Determine the impulse response of a filter matched to this signal and sketch it as a function of time. Plot the matched filter output, where s (t ) is the input, as a function of time. What is the peak value of the output?
Qu. 1 solution (a) The matched filter impulse response is h(t ) = s (T − t ) :
(b) The output of the matched filter is given by: y (t ) =
∫
T
0
s (τ )h(t − τ )d τ , that is we multiply s (τ )
by shifted versions of h( −τ ) (from h( −τ ) to h(T − τ ) ) and integrate the product: s (τ )
τ
h(t − τ )
t = 0 τ
With the result:
3T/2
(c) The peak output value of A2T 4 occurs at t = T .
2.
A PCM system using polar NRZ signalling operates just above the error threshold with an − average probability of error equal to 10 6 . Suppose that the signalling rate is doubled. Find the new value of the average probability of error. You may use the upper bound approximation 2 2 exp( −u ) exp( −u ) erfc(u ) ≤ for evaluating the complementary error function and erfc(u ) ≤ π u π for evaluating its inverse.
Qu. 2 solution
In PCM or binary PAM we have that Pe
=
1 2
⎛ E b ⎞ ⎟ ⎜ N ⎟ where E b ⎝ O ⎠
erfc⎜
=
T
∫ A dt = A T . 0
2
2
b
If the signalling rate is doubled then the bit duration, T b , is reduced by half and hence E b is also reduced by half. Let u = E b N 0 then we know that Pe u≈ Pe
=
− ln
1 2
(2(10
−6
)
)
π =
= 10
−6
=
1 2
erfc(u ) ≈
exp( −u 2 ) 2
and hence can state that
π
3.54 . With the signalling rate doubled, the new value of Pe is:
⎛ E b 2 ⎞ 1 u ⎞ 1 ⎟ = erfc⎛ ⎜ ⎟ = erfc(2.503) ≈ 2.14(10 4 ) ⎜ N ⎟ 2 ⎝ 2 ⎠ 2 0 ⎠ ⎝
erfc⎜
−
Note that using the MATLAB erfcinv( ) we get 3.36 and using both the MATLAB erfcinv() and erfc() functions gives the more accurate answer as 3.88(10 −4) . So care needs to be exercised when using the approximations to the complementary error function and its inverse.
3.
A continuous-time signal is sampled and then transmitted as a PCM signal. The random variable at the input of the decision device in the receiver has a variance of 0.01 volts2 . (a) Assuming the use of polar NRZ signalling, determine the pulse amplitude that must be 8 transmitted for the average error rate not to exceed 1 bit in 10 bits.
6
(b)
If the added presence of interference causes the error rate to increase to 1 bit in 10 bits, what is the variance of the interference?
Qu. 3 solution The bit error probability for PAM is Pe for PCM we can state that T b (a)
= t S
Hence we can write u =
We know that With Pe
= 10
−8
2 σ =
=
1 2
⎛ E b ⎞ ⎟ where E b = A2T b and σ 2 ⎜ N ⎟ ⎝ O ⎠
erfc⎜
=
N 0
=
2t S
N 0
2T b
since
.
E b N 0
A
=
so A = 2σ u .
2σ
0.01 ⇒ σ = 0.1 , so we need to find u.
we have u = erfc (2(10 )) and using the approximation erfc(u ) ≤ -1
−8
2
exp( −u )
we can
π
solve u ≈
− ln
(2(10
−8
)
)
π =
4.14 , and using the MATLAB erfcinv() gives u = 3.97 and hence
2 (0.1)(3.97) = 0.561 volts.
A =
2σ u =
(b)
If now Pe
= 10
−6
then u ≈
− ln
(2(10
−6
)
)
π =
⎛ A ⎞ 2 u = 3.36 and hence we have that σ T = ⎜ ⎟ ⎝ 2u ⎠ variance of interference is
4.
2 2 2 σ i = σ T − σ =
2
3.54 , and using the MATLAB erfcinv() gives
⎛ 0.561 ⎞ ⎟ =⎜ ⎜ 2 (3.36) ⎟ ⎝ ⎠
2
2
= 1.394(10 3
1.394(10 − ) − 0.01 = 3.939(10 − ) volts
−2
) and the added
2
A computer puts out binary data at the rate of 56 kb/s. The computer output is transmitted using a baseband binary PAM system that is designed to have a raised-cosine spectrum. Determine the transmission bandwidth required for each of the following rolloff factors: α = 0.25, 0.5, 0.75, 1.0 .
Qu. 4 solution For the raised-cosine spectrum we have BT
=
2W − f 1
Since Rb
=
28000 (1 + α ) . So:
= 56000
then W = 28000 and BT
= W (1 + α )
α
BT
0.25 0.5 0.75 1.0
35 kHz 42 kHz 49 kHz 56 kHz
where W =
Rb
2
=
1 2T b
.
5.
Repeat the previous question given that each set of successive binary digits in the computer output are coded into one of eight possible amplitude levels and the resulting signal is transmitted using an eight-level (8-ary) baseband PAM system designed to have a raisedcosine spectrum.
Qu. 5 solution The use of eight amplitude levels ensures that 3 bits can be transmitted per PAM pulse. Thus the pulse duration or symbol duration will be increased by a factor of 3. This means that the channel bandwidth, W , can be reduced by a factor of 3 and the bandwidths in the previous question will be reduced to (1/3) of their binary PAM values. For example for α = 0.5 the transmission bandwidth is now (42/3) = 14 kHz.
