EE 661: Modulation Theory Solutions to Homework 6

Transcription

1 EE 66: Modulation Theory Solutions to Homework 6. Solution to problem. a) Binary PAM: Since 2W = 4 KHz and β = 0.5, the minimum T is the solution to (+β)/(2t ) = W = Hz. Thus, we have the maximum symbol rate /T = 2667 symbols/sec. The bit rate for binary modulation is then given by R b = (log 2 2) (/T ) = 2667 bits/sec. b) Similarly to the procedures in a), we have R b = 5333 bits/sec. c) Similarly to the procedures in a), we have R b = 8000 bits/sec. d) As discussed in previous homework 3, the minimum frequency separation for orthogonal constellation, denoted by f min, is equal to /(2T ) for coherent detection. For 4-FSK, we have f min = 2W/4 = 000 Hz, resulting in T = /(2 f min ) = sec. Then, the bit rate is given by R b = (/T ) log 2 4 = 4000 bits/sec. e) As discussed in previous homework 3, the minimum frequency separation for orthogonal constellation, denoted by f min, is equal to /T for coherent detection. For 4-FSK, we have f min = 2W/4 = 000 Hz, resulting in T = / f min = 0 3 sec. Then, the bit rate is given by R b = (/T ) log 2 4 = 2000 bits/sec. 2. Solution to problem 2. Without loss of generality, we set T = sec and assume that the discrete sampled (with sampling rate 50/T ) pulse shaper g(t) has unit energy, i.e., i= i= g(50i/t ) =. We plot the output of matched filter and the function y n I n 2 / I n 2 as follows: Fig.

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3 3. Solution to problem 3. The average SNR is given by SNR avg = N 0 + x 2 (τ)e s ; () x 2 ( nt + τ) n= n 0 the worst-case SNR is given by SNR worst = N 0 + n= n 0 x 2 (τ)e s x( nt + τ) 2, (2) where sinc(πt/t ) cos(πβt/t ) x(t) = g(t) g(t) = 4β 2 t 2, for RRC pulse; /T 2 sinc(πt/t ), for sinc pulse. (3) The plot of SNR as a function of τ is given in Fig. 3. Fig. 3 As shown in Fig. 3, RRC shaper does result in much higher worst-case SNR than the sinc shaper. However, the superiority of the average SNR is not that significant. We do need to give credits to RRC shaper for causing less SNR loss when there is timing offset. However, the main reasons for using RRC shaper are as follows: a) The sinc shaper is hard to implement in practical systems due to the sharp edge in frequency domain. b) The amplitude decaying speed of the RRC shaper is much higher than that of the sinc shaper. Thus, RRC shaper can be implemented with shorter filters, significantly reducing the complexity and processing delay. 3

4 4. Solution to problem 4. Observing the output of the matched filter y(t) at t = τ, y(τ) = k = k a k g(t kt ) g T (t τ) t=τ (4) a k sinc(2w(t kt )) sinc(2w(t τ)) t=τ (5) = k sin(2πw(τ kt )) 2πw(τ kt ) (6) Assuming the Nyquist criterion is met, T = 2w and hence y(τ) = a 0 sinc(2wτ) + sin(2πwτ) π k 0 ( ) k a k 2wτ k } {{ } ISI (7) As seen, the series in the ISI term is a divergent series. Furthermore if a k = ( ) k, the ISI would be infinite. 4

5 5. Solution to problem 5. Denote the transmitter frequency response and the receiver frequency response by G T (f) and G R (f), respectively. Then, the overall frequency response after matched filter is given by X(f) = G T (f)c(f)g R (f). (8) Assume that the output of matched filter have raised-cosine spectrum. For β = 0.5, we have T, { [ ( )]} if f T X(f) = 2 + cos 2πT f 4T, if 4T < f 3 (9) 0, otherwise. Then, the ISI-free system needs to satisfy T cos 2πfT X(f) G T (f) = C(f) = T 2, if f + sin 2πT f cos 2πfT, if 4T < f 3 0, otherwise. (0) Without loss generality, we set G T (f) = G R (f) = G T (f), () which corresponds to a real time-domain signals. Also, G T (f) is plotted in Fig. 4. Fig. 4 G T (f) versus f, where T is set equal to. 5

6 6. Solution to problem 6. The diagrams for (a) and (b) are plotted in Fig. 5. Fig. 5 Diagrams for ISI-free QPSK Transmitters. 6