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Digital Signal Processing Lab Lab Manual and Workbook
Academic Year: _________ Semester: _______ Name: _________________ Reg. No.: ________ Department: ____________ Class: __________
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Gokaraju Rangaraju Institute of Engineering and Technology
Certificate
This is to certify that this is a bonafide record of practical work done by Name ____________________, ____________________, Reg. No. ________________, ________________, in the laboratory during the academic year ____________.
Signature of Faculty
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Contents C6713-based Experiments 1 2 3 4 5 6
Introduction to the TMS 320C6713 DSP Processor Linear convolution Circular convolution FIR filters IIR filters Echo with variable delay and feedback
7 11 15 19 23 27
MATLAB-based Experiments 1 2 3 4 5 6 7 8
31-Linear and circular convolution 32-Filtering via DFT 33-Picket fence effect in DFT 41-IIR low-pass filter frequency response and simulation 43-Analog Butterworth low-pass filter frequency response 47-Design of analog Butterworth low-pass filter 48-Design of digital Butterworth low-pass filter 51-FIR low-pass filter
29 33 41 51 59 67 73 81
Extra-credit Experiments 1 2 3 4
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E1-Generation of DTMF E2-Implementation of Decimation E3-Implementation of Interpolation E4-Implementation of I/D Rate Conversion
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1. Introduction to the TMS 320C6713 DSP Processor DSK Board The DSK package is powerful, yet relatively inexpensive, with the necessary hardware and software support tools for real-time signal processing. It is a complete DSP system. The DSK board, with an approximate size of 5 X 8 in., includes the C6713 floating-point digital signal processor and a 32-bit stereo codec TLV320AIC23 (AIC23) for input and output. The onboard codec AIC23 uses a sigma – delta delta technology that provides ADC and DAC. It connects to a 12-MHz system clock. Variable sampling rates from 8 to 96 kHz can be set readily. A daughter card expansion is also provided on the DSK board. Two 80-pin connectors provide for external peripheral and external memory interfaces with light-emitting diodes (LEDs) and liquid-crystal displays (LCDs) for spectrum display. The DSK board includes 16MB (megabytes) of synchronous dynamic random access memory (SDRAM) and 256kB (kilobytes) of flash memory. Four connectors on the board provide input and output: MIC IN for microphone input, LINE IN for line input, LINE OUT for line output, and HEADPHONE for a headphone output (multiplexed with line output). The status of the four user dip switches on the DSK board can be read from a program and provides the user with a feedback control interface. The DSK operates at 225MHz.Also onboard the DSK are voltage regulators that provide 1.26 V for the C6713 core and 3.3 V for its memory and peripherals.
(a)
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(b) FIGURE. TMS320C6713-based DSK board: ( a ) board; (b ) diagram
TMS320C6713 Digital Signal Processor The TMS320C6713 (C6713) is based on the VLIW architecture, which is very well suited for numerically intensive algorithms. The internal program memory is structured so that a total of eight instructions can be fetched every cycle. For example, with a clock rate of 225MHz, the C6713 is capable of fetching eight 32-bit instructions every 1/(225 MHz) or 4.44 ns. Features of the C6713 include 264 kB of internal memory (8kB as L1P and L1D Cache and 256kB as L2 memory shared between program and data space), eight functional or execution units composed of six arithmetic-logic units (ALUs) and two multiplier units, a 32-bit address bus to address 4 GB (gigabytes), and two sets of 32-bit general purpose registers. The C67xx (such as the C6701, C6711, and C6713) belong to the family of the C6x floating point processors, whereas the C62xx and C64xx belong to the family of the C6x fixed-point processors. The C6713 is capable of both fixed- and floating point processing.
Code Composer Studio CCS provides an IDE to incorporate the software tools. CCS includes tools for code generation, such as a C compiler, an assembler, and a linker. It has graphical capabilities and supports real-time debugging. It provides an easy-to-use software tool to build and debug programs. The C compiler compiles a C source program with extension .c to produce an assembly source file with extension.asm. The assembler assembles an.asm source file to produce a machine language object file with extension.obj. The linker combines object files and object libraries as input to produce an executable file with extension.out . This executable file represents a linked common object file format (COFF), popular in Unix-based systems and adopted by several makers of digital signal processors. This executable file can be loaded and run directly on the C6713 processor. A linear optimizer optimizes this source file to create an assembly file with extension .asm (similar to the task of the C compiler). To create an application project, one can “add” the appropriate files to the project. Compiler/linker options can readily be specified. A number of debugging features are available, DSP Lab
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including setting breakpoints and watching variables; viewing memory, registers, and mixed C and assembly code; graphing results; and monitoring execution time. One can step through a program in different ways (step into, over, or out). Through the joint team action group (JTAG), communication with on-chip emulation support occurs to control and monitor program execution. The C6713 DSK board includes a JTAG interface through the USB port.
CCS Installation and Support Use the USB cable to connect the DSK board to the USB port on the PC. Use the 5-V power supply included with the DSK package to connect to the +5-V power connector on the DSK to turn it on. Install CCS with the CD-ROM included with the DSK, preferably using the c:\C6713 structure. The CCS icon should be on the desktop as “C6713DSK CCS” and is used to launch CCS. The code generation tools (C compiler, assembler, linker) are used with CCS version 2.x. CCS provides useful documentations included with the DSK package on the following (see the Help icon): 1. Code generation tools (compiler, assembler, linker, etc.) 2. Tutorials on CCS, compiler, RTDX 3. DSP instructions and registers 4. Tools on RTDX, DSP/basic input/output system (DSP/BIOS), and so on. An extensive amount of support material ( pdf files) is included with CCS. There are also examples included with CCS within the folder c:\C6713\examples. They illustrate the board and chip support library files, DSP/BIOS, and so on. CCS Version 2.x was used to build and test the examples included in this book. A number of files included in the following subfolders/directories within c:\C6713 can be very useful: 1. myprojects: a folder supplied only for your projects. All the folders in the accompanying book CD should be placed within this subdirectory. 2. bin: contains many utilities. 3. docs: contains documentation and manuals. 4. c6000\cgtools: contains code generation tools. 5. c6000\RTDX : contains support files for real-time data transfer. 6. c6000\bios: contains support files for DSP/BIOS. 7. examples: contains examples included with CCS. 8. tutorial : contains additional examples supplied with CCS.
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Useful Types of Files You will be working with a number of files with different extensions. They include: 1. file.pjt: to create and build a project named file 2. file.c: C source program 3. file.asm: assembly source program created by the user, b y the C compiler, or by the linear optimizer 4. file.sa: linear assembly source program. The linear optimizer uses file.sa as input to produce an assembly program file.asm 5. file.h: header support file 6. file.lib: library file, such as the run-time support library file rts6700.lib 7. file.cmd: linker command file that maps sections to memory 8. file.obj: object file created by the assembler 9. file.out: executable file created by the linker to be loaded and run on the C6713 processor 10. file.cdb: configuration file when using DSP/BIOS
Support Files The following support files located in the folder support (except the library files) are used for most of the examples. 1. C6713dskinit.c : contains functions to initialize the DSK, the codec, the serial ports, and for I/O. It is not included with CCS. 2. C6713dskinit.h : header file with function prototypes. Features such as those used to select the mic input in lieu of line input (by default), input gain, and so on are obtained from this header file. 3. C6713dsk.cmd : sample linker command file. This generic file can be changed when using external memory in lieu of internal memory. 4. Vectors_intr.asm : a modified version of a vector file included with CCS to handle interrupts. Twelve interrupts, INT4 through INT15, are available, and INT11 is selected within this vector file. They are used for interrupt-driven programs. 5. Vectors_poll.asm : vector file for programs using polling. 6. rts6700.lib, dsk6713bsl.lib, csl6713.lib : run-time, board, and chip support library files, , respectively. These files are included with CCS and are located in C6000\cgtools\lib C6000\dsk6713\lib , and c6000\bios\lib , respectively.
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2. Linear Convolution Aim:
To verify linear convolution. Equipments Required: Operating System Constructor Software
Windows XP Simulator CStudio3 & MATLAB 7.5
Theory:
Convolution is a formal mathematical operation, just as multiplication, addition, and integration. Addition takes two numbers and produces a third number, while convolution takes two signals and produces a third signal. Convolution is used in the mathematics of many fields, such as probability and statistics. In linear systems, convolution is used to describe the relationship between three signals of interest: the input signal, the impulse response, and the output signal.
In this equation, x1(k), x2(n-k) and y (n) represent the input to and output from the system at time n. Here we could see that one of the inputs is shifted in time by a value every time it is multiplied with the other input signal. Linear Convolution is quite often used as a method of implementing filters of various types. Program for:
//Linear convolution program in c language using CCStudio #include int x[15],h[15],y[15]; main() { int i,j,m,n; printf("\n enter value for m"); scanf("%d",&m); printf("\n enter value for n"); scanf("%d",&n); printf("Enter values for i/p x(n):\n"); for(i=0;i
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for(i=0;i
//displaying the o/p
Result: enter value for m4 enter value for n4 Enter values for i/p 1234 Enter Values for n 1234 The Value of output y[0]=1 The Value of output y[1]=4 The Value of output y[2]=10 The Value of output y[3]=20 The Value of output y[4]=25 The Value of output y[5]=24 The Value of output y[6]=16
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3. Circular Convolution Aim:
To verify circular convolution. Equipment Required: Operating System Constructor Software
Windows XP Simulator CCStudio 3 & MATLAB 7.5
Theory: Circular convolution is another way of finding the convolution sum of two input signals. It resembles the linear convolution, except that the sample values of one of the input signals is folded and right shifted before the convolution sum is found. Also note that circular convolution could also be found by taking the DFT of the two input signals and finding the product of the two frequency domain signals. The Inverse DFT of the product would give the output of the signal in the time domain which is the circular convolution output. The two input signals could have been of varying sample lengths. But we take the DFT of higher point, which ever signals levels to. For eg. If one of the signals is of length 256 and the other spans 51 samples, then we could only take 256 point DFT. So the output of IDFT would be containing 256 samples instead of 306 samples, which follows N1+N2 – 1 where N1 & N2 are the lengths 256 and 51 respectively of the two inputs. Thus the output which should have been 306 samples long is fitted into 256 samples. The256 points end up being a distorted version of the correct signal. This process is called circular convolution. Program: /* program to implement circular convolution */ #include int m,n,x[30],h[30],y[30],i,j, k,x2[30],a[30]; void main() { printf(" Enter the length of the first sequence\n"); scanf("%d",&m); printf(" Enter the length of the second sequence\n"); scanf("%d",&n); printf(" Enter the first sequence\n"); for(i=0;in) /* Pad the smaller sequence with zero*/ { for(i=n;i
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m=n; } y[0]=0; a[0]=h[0]; for(j=1;j
/*folding h(n) to h(-n)*/
/*Circular convolution*/ for(i=0;i
/*displaying the result*/;
OUTPUT:
Enter the length of the first sequence 4 Enter the length of the second sequence 3 Enter the first sequence 1234 Enter the second sequence 123 The circular convolution is 18 16 10 16
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4. FIR Filters Aim:
To verify FIR filters. Equipment Required: Operating System Constructor Software
Windows XP Simulator CCStudio 3 & MATLAB 7.5
Theory:
A Finite Impulse Response (FIR) filter is a discrete linear time-invariant system whose output is based on the weighted summation of a finite number of past inputs. An FIR transversal filter structure can be obtained directly from the equation for discrete-time convolution.
In this equation, x(k) and y(n) represent the input to and output from the filter at time n. h(n-k) is the transversal filter coefficients at time n. These coefficients are generated by using FDS (Filter Design Software or Digital filter design package). FIR – filter is a finite impulse response filter. Order of the filter should be specified. Infinite response is truncated to get finite impulse response, placing a window of finite length does this Types of windows available are Rectangular, Barlett, Hamming, Hanning, Blackmann window etc. This FIR filter is an all zero filter. Program:
#include #include #define pi 3.1415 int n,N,c; float wr[64],wt[64]; void main() { printf("\n enter no. of samples,N= :"); scanf("%d",&N); printf("\n enter choice of window function\n 1.rect \n 2. triang \n c= :"); scanf("%d",&c); printf("\n elements of window function are:"); switch(c) { case 1: for(n=0;n<=N-1;n++) { wr[n]=1; printf(" \n wr[%d]=%f",n,wr[n]); } break; DSP Lab
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case 2: for(n=0;n<=N-1;n++) { wt[n]=1-(2*(float)n/(N-1)); printf("\n wt[%d]=%f",n,wt[n]); } break; } } Result:
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5. IIR Filters Aim:
To design and implement IIR (LPF/HPF) filters. Equipment Required: Operating System Construct Software
Windows XP Simulator CCStudio 3 & MATLAB 7.5
Theory: The IIR filter can realize both the poles and zeroes of a system because it has a rational transfer function, described by polynomials in z in both the numerator and the denominator
∑ ∑ ∑ ∑
The difference equation for such a s ystem is described by the following: M and N are order of the two polynomials bk and ak are the filter coefficients. These filter coefficients are generated using FDS (Filter Design software or Digital Filter design package).IIR filters can be expanded as infinite impulse response filters. In designing IIR filters, cutoff frequencies of the filters should be mentioned. The order of the filter can be estimated using butter worth polynomial. That’s why the filters are named as butter worth filters. Filter coefficients can be found and the response can be plotted. C-Program iirfilters:
#include #include int i,w,wc,c,N; float H[100]; float mul(float, int); void main() { printf("\n enter order of filter "); scanf("%d",&N); printf("\n enter the cutoff freq "); scanf("%d",&wc); printf("\n enter the choice for IIR filter 1. LPF 2.HPF "); scanf("%d",&c); switch(c)
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{ case 1: for(w=0;w<100;w++) { H[w]=1/sqrt(1+mul((w/(float)wc),2*N)); printf("H[%d]=%f\n",w,H[w]); } break; case 2: for(w=0;w<=100;w++) { H[w]=1/sqrt(1+mul((float)wc/w,2*N)); printf("H[%d]=%f\n",w,H[w]); } break; } } float mul(float a,int x) { for(i=0;i
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6. Echo with variable delay and feedback AIM:
To verify echo with variable delay and feedback EQUIPMENTS:
Operating System Constructor Software
Windows XP Simulator CCStudio 3
PROGRAM:
#include "DSK6713_AIC23.h" // codec support Uint32 fs=DSK6713_AIC23_FREQ_8KHZ; // set sampling rate #define DSK6713_AIC23_INPUT_MIC 0x0015 #define DSK6713_AIC23_INPUT_LINE 0x0011 Uint16 inputsource=DSK6713_AIC23_INPUT_MIC;// select input #define BUF_SIZE 8000 // this sets maximum length of delay short input,output,delayed; short buffer[BUF_SIZE]; // storage for previous samples int i; //index into buffer interrupt void c_int11() // interrupt service routine { input = input_left_sample();// read new input sample from ADC delayed = buffer[i]; // read delayed value from buffer output = input + delayed; // output sum of input and delayed values buffer[i] = input // store new input and a fraction //of the delayed value in buffer if(++i >= BUF_SIZE)//test for end of buffer output_left_sample(output) return; //return from ISR } void main() { for(i=0 ; i
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Lab-31 Linear and circular convolution and DFT Objective To perform linear and circular convolution of sequences and explore their relationship to polynomial multiplication.
( ) ( ) Background The formulae for linear and circular convolution are:
To Do A. Evaluate, by hand, the linear convolution of the sequences . Compare with the MATLAB solution. B. Perform the circular convolution of the sequences compare with the MATLAB solution.
and
C. Evaluate, by hand, the linear convolution of the sequences convolution. Compare with the MATLAB solution. D. Verify .
, or equivalently,
and
by hand (matrix method) and
and
using circular
=
E. [Polynomial multiplication] Multiply the polynomials
Put the result in the form
Give a general expression for the coefficients by the convolution of the coefficient sequences
and
and show that the sequence . Verify for
is given
A1. Linear convolution by hand. A2. Linear convolution, using the conv function. In the subplots we have zero-padded all sequences so as, firstly, to extend the plots and, secondly, to line up all sequences on a matching time scale. Whether we convolve the zero-padded or non-zero-padded sequences the result is the same.
x1n = [1, 2, 2, 1, 1]; x2n = [3, 2, 1]; nz=2; %Extra zeros yn = conv(x1n, x2n) , % Linear convolution ynz=[yn, zeros(1, nz)];% Zeros appended to yn to extend the plot lynz= length(yn) + nz; % Length of yn = length(x1n) + length(x2n) - 1; e1=lynz - length(x1n); x1z=[x1n, zeros(1, e1)]; % Zero-pad for plotting purpose e2=lynz - length(x2n); x2z=[x2n, zeros(1, e2)]; % Zero-pad for plotting purpose m = 0: lynz-1; subplot (2, 2, 1), stem(m, x1z); title ('Sequence x1(n)'), xlabel ('n'), ylabel('x1(n)'); DSP Lab
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subplot (2, 2, 2), stem(m, x2z); title ('Sequence x 2(n)'), xlabel ('n'), ylabel('x2(n)'); subplot (2, 2, 3), stem(m, ynz); % Bottom left plot title ('Sequence y(n)'), xlabel ('n'), ylabel('y(n)'); The result of convolution is given by yn = [3 8 11 9 7 3 1] Q. Would the result be affected if an arbitrary number of zeros were appended to the original sequences prior to convolution?
B1. Circular convolution by hand.
B2. Circular convolution in MATLAB. The sequences are We shall need to zero-pad the shorter sequence to the same length as the longer one. Thus .
.
x1 = [1, 2, 2, 1, 1], %Row vector % x2m is a matrix made up of circularly shifted rows of x2z x2m = [3, 0, 0, 1, 2; 2, 3, 0, 0, 1; 1, 2, 3, 0, 0; 0, 1, 2, 3, 0; 0, 0, 1, 2, 3], %5 x 5 Matrix ycn= (x2m*x1')', %As calculated (x2m*x1') is a column vector
x1 =[1 2 2 1 1] x2m =[3 0 0 1 3] ycn =[6 9 11 9 7]
2; 2
3
0
0
1; 1
2
3
0
0; 0
1
2
3
0; 0
0
1
2
( ) ( )
Q. What function(s) does MATLAB offer to perform circular convolution? Q. Write and execute a MATLAB program which, given the input sequences perform the circular convolution using the matrix method.
and
, will
C1. Circular convolution of zero-padded sequences to get the result of linear convolution. The length of the sequences and are, respectively, and . The length of the result of linear convolution, , is . Therefore, we zero-pad and each to a length of 7 giving and . We now perform circular convolution of and . The result
will be the same as the linear convolution of
and
or of . Thus
and
. In other words,
x1 = [1, 2, 2, 1, 1, 0, 0], %Row vector % x2m is a matrix made up of circularly shifted rows of x2z x2m = [3 0 0 0 0 1 2; 2 3 0 0 0 0 1; 1 2 3 0 0 0 0; 0 1 2 3 0 0 0; 0 0 1 2 3 0 0; 0 0 0 1 2 3 0; 0 0 0 0 1 2 3], %7 x 7 Matrix ycn= (x2m*x1')', %As calculated (x2m*x1') is a column vector
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x1 = [1 2 2 1 1 0 0] x2m =[3 0 0 0 0 1 2; 2 3 0 3 0 0 0; 0 0 1 2 3 0 0; 0 ycn = [3 8 11 9 7 3 1]
0 0
0 0
0 1
1; 1 2 3 2 3 0; 0
0 0
0 0
0 0
0; 0 1 2 1 2 3]
C2. Linear convolution of the original (unpadded) sequences, sequences, and same result.
and , or of the zero-padded , should, of course, yield the
x1n = [1, 2, 2, 1, 1, 0, 0]; x2n = [3, 2, 1, 0, 0, 0, 0]; nz=2; %Extra zeros yn = conv(x1n, x2n) , % Linear convolution ynz=[yn, zeros(1, nz)];% Zeros appended to yn to extend the plot lynz= length(yn) + nz; % Length of yn = length(x1n) + length(x2n) - 1; e1=lynz - length(x1n); x1z=[x1n, zeros(1, e1)]; % Zero-pad for plotting purpose e2=lynz - length(x2n); x2z=[x2n, zeros(1, e2)]; % Zero-pad for plotting purpose m = 0: lynz-1; subplot (2, 2, 1), stem(m, x1z); % Top left plot title ('Sequence x1(n)'), xlabel ('n'), ylabel('x1(n)'); subplot (2, 2, 2), stem(m, x2z); % Top right plot title ('Sequence x2(n)'), xlabel ('n'), ylabel('x2(n)'); subplot (2, 2, 3), stem(m, ynz); % Bottom left plot title ('Sequence y(n)'), xlabel ('n'), ylabel('y(n)'); yn = [3
8
11
9
7
3
1
0
0
0
0
0
0]
= or, equivalently, D1. Verify . The sequences are and shorter sequence to the same length as the longer one. Thus MATLAB program
1. Computes the DFT’s of and resulting in 2. Multiplies the DFT’s resulting in 3. Computes the IDFT of the product, resulting in
=
. We shall need to zero-pad the . The following
and
x1n = [1, 2, 2, 1, 1]; x2n = [3, 2, 1, 0, 0]; X1k=fft(x1n) , X2k=fft(x2n) , Yk=X1k.*X2k, yn=ifft(Y)
X1k =[7 X2k =[6 Yk =[42 yn =[6
(-0.5000 - 1.5388i) (-0.5000 + 0.3633i) (-0.5000 - 0.3633i) (-0.5000 + 1.5388i)] (2.8090 - 2.4899i) (1.6910 - 0.2245i) (1.6910 + 0.2245i) (2.8090 + 2.4899i)] (-5.2361 - 3.0777i) (-0.7639 + 0.7265i) (-0.7639 - 0.7265i) (-5.2361 + 3.0777i)] 9 11 9 7]
E. Polynomial multiplication.
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Lab-32 Filtering through DFT/FFT Objective To perform digital filtering in the frequency domain using DFT.
may be filtered in the time domain by implementing the filter difference Background The signal equation. Alternatively, where it is possible to process in blocks or off-line, we may work in the frequency domain: we first find the DFT, . Next we find ( f for “filtered”) by setting to zero selected components of while preserving the symmetry properties of about the mid point, , of the sequence. We then find , the IDFT of , which should be a filtered version of . Thus has been filtered entirely in the frequency domain. Consider a signal consisting of two frequency components, a 2 Hz and a 4 Hz, from which we would like to filter out, say, the 2 Hz component. Given the signals and , the signal is sampled at 16 Hz ( =1/16) resulting in the discrete-time sequence . Then we find its 16-point DFT, i.e., the values. We then filter the signal in the frequency domain: remove the DFT values that correspond to 2 Hz. The resulting DFT, , when inverse-transformed should give a sequence, , that contains only the 4 Hz component.
When sampled at 16 Hz, the 2 Hz and 4 Hz components have digital frequencies of 1/8 cycles/sample and 1/4 cycles/sample. To Do A. Demonstrate filtering in the frequency domain. A1. Sketch the continuous-time and discrete-time signals using 16 Hz. A2. Sketch the 16-point DFT of the 2-Hz signal. A3. Sketch the 16-point DFTof the composite (2-Hz + 4-Hz) signal. A4. Filter out the 4 Hz component.
sampled at
B. Generate the continuous-time signal
from the discrete-time sequence
.
is to be sampled at 16 C. Repeat A and B if the signal Hz and filtered using the DFT method so as to eliminate all frequencies outside the range .
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A1. Sketch the signals period. Sketch the sequences
,
, and
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over a 1-second .
In MATLAB the following segment plots the three functions
.
t = 0:1/160:1.0; x1t = cos(2*pi*2*t); x2t = cos(2*pi*4*t); xt = x1t + x2t; % subplot(3,1,1), plot(t,x1t); xlabel('t, sec.'), ylabel('x1(t)'); title('x1(t) = cos 2 \pi2t'); % subplot(3,1,2), plot(t,x2t); xlabel('t, sec.'), ylabel('x2(t)'); title('x2(t) = cos 2 \pi4t'); % subplot(3,1,3), plot(t,xt); xlabel('t, sec.'), ylabel('x(t)'); title('x(t) = cos 2\pi2t + cos 2\pi4t');
x1(t) = cos 22t 1 ) t ( 1 x
0 -1
0
0.1
0.2
0.3
0.4
0.5 0.6 t, sec. x2(t) = cos 24t
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5 0.6 0.7 t, sec. x(t) = cos 22t + cos 24t
0.8
0.9
1
0
0.1
0.2
0.3
0.8
0.9
1
1 ) t ( 2 x
0 -1
2 ) t ( x
0 -2
0.4
0.5 t, sec.
0.6
0.7
√ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √
The sequence values are:
Note that there are 16 samples per second. The 16-sample record above occupies a period of one second. In MATLAB the following segment plots the sequences , and . DSP Lab
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%t = 0: 1/160: 1.0; x1t = cos(2*pi*2*t); x2t = cos(2*pi*4*t); xt = x 1t + x2t; % n = 0:16; x1n = cos(pi*n/4); x2n = cos(pi*n/2); xn = x1n + x2n, % subplot(3,1,1), stem(n,x1n); xlabel('n'), ylabel('x1(n)'); title('x1(n) = cos(\pin/4)'); % subplot(3,1,2), stem(n,x2n); xlabel('n'), ylabel('x2(n)'); title('x2(n) = cos(\pin/2)'); % subplot(3,1,3), stem(n,xn); xlabel('n'), ylabel('x(n)'); title('x(n) = cos(\pin/4)+ cos(\pin/2)'); The sequence is -1 0.7071}
= {2 0.7071 -1 -0.7071 0 -0.7071 -1 0.7071 2 0.7071 -1 -0.7071 0 -0.7071
x1(n) = cos(n/4) 1 ) n ( 1 x
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Q. Indicate the digital frequencies, record length o f sequence and the frequency resolution of the corresponding DFT.
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A2. Sketch the 16-point DFT
. Note
for
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In this case the DFT is real valued for all . Accordingly, all phase angle values should be zero, that is, . However, … o If the input is listed explicitly by specifying all 16 values MATLAB computes all the phase angles correctly (as zero). If the input sequence is implicitly specified by a statement like “n = 0:15; x1n = o cos(pi*n/4)” then the MATLAB output shows non-zero phase at several points. We shall tentatively attribute this to finite precision effects within MATLAB. Thus two sets of MATLAB plots for the magnitude and phase angle of are shown below. First the explicit listing: Input specified as x1n = [1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2), 1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2)] Fs = 16; N = 16; % Sampling rate & Number of points % Specify input sequence values explicitly x1n = [1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2), 1, 1/sqrt(2), 0, -1/sqrt(2), -1, 1/sqrt(2), 0, 1/sqrt(2)], % Magnitude and Phase of X1k X1k = fft(x1n); MX1k = abs(X1k), PX1k = angle(X1k), n = 0: (N-1); F = (Fs/N)*n; subplot(2,1,1), stem(F, abs(X1k)); axis([0 16 0 10]), xlabel('Frequency, Hz'), ylabel('|X1(k)|'); title ('Magnitude of X1(k)'); subplot(2,1,2), stem(n, angle(X1k)); axis([0 16 -4 4]), xlabel('k'), ylabel('
| ) k ( 1 X |
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The 2 Hz component is indicated by the pair and . Recall properties of symmetry about the midpoint of the sequence . Recall also that for a sampling rate of 16 Hz the maximum allowed frequency is 8 Hz.
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√ √ √ √ √ √ √ √
A3. Sketch the 16-point DFT
for
for
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The 16-point sequence is
The magnitude and phase of are plotted below. Note that is specified by an explicit list. (Implicit specification with “n = 0:15; xn = cos(pi*n/4) + cos(pi*n/2)” is prob lematic.) Fs = 16; N = 16; % Sampling rate & Number of points % Specify input sequence values explicitly xn = [2, 1/sqrt(2), -1, -1/sqrt(2), 0, -1/sqrt(2), -1, 1/sqrt(2), 2, 1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2), -1, 1/sqrt(2)], % MXk = magnitude of Xk, PXk = phase angle of Xk Xk = fft(xn), MXk = abs(Xk), PXk = angle(Xk), % k = 0: (N-1); F = (Fs/N)*k; subplot(2,1,1), stem(F, abs(Xk)); axis([0 16 0 10]), xlabel('Frequency, Hz'), ylabel('|X(k)|'); title ('Magnitude of X(k)'); subplot(2,1,2), stem(k, angle(Xk)); axis([0 16 -4 4]), xlabel('k'), ylabel('
| ) k ( X |
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The 2 Hz component is indicated by the pair the 4 Hz component by the pair and maximum allowed frequency is 8 Hz.
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8 k
and (symmetry considerations) and . Recall that for a sampling rate of 16 Hz the
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{ }
A4. [Filtering] Filter the sequence ) is removed. From the resulting
find and plot
Solution In terms of the DFT, , this means setting both symmetry properties of the DFT. The resulting DFT sequence,
We next find
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so that the 4 Hz component (the .
and to zero to preserve the , is given by
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 0 8 0 8 0 0 0 0 0 0 0 8 0 8 0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 8 0
the inverse DFT of
:
Xfk = [0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0], xfn = ifft(Xfk), N=length(Xfk), n = 0: (N-1); stem(n, xfn); xlabel('n'), ylabel('xf(n)'); title ('Filtered signal xf(n)');
√ √ √ √ √ √ √ √
The output from MATLAB gives the filtered version o f the discrete-time sequence as xfn = [1 0.7071 0 -0.7071 -1 -0.7071 0 0.7071 1 0.7071 0 -0.7071 -1 -0.7071 0 0.7071] It is seen that is identical to the original 2 Hz component, :
Filtered signal xf(n)
1 0.8 0.6 0.4 0.2 ) n ( f x
0 -0.2 -0.4 -0.6 -0.8 -1
Q. Repeat with either
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0
or
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n
(but not both) set to zero and explain the outcome.
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from the discrete-time sequence in B above. B. Generate the continuous-time signal Given the sequence, , the operation of D/A conversion followed b y low pass filtering should produce the continuous-time signal ( ). This may be viewed roughly as “connecting the dots,” resulting in a continuous-like continuous-like graph. In the following MATLAB segment the subplot (2, 1, 1) uses the plot function to connect the dots resulting in the function th e continuous-looking graph of while while subplot (2, 1, 2) simply calibrates the horizontal axis in terms of time in seconds and re-labels the diagram accordingly. In either plot the same 16 data points are connected. Fs = 16; % Sampling frequency Xfk = [0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0], xfn = ifft(Xfk), N=length(Xfk), n = 0: (N-1); % subplot(2, 1, 1), plot(n, xfn); axis([0 16 -1 1]); xlabel('n'), ylabel('xf(n)'); title ('Filtered signal xf(n)'); % subplot(2,1,2), plot(n/Fs, xfn); xlabel('Time, sec'), ylabel('xf(t)'); title ('Filtered signal xf(t)'); Filtered signal xf(n) 1 0.5 ) n ( f x
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Q. Compare filtering in the time domain with filtering in the frequency domain.
C. Repeat A and B if the signal is to be sampled at 16 Hz and filtered using the DFT method so as to eliminate all frequencies outside ou tside the range .
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Lab-33 Picket fence effect in DFT
∑
Objective This Lab deals with the Fourier spectra of finite-duration sequences using the signal to demonstrate the picket fence effect in DFT.
is is given by Background The DTFT of the -point sequence . The DFT of the sequence is obtained by taking samples of the DTFT:
,
We may either use f f t to compute the DFT or use freqz to to compute the DTFT and take samples of it to arrive at the DFT, the latter approach being more insightful. The resulting set of DFT values, , thus depends on the particular set of frequencies where is sampled. The spectrum is is, is, in effect, viewed through a picket fence. Therefore when a DFT is computed it is possible that spectral components located at frequencies other than the selected set may be missed (obscured by the pickets). This is known as the picket-fence effect . To minimize the picket-fence effect we increase t he number of DFT points while holding h olding the (time-domain) sampling rate fixed. This increases the record length and hence the spectral resolution of the DFT.
To Do A. For the -point sequence obtained by sampling the signal obtained find the DTFT ( ) and the DFT ( ). ). A1. DTFT of 8-point sequence and 8-point DFT A2. The 8-point DFT using the function f f t A3. DTFT of 16-point zero-padded sequence A4. 16-point DFT with zero-padding using the function f f t A5. 16-point DFT without zero-padding using the function f f t
at 16 Hz starting
B. Repeat A using a sampling frequency of 12 Hz.
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] [ [ ] []
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, when sampled at 16 Hz, gives . The samples in the A1. The signal 8-point sequence = {1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2)} are spaced (1/16)-second apart over a record length of 0.5 sec. (one cycle of the cosine). The average value of the sequence (the dc component) is zero. The DTFT, , of the 8-point sequence is a continuous function of :
The numerator and denominator coefficients are = and . The DTFT is computed using the function freqz and plotted below (Xw). Further, the DTFT is sampled at 8 points (Xw2) resulting in the 8-point DFT ( , spaced 2 Hz apart). In the two sub-plots the horizontal axes are calibrated both in Hz and in the frequency index, . % DTFT of ={1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2)} Fs= 16; N= 8; F0= Fs/N; w0= 2*pi/N; %Sampling freq; No. of points; Resolution b=[1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2)]; %Numerator coefficients a=[1]; %Denominator coefficients w=0: pi/256: 2*pi; k=w/w0; F=k*F0; Xw=freqz(b, a, w); %DTFT w2=0: 2*pi/N: 2*pi; k2=w2/w0; F2=k2*F0; Xw2=freqz(b, a, w2); %DFT subplot(2, 1, 1), stem(F2, abs(Xw2), 'k'); hold on plot(F, abs(Xw)); legend ('Samples of DTFT', 'DTFT', 'Location', 'North' ); xlabel('F, Hz'), ylabel('|X(k)|'); grid; title('Magnitude'); hold off subplot(2, 1, 2), plot(k, angle(Xw)); title('Phase'); xlabel('k'), ylabel('
X: 2 Y: 4 | ) k ( X |
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A2. Compute the 8-point DFT using the function fft and compare with the result of A1.
Fs= 16; N= 8; F0= Fs/N; %Sampling freq; No. of points; Resolution xn = [1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2)]; Xk = fft(xn), MXk = abs(Xk); AXk = angle(Xk); %DFT, Ma gnitude and Phase % k = 0:7; F = k*F0; % subplot(2,1,1), stem(F, abs(Xk)); axis([0 16 0 5]) xlabel('Frequency, Hz (= k*F0)'), ylabel('|X(k)|'); title ('Magnitude of X(k)'); subplot(2,1,2), stem(k, angle(Xk)); xlabel('k'), ylabel('
The table below shows that the 8-point DFT contains a component at 2 Hz, ( 1 and 7, based on symmetry properties), that is and . The DFT values are all real numbers symmetrically disposed about , the center of symmetry. The frequency resolution is Hz. Note that with a sampling frequency of 16 Hz the maximum frequency possible is 8 Hz corresponding to . For easier comparison the plots go up to 8.
k =0 ↓ Hz→ 0 {0 {0 {0
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Magnitude of X(k) 5 4 | ) 3 k ( X |
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Q. What frequency components do you expect to see in the DFT?
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A3. [DTFT of the 16-point zero-padded sequence] The 8-sample sequence of the 2-Hz cosine, (sampled at 16 Hz) is concatenated with 8 zeros: = {1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2), 0, 0, 0, 0, 0, 0, 0, 0} Zero-padding improves the spectral resolution by increasing the reco rd length. Demonstrate that this 16 point sequence’s DTFT is identical to that of the 8-point sequence in A1. Take 16 samples of this DTFT and compare the resulting 16-pont DFT to the 8-point DFT.
% DTFT of xnz, the zero-padded sequence % xnz ={1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2), zeros(1, 8)} Fs= 16; N= 16; %Sampling frequency & No. of points F0= Fs/N; w0= 2*pi/N; %Resolution xn=[1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2)], b=[xn, zeros(1, 8)], a=[1]; %Numerator & denominator coefficients w=0: pi/256: 2*pi; k=w/w0; F=k*F0; Xw=freqz(b, a, w); %DTFT w2=0: 2*pi/N: 2*pi; k2=w2/w0; F2=k2*F0; Xw2=freqz(b, a, w2); %DFT subplot(2, 1, 1), stem(F2, abs(Xw2), 'k'), hold on, plot(F, abs(Xw)); legend ('Samples of DTFT', 'DTFT', 'Location', 'North' ), legend boxoff; xlabel('F, Hz'), ylabel('|X(k)|'); grid; title('Magnitude'), hold off subplot(2, 1, 2), plot(k, angle(Xw)); title('Phase'); xlabel('k'), ylabel('
A4. [16-point DFT with zero-padding] The MATLAB function fft(xn, 16) will zero-pad the sequence to a length of 16 and compute the DFT of the resulting 16-point sequence. Zero-padding improves the spectral resolution by increasing the record length. The 8-sample sequence of the 2-Hz cosine, sampled at 16 Hz, is appended with 8 zeros: = {1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2), 0, 0, 0, 0, 0, 0, 0, 0} Find the 16-point DFT using the function f ft . The frequency resolution is .
⁄ ⁄
Fs= 16; N= 16; F0= Fs/N; %Sampling freq; No. of points; Resolution xn=[1, 1/sqrt(2), 0, -1/sqrt(2), -1, -1/sqrt(2), 0, 1/sqrt(2)], xnz=[xn, zeros(1, 8)], %Zero-padded sequence n = 0:15; subplot(3,1,1), stem(n, xnz); xlabel('n'), ylabel('x(n)'); title ('Zero-padded sequence xnz'); Xk = fft(xn, 16), %x(n) is zero-padded to a length of 16 MXk = abs(Xk), AXk = angle(Xk), % Magnitude & Phase angle of Xk k = 0:15; F=k*F0; subplot(3,1,2), stem(F, abs(Xk)); axis([0 16 0 5]) xlabel('Frequency, Hz (= k*F0)'), ylabel('|X(k)|'); title ('Magnitude of X(k)'); subplot(3,1,3), stem(k, angle(Xk)); axis([0 16 -2 2]) xlabel('k'), ylabel('
Q. What kind of symmetry does the sequence Q. List the MATLAB-generated DFT sequence
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possess? and identify the 2 Hz component.
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A5. [16-point DFT without zero-padding] Given all 16 sample values of a 2-Hz cosine, , obtained at a sampling rate of 16 Hz find its 16-point DFT using f ft . The frequency resolution of the DFT is . In the MATLAB segment below the sequence values are specified implicitly (by a statement such as “t = 0: 1/16: 15/16; xn = cos (2*pi*2*t)” or “n = 0: 1: 15; xn = cos (2*pi*2*n/Fs)”. Although the MATLAB-generated DFT, , is correct, the phase values, , are incorrect (seems to happen when is implicitly specified).
N=16; Fs=16; F0= Fs/N; %N points sampled at Fs Hz & Resolution t = 0: 1/16: 15/16; xn = cos (2*pi*2*t), %Sequence specified implicitly n = 0:15; subplot(3,1,1), stem(n, xn); xlabel('n'), ylabel('x(n)'); title ('Sequence x(n)'); % Xk = fft(xn), MXk = abs(Xk), AXk = angle(Xk), % Magnitude & Phase of Xk k = 0:15; F=k*F0; subplot(3,1,2), stem(F, abs(Xk)); axis([0 16 0 10]) xlabel('Frequency, Hz (= k*F0)'), ylabel('|X(k)|'); title ('Magnitude of X(k)'); subplot(3,1,3), stem(k, angle(Xk)); axis([0 16 -5 5]) xlabel('k'), ylabel('
The DFT (reproduced from the MATLAB output) is X(k) = {0, (-0+0i), (8-0i), (0-0i), (0-0i), (0-0i), (0-0i), (0+0i), -0, (0-0i), (0+0i), (0+0i), (0+0i), (0+0i), (8+0i), (-0-0i)}
Q. What kind of symmetry does the sequence possess? Q. List the MATLAB-generated DFT sequence and identify the 2 Hz component. Q. Examine why the MATLAB-generated phase values are incorrect.
⁄ ⁄ [ ] [ ] []
at 12 Hz starting at B1. An -point sequence is obtained by sampling the signal (satisfies the sampling theorem). Demonstrate the picket fence effect by calculating the DTFT and from it the -point DFT for ranging from 6 to 16. Compare the results. (1) 6- point DFT. The statement “n = 0: 1: 5; xn = cos (pi*n/3),” produces the sequence . The average value of the sequence (the dc component) is zero. The samples are spaced (1/12)-second apart and the record length is 6(1/12) = 0.5 sec. The DTFT, , of the 6-point sequence is a continuous function of :
The numerator and denominator coefficients are = and . The DTFT is computed using the function freqz and plotted below (Xw). Xw2 represents the 6 samples (of the DTFT) which make up the 6-point DFT ( , spaced 12/6 = 2 Hz apart). In the two sub-plots the horizontal axes are calibrated both in Hz and in the frequency index, . % DTFT of ={1, 1/2, -1/2, -1, -1/2, 1/2} Fs= 12; N= 6; F0= Fs/N; w0= 2*pi/N; %Sampling freq; No. of points; Resolution b=[1, 1/2, -1/2, -1, -1/2, 1/2]; %Numerator coefficients a=[1]; %Denominator coefficients
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w=0: pi/256: 2*pi; k=w/w0; F=k*F0; Xw=freqz(b, a, w); %DTFT w2=0: 2*pi/N: 2*pi; k2=w2/w0; F2=k2*F0; Xw2=freqz(b, a, w2); %6 points subplot(2, 1, 1), stem(F2, abs(Xw2), 'k'); hold on plot(F, abs(Xw)); legend ('Samples of DTFT', 'DTFT', 'Location', 'North' ); xlabel('F, Hz'), ylabel('|X(k)|'); grid; title('Magnitude'); hold off subplot(2, 1, 2), plot(k, angle(Xw)); title('Phase'); xlabel('k'), ylabel('
⁄ ⁄ ⁄ ⁄ ⁄ ] [ [ ] [] Q. What kind of symmetry does the sequence possess? Q. List the DFT sequence and identify the 2 Hz component.
(2) 8- point DFT. The statement “n = 0: 1: 7; xn = cos (pi*n/3),” produces the sequence . Unlike in the 6-point DFT the average value of the sequence (the dc component) in this case is not zero. The samples are spaced (1/12)-second apart and the record length is 8(1/12) = 2/3 sec. The DTFT, , of the 8-point sequence is a continuous function of :
The numerator and denominator coefficients are = and . The DTFT is computed using the function freqz and plotted below (Xw). Xw2 represents the 8 samples (of the DTFT) which make up the 8-point DFT ( , spaced 12/8 = 1.5 Hz apart). In the two sub-plots the horizontal axes are calibrated both in Hz and in the frequency index, . % DTFT of ={1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2} Fs= 12; N= 8; F0= Fs/N; w0= 2*pi/N; %Sampling freq; No. of points; Resolution b=[1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2]; %Numerator coefficients a=[1]; %Denominator coefficients w=0: pi/256: 2*pi; k=w/w0; F=k*F0; Xw=freqz(b, a, w); %DTFT w2=0: 2*pi/N: 2*pi; k2=w2/w0; F2=k2*F0; Xw2=freqz(b, a, w2); %6 points subplot(2, 1, 1), stem(F2, abs(Xw2), 'k'); hold on plot(F, abs(Xw)); legend ('Samples of DTFT', 'DTFT', 'Location', 'North' ); xlabel('F, Hz'), ylabel('|X(k)|'); grid; title('Magnitude'); hold off subplot(2, 1, 2), plot(k, angle(Xw)); title('Phase'); xlabel('k'), ylabel('
Q. What kind of symmetry does the sequence possess? Q. List the DFT sequence and identify the 2 Hz component.
(3) 12- point DFT. The statement “n = 0: 1: 11; x = cos (pi*n/3),” produces the sequence . The average value of the sequence (the dc component) in this case is zero. The samples are spaced (1/12)-second apart and the record length is 12(1/12) = 1 sec. The DTFT, , of the 12-point sequence is a continuous function of :
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The numerator and denominator coefficients are = and . The DTFT is computed using the function freqz and plotted below (Xw). Xw2 represents the 12 samples (of the DTFT) which make up the 12-point DFT ( , spaced 12/12 = 1 Hz apart). In the two sub-plots the horizontal axes are calibrated both in Hz and in the frequency index, .
% DTFT of ={1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2, -1/2, -1, -1/2, 1/2} Fs= 12; N= 12; F0= Fs/N; w0= 2*pi/N; %Sampling freq; No. of points; Resolution b=[1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2, -1/2, -1, -1/2, 1/2]; %Numerator coefficients a=[1]; %Denominator coefficients w=0: pi/256: 2*pi; k=w/w0; F=k*F0; Xw=freqz(b, a, w); %DTFT w2=0: 2*pi/N: 2*pi; k2=w2/w0; F2=k2*F0; Xw2=freqz(b, a, w2); %6 points subplot(2, 1, 1), stem(F2, abs(Xw2), 'k'); hold on plot(F, abs(Xw)); legend ('Samples of DTFT', 'DTFT', 'Location', 'North' ); xlabel('F, Hz'), ylabel('|X(k)|'); grid; title('Magnitude'); hold off subplot(2, 1, 2), plot(k, angle(Xw)); title('Phase'); xlabel('k'), ylabel('
] [ [ ] [] Q. What kind of symmetry does the sequence possess? Q. List the DFT sequence and identify the 2 Hz component.
(4) 16- point DFT. The statement “n = 0: 1: 15; x = cos (pi*n/3),” produces the sequence . The average value of the sequence (the dc component) in this case is zero. The samples are spaced (1/12)-second apart and the record length is 16(1/12) = 1.333 sec. The DTFT, , of the 16-point sequence is a continuous function of :
The numerator and denominator coefficients are = and . The DTFT is computed using the function freqz and plotted below (Xw). Xw2 represents the 16 samples (of the DTFT) which make up the 16-point DFT ( , spaced 12/16 = 3/4 Hz apart). In the two sub-plots the horizontal axes are calibrated both in Hz and in the frequency index, . % DTFT of ={1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2, -1/2, -1} Fs= 12; N= 16; F0= Fs/N; w0= 2*pi/N; %Sampling freq; No. of points; Resolution b=[1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2, -1/2, -1]; %Numerator coefficients a=[1]; %Denominator coefficients w=0: pi/256: 2*pi; k=w/w0; F=k*F0; Xw=freqz(b, a, w); %DTFT w2=0: 2*pi/N: 2*pi; k2=w2/w0; F2=k2*F0; Xw2=freqz(b, a, w2); %6 points subplot(2, 1, 1), stem(F2, abs(Xw2), 'k'); hold on plot(F, abs(Xw)); legend ('Samples of DTFT', 'DTFT', 'Location', 'North' ); xlabel('F, Hz'), ylabel('|X(k)|'); grid; title('Magnitude'); hold off subplot(2, 1, 2), plot(k, angle(Xw)); title('Phase'); xlabel('k'), ylabel('
Q. What kind of symmetry does the sequence DSP Lab
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⁄ ⁄ ⁄⁄ ⁄ Q. List the DFT sequence
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and identify the 2 Hz component.
is sampled at 12 Hz (satisfies the sampling theorem). Using the B2. The signal MATLAB function f ft calculate the -point DFT for = 6, 8 and 12. Compare the results and illustrate the picket fence effect.
(1) 6- point DFT. The statement “n = 0: 1: 5; xn = cos (pi*n/3),” produces the sequence . The average value of the sequence (the dc component) is zero. The frequency resolution is = 12/6 = 2 Hz. The DFT is which does show a component at 2 Hz.
N=6; Fs=12; F0=Fs/N; %N points sampled at Fs Hz & Resolution xn = [1, 1/2, -1/2, -1, -1/2, 1/2], %List the sequence values explicitly n = 0: 5; %Plot of sequence subplot(3,1,1), stem(n, xn); xlabel('n'), ylabel('x(n)'); title ('6-point sequence x(n)'); % Xk = fft(xn), MXk = abs(Xk), AXk = angle(Xk), % Magnitude & Phase of X(k) k = 0:5; F=k*F0; subplot(3,1,2), stem(F, abs(Xk)); axis([0 12 0 4]) xlabel('Frequency, Hz (= k*F0)'), ylabel('|X(k)|'); title ('Magnitude of X(k)'); subplot(3,1,3), stem(k, angle(Xk)); axis([0 6 -1 1]) xlabel('k'), ylabel('
Q. What kind of symmetry does the sequence possess about its midpoint ( )? Q. List the MATLAB-generated DFT sequence and identify the 2 Hz component.
⁄ ⁄ ⁄ ⁄ ⁄
⁄ ⁄
(2) 8- point DFT. The statement “n = 0: 1: 7; xn = cos (pi*n/3),” produces the sequence . Unlike in the 6-point case the average value of the sequence (the dc component) in this case is not zero. The frequency resolution is Hz. The DFT is {1.5, (2.5607 + j2.5607), -j1.5, (0.4393 - j0.4393), 0.5, (0.4393 + j0.4393), j1.5, (2.5607 - j2.5607)}. The frequency resolution being 1.5 Hz and since the DFT spectral components only occur at integer multiples of 1.5 Hz it is not possible to know that there is a spectral component at 2 Hz. This is akin to looking at a scene through a picket fence. You only see what is visible through the gaps between the pickets, the rest being blocked from view by the pickets. It is as though the 2 Hz component is “blocked from view” by a picket. N=8; Fs=12; F0=Fs/N; %N points sampled at Fs Hz & Resolution xn = [1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2], %List sequence v alues explicitly n = 0:7; %Plot of sequence subplot(3,1,1), stem(n, xn); xlabel('n'), ylabel('x(n)'); title ('8-point sequence x(n)'); % Xk = fft(xn), MXk = abs(Xk), AXk = angle(Xk), % Magnitude & Phase of X(k) DSP Lab
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k = 0:7; F=k*F0; subplot(3,1,2), stem(F, abs(Xk)); axis([0 12 0 4]) xlabel('Frequency, Hz (= k*F0)'), ylabel('|X(k)|'); title ('Magnitude of X(k)'); subplot(3,1,3), stem(k, angle(Xk)); axis([0 8 -2 2]) xlabel('k'), ylabel('
⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ Q. What kind of symmetry does the sequence possess about its midpoint ( )? Q. Identify the 2 Hz component. What is your best estimate of the frequency cont ent of Q. How do you explain the non-zero ?
?
(3) 12- point DFT. The statement “n = 0: 1: 11; xn = cos (pi*n/3),” produces the sequence . Like in the 6-point case the average value of the sequence (the dc component) in this case is zero. The frequency resolution is Hz. The DFT is {0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 6, 0} which does show a component at 2 Hz. N=12; Fs=12; F0=Fs/N; %N points sampled at Fs Hz & Resolution xn = [1, 1/2, -1/2, -1, -1/2, 1/2, 1, 1/2, -1/2, -1, -1/2, 1/2], %List explicitly n = 0:11; %Plot of sequence subplot(3,1,1), stem(n, xn); xlabel('n'), ylabel('x(n)'); title ('12-point sequence x(n)'); % Xk = fft(xn), MXk = abs(Xk), AXk = angle(Xk), % Magnitude & Phase of X(k) k = 0:11; F=k*F0; subplot(3,1,2), stem(F, abs(Xk)); axis([0 12 0 14]) xlabel('Frequency, Hz (= k*F0)'), ylabel('|X(k)|'); title ('Magnitude of X(k)'); subplot(3,1,3), stem(k, angle(Xk)); axis([0 12 -12 12]) xlabel('k'), ylabel('
Q. What kind of symmetry does the sequence possess about its midpoint ( )? Q. List the MATLAB-generated DFT sequence and identify the 2 Hz component. Q. How do you explain that compared to case of ?
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Lab-41 Frequency response and time-domain simulation of IIR-LP filter
() (()) ⁄ || || || | || || || | || Objective To examine and simulate the IIR filter sampling rate of 150 Hz.
Background The impulse response of the system
transfer function is
which operates at a
is
. The frequency response is
.
It is useful to find the upper and lower bounds for the gain at dc and at parameter . The magnitude and phase are:
The dc gain,
For have
For
at
we have 1
. The
as a function of the
, is
. Similarly, at
, the high end of the frequency scale, we
we have 1 . In calibrating the horizontal axis in terms of the cyclic frequency, , we use the relation from which the point corresponds to .
To Do A. Compute the frequency response b y hand at a few critical points and plot it. Compare with MATLAB. Calibrate the frequency axis in both digital and analog units ( rad/sample and Hz).
as the sum of two cosinusoids of frequencies 2 Hz and 75 Hz and appropriate B. Construct a signal relative amplitudes. Generate time domain plots of and and comment on the gain (or attenuation) of each frequency component. C. In B, at the low frequency end we use either dc or something very close to it. At the high end, from the viewpoint of a successful demo, we choose 75 Hz (the highest possible) which, since it has just 2 samples per cycle, keeps the diagram least cluttered. Further, we use cosines rather than sines. Repeat with the 75 Hz cosine replaced with a 75 Hz sine. Comment on the results. D. Repeat A if the sampling frequenc y were 3 kHz (instead of 150 Hz). In what way does this affect the plots you obtained above?
E. Indicate what kind of filter is represented by comparison with that in A. [Cascade, gain is squared, band width is reduced.]
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|| √ | | | || ⁄ | || ( )
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A1. Hand calculation For
we have
transfer function is
with
. The
. The frequency response is
magnitude and phase are
. The
and
.
The value of
have
at is and . Similarly, at and . In calibrating the horizontal axis in terms of the cyclic frequency, , we use the relation from which the point corresponds to .
we
Q. What is the gain at 2 Hz? At 75 Hz?
A2. Plots of and may be generated by the MATLAB function freqz . The frequency range of interest is 0 to rad., corresponding to 0 to 150 Hz. The MATLAB format is
Here the vectors and specify, respectively, the numerator and denominat or coefficients. In our example , , and . In the MATLAB segment below we have used ‘w’ for and . Both and the phase, , are in radians. %Frequency response Fs=150; b = [1]; a = [1, -0.5]; %Numerator & denominator coefficients w = -pi: pi/256: pi; F = Fs*w/(2*pi); % Digital & analo g frequency [Hw] = freqz(b, a, w) ; subplot(2, 1, 1), plot(F, abs(Hw)); xlabel('Frequency, Hz'), ylabel('Magnitude of H(\omega)'); grid subplot(2, 1, 2), plot(F, angle(Hw)); xlabel('Frequency, Hz'), ylabel('Phase of H(\omega)'); grid
2 ) ( H f 1.5 o e d u t i n 1 g a M
0.5 -80
X: 17.29 Y: 1.413
-60
-40
-20
0 20 Frequency, Hz
40
60
80
-60
-40
-20
0 20 Frequency, Hz
40
60
80
1 ) 0.5 ( H f o 0 e s a h P -0.5
-1 -80
Q. Indicate what frequencies you would consider as low (or high or medium) frequencies. Does this system approximate a low pass or band pass etc. filter? Q. What is the (3 dB) bandwidth? Q. What is the gain at 2 Hz? At 75 Hz? DSP Lab
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Q. Indicate if the phase is linear or nonlinear. Q. Test and indicate in what way the system
is different.
B1. Let the input consist of a low frequency and a high frequency component: and . Plot , and over a quarter cycle of the lower frequency.
% Continuous-time plots of x(t) A=5; F1=2; B=1; F2=75; t = 0: 0.001: 0.2; x1 = A*cos (2*pi*F1*t); x2 = B*cos (2*pi*F2*t); %2 Hz and 75 Hz xt = x1 +x2; % 2 Hz + 75 Hz % Choose a multiplot so that all waveforms are on the same scale plot(t, x1, 'b', t, x2, 'r', t, xt, 'k'); title('Continuous-time plots'); legend ('2 Hz', '75 Hz', '2 Hz + 75 Hz'); xlabel ('Time, t, seconds'), ylabel('x1, x2 and x(t)'); Continuous-time plots 6 2 Hz 75 Hz 2 Hz + 75 Hz
4
2 ) t ( x d n a 2 x , 1 x
0
-2
-4
-6
0
0.02
0.04
0.06
0.08 0.1 0.12 Time, t, seconds
0.14
0.16
0.18
0.2
B2. In the following MATLAB segment we show the input, both continuous and discrete time, .
% Plots of x(t) and x(n) A=5; F1=2; B=1; F2=75; Fs=150; t= 0: 0.001: 0.4; % 0.4 sec interval so continuous & discrete time scales align xt = A*cos (2*pi*F1*t) + B*cos (2*pi*F2*t); % 2 Hz + 75 Hz % n = 0: 1: 60; % Choose n = 60 so continuous & discrete time scales align x = A*cos(2*pi*F1*n/Fs) + B*cos(2*pi*F2*n/Fs); % 2 Hz + 75 Hz subplot(2, 1, 1), plot(t, xt, 'b'); legend ('2 Hz + 75 Hz'); title ('Continuous-time signal x(t)');
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xlabel ('Time, t, seconds'), ylabel('x(t)'); subplot(2, 1, 2), stem(n, x, 'b*'); legend ('(1/75) rad + (1/2) rad'); title ('Discrete-time signal x(n)'); xlabel ('n'), ylabel('x(n)'); Continuous-time signal x (t) 10 2 Hz + 75 Hz 5 ) t ( x
0 -5 -10 0
0.05
0.1
0.15 0.2 0.25 Time, t , s econds Discrete-time signal x(n)
0.3
0.35
0.4
10 (1/75) rad + (1/2) rad 5 ) n ( x
0 -5 -10 0
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B3. In the following MATLAB segment we show the operation of the filter. Both sequences are shown. The low frequency component (2 Hz) is amplified and the high frequency component (75 Hz) is attenuated.
% Plots of input x(n) and low-pass filter output y(n) A=5; B=1; F2=75; Fs=150; b = [1]; a = [1, -0.5]; %Numerator and denominator coefficients n = 0: 1: 60; x = A*cos(2*pi*F1*n/Fs) + B*cos(2*pi*F2*n/Fs); % dc + 75 Hz y = filter(b, a, x) ; subplot(2, 1, 1), stem(n, x, 'b*'); axis([0, 60, -20, 10]); legend ('2 Hz + 75 Hz '); title ('Discrete-time signal x(n)'); xlabel ('n'), ylabel('x(n)'); subplot(2, 1, 2), stem(n, y, 'ko'); axis([0, 60, -20, 10]); legend ('Low pass filtered'); title ('Low pass filter output y(n)'); xlabel ('n'), ylabel('y(n)'); Discrete-time signal x(n) 10 2 Hz + 75 Hz ) n ( x
0
-10
-20
0
10
20
30 40 n Low pass filter output y(n)
50
60
10 Low pass filtered 0 ) n ( y
-10
-20
0
10
20
30 n
40
50
60
Q. On an approximate basis verify if the output amplitudes of the 2 and 75 Hz components agree with the gain figures obtained in A.
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. B4. In B2 replace the 2 Hz cosine with dc (the lowest frequenc y). Plot both sequences Use the data cursor and verify that the dc is amplified and the high frequency component (75 Hz) is attenuated. % Plots of input x(n) and low-pass filter output y(n) A=1; B=1; F2=75; Fs=150; b = [1]; a = [1, -0.5]; %Numerator and denominator coefficients n = 0: 1: 60; x = A + B*cos(2*pi*F2*n/Fs); % dc + 75 Hz y = filter(b, a, x); plot(n, x, 'b*', n, y, 'ko'); legend ('x(n)', 'y(n)'); xlabel ('n'), ylabel('x(n) and y(n)');
3 x(n) y(n) 2.5
2 ) n ( y d n a ) n ( x
1.5
1
0.5
0
0
10
20
30 n
40
50
60
Q. Identify the input and output by connecting the dots. On an approximate basis verify if the output amplitudes of the dc and 75 Hz components agree with the gain figures obtained in A. Ans. Eyeballing the above diagram, the dc level of the input (blue) is 1 while that of the output (black) is about 2 which agrees with a dc gain of 2 calculated earlier. In a similar fashion the amplitude of the 75 Hz at the input (blue) is 1 while that of the output (black) is about 0.7 which agrees with calculated earlier. Q. Try the demo with sines instead of cosines. Wh y does the demo fail if we use sines instead of cosines?
||
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C1. Repeat B4 with the cosine replaced with a sine.
% Plots of input x(n) and low-pass filter output y(n) A=1; B=1; F2=75; Fs=150; b = [1]; a = [1, -0.5]; %Numerator and denominator coefficients n = 0: 1: 60; x = A + B*sin(2*pi*F2*n/Fs); % dc + 75 Hz y = filter(b, a, x); plot(n, x, 'b*', n, y, 'ko'); legend ('x(n)', 'y(n)'); xlabel ('n'), ylabel('x(n) and y(n)');
2.2 x(n) y(n)
2
1.8 ) n ( y d n a ) n ( x
1.6
1.4
1.2
1
0.8
0
10
20
30 n
40
50
60
Q. Identify the input and output by connecting the dots. Q. Why does the demo fail if we use a sine instead of a cosine (of 75 Hz)? Reconcile with the sampling theorem. Q. On an approximate basis verify if the output amplitude of the dc component agrees with the gain figures obtained in A. D1. Repeat A2 with sampling frequency equal to 3000 Hz.
Q. In what way does changing the sampling frequency from 150 Hz to 3000 Hz affect the magnitude and phase versus frequency plots? Would the vertical axis have to be relabeled? Q. What is the effect of changing the sampling frequency from 150 Hz to 3000 Hz on the time-domain demonstration? Q. Repeat for a sampling frequency of 8000 Hz.
E1. The frequency response is
At dc:
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. The magnitude and phase are
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Lab-43 Frequency response of analog Butterworth low-pass filter Objective Analog Butterworth filter transfer functions are readily available and are the starting point for some IIR digital filter design. In this Lab you will learn low -pass (LP) filter design through the use of MATLAB functions such as butter(N,Wn,'s') , where N is the filter order, Wn is the cut-off frequency in rad/second, and 's' signifies analog design.
Background The transfer function and frequency response of the RC low pass filter below are given by
and
, respectively. The dc gain (at
maximum gain. The 3dB cut-off frequency or half-power point is
+
Input =
rad/sec.
) is 1 and is the
| | +
Output =
–
–
1
0.707
0
, rad/sec
Nomenclature Further terminology of filter transfer functions is illustrated in figure below using a low pass filter characteristic. Here and are the pass-band edge frequency and the corresponding gain and and are the stop-band edge frequency and the corresponding gain.
|| Pass band
Transition band
Stop band
||
rad/sec
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, we would have Normalized filter Adjusting the values of and so that . This is a first order normalized low-pass Butterworth filter with a cut-off frequency rad/sec and a “dc” gain of 1 at . It serves as a prototype in further design.
Frequency transformation and magnitude scaling The normalized LP filter could be transformed to a different LP filter with an arbitrary cut-off frequency of rad/sec by the low pass to low pass transformation . Further, in addition to scaling on the frequency axis, the gain of this filter could be scaled by an arbitrary multiplier, . For example, if we desire a cut-off frequency of, say, rad/sec, and a dc gain of, say, , the required filter is given by
The cut-off frequency, rad/sec, is where the magnitude drops to to . The magnitude at other frequencies will be scaled proportionately.
of the maximum, that is,
To Do A. [First order, normalized] Frequency response of the first order normalized low-pass Butterworth filter. A1. Frequency response A2. Frequency response: frequency axis calibrated in Hz A3. Frequency response: the plot function instead of fplot
B. [Low-pass to low-pass transformation, band-width 500 Hz] The first order low-pass Butterworth filter with a cut-off frequency of 500 Hz. B1. Frequency response: writing explicitly B2. Frequency response: the freqs function plus the filter coefficient list B3. Frequency response: the freqs function plus the coefficients via the butter function B4. Frequency response: the freqs function + the butter function; dB magnitude
C. [Fifth order, normalized] Design of normalized fifth order LP analog Butterworth filter. C1. Fifth order: the butter function D. [Fifth order, band-width 500 Hz] Design of fifth order LP analog Butterworth filter with a cut-off frequency of 500 Hz. D1. Fifth order: the butter function
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⁄ ⁄ ⁄
A. [First order, normalized] Plot the frequency response of the first order normalized low-pass Butterworth filter . A3. [Frequency response: the plot function instead of fplot ] Use the function plot to obtain the frequency response of the first order normalized filter with the frequency axis calibrated in Hz. . The corresponding MATLAB plots are Solution The frequency response is shown below using the function plot . The step size along the frequency axis can be controlled more directly than in fplot (by changing N in F = -L: L/N: L). Omega (or F) is a vector , consequently we use “./” instead of “/” etc.
%L=6*pi; N=512; Omega = -L: pi/N: L; H = 1./(1.+ j .*Omega); F=Omega/(2*pi); L= 3; N=512; F = -L: L/N: L; H = 1./(1.+ (j*2*pi) .*F); %Omega=2*pi*F subplot(2, 1, 1), plot(F, abs(H), 'k'); xlabel ('Frequency, Hz'), ylabel('|H(F)|'); grid; title ('Magnitude') subplot(2, 1, 2), plot(F, angle(H), 'k'); xlabel ('Frequency, Hz'), ylabel('Phase of H(F)'); grid; title ('Phase') Magnitude 1
X: 0.1582 Y: 0.7092
| ) F ( 0.5 H |
0 -3
-2
-1
0 Frequency, Hz Phase
1
2
3
-2
-1
0 Frequency, Hz
1
2
3
2 ) 1 F ( H f o 0 e s a h P -1
-2 -3
Q. Use help plot and explain how its arguments are used. Q. At how many points along the frequency axis does MATLAB compute the frequency response function? Use the data cursor to verify. Are these points uniformly spaced? Q. Does plot give any better control (than fplot ) over the resolution along the frequency axis? Q. With the plot function is there some way by which you can place the data cursor exactly at i.e., ? (Can you control the horizontal spacing of the data points?
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⁄
B. [Low-pass to low-pass transformation] Transform the first order normalized low-pass Butterworth filter , to obtain a first order low-pass Butterworth filter with a cut-off frequenc y of 500 Hz.
⁄
B2. [Frequency response: the freqs function plus the filter coefficient list] Solution In prior work the frequency response was generated by spelling out the function in full in the MATLAB program. This can be tedious as the order of the filter increases and the response function gets quite involved. In this example we avoid writing out the frequency response function by using the freqs function along with a list of the filter coefficients. From prior work the first order filter with a cut-off frequency of 500 Hz is given by
The numerator and denominator coefficients are and are used as such by the MATLAB program below.
and
Fc= 500; L= 2*pi*3*Fc; % Cut-off freq. in Hz & plotting limit b=[0, 2*pi*500]; a=[1, 2*pi*500]; %Coefficients % Omega = -L: L/512: L; F= Omega/(2*pi); % Analog frequency range H = freqs(b, a, Omega) ; subplot(2, 1, 1), plot(F, abs(H), 'k'); xlabel ('Frequency, Hz'), ylabel('|H(F)|'); st grid; legend ('Magnitude', 'Location', 'NorthWest'); title ('1 order Butterworth LP'); subplot(2, 1, 2), plot(F, angle(H), 'k'); xlabel ('Frequenc y, Hz'), ylabel('Phase of H(F)'); st grid; legend ('Phase'); title ('1 order Butterworth LP');
Q. Use help fr eqs and explain how its arguments are used. Q. Determine the dc gain and half-power bandwidth (in rad/sec and Hz) from the magnitude plot. Q. Write an expression for so that it has a dc gain of 4 and redo the plots. Q. What is the effect of magnitude scaling (changing dc gain from 1 to 4) on the phase plot?
B3. [Frequency response: the freqs function plus the coefficients via the butter function] function along with a list Solution In prior work we generated the frequency response by using the freqs of the filter coefficients. On the one hand the syntax for specifying the list of coefficients is simpler than that of spelling out the frequency response function itself, but, on the other hand, the coefficients still have to be derived by hand. In this example the function butter generates the coefficients which are then used by freqs .
Fc= 500; L= 2*pi*3*Fc; % Cut-off freq. in Hz & plotting limit %b=[0, 2*pi*500]; a=[1, 2*pi*500]; %Coefficients [b, a] = butter(1, 2*pi*Fc, 's') % Design filter of order 1 and cut-off = 2πFc rad/sec. % Omega = -L: L/256: L; F= Omega/(2*pi); % Analog frequency range H = freqs(b, a, Omega) ; subplot(2, 1, 1), plot(F, abs(H), 'k'); xlabel ('Frequency, Hz'), ylabel('|H(F)|'); st grid; legend ('Magnitude', 'Location', 'NorthWest'); title ('1 order Butterworth LP'); subplot(2, 1, 2), plot(F, angle(H), 'k'); xlabel ('Frequenc y, Hz'), ylabel('Phase of H(F)'); st grid; legend ('Phase'); title ('1 order Butterworth LP'); DSP Lab
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The program generates coefficients (same as in prior work): b = [0
3141.6]; a = [1
3141.6].
1st order Butterworth LP 1 Magnitude
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0.8 | ) F ( 0.6 H |
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-1000
-500
0 500 Frequency, Hz 1st order Butterworth LP
1000
1500
2 Phase ) 1 F ( H f o 0 e s a h P -1
-2 -1500
-1000
-500
0 Frequency, Hz
500
1000
1500
Q. Use help butter and explain how its arguments are used. Q. Determine the dc gain and half-power bandwidth (in rad/sec and Hz) from the magnitude plot. Q. Generate the coefficients and write down the transfer function for a filter of order 1, dc gain 4 and half-power point of 500 Hz. nd Q. Generate the coefficients and write down the transfer function of a normalized 2 order filter. Q. Generate the coefficients and write down the transfer function of a filter of order 3 and cut-off frequency 5 rad/sec.
B4. [Frequency response: the freqs function + the butter function; dB magnitude] Plot the absolute magnitude, decibel magnitude and phase versus frequency. Solution This is a repeat of prior work with the dB magnitude as an important alternative to the absolute magnitude. Here again the function butter supplies the coefficients to freqs which generates the frequency response.
Fc= 500; L= 2*pi*3*Fc; % Cut-off freq. in Hz & plotting limit [b, a] = butter(1, 2*pi*Fc, 's') % Design filter of order 1 and cut-off = 2πFc rad/sec. %b=[0, 2*pi*500]; a=[1, 2*pi*500]; %Coefficients Omega = -L: L/512: L; F= Omega/(2*pi); % Analog frequency range H = freqs(b, a, Omega) ; M=abs(H); dBM=20*log10(M) ; Phase= angle(H) ; subplot(3, 1, 1), plot(F, M, 'k'); xlabel ('Frequency, Hz'), ylabel('|H(F)|'); st grid; legend ('Abs. Magnitude'); legend boxoff; title ('1 order Butterworth LP'); subplot(3, 1, 2), plot(F, dBM, 'k'); xlabel ('Frequency, Hz'), ylabel('20 log |H(F)|'); st grid; legend ('dB Magnitude'); legend boxoff; title ('1 order Butterworth LP'); subplot(3, 1, 3), plot(F, Phase, 'k') ; xlabel ('Frequency, Hz'), ylabel('Phase of H(F)'); DSP Lab
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grid; legend ('Phase'); legend boxoff; title ('1 order Butterworth LP'); 1st order Butterworth LP 1 Abs. Magnitude
| ) F ( 0.5 H |
0 -1500
| ) 0 F ( H | g -5 o l 0 2 -10
-1500
) 2 F ( H f o 0 e s a h P -2
-1500
X: 501 Y: 0.7064
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0 500 Frequency, Hz 1st order Butterworth LP
1000
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dB Magnitude
-1000
-500
0 500 Frequency, Hz 1st order Butterworth LP
1000
1500
Phase
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0 Frequency, Hz
500
1000
1500
Q. On the dB magnitude plot determine the dc gain and half-power bandwidth (in rad/sec and Hz). Q. Write an expression for so that it has a dc gain of 4 and redo the plots. Q. What is the effect of magnitude scaling (changing dc gain from 1 to 4) on the phase plot? Q. Generate the coefficients and write down the transfer function for a filter of order 2, dc gain 0 dB and cut-off frequency 500 Hz. Q. Generate the coefficients and write down the transfer function of a filter of order 3 and dc gain 10 dB.
C. [Fifth order, normalized] Design a normalized fifth order LP analog Butterworth filter. Give the transfer function and plot the frequency response of the filter. D. [Fifth order] Design a fifth order LP analog Butterworth filter with a cut-off frequency of 500 Hz. Plot the absolute magnitude, decibel magnitude and phase versus frequency. Give the transfer function. D1. [Fifth order: the freqs function + the butter function; dB magnitude] which generates the frequency Solution Here again the function butter supplies the coefficients to freqs response.
format short e % To display coefficients in full! Fc= 500; L= 2*pi*3*Fc; % Cut-off freq. in Hz & plotting limit [b, a] = butter(5, 2*pi*Fc, 's') % Design filter of order 5 and cut-off = 2πFc rad/sec. Omega = -L: L/512: L; F= Omega/(2*pi); % Analog frequency range H = freqs(b, a, Omega) ; M=abs(H); dBM=20*log10(M) ; Phase= angle(H) ; subplot(3, 1, 1), plot(F, M, 'k'); xlabel ('Frequency, Hz'), ylabel('|H(F)|'); grid; DSP Lab
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legend ('Abs. Magnitude', 'Location', 'NorthWest'); title ('5 order Butterworth LP'); subplot(3, 1, 2), plot(F, dBM, 'k'); xlabel ('Frequency, Hz'), ylabel('20 log |H(F)|'); grid; th legend ('dB Magnitude', 'Location', 'NorthWest'); title ('5 order Butterworth LP'); subplot(3, 1, 3), plot(F, Phase, 'k') ; xlabel ('Frequency, Hz'), ylabel('Phase of H(F)'); grid; th legend ('Phase'); title ('5 order Butterworth LP'); b = [0 0 0 0 0 3.0602e+17]; a = [1 1.0166e+4 5.1678e+7 1.6235e+11 3.1522e+14 3.0602e+17] 5th order Butterworth LP 1 Abs. Magnitude
| ) F ( 0.5 H |
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Q. On the dB magnitude plot determine the dc gain and half-power bandwidth (in rad/sec and Hz). Q. Write an expression for so that it has a dc gain of 4 and redo the plots. Q. What is the purpose of for mat short e ? What happens if it is removed?
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Lab-47 Design of Analog Butterworth LP filter
Objective To design an analog Butterworth filter to given specifications,
and
.
Background The standard design procedure for an analog Butterworth filter starts with specifying the coordinates of two points on the Butterworth characteristic, say, and . The ’s are analog frequencies (rad/sec) and the ’s are the corresponding gains in dB. The parameters are shown in the dB magnitude characteristic below.
0
| |
We then determine the order, , and natural (3-dB) frequency, specifications from the following design equations:
, of the filter that satisfies the
The relevant MATLAB functions are shown in the left side column in Table below. The connection to MATLAB symbols (the ’s and the ’s) is as follows: & MATLAB Functions Analog, in rad/sec. Digital, in rad/sample [N, Wn] = buttord(Wp, Ws, Rp, Rs, 's') [N, Wn] = buttord(Wp, Ws, Rp, Rs) Wp, Ws are in rad/sec. Wp, Ws are normalized [b, a] = butter(N, Wn, 's') [b, a] = butter(N, Wn) Wn is in rad/sec. Wn is normalized H = freqs(b, a, W) H = freqz(b, a, W) W is in rad/sec. W is in rad/sample
The pass-band and stop-band edge frequencies are denoted and associated attenuations. The parameter 's' denotes analog design.
with
and
being the
To Do Design a low-pass analog Butterworth filter with (a) – 3.01 dB cut-off frequency of 500Hz, (b) monotonic stop and pass bands, (c) magnitude of frequency response down at least 15 dB at 750 Hz.
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A1. Filter order and natural frequency. Determine the filter order (N) and natural frequency (Wn) of the LP filter such that Pass band edge frequency = Wp = rad/sec, attenuation = Rp = 3.01dB Stop band edge frequency = Ws = rad/sec, attenuation = Rs = 15 dB
| | 0
th
In the following MATLAB segment “[N, Wn] = buttord (.)” computes N = 5, Wn = 3346.9, which is 5 order with a natural frequency of 3346.9 rad/sec. And “[b, a] = butter (.)” gives the filter coefficients (requires “format short e” in order to display the co efficients in full) b = [0 0 0 0 0 4.1994e+17] a = [1 1.0831e+4 5.8651e+7 1.9630e+11 4.0604e+14 4.1994e+17] The rest of the program produces the frequency response plots. (Note that the phase curve has wraparound and looks as though it is periodic, but is not. This may be demonstrated by plotting the frequency response over a wider range, say, of 3 kHz.) format short e % To display coefficients in full! F1=500 ; Wp=2*pi*F1, Rp = 3.01, %Pass band edge spec F2=750; Ws = 2*pi*F2, Rs = 15, % Stop band edge spec [N, Wn] = buttord(Wp, Ws, Rp, Rs, 's') % Determine N and Wn [b, a] = butter(N, Wn, 's') % Determine filter coefficients % Fmax=1000; L=2*pi*Fmax; Omega = -L: L/1024: L; % Analog frequency range H = freqs(b, a, Omega); subplot(3, 1, 1), plot(Omega/(2*pi), abs(H), 'k'); %Absolute value xlabel ('Frequency, Hz'), ylabel('|H(Omega)|'); grid; title ('Absolute Magnitude, Low Pass') subplot(3, 1, 2), plot(Omega/(2*pi), 20*log10(abs(H)), 'k'); %db = 20*log10(abs(H)) xlabel ('Frequency, Hz'), ylabel('20 log10 |H(Omega)|'); grid; title ('dB Magnitude, Low Pass') subplot(3, 1, 3), plot(Omega/(2*pi), angle(H), 'k'); xlabel ('Frequency, Hz'), ylabel('Phase of H(Omega)'); grid; title ('Phase, Low Pass')
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Absolute Magnitude, Low Pass | 1 ) a g e m0.5 O ( H |
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| ) a g e 0 m O ( H | -20 0 1 g o -40 l 0 -1000 2 ) a g e 5 m O ( H 0 f o e s -5 a h -1000 P
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From the dB magnitude plot, using the data cursor it can be seen that the attenuation at 500 Hz is 1.85 dB (which is better than 3.01 dB). Thus the design exceeds the specification at the pass-band edge frequency (and meets the spec exactly at the stop-band edge frequency, that is, the attenuation at 750 Hz is exactly 15 dB).
Q. Using help fr eqs write down the transfer function . The filter coefficients are b = = [0 0 0 0 0 4.1994e+17] a= = [1 1.0831e+4 5.8651e+7 1.9630e+11 4.0604e+14 4.1994e+17]
[[ ]]
The MATLAB notation gives
Q. Comment on the pole-zero plot. In MATLAB the transfer function is specified as a ratio of po lynomials in
The numerator coefficients, specified as two vectors below.
and the denominator coefficients
.
are
%Pole-zero plot b = [0 0 0 0 0 4.1994e+17]; DSP Lab
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a = [1 1.0831e+4 5.8651e+7 1.9630e+11 4.0604e+14 4.1994e+17]; zplane (b, a);
The 5 poles are on the Butterworth circle in the left-half s-plane (the real pole is located at -3348 where the data cursor is). The radius of the circle is the natural frequency (=3348).
A2. Design by hand calculation. The order, cutoff frequency and transfer function of the analog Butterworth filter that satisfies the specification
⌈⌉ √ are obtained as follows:
This agrees with the MATLAB solution. Next, determine the cutoff frequency that corresponds to the given specifications and the order N = 5 determined above. We choose to match the spec exactly at and exceed the spec at , (as does the MATLAB solution) in which case:
or 532.7 Hz. Thus we have the order (= 5) and cut-off frequency (= 3346.9 rad/sec) of the desired filter. To th arrive at its transfer function we start with the normalized 5 order filter whose cut-off frequency we know is 1 rad/sec and whose transfer function we can get from a standard table or construct from the th Butterworth circle. The next step is to apply to the normalized 5 order Butterworth filter the low-pass to low-pass transformation so that it will have a cut-off frequency of 3346.9 rad/sec. th The normalized 5 order filter is
The required filter is thus given by
After a modest amount algebra to simplify the expression for the obtained here is the same as the MATLAB-generated DSP Lab
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, it should be possible to show that in A1. We can skip the algebra but GRIET-ECE
it is fairly straight forward to see that the numerator of the , which is the same as in A1.
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obtained here should be
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Lab-48 Design of Digital Butterworth LP filter
Objective To design a digital Butterworth filter to given analog specifications,
and
.
Background In the standard design procedure for an IIR filter, based on the bilinear transformation, the filter’s specifications may be digital : ( and ), or, where an structure is involved, the specs may be analog : ( and ). In the latter case the specs are first converted to digital using , where is the sampling interval, the ’s are analog frequencies (rad/sec), the ’s are digital frequencies and the ’s are the corresponding gains in dB. The next steps are: 1. The digital specs are converted to analog using pre-warping. The pre-warped specs may be denoted ( and ). 2. Using the pre-warped specs the required analog filter of the proper order and natural frequency is designed. 3. Finally, the analog filter is transformed to digital using bilinear transformation.
In Step 2 we follow the standard design procedure for an analog Butterworth filter using the pre-warped specs ( and ). We determine the order, , the natural (3-dB) frequency, , and the transfer function of the prototype filter using the MATLAB functions shown in the left side column in Table below. This is followed by the bilinear transformation of Step 3. Alternatively, in MATLAB we may use the functions shown in the right side column in Table and thus dispense with Steps 1 through 3. Note that Wp, Ws and Wn are normalized digital frequencies.
MATLAB Functions Analog, in rad/sec. Digital, in rad/sample [N, Wn] = buttord(Wp, Ws, Rp, Rs, 's') [N, Wn] = buttord(Wp, Ws, Rp, Rs) Wp, Ws are in rad/sec. Wp, Ws are normalized [b, a] = butter(N, Wn, 's') [b, a] = butter(N, Wn) Wn is in rad/sec. Wn is normalized H = freqs(b, a, W) H = freqz(b, a, W) W is in rad/sec. W is in rad/sample
To Do A. Design a digital Butterworth filter that when used in an structure gives an equivalent low-pass analog filter with (a) – 3.01 dB cut-off frequency of 500 Hz, (b) monotonic stop and pass bands, (c) magnitude of frequency response down at least 15 dB at 750 Hz. The sampling rate is 2000 samples/sec.
B. How would the design be affected if the digital filter were specified in terms of the digital specification ( and )?
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and A1. Design the digital Butterworth filter. Convert the analog frequencies to digital frequencies and using the relation . Here the sampling rate is . , attenuation = Rp = 3.01 dB , attenuation = Rp = 15 dB
0
| |
Furthermore, in the MATLAB functions “ buttord (Wp, Ws, Rp, Rs)” and “butter(N, Wn)” the Wp, Ws and Wn are normalized digital frequencies. The normalized frequenc y is defined as . Thus the specifications become Pass-band edge frequency = Wp = Stop-band edge frequency = Ws =
, attenuation = Rp = 3.01dB , attenuation = Rs = 15 dB
We next determine the order and natural (3-dB) frequency of the digital Butterworth filter that meets the above specifications. In the following MATLAB segment the statement “[N, Wn] = buttord (Wp, Ws, Rp, Rs)” computes the filter order and natural frequency, respectively, as N = 2 and Wn = 0.5083. The corresponding filter coefficients are generated by “[b, a] = butter (N, Wn)” as b =[0.3005 0.6011 0.3005] a =[1 0.0304 0.1717] The rest of the program produces the frequency response plots which are seen to be periodic. In the MATLAB function “freqz(b, a, w)” the w is just digital (not normalized) frequency. Fs = 2000; F1=500; F2=750; %Sampling rate & analog edge frequencies Wp=2*F1/Fs, Rp = 3.01, %Digital normalized pass-band edge frequency & attenuation Ws =2*F2/Fs, Rs = 15, %Digital normalized stop-band edge frequency & attenuation [N, Wn] = buttord(Wp, Ws, Rp, Rs) %Determine order (N) and natural frequenc y (Wn) [b, a] = butter(N, Wn) %Determine filter coefficients % % Plot frequency response DSP Lab
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w = -3*pi: pi/256: 3*pi; H = freqz(b, a, w); subplot(3, 1, 1), plot(w, abs(H), 'k'); %Absolute value xlabel ('Digital Frequency, \omega, rad/sample'), ylabel('|H(\omega)|'); grid; title ('Absolute Magnitude, Low Pass') subplot(3, 1, 2), plot(w, 20*log10(abs(H)), 'k'); %db = 20*log10 (abs(H)) xlabel ('Digital Frequency, \omega, rad/sample'), ylabel('20 log10 |H(\omega)|'); grid; title ('dB Magnitude, Low Pass') subplot(3, 1, 3), plot(w, angle(H), 'k'); xlabel ('Digital Frequency, \omega, rad/sample'), ylabel('Phase of H(\omega)'); grid; title ('Phase, Low Pass')
Absolute Magnitude, Low Pass 1 | ) 0.5 ( H |
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[[ ]] ( ) ) (
From the dB magnitude plot, using the data cursor it can be seen that the attenuation at rad/sec is 2.791 dB which exceeds the specified 3.01 dB at the pass-band edge frequency. Further, the design meets the spec exactly at the stop-band edge frequency, that is, the attenuation at rad/sec is exactly 15 dB). Q. Using help fr eqz write down the transfer function Answer The filter coefficients are b = = [0.3005 0.6011 0.3005] a= = [1 0.0304 0.1717] According to the MATLAB notation
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and the difference equation.
, so that
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Q. Comment on the pole-zero plot. Answer In MATLAB the transfer function is specified as a ratio of pol ynomials in
The numerator coefficients, specified as two vectors below.
and the denominator coefficients
.
are
%Pole-zero plot b = [0.3005 0.6011 0.3005]; a = [1 0.0304 0.1717]; zplane (b, a); The two poles are inside the unit circle, so the filter is stable.
A2. Repeat the plots of A1 with the frequency axis relabeled in Hz over the range The sampling frequency is Hz. The cyclic frequency is given by
.
Fs = 2000; F1=500; F2=750; %Sampling rate & analog edge frequencies Wp=2*F1/Fs, Rp = 3.01, %Digital normalized pass-band edge frequency & attenuation Ws =2*F2/Fs, Rs = 15, %Digital normalized stop-band edge frequency & attenuation [N, Wn] = buttord(Wp, Ws, Rp, Rs) %Determine order (N) and natural frequenc y (Wn) [b, a] = butter(N, Wn) %Determine filter coefficients % % Plot frequency response w = -1*pi: pi/256: 1*pi; H = freqz(b, a, w); F= w*Fs/(2*pi); subplot(3, 1, 1), plot(F, abs(H), 'k'); %Absolute value xlabel ('Frequency, Hz'), ylabel('|H(\omega)|'); grid; title ('Absolute Magnitude, Low Pass') % subplot(3, 1, 2), plot(F, 20*log10(abs(H)), 'k'); %db = 20*log10(abs(H)) xlabel ('Frequency, Hz'), ylabel('20 log10 |H(\omega)|'); grid; title ('dB Magnitude, Low Pass') % subplot(3, 1, 3), plot(F, angle(H), 'k'); xlabel ('Frequency, Hz'), ylabel('Phase of H(\omega)'); grid; title ('Phase, Low Pass')
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Absolute Magnitude, Low Pass 1 | ) 0.5 ( H |
0 -1000 | ) 0 ( H | 0 -50 1 g o l 0 -100 2
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⌈⌉ √ √ √ √
A3. [Bilinear transformation] Design the digital Butterworth filter by hand calculation. The analog frequencies and have previously been converted to digital frequencies and using the relation , resulting in the digital specification:
These digital specs are pre-warped to analog specs using the relation bilinear transformation. The pre-warping step gives (use
that applies to
)
so that the analog specification to which we design the Butterworth filter is:
Plugging in numerical values, the filter order is
Choosing to match the specs exactly at the stop-band edge frequency (as does MATLAB)
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√ √ √ ) ( | () ( )
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Note that in this case . Another way of saying this is that this design exceeds the specification at the pass-band edge frequency and meets it exactly at the stop-band edge. Therefore, the required pre-warped , normalized (unit bandwidth), analog Butterworth filter of order 2 (using the Butterworth Table or the Butterworth circle) is
Since we need a cut-off frequency of rad/sec, we next use the low-pass to low-pass transformation in order to move the cut-off frequency from 1 to 2.0527 rad/sec.
See Aside-1 in box below for MATLAB’s way of arriving at the above
.
Finally, the digital Butterworth filter is obtained by substituting
above
, with
, in the
:
Q. Where have we seen this before? What is MATLAB’s way of arriving at this
?
**************** Aside-1 The
, with
, arrived at above is done in
MATLAB by the following segment: format short e % To display coefficients in full! Wp= 2, Rp = 3.01, %Pass-band edge specification Ws = 4.828, Rs = 15, % Stop-band edge spec [N, Wn] = buttord(Wp, Ws, Rp, Rs, 's') % Determine N and Wn [b, a] = butter(N, Wn, 's') % Determine filter coefficients which generates the result: (Wp = 2, Rp = 3.01; Ws = 4.828, Rs = 15) N = 2, Wn = 2.0524 – Order & natural frequency b = [0 0 4.2122], a = [1 2.9025 4.2122] – Filter coefficients This MATLAB output is in agreement with the ha nd-calculated to within 2 or 3 decimal places.
**************** Aside-2 The
arrived at above is obtained in a more merciful
fashion by the following MATLAB segment: DSP Lab
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Fs = 2000; F1=500; F2=750; %Sampling rate & analog edge frequencies Wp=2*F1/Fs, Rp = 3.01, %Digital normalized pass-band edge frequency & attenuation Ws =2*F2/Fs, Rs = 15, %Digital normalized stop-band edge frequency & attenuation [N, Wn] = buttord(Wp, Ws, Rp, Rs) %Determine order (N) and natural frequenc y (Wn) [b, a] = butter(N, Wn) %Determine filter coefficients which generates the result: (Wp = 0.5, Rp = 3.01; Ws = 0.75, Rs = 15) – Normalized digital frequencies N = 2, Wn = 0.50826 – Order & (normalized) natural frequency b = [0.30053 0.60106 0.30053], a = [1 0.030385 0.17174] – Filter coefficients This MATLAB output is in agreement with the partial ha nd-calculated above. ****************
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Lab-51 Frequency response and time-domain simulation of FIR-LP filter
( ) || || Objective Examine and simulate the FIR filter
which computes a moving
average (a smoothing operation) at a sampling frequency of 150 Hz. Background The frequency response is
so that
The zero crossings of the magnitude function occur where , or . In other words, a frequency of rad./sample ( cycle/sample) is totally stopped (filtered out) by the filter. The corresponding digital signal is . The underlying continuous-time signal,
use the relation that the point
(or
frequency is 16 Hz, then
, depends on the sampling frequency. We
) corresponds to
. If, for example, the sampling
, and a frequency of 16/3 Hz will be totally filtered out.
If the sampling frequency is 150 Hz, then
, and a frequency of 50 Hz will be
eliminated. In calibrating the horizontal axis in terms of the cyclic frequency, , we use the relation . As goes from 0 to , goes from 0 to 1 and goes from 0 to
.
To Do A. Plot the frequency response by hand and compare with MATLAB. Calibrate the frequency axis in digital units ( rad/sample) as well as analog units (Hz). B. Find the one frequency that is totally filtered out. Demonstrate with time domain plots.
C. Evaluate by hand the output signal for where the input is . Take . Explain the result in terms of the frequenc y response computed in A.
D. Construct a signal as the sum of two cosinusoids of frequencies 2 Hz and 75 Hz and appropriate relative amplitudes. Generate time domain plots of and and comment on the gain (or attenuation) of each frequency component.
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A1. Hand calculation.
|| ( )
A2. Plots of and may be generated by the MATLAB function freqz . The frequency range of interest is 0 to rad., corresponding to 0 to 150 Hz. The MATLAB format is
The vectors and are the numerator and denominator coefficients, respectively. In our example , and . In the MATLAB segment below we have used ‘w’ for and . Both and the phase, , are in radians. %Frequency response b = [1/3, 1/3, 1/3]; a = [1]; % Numerator & denominator coefficients Fs=150; w = -pi: pi/256: pi; F = Fs*w/(2*pi); % Digital & analog frequency [Hw] = freqz(b, a, w) ; subplot(2, 1, 1), plot(F, abs(Hw)); xlabel('Frequency, Hz'), ylabel('Magnitude of H(\omega)'); grid subplot(2, 1, 2), plot(F, angle(Hw)); xlabel('Frequency, Hz'), ylabel('Phase of H(\omega)'); grid
1 ) ( H f o e d 0.5 u t i n g a M
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Q. Indicate what frequencies you would consider as low (or high or medium) frequencies. Does this system approximate a low pass or band pass etc. filter? What is the (3 dB) bandwidth? Q. Indicate if the phase is linear or nonlinear. Explain the discontinuity in the plot. Q. How will it affect the plots you obtained above if the sampling frequency were 3 kHz (instead of 150 Hz)? Would the vertical axis have to be relabeled? On the digital frequency scale, the zero crossing of the magnitude plot stays the same at . If the sampling frequency is 3000 Hz, then
corresponds to
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B1. If the sampling frequency is 150 Hz, then a frequency of 50 Hz will be eliminated.
B2. Consider a signal sampled at 150 Hz. The corresponding is the input to the moving average filter. The following MATLAB segment plots .
and
% Continuous-time plot of x(t) % Choose 0.2 sec interval so that continuous and discrete time scales align Fs=150; F2=50; %Sampling frequency & 50 Hz t = 0: 1/500: 0.2; xt = cos (2*pi*F2*t); %50 Hz subplot(3, 1, 1), plot(t, xt, 'b'); xlabel ('Time, t, seco nds'), ylabel('x(t)'); % Discrete-time plots. % Choose 30 samples so that continuous and discrete time scales align n=0:1:30; x=cos(2*pi*F2*n/Fs); b=[1/3, 1/3, 1/3]; a=[1]; %Filter coefficients y=filter(b, a, x); subplot(3, 1, 2), stem(n, x, 'b*'); title('Input x(n)=Cosine(2\pin/3)'); xlabel('n'), ylabel('x(n)') subplot(3, 1, 3), stem(n, y, 'ko'); title('Output y(n)=[x(n)+x(n-1)+x(n-2)]/3'); xlabel('n'), ylabel('y(n)');
1 ) t ( x
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Output y (n)=[x(n)+x(n-1)+x(n-2)]/3 0.5 ) n ( y
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20
Q. On an approximate basis verify if the output amplitudes of the 50 Hz component agrees with the gain figures obtained in A.
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B3. As an extension of above example, consider a signal consisting of a 2 Hz desirable component plus a noise component of 50 Hz (with a smaller amplitude), sampled at 150 Hz: . The corresponding is the input to the moving
average filter
. It is expected that the 2 Hz will be passed through while
the 50 Hz will be eliminated. Using MATLAB plot plots below.
. We show the continuous-time
% Continuous-time plots of x(t) t = 0: 1/500: 0.4; x2Hz = 5*cos (2*pi*2*t); %2 Hz x50Hz = 1*cos (2*pi*50*t); %50 Hz xt = x2Hz +x50Hz; %2 Hz + 50 Hz % Choose a multiplot so that all waveforms are on the same scales plot(t, x2Hz, 'b', t, x50Hz, 'r', t, xt, 'k'); title('Continuous-time plots'); legend ('2 Hz', '50 Hz', '2 Hz + 50 Hz'); xlabel ('Time, t, seconds'), ylabel('x2Hz, x50Hz and x(t)'); Continuous-time plots 6 2 Hz 50 Hz 2 Hz + 50 Hz
4
) t ( 2 x d n a z H 0 0 5 x , z H 2 x -2
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-6
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B4. In the following MATLAB segment we show b oth
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.
% Plots of x(t) and x(n) % Choose 0.4 sec interval so that continuous and discrete time scales align A=5; F1=2; B=1; F2=50; Fs=150; t = 0: 1/500: 0.4; xt = A*cos (2*pi*F1*t) + B*cos (2*pi*F2*t); % 2 Hz + 50 Hz % Choose n=60 so that continuous and discrete time scales align n = 0: 1: 60; xn = A*cos(2*pi*F1*n/Fs) + B*cos(2*pi*F2*n/Fs); subplot(2, 1, 1), plot(t, xt, 'b'); legend ('2 Hz + 50 Hz'); title ('Continuous-time signal x(t)'); xlabel ('Time, t, seconds'), ylabel('x(t)'); subplot(2, 1, 2), stem(n, xn, 'b*'); legend ('(1/75) rad + (1/3) rad'); title ('Discrete-time signal x(n)'); xlabel ('n'), ylabel('x(n)'); Continuous-time signal x(t) 10 2 Hz + 50 Hz 5 ) t ( x
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0.3
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10 (1/75) rad + (1/3) rad 5 ) n ( x
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B5. We show the operation of the filter in the following MATLAB segment. Both sequences are plotted on the same scale so the smoothing action of the filter is easily discernible: The 50 Hz component is totally removed. [Note: The signal magnitudes are such that MATLAB, o n its own, would use two different scales for In order to force MATLAB to use the same scale for both we have used the axis() command.]
% Plots of input x(n) and moving average filter output y(n) A=5; F1=2; B=1; F2=50; Fs=150; b = [1/3, 1/3, 1/3]; a = [1]; %Filter coefficients n = 0: 1: 60; x = A*cos(2*pi*F1*n/Fs) + B*cos(2*pi*F2*n/Fs); y = filter(b, a, x) ; subplot(2, 1, 1), stem(n, x, 'b*'); axis([0, 60, -7, 7]); legend ('(1/75) rad + (1/3) rad'); title ('Discrete-time signal x(n)'); xlabel ('n'), ylabel('x(n)'); subplot(2, 1, 2), stem(n, y, 'ko'); axis([0, 60, -7, 7]); legend ('(1/3) rad is absent'); title ('Moving average filter output y(n)'); xlabel ('n'), ylabel('y(n)'); Discrete-time signal x(n) (1/75) rad + (1/3) rad
5 ) n ( x
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30 40 n Moving average filter output y(n)
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(1/3) rad is absent
5 ) n ( y
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Q. On an approximate basis verify if the output amplitudes of the 2 and 50 Hz components agree with the gain figures obtained in A. Q. What is the effect, on the time-domain demonstration, of changing the sampling frequenc y from 150 Hz to, say, 3000 Hz or 8000 Hz?
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( )
C1. Evaluate by hand.
0 1 2 3 4 5 6 …
1 -0.5 -0.5 1 -0.5 -0.5 1 …
=
0 0 …
Q. Write your own MATLAB program that will implement the above evaluation and is, in effect, the equivalent of the MATLAB function filter() . Q. In the context of the 3-term moving average comment on the kind of filter represented by
⁄ ⁄
C2. The three-term moving average filter has an impulse response of . Indicate what kind of filter (low-pass or high-pass etc.) the following are. In each case give the difference equation and indicate the type of symmetry (of the impulse response). 1. 2. 3. h(n) = {0.25, 0.5, – 0.25} 4. h(n) = {0.25, 0, – 0.25}
Q. Comment on shape of magnitude plot for average of odd vs. average of even number of terms. For instance consider a 4-term average or 5-term average.
D1. Let the input consist of a 2 Hz and a 75 Hz, the latter being the maximum permissible with a sampling frequency of 150 Hz. . The corresponding continuoustime plots are given below:
% Continuous-time plots of x(t) A=5; F1=2; B=1; F2=75; Fs=150; t = 0: 0.001: 0.2; x2Hz = A*cos (2*pi*F1*t); % 2 Hz x75Hz = B*cos (2*pi*F2*t); % 75 Hz xt = x2Hz +x75Hz; % 2 Hz + 75 Hz % Choose a multiplot so that all waveforms are on the same scale plot(t, x2Hz, 'b', t, x75Hz, 'r', t, xt, 'k'); title('Continuous-time plots'); legend ('2 Hz', '75 Hz', '2 Hz + 75 Hz'); xlabel ('Time, t, seconds'), ylabel('x2Hz, x75Hz and xt'); DSP Lab
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D2. In the following MATLAB segment we show the input, both continuous and discrete time, .
% Plots of x(t) and x(n) % Choose 0.4 sec interval so that continuous and discrete time scales align A=5; F1=2; B=1; F2=75; Fs=150; t = 0: 0.001: 0.4; xt = A*cos (2*pi*F1*t) + B*cos (2*pi*F2*t); % 2 Hz + 75 Hz % Choose n = 60 n = 0: 1: 60; xn = A*cos(2*pi*F1*n/Fs) + B*cos(2*pi*F2*n/Fs); % 2 Hz + 75 Hz subplot(2, 1, 1), plot(t, xt, 'b'); legend ('2 Hz + 75 Hz'); title ('Continuous-time signal x(t)'); xlabel ('Time, t, seconds'), ylabel('x(t)'); subplot(2, 1, 2), stem(n, xn, 'b*'); legend ('(1/75) rad + (1/2) rad'); title ('Discrete-time signal x(n)'); xlabel ('n'), ylabel('x(n)');
D3. In the following MATLAB segment we show the operation of the filter. Both sequences are shown. The low frequenc y component (2 Hz) suffers much less attenuation than the high frequency component (75 Hz).
% Plots of input x(n) and high pass filter output y(n) A=5; F1=2; B=1; F2=75; Fs=150; b = [1/3, 1/3, 1/3]; a = [1]; %Filter coefficients n = 0: 1: 60; xn = A*cos(2*pi*F1*n/Fs) + B*cos(2*pi*F2*n/Fs); % 2 Hz + 75 Hz yn = filter(b, a, x) ; subplot(2, 1, 1), stem(n, xn, 'b*'); axis([0, 60, -10, 10]); legend ('2 Hz + 75 Hz '); title ('Discrete-time signal x(n)'); xlabel ('n'), ylabel('x(n)'); subplot(2, 1, 2), stem(n, yn, 'ko'); axis([0, 60, -10, 10]); legend ('Low pass filtered'); title ('Low pass filter output y(n)'); xlabel ('n'), ylabel('y(n)'); Q. On an approximate basis verify if the output amplitudes of the 2 and 75 Hz components agree with the gain figures obtained in A. Q. What is the effect, on the time-domain demonstration, of changing the sampling frequenc y from 150 Hz to, say, 3000 Hz or 8000 Hz?
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Extra-credit Experiments
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Lab-E1 Generation of DTMF signals Objective In this Lab you will generate dual tone multi frequency signals used in the modern telephone. Background A DTMF signal consists of the sum of two sinusoids - or tones - with frequencies taken from two mutually exclusive groups. These frequencies we re chosen to prevent any harmonics from being incorrectly detected by the receiver as some other DTMF frequency. Each pair of tones contains one frequency of the low group (697 Hz, 770 Hz, 852 Hz, 941 Hz) and one frequency of the high group (1209 Hz, 1336 Hz, 1477Hz) and represents a unique symbol. The frequencies allocated to the push buttons of the telephone pad are shown below:
Frequency 697 Hz 770 Hz 852 Hz 941 Hz
1209 Hz 1336 Hz 1477 Hz ABC DEF 1 2 3 GHI JKL MNO 4 5 6 PRS TUV WXY 7 8 9 * 0 #
symbol = {'1','2','3','4','5','6','7','8','9','*','0','#'}; lfg = [697 770 852 941]; % Low frequency group hfg = [1209 1336 1477]; % High frequency group f = []; for c=1:4, for r=1:3, f = [ f [lfg(c);hfg(r)] ]; end end >> Fs = 8000; % Sampling frequency 8 kHz N = 800; % Tones of 100 ms t = (0:N-1)/Fs; % 800 samples at Fs pit = 2*pi*t; tones = zeros(N, size(f,2)); for toneChoice=1:12, % Generate tone tones(:,toneChoice) = sum(sin(f(:,toneChoice)*pit))'; % Plot tone subplot(4,3,toneChoice),plot(t*1e3,tones(:,toneChoice)); title(['Symbol "', symbol{toneChoice},'": [',num2str(f(1,toneChoice)),',',num2str(f(2,toneChoice)),']']) set(gca, 'Xlim', [0 25]); ylabel('Amplitude'); if toneChoice>9, xlabel('Time (ms)'); end end set(gcf, 'Color', [1 1 1], 'Position', [1 1 1280 1024]) annotation(gcf,'textbox', 'Position',[0.38 0.96 0.45 0.026],... 'EdgeColor',[1 1 1],... 'String', '\bf Time response of each tone of the telephone pad', DSP Lab
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'FitHeightToText','on'); Output:
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Lab-E2 Implementation of decimation Objective In this Lab you will convert the sampling rate of a discrete time signal from its original higher sampling rate to a new lower rate. Background
clf; M = input('Down-sampling factor = '); n = 0:99; x = sin (2*pi*0.043*n) + sin (2*pi*0.031*n); y = decimate(x, M); subplot(2,1,1); stem(n, x(1:100)); title('Input Sequence'); xlabel('Time index n'); ylabel('Amplitude'); subplot(2,1,2); m = 0 : (100/M)-1; stem(m, y(1:100/M)); title('Output Sequence'); xlabel('Time index n'); ylabel('Amplitude');
Ouput:
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Lab-E3 Implementation of interpolation Objective In this Lab you will convert the sampling rate of a discrete time signal from its original lower sampling rate to a new higher rate. Background
% Illustration of Interpolation Process clf; L = input('Up-sampling factor ='); % Generate the input sequence n = 0:49; x = sin(2*pi*0.043*n) + sin(2*pi*0.031*n); % Generate the interpolated output sequence y = interp(x,L); % Plot the input and the output sequences subplot(2,1,1); stem(n,x(1:50)); title('Input Sequence'); xlabel('Time index n'); ylabel('Amplitude'); subplot(2,1,2); m = 0:(50*L)-1; stem(m,y(1:50*L)); title('Output Sequence'); xlabel('Time index n'); ylabel('Amplitude'); Output:
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Lab-E4 Implementation of I/D sampling rate converters Objective In this Lab you will convert the sampling rate of a discrete time signal by an integer factor. Background
% Illustration of Down-Sampling by an Integer Factor % clf; n = 0: 49; m = 0: 50*3 - 1; x = sin(2*pi*0.042*m); y = x([1: 3: length(x)]); subplot(2,1,1) stem(n, x(1:50)); axis([0 50 -1.2 1.2]); title('Input Sequence'); xlabel('Time index n'); ylabel('’Amplitude'); subplot(2,1,2) stem(n, y); axis([0 50 -1.2 1.2]); title('Output Sequence'); xlabel('Time index n'); ylabel('Amplitude'); Output:
% Illustration of Up-Sampling by an Integer Factor % clf; n = 0:50; x = sin(2*pi*0.12*n); DSP Lab
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y = zeros(1, 3*length(x)); y([1: 3: length(y)]) = x; subplot(2,1,1) stem(n,x); title('Input Sequence'); xlabel('Time index n'); ylabel('Amplitude'); subplot(2,1,2) stem(n,y(1:length(x))); title('Output Sequence'); xlabel('Time index n'); ylabel('Amplitude'); Output:
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