Wireless Technology Questions1_answer

DKV/MET

Wireless Technology Questions –1 03-2-2007 Text: William Stalling Chap 1-6

Assignment -

To be submitted on 17-2-2007

1. Chap 2 Fundamentals a.

Problems

Q 7 What is the channel capacity for a teleprinter channel with a 300Hz bandwidth and a signal to 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 encodes a 4-bit 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 on the bit rate of a channel based on two different approaches. How are they related?

Ans. Nyquist analyzed the theoretical capacity of a noiseless channel; therefore, in that case, the signaling rate is limited solely by channel bandwidth. Shannon addressed the question of what signaling rate can be achieved over a channel with a given bandwidth, a given signal power, and in the presence of noise. Q10 Given a channel with an intended capacity of 20 Mbps, the bandwidth of the cahnnel is 3 MHz. What signal-to-noise ratio is required to achieve this capacity?

Ans.

C = B log2 (1 + SNR) 20 × 106 = 3 × 106 × log2(1 + SNR) log2(1 + SNR) = 6.67 1 + SNR = 102 SNR = 101

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DKV/MET Q11 Show that doubling the transmission frequency or doubling the distance between transmitting antenna and the receiving antenna attenuates the power received by 6 dB.

Ans. From Equation 2.1, we have LdB = 20 log (4πd/λ) = 20 log (4πdf/ v), where λf  = v (see Question 4). If we double either d or f, we add a term 20 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. 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 with a power level of 10mW is inserted onto a transmission line and the measured power some distance away is 5mW, the loss can be expressed as L dB = 10 log(10/5) = 10(0.3) = 3 dB

2.

Decibels are useful in determining the gain or loss over a series of transmission elements. Consider a series in which the input is at a low power level of 4 mW, the first lement is a transmission line with a 12 dB loss, the second element is an amplifier with a 35 dB gain, and the third element is a transmission line with a 10 dB loss. GdB = 13 = 10 log(Pout / 4 mW) Pout = 4 X 101.3 mW = 79.8 mW

3.

Power of 1000 W is 30 dBW, and power of 1 mW is –30 dBW.

Review Questions Q1 What are some major advantages and disadvantages of microwave transmission.

Ans. A continuous or analog signal is one in which the signal intensity varies in a smooth fashion over time while a discrete or digital signal is one in which the signal intensity maintains one of a finite number of constant levels for some period of time and then changes to another constant level. Q2 Explain how synchronous time division multiplexing (TDM) works.

Ans. Amplitude, frequency, and phase are three important characteristics of a periodic signal.

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2. Chap 5 Antennas a. Review Questions Q1. What two functions are performed by antenna?

The two functions of an antenna are: (1) For transmission of a signal, radiofrequency electrical energy from the transmitter is converted into electromagnetic energy by the antenna and radiated into the surrounding environment (atmosphere, space, water); (2) for reception of a signal, electromagnetic energy impinging on the antenna is converted into radio-frequency electrical energy and fed into the receiver. Q2. What is isotropic antenna?

An isotropic antenna is a point in space that radiates power in all directions  equally. Q4. What is the advantage of parabolic reflective antenna?

A parabolic antenna creates, in theory, a parallel beam without dispersion. In  practice, there will be some beam spread. Nevertheless, it produces a highly  focused, directional beam. Q5. What factors determine antenna gain?

Effective area and wavelength. Q6. What is the primary cause of signal loss in satellite communication?

Free space loss. Q7. Name and briefly define four types of noise.

Thermal noise is due to thermal agitation of electrons. Intermodulation noise  produces signals at a frequency that is the sum or difference of the two original  frequencies or multiples of those frequencies. Crosstalk is the unwanted coupling  between signal paths. Impulse noise is noncontinuous, consisting of irregular  pulses or noise spikes of short duration and of relatively high amplitude. Q9. What is fading?

The term fading refers to the  time variation of received signal power caused by  changes in the transmission medium or path(s). Q10. What is the difference between diffraction and fading?

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DKV/MET Diffraction occurs at the edge of an impenetrable body that is large compared to  the wavelength of the radio wave. The edge in effect become a source and waves  radiate in different directions from the edge, allowing a beam to bend around an  obstacle. If the size of an obstacle is on the order of the wavelength of the signal  or less, scattering occurs. An incoming signal is scattered into several weaker  outgoing signals in unpredictable directions. 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 about  one­half a wavelength. Slow fading refers to changes in signal strength between  a transmitter and receiver as the distance between the two changes by a larger  distance, well in excess of a wavelength. Q12. What is the difference between flat and selective fading?

Flat fading, or nonselective fading, is that type of fading in which all frequency  components of the received signal fluctuate in the same proportions  simultaneously. Selective fading affects unequally the different spectral  components of a radio signal. Q13. Name and briefly define three diversity techniques.

Space diversity involves the physical transmission path and typical refers to the  use of multiple transmitting or receiving antennas. With frequency diversity, the  signal is spread out over a larger frequency bandwidth or carried on multiple  frequency carriers. Time diversity techniques aim to spread the data out over  time so that a noise burst affects fewer bits. b. Problems Q 9 Assume that two antennas are half-wave dipoles and each has a directive gain of 3 dB. If the transmitted power is 1 W and the two antennas are separated by a distance of 10 Km, what is the received power? Assume that the antennas are aligned so that the directive gain numbers are correct and the frequency used is 100MHz.

         Ans. We have Pr

= [(Pt) (Gt) (Gr) (c)2]/(4πfd)2

= [(1) (2) (2) (3 × 108)2]/[(16) (π)2 (3 × 108)2 (104)2] = 0.76 × 10–9 W Q10 Suppose a transmitter produces 50 W of power. a. Express the transmitted power in units of dBm and dBW. b. If the transmitter’s power is applied to a unity gain antenna with a 900 MHz carrier frequency, what is the received power in dBm at a free space distance of 100 m?

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DKV/MET a. From Appendix 2A, PowerdBW = 10 log (PowerW) = 10 log (50) = 17 dBW PowerdBm = 10 log (PowermW) = 10 log (50,000) = 47 dBm 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 transmission over analog media? Ans. Modulation involves operation on one or more of the tree characteristics of the carrier signal: amplitude, frequency and phase. There are three basic encoding or modulation techniques for transforming digital data into analog signals: Amplitude Shift Keying (ASK), Frequency Shift Keying (FSK) and Phase Shift Keying (PSK). Draw Fig 6.2 (ASK, BFSK, BPSK)

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DKV/MET Q1. What is differential encoding?

In differential encoding, the signal is decoded by comparing the polarity of adjacent signal elements rather than determining the absolute value 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 enjoyed by digital transmission over analog transmission. Problems Q6. Why should PCM be preferable to DM for encoding analog signals that represent digitized data? As was mentioned in the text, analog signals in the voice band that represent digital data have more high frequency components than analog voice signals. These higher components cause the signal to change more rapidly over time. Hence, DM will suffer from a high level of slope overload noise. PCM, on the other hand, does not estimate changes in signals, but rather the absolute value 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 considered logical extension of QPSK. Two different signals (carrier) are sent simultaneously with a phase shift of 90 degree with respect to one another. Each carrier is ASK modulated. The two different signals are simultaneously transmitted over the same medium. At the receiver, the two signals are demodulated and the results combined to produce the original signal. If two level ASK is used, then each of the streams can be in one of two states and the combined stream can be in one of 4 = 2x2 states. This is essentially QPSK. If four-level ASK is used, then the combined stream can be in one of 16=4x4 states. S(t) = d1 cos 2π f t + d2 sin 2π f t

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Chap 7 Spread Spectrum Review Questions 1. What is the relationship between the bandwidth of a signal before and after it has been encoded using spread spectrum? The bandwidth is wider after the signal has been encoded using spread spectrum. 2. List three benefits of using spread spectrum. (1) We can gain immunity from various kinds of noise and multipath distortion. (2) It can also be used for hiding and encrypting signals. Only a recipient who knows the spreading code can recover the encoded information. (3) Several users can independently use the same higher bandwidth with very little interference, using code division multiple access (CDMA). 3. What is frequency hopping spread spectrum? With frequency hopping spread spectrum (FHSS), the signal is broadcast over a 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 the difference between slow and fast FHSS. Slow FHSS = multiple signal elements per hop; fast FHSS = multiple hops per signal element. 5. What is direct sequence spread spectrum? With direct sequence spread spectrum (DSSS), each bit in the original signal is represented by multiple bits in the transmitted signal, using a spreading code. 6. What is the relationship between the bit rate of a signal before and after it has been encoded using DSSS? For an N-bit spreading code, the bit rate after spreading (usually called the chip rate) is N times the original bit rate. 7. What is CDMA? CDMA allows multiple users to transmit over the same wireless channel using spread spectrum. Each user uses a different spreading code. The receiver picks out one signal by matching the spreading code. 8. Explain the difference between auto-correlation and cross correlation. Autocorrelation is the correlation or comparison of a sequence with all phase shifts of itself. Cross-correlation is the comparison is made between two sequences from different sources rather than a shifted copy of a sequence with itself. QA

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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 dividing 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, it 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 not all 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 least a distance of 5 from all other codewords. Then for those particular codewords, but 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 output of n bits for each incoming k bits. The current output of n bits is a function of the last K × k input bits. Q8. What is trellis in context of a convolution code? A trellis is a diagram that shows the state transitions over time in a convolutional code. Q9. What two key elements comprise error control? Detection of errors and retransmission of frames that are received in error. Q10. Explain how Go-back N ARQ work. Go-back-N ARQ is a form of error control in which a destination station sends a negative acknowledgment (NAK) when it receives an error. The source station receiving the NAK will retransmit the frame in error plus all succeeding frames transmitted in the interim. Problems Q1. What is the purpose of using modulo-2 arithmetic rather than binary arithmentic in computing an FCS? QA

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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 in 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 applicationlevel procedures. It is basically an unreliable service, delivery and duplicate protection are not guaranteed.. However, this reduces the overhead of the protocol and may be adequate in many cases. For example, - Inward data collection where loss of occasional data unit may not cause problem. Request-Response in transaction service where application can handle error. - Run-time applications such as voice video involving a degree of redundancy or realtime requirement. UDP sits on top of IP. Because it is connectionless, UDP may have very little to do. Essentially it adds port-addressing capability to IP.

UDP provides the source and destination port addresses and a checksum that covers the data field. These functions would not normally be performed by protocols above the transport layer. Thus UDP provides a useful, though limited, service.

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