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Wireless Technology Questions –1 03-2-2007 Text: William Stalling Chap 1-6
Assignment -
To be submitted on 17-2-2007
1. Chap hap 2 Fund Fundam amen enta tals ls a.
Problems
Q 7 What is the channel capacity capacity for a teleprinte teleprinterr channel with a 300Hz bandwidth bandwidth and a noise ratio of 3 dB?
Ans. Using Shannon's equation: C = B log2 (1 + SNR) We have W = 300 Hz
(SNR)dB = 3
Therefore, SNR = 100.3 C = 300 log2 (1 + 100.3) = 300 log2 (2.995) = 474 bps
Q 8a A digital signaling system is required to operate at 9600 bps. If a signal element encode word, what is the minimum required bandwidth of the channel?
Ans. Using Nyquist's equation: C = 2B log2M We have C = 9600 bps a. log2M = 4, because a signal element encodes a 4-bit word Therefore, C = 9600 = 2B × 4, and B = 1200 Hz Q 9 Study the works of Shannon & Nyquist on channel capacity. Each places an upper limit rate of a channel based on two different approaches. How are they related?
Master your semester with Scribd Ans. Nyquist analyzed the theoretical capacity Read of a up noiseless channel; Free Foron 30 Days Sign to vote this title therefore, in case,Times the signaling rate is limited solely by channel bandwidth. addresse & The New York useful Useful NotShannon
anytime. question of what signaling rate can be achieved overCancel a channel with a given band Special offer for students: Only $4.99/month.
given signal power, and in the presence of noise.
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Q11 Show that doubling the transmission frequency or doubling the distance between tra antenna and the receiving antenna attenuates the power received by 6 dB.
Ans. From Equation 2.1, we have LdB = 20 log (4 π d/λ ) = 20 log (4 π df /
f = v (see Question 4). If we double either d or f where λ f = or f , we add a term log(2), which is approximately 6 dB. Q13 If an amplifier has a 30 dB voltage gain, what voltage ratio does the gain represent?
Ans. For a voltage ratio, we have NdB = 30 = 20 log(V2/V1) V2/V1 = 1030/20 = 101.5 = 31.6 Q14. Q14. An amplifier has an output of 20 W. What is its output in dBW?
Ans. Power (dBW) = 10 log (Power/1W) = 10 log20 = 13 dBw Worked examples – p.42
1. If a signal signal with a power power level of 10mW is insert inserted ed onto a transm transmiss ission ion line and the m power some distance away is 5mW, the loss can be expressed as L dB = 10 log(10/5) = 10(0.3) = 3 dB
2. Deci Decibe bels ls are are use usefu full in det deter ermi mini ning ng the the gai gain n or los loss s over over a ser serie ies s of tra trans nsmi miss ssio io Consider a series in which the input is at a low power level of 4 mW, the first lement is a tran line with a 12 dB loss, the second element is an amplifier with a 35 dB gain, and the third elem transmission line with a 10 dB loss. GdB = 13 = 10 log(Pout / 4 mW)
Master your semester Pwith Scribd = 4 X 10 mW = 79.8 mW Read Free Foron 30this Days Sign up to vote title 3. Times Powe ower of 10 1000 W is 30 30 dB dBW, and po power of of 1 Useful mW mW is is –3 –3 0 dB dNot BW.useful & The New York out
Special offer for students: Only $4.99/month. Review Questions
1.3
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2. Chap hap 5 Antenna ennas s a.
Review Questions
Q1. What two functions are performed by antenna?
The two funct function ions s of an antenn antenna a are: are: (1) For transm transmiss ission ion of a signa signa frequency electrical energy from the transmitter is converted into electroma energy by the antenna and radiated into the surrounding environment (atmos space, water); (2) for reception of a signal, electromagnetic energy impinging antenna is converted into radio-frequency electrical energy and fed into the rec Q2. Q2. What is isotropic antenna?
An isotropic antenna is a point in space that radiates power in all direction equally. Q4. Q4. What is the advantage of parabolic reflective antenna?
A parabolic antenna creates, in theory, a parallel beam without dispersion. practice, there will be some beam spread. Nevertheless, it produces a highl focused, directional beam. Q5. Q5. What factors determine antenna gain?
Effective area and wavelength. Q6. Q6. What is the primary cause of signal loss in satellite communication?
Free space loss.
Master your semester with Scribd Q7. Q7. Name and briefly define four types of noise. & The New York Times
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Useful Not useful Cancel anytime. Special offer for students: Only $4.99/month. Thermal noise is due to thermal agitation of electrons. Intermodulation Intermodulation produces signals at a frequency that is the sum or difference of the two orig
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Diffraction occurs at the edge of an impenetrable body that is large compa the wavelength of the radio wave. The edge in effect become a source and radiate in different directions from the edge, allowing a beam to bend arou obstacle. If the size of an obstacle is on the order of the wavelength of the s or less, scattering occurs. An incoming signal is scattered into several weak outgoing signals in unpredictable directions. Q11. Q11. What is the difference between fast and slow fading?
Fast fading refers to changes in signal strength between a transmitter and receiver as the distance between the two changes by a small distance of abo onehalf a wavelength. Slow fading refers to changes in signal strength be a transmitter and receiver as the distance between the two changes by a lar distance, well in excess of a wavelength. Q12. Q12. What is the difference between flat and selective fading?
Flat fading, or nonselective fading, is that type of fading in which all frequ components of the received signal fluctuate in the same proportions simultaneously. Selective fading affects unequally the different spectral components of a radio signal. Q13. Q13. Name and briefly define three diversity techniques.
Space diversity involves the physical transmission path and typical refers use of multiple transmitting or receiving antennas. With frequency divers signal is spread out over a larger frequency bandwidth or carried on multi frequency carriers. Time diversity techniques aim to spread the data out o time so that a noise burst affects fewer bits.
Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title b. Problems & The New York Times Useful Not useful Q 9 Assume that two antennas are half-wave dipoles and each has a directive gain of 3 d Cancel anytime.
Special offer for students: Only $4.99/month. transmitted power is 1 W and the two antennas are separated by a distance of 10 Km, wh
received power? Assume that the antennas are aligned so that the directive gain numbers a
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b. Using Equation (5.2), LdB = 20 log(900 × 106) +20 log (100) – 147.56 = 120 + 59.08 +40 – 147.56 = 71.52
Therefore, received power in dBm = 47 – 71.52 = –24.52 dBm
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3.
Chap 6
Encoding
Q. What are the different modulations that can be used to encode digital data for transmis analog media?
Ans. Modulation involves operation on one or more of the tree characteristics of the carrie amplit amplitude, ude, freque frequency ncy and phase. phase. There There are three three basic basic encodi encoding ng or modula modulatio tion n tec transforming digital data into analog signals: Amplitude Shift Keying (ASK), Frequency Shi (FSK) and Phase Shift Keying (PSK). Draw Fig 6.2 (ASK, BFSK, BPSK)
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Q1. What is differential encoding?
In differential encoding, the signal is decoded by comparing the polarity of adjace signal elements rather than determining the absolute v alue of a signal element. Q3 Indicate three major advantages of digital transmission over analog transmission.
Cost, capacity utilization, and security and privacy are three major advantages en digital transmission over analog transmission.
Problems Q6. Why should PCM be preferable to DM for encoding analog signals that repr digitized data?
As was mentioned in the text, analog signals in the voice band that represe digital data have more high frequency components than analog voice signa These higher components cause the signal to change more rapidly over tim Hence, DM will suffer from a high level of slope overload noise. noise. PCM, on t other hand, does not estimate changes in signals, but rather the absolute v of the signal, and is less affected than DM. Q. What is QAM technique? Show diagrams of QAM modulator and demodulator.
Quadrature Amplitude Modulation is a combination of ASK and PSK. It can be co logical extension of QPSK.
Two different signals (carrier) are sent simultaneously with a phase shift of 90 deg respect respect to one another another.. Each Each carrie carrierr is ASK modulated modulated.. The two differen differentt simultaneously transmitted over the same medium. At the receiver, the two sig demodulated and the results combined to produce the original signal.
Master your semester with Scribd If two level ASK is used, then each of the streams can be in one of two states an Read Free Foron 30this Days Sign up to vote title combined stream can be in one of 4 = 2x2 states. This is essentially QPSK. If fo & The New York Times Useful Not useful ASK is used, then the combined stream can be in one of 16=4x4 states. Special offer for students: Only $4.99/month.
S(t) = d1 cos 2π f t + d2 sin 2 π f t
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Chap 7 Spread Spectrum Review Questions
1. What is the relation relationship ship between between the bandwidth bandwidth of a signal before before and and after i been encoded using spread spectrum? The bandwidth is wider after the signal has been encoded using spread spectrum.
2. List List three three benefi benefits ts of using using spre spread ad spectr spectrum. um. (1) We can gain immunity from various kinds of noise and multipath distorti It can also be used for hiding and encrypting signals. Only a recipient who k the spreading code can recover the encoded information. (3) Several user independently use the same higher bandwidth with very little interference, u code division multiple access (CDMA).
3. What What is freque frequency ncy hoppi hopping ng spread spread spect spectrum rum?? With frequency hopping spread spectrum (FHSS), the signal is broadcast o seemingly random series of radio frequencies, hopping from frequency to frequency at fixed intervals. A receiver, hopping between frequencies in synchronization with the transmitter, picks up the message. 4. Explain Explain the the differen difference ce between between slow slow and fast fast FHSS. FHSS. Slow FHSS = multiple signal elements per hop; fast FHSS = multiple hops per signal element.
5. What What is direc directt sequen sequence ce sprea spread d spectr spectrum? um? With direct sequence spread spectrum (DSSS), each bit in the original sign represented by multiple bits in the transmitted signal, using a spreading cod
Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title it ha 6. What is the relation relationship ship between between the bit bit rate of a signal signal before before and after after & The New York Times Useful Not useful encoded using DSSS? Special offer for students: Only $4.99/month.
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spreading code, the bit rate after spreading (usually called the
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Chap 8 Coding and Error Control Review Questions
Q2. What is CRC? An error detecting code in which the code is the remainder resulting from d the bits to be checked by a predetermined binary number.
Q3. Why would you expect a CRC to detect more errors than a parity bit? The CRC has more bits and therefore provides more redundancy. That is, i provides more information that can be used to detect errors. Q4. List three different ways to describe CRC algorithm. Modulo 2 arithmetic, polynomials, and digital logic.
Q5. Is it possible to design an ECC that will correct some double bit errors but n double bit errors? Why or why not? It is possible. You could design a code in which all codewords are at least a distance of 3 from all other codewords, allowing all single-bit errors to be corrected. Suppose that some but not all codewords in this code are at leas distance of 5 from all other codewords. Then for those particular codeword not the others, a double-bit error could be corrected. Q6. In an (n, k) block ECC, what do n and k represent? An (n, k ) block code encodes k data bits into n-bit codewords.
Q7. In an (n, k, K) convolution code, what do n, k, and K represent? An (n, k , K ) code processes input data k bits at a time and produces an ou n bits for each incoming k bits. The current output of n bits is a function of K × k input bits. Read Free Foron 30this Days Sign up to vote title
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Special offer for students: Only $4.99/month. A trellis is a diagram that shows the state transitions over time in a convolu
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DKV/MET Any arithmetic scheme will work if applied in exactly the same way to the forward and reverse process. The modulo 2 scheme is easy to implement circuitry. It also yields a remainder one bit smaller than binary arithmetic. Chap 4 Problems. Q8 . Describe UDP protocol. Why can’t a user program directly access IP?
UDP is a transport level protocol. UDP provides connectionless service fro appli level procedures. It is basically an unreliable service, delivery and duplicate prot are not guaranteed.. However, this reduces the overhead of the protocol and may adequate in many cases. For example, - Inward Inward data collection collection where loss of occasio occasional nal data data unit may not not cause cause pr Request-Res Request-Response ponse in transaction transaction service service where applicatio application n can can handle erro - Run-time Run-time applic applications ations such as voice voice video video involvi involving ng a degree of redundanc redundanc time requirement. UDP sits on top of IP. Because it is connectionless, UDP may h ave very little to Essentially it adds port-addressing capability to IP.
UDP provides the source and destination port addresses and a checksum that cove data field. These functions would not normally be performed by protocols above t transport layer. Thus UDP provides a useful, though limited, service.
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