Digital Communication Systems Third year communications Midterm exam (15 points)

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Mitchell Wilkerson
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1 Name: Section: BN: Digital Communication Systems Third year communications Midterm exam (15 points) May 2011 Time: 1.5 hours 1- Determine if the following sentences are true of false (correct answer 0.5 point, wrong or no answer 0).: 1 Any Random process consists of infinite number of RVs. One random variable at True each time instant 2 If X(t)=cos(2*pi*t+theta) is a R.P., where theta takes two values (0,pi) with equal False probabilities. Then X(t=1/4) is a R.V. that can take two values only with equal probabilities. [It will only take one value, which is 0] 3 The mean of X(t) in (2) is cos(2*pi*t+pi/2) [The mean is 0] False 4 X(t) in (2) is stationary [non stationary as the pdf of the R.V at different time False instants is not the same] 5 Any random process should start at inf and end at +inf [The temperature False random process that we take in the first lecture] 6 Power Spectral Density (PSD) exists for all R.Ps [only for stationary R.P] False 7 Any periodic stationary random process should have tones in its PSD True 8 When a stationary R.P passes by a time variant system, the output R.P will still be False stationary. [because it is time variant, not time invariant] 9 Matched filter (M.F.) is used to remove the effect of noise from the signal [it only maximizes the SNR after the filter, but there no filter that removes the effect of noise] False 10 When the M.F. of the signal is multiplied by a real number, it will still be a True matched filter 11 For the same received signal to noise ratio, the M.F. is the best filter that will True result in the minimum BER. [because it maximizes the SNR after the filter] 12 Intersymbol interference is the interference due to the transmission of two False symbols on the real and imaginary components at the same time [wrong definition] 13 If there is infinite BW channel, no external interference and no noise effect, then False the maximum symbol rate on this channel is 1M symbols/second [no limit] 14 The power of unfiltered AWGN is infinity True 15 When sending a raised cosine signal from the transmitter through a channel with False δ(t) impulse response and receiving it with a MF, the system will have zero ISI 1 / 5

2 [square root raised cosine should be transmitted. The conv between raised cosine at the transmitter and its MF at the receiver will not give zero ISI] 16 When sending a sinc signal with zero crossings at multiples of the symbol duration from the transmitter through a channel with impulse response = δ(t)+ δ(t- T/4) and receiving it with a M.F., the system will have zero ISI [channel effect should be removed by an equalizer] False 2 / 5

3 2- Fill in the missing parts of the code (2.5 points, any mistake = -0.5 points.) : It is required to generate an ensemble that consists of 500 waveforms, each containing 100 bits, for the polar NRZ line code and compute its statistical mean. The sampling time of the waveforms is 10ms, and the bit duration is 100ms. %% Polar NRZ code clear num_samples_per_bit = 10; % the sampling time is 10 ms, and the bit duration is 70 ms num_bits = 100; % number of bits per waveform total_samples = num_samples_per_bit * (num_bits);% total number of samples in the waveform s1=ones(1,num_samples_per_bit); % logic '1' is mapped to +1, but this will be repeated 7 times to be sent as a symbol s0=-ones(1,num_samples_per_bit);% logic '0' is mapped to -1, but this will be repeated 7 times to be sent as a symbol %%% generate num_bits symbols tot_syms = []; num_waveforms = 500; % generate the waveforms with random initial shift for v=1:num_waveforms % loop over all waveforms random_shift = randint(1,1,[0 num_samples_per_bit-1]); % the random initial shift syms_to_transmit = []; for u=1:(num_bits+1) if(rand>0.5) s = s1; else s=s0; end; %send s1 or s0 with probability 50% if(u==1) % for the first bit, shift the transmitted symbol to the left by random_shift samples s=s(random_shift+1:end);% take the last 7-random_shift samples elseif(u==num_bits)% at the last bit, take only the first random_shift samples s=s(1:random_shift);% take the first random_shift samples end; syms_to_transmit = [syms_to_transmit s]; end tot_syms=[tot_syms ;syms_to_transmit]; end % for copyright purposes, part of the previous code was taken from a report % from last year's students and edited in it. ensemble_mean = sum(tot_syms)/num_waveforms; 3 / 5

4 3- Answer the following questions: 1- (3 points) Can these signals be the output of a signal passed by its M.F. (ignore the noise effect)? The signal has a symbol duration = T. Justify your answer. A 0 2T t No, for few reasons: 1- The output of the MF is the result of the convolution between two energy limited signals and it is not possible to have sharp edges at 0 and 2T 2- The M.F freq response is { H * (f)exp(-j2πft) }, Hence the freq response of the output of the M.F is H(f) 2 exp(- j2πft). That means that, by ignoring the linear phase which only represents the delay of the output signal by (T), the spectrum of the output of the matched filter should be +ve, because the term H(f) 2 is +ve. However, the response of the previous rect pulse is sinc in the frequency domain, which has both +ve and ve parts. A 0 T 2T t No, for few reasons: 1- The output of the MF is the result of the convolution between two energy limited signals and it is not possible to have sharp edges at 0 2- The M.F freq response is { H * (f)exp(-j2πft) }, Hence the result of the output of the M.F is H(f) 2 exp(-j2πft). That means that, by ignoring the linear phase which only represents the delay of the output signal by (T), the spectrum of the output of the matched filter should be even, because, for real signals, H(f) = H * (-f). Hence H(f) 2 = H(-f) 2, which means that it is both real and even. So, the output of the matched filter in the time domain should also be real and even function, shifted by (T). The function shown in the figure is not an even function shifted by (T), so it is not possible to be the output of a matched filter Try any real g(t) in MATLAB that has any shape and convolve it with its matched filter to verify that the output of the matched filter is an even function shifted at (T) 4 / 5

5 2- (2.5 points) An analog signal of bandwidth of 10 khz is sampled at a rate of 24 khz, quantized into 256 levels and coded using M-ary multi-amplitude pulses satisfying the Nyquist criterion with a roll-off factor 0.2. A 30 khz one sided bandwidth total of 60 khz in both positive and negative frequencies is available to transmit the data. Determine the smallest acceptable M? You should be able to solve this easy question! Note that the 10kHz at the first sentence will not be used Try to solve it! I guess you should understand this hint! 5 / 5

QUESTION BANK SUBJECT: DIGITAL COMMUNICATION (15EC61)

QUESTION BANK SUBJECT: DIGITAL COMMUNICATION (15EC61) Module 1 1. Explain Digital communication system with a neat block diagram. 2. What are the differences between digital and analog communication systems?