We use cookies to give you the best experience possible. By continuing we’ll assume you’re on board with our cookie policy

See Pricing

Hire a Professional Writer Now

The input space is limited by 250 symbols

Choose 3 Hours or More.
Back
2/4 steps

How Many Pages?

Back
3/4 steps

Back
Get Offer

# Solution for Communication Systems Essay

Hire a Professional Writer Now

The input space is limited by 250 symbols

Write my paper

Solutions Manual for: Communications Systems, 5th edition by Karl Wiklund, McMaster University, Hamilton, Canada Michael Moher, Space-Time DSP Ottawa, Canada and Simon Haykin, McMaster University, Hamilton, Canada Published by Wiley, 2009. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Chapter 2 2. 1 (a) g (t ) = A cos(2? f c t ) fc = 1 T ? ?T T ? t?? , ? ? 2 2? We can rewrite the half-cosine as: ? t ? A cos(2? f c t ) ? rect ? ? ? T ? Using the property of multiplication in the time-domain: G ( f ) = G1 ( f ) ? G2 ( f ) 1 sin(? fT ) [? ( f ? fc ) + ? ( f + fc )] ? AT ? fT 2 Writing out the convolution: ?

AT ? sin(?? T ) ? G( f ) = ? ? ? [? (? ? ( f + f c ) + ? (? ? ( f ? f c ) ] d ? 2 ? ?? T ? ?? = A ? sin(? ( f + f c )T ) sin(? ( f ? f c )T ) ? + ? ? f + fc f ? fc 2? ? ? ? ? A ? cos(? fT ) cos(? fT ) ? = ? ? 1 ? 2? ? f ? 1 ? f+ 2T 2T ? ? = (b)By using the time-shifting property: T exp(? j 2? ft0 ) g (t ? t0 ) t0 = 2 ? ? A ? cos(? fT ) cos(? fT ) ? G( f ) = ? ? exp(? j? fT ) ? 1 ? 2? ? f ? 1 ? f+ ? 2T 2T ? fc = 1 2T Copyright © 2009 John Wiley & Sons, Inc.

Don't use plagiarized sources. Get Your Custom Essay on
Solution for Communication Systems
Just from \$13,9/Page

All Rights Reserved. (c)The half-sine pulse is identical to the half-cosine pulse except for the centre frequency and time-shift. fc = 1 2Ta ? cos(? fTa ) cos(? fTa ) ? ? ? ? (cos(? fTa ) ? j sin(? fTa )) f + fc ? ? f ? fc A ? cos(2? fTa ) cos(2? fTa ) sin(2? fTa ) sin(2? fTa) ? = ? +j ? j ? ? 4? ? f ? f c f + fc f ? fc f + fc ? A 2? A ? exp(? j 2? fTa ) exp(? j 2? fTa ) ? ? ? ? 4? ? f ? fc f + fc ? G( f ) = = (d) The spectrum is the same as for (b) except shifted backwards in time and multiplied by -1. ? ? A ? cos(? fT ) cos(? fT ) ? ? ? exp( j? fT ) G( f ) = ? 1 ? 2? ? f ? 1 ? f+ 2T 2T ? ? ? ? A ? exp( j 2? fT ) exp( j 2? fT ) ? = ? ? ? 1 4? ? f ? 1 ? f+ 2T 2T ? ? (e) Because the Fourier transform is a linear operation, this is simply the summation of the results from (b) and (d) ? A ? exp( j 2? fT ) + exp(? j 2? fT ) exp( j 2? fT ) + (? j 2? fT ) ? ? G( f ) = ? ? 1 1 4? ? ? f? f+ 2T 2T ? ? ? ? A ? cos(2? fT ) cos(2? fT ) ? = ? ? 1 ? 2? ? f ? 1 ? f+ 2T 2T ? ? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 2 g (t ) = exp(? t ) sin(2? f c t )u(t ) = ( exp(? t )u(t ) )( sin(2? f c t ) ) ? 1 ? 1 ? ? (? ( f ? f c ) ? ? ( f + f c ) ) ? 1 + j 2? f ? 2 j ? ? 1 ? 1 1 = ? ? ? 2 j ? 1 + j 2? ( f ? f c ) 1 + j 2? ( f + f c ) ? ? G( f ) = 2. 3 (a) g (t ) = g e (t ) + g o (t ) 1 [ g (t ) + g (? t )] 2 ? t ? g e (t ) = Arect ? ? ? 2T ? g e (t ) = 1 [ g (t ) ? g (? t )] 2 ? ? 1 ? ? 1 ? t ? 2T ? ?t+ T ? rect ? 2 g o (t ) = A ? rect ? ? ? ? T ? ? T ? ? ? ? ? g o (t ) = ?? ?? ?? ?? ? ?? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. (b) By the time-scaling property g(-t) Ge ( f ) = G(-f) 1 [G ( f ) + G (? f ) ] 2 1 = [sinc( fT ) exp(? j 2? fT ) + sinc( fT ) exp( j 2? fT ) ] 2 = sinc( fT ) cos(? fT ) Go ( f ) = 1 [G ( f ) ? G (? f )] 2 1 = [sinc( fT ) exp(? j 2? fT ) ? sinc( fT ) exp( j 2? fT ) ] 2 = ? jsinc( fT ) sin(? fT ) Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 4. We need to find a function with the stated properties. We can verify that: G ( f ) = ? sgn( f ) + ju( f ? W ) ? ju(? f ? W ) meets the stated criteria. By duality g(f) G(-t) g (t ) = ?1 1 ? j ? ? (t ) ? ? exp(? j 2? Wt ) ? j 2? t ? ?2 1 sin(2? Wt ) = +j 2? t ? t 1 + ? t ? ? u2 ? exp ? ? 2 ? du ? ? ? ? ? t ? T 1 1 t +T ?1 1 ? j ? ? (t ) ? ? exp( j 2? Wt ) j 2? t ? ?2 2. 5 g (t ) = = ? t ? T ? 0 h(? )d? + 1 t +T ? ? h(? )d? 0 dg (t ) 1 1 = ? h(t ? T ) + h(t + T ) dt ? ? By the differentiation property: ? dg (t ) ? F? ? = j 2? fG ( f ) ? dt ? 1 = [ H ( f ) exp( j 2? f ? ) ? H ( f ) exp(? j 2? f ? )] ? = 2j ? H ( f ) sin(2? f ? ) But H ( f ) = ? exp(?? f 2? 2 ) 1 exp(?? f 2? 2 ) sin(2? fT ) ?

All Rights Reserved. 2. 9 a) Let h(t ) = g1 (t ) ? g 2 (t ) dh(t ) j 2? fH ( f ) dt = j 2? fG1 ( f )G2 ( f ) = ( j 2? fG1 ( f ) ) G2 ( f ) ? dg1 (t ) ? ? dt ? ? g 2 (t ) ? ? d ? dg (t ) ? ? [ g1 (t ) ? g 2 (t )] = ? 1 ? ? g 2 (t ) dt ? dt ? ( j 2? fG1 ( f ) ) G2 ( f ) b) ?? ? g (t ) ? g (t )dt 1 2 t 1 G (0)G2 (0) ? ( f ) G1 ( f )G2 ( f ) + 1 j 2? f 2 ? 1 ? ? G (0) ? =? G1 ( f ) ? G2 ( f ) + ? 1 ? ( f ) ? G2 ( f ) ? 2 ? ? j 2? f ? ? 1 ? G (0) =? G1 ( f ) + 1 ? ( f ) ? G2 ( f ) 2 ? j 2? f ? t ? t ? ? ? g1 (t ) ? g 2 (t )dt = ? ? g1 (t ) ? ? g 2 (t ) ?? ? ?? ? Y( f ) = 2. 10. ?? ? X (? ) X ( f ?? )d? t X (? ? 0 if ? ? W X ( f ?? ) ? 0 if f ?? ? W ( f ?? ) ? W for f ? W +? when ? ? 0 and ? ? W ( f ?? ) ? ?W for f ? ?W +? when ? ? 0 and ? ? ? W ? ( f ?? ) ? W for 0 ? ? ? W when f ? 2W ( f ?? ) ? ?W for -W ? ? ? 0 when f ? ?2W ? Over the range of integration [ ? W ,W ] , the integral is non-zero if f ? 2W Copyright © 2009 John Wiley , Inc. All Rights Reserved. 2. 11 a) Given a rectangular function: g (t ) = always equal to 1, and the height is 1/T. 1 ? t ? rect ? ? , for which the area under g(t) is T ? T ? 1 ? t? rect ? ? T ? T ? sinc( fT ) Taking the limits: 1 ? t? lim rect ? ? = ? (t ) T >0 T ? T ? lim sinc( fT ) = 1 T >0 T b) g (t ) = 2Wsinc(2Wt ) 2Wsinc(2Wt ) ? f ? rect ? ? ? 2W ? W >? lim 2Wsinc(2Wt ) = ? (t ) ? 2 ? lim rect ? ? =1 ? 2W ? W >? 2. 12. G( f ) = 1 1 + sgn( f ) 2 2 By duality: 1 1 ? ( ? t ) ? j 2? t 2 j 1 ? g (t ) = ? (t ) + 2 2? t G( f ) Copyright © 2009 John Wiley , Inc. All Rights Reserved. 2. 13. a) By the differentiation property: 2 ( j 2? f ) G ( f ) = ? ki exp(? j 2? fti ) i ?G( f ) = ? 1 4? 2 f 2 ? k exp(? j 2? ft ) i i i b)the slope of each non-flat segment is: ± A tb ? t a ? 1 ?? A ? G( f ) = ? ? 2 2 ? ? ? [ exp( j 2? ftb ) ? exp( j 2? fta ) ? exp( j 2? fta ) + exp( j 2? tb ) ] ? 4? f ? ? tb ? ta ? A =? 2 2 [cos(2? ftb ) ? cos(2? fta )] 2? f ( tb ? ta ) 1 But: sin(? f (tb ? ta )) sin(? f (tb + ta )) = [ cos(2? fta ) ? cos(2? ftb ) ] by a trig identity. 2 A ? G( f ) = 2 2 [sin(? f (tb ? ta )) sin(? f (tb + ta ))] ? f (tb ? ta ) 2. 14 a) let g(t) be the half cosine pulse of Fig. P2. 1a, and let g(t-t0) be its time-shifted counterpart in Fig. 2. 1b ? = G ( f )G* ( f ) = G( f ) 2 ( G ( f ) exp(? j 2? ft0 ) ) ( G* ( f ) exp( j 2? ft0 ) ) = ( G ( f ) exp(? j 2? ft0 ) ) ( G* ( f ) exp( j 2? ft0 ) ) = G ( f ) exp(? j 2? ft0 ) exp( j 2? ft0 ) 2 G( f ) 2 Copyright © 2009 John Wiley , Inc.

All Rights Reserved. 2. 14 b)Given that the two energy densities are equal, we only need to prove the result for one. From before, it was shown that the Fourier transform of the half-cosine pulse was: AT 1 [sinc(( f + fc )T ) + sinc(( f ? f c )T )] for fc = 2 2T After squaring, this becomes: sin(? ( f + f c )T ) sin(? ( f ? f c )T ) ? A2T 2 ? sin 2 (? ( f + f c )T ) sin 2 (? ( f ? f c )T ) + +2 ? ? 4 ? (? ( f + f c )T ) 2 (? ( f ? f c )T ) 2 ? 2T 2 ( f + f c )( f ? f c ) ? The first term reduces to: ?? ? sin 2 ? ? fT + ? 2 cos 2 (? fT ) 2 ? cos (? fT ) ? = = 2 2 2 2 2 ?? ? ? ? T ( f + fc ) ? ? ? ? fT + ? ? ? fT + ? ? 2? ? ? The second term reduces to: ?? ? sin 2 ? ? fT ? ? cos 2 (? fT ) 2? ? = 2 2 2 2 ?? ? T ( f ? fc ) ? ? fT ? ? ? 2? ? The third term reduces to: sin(? ( f + f c )T ) sin(? ( f ? f c )T ) cos(? ) ? cos 2 (2? fT ) = 1 ? ? 2T 2 ( f + f c )( f ? f c ) ? ? 2T 2 ? f 2 ? 2 ? 4T ? ? ? 1 ? cos(2? fT ) = 1 ? ? ? 2T 2 ? f 2 ? 2 ? 4T ? ? 2 cos 2 (? fT ) =? 1 ? ? ? 2T 2 ? f 2 ? 2 ? 4T ? ? Summing these terms gives: ? ? ? 2 2 2 2 ? 2 cos (? fT ) A T ? cos (? fT ) cos (? fT ) ? + ? 2 2 2 1 ?? 1 ?? 4? 2T 2 ? ? ? 1 ? 1 ? ? ? f+ ?? f ? ? ?? f + ? ?f? ? 2T ? ? 2T ? ? ? 2T ? 2T ? ? ?? ? 2 Copyright © 2009 John Wiley , Inc.

All of the other product components can be discarded. (a) Given the sum of the modulated carrier waves, the individual message signals are extracted by multiplying the signal with the required carrier. For m1(t), this results in the conditions: cos(? 1 ) + cos( ? 1 ) = 0 cos(? 2 ) + cos( ? 2 ) = 0 cos(? 3 ) + cos( ? 3 ) = 0 ?? i = ? i ± ? For the other signals: m2 (t ) : cos(?? 1 ) + cos(? ?1 ) = 0 cos(? 2 ? ?1 ) + cos( ? 2 ? ?1 ) = 0 cos(? 3 ? ?1 ) + cos( ? 3 ? ?1 ) = 0 Similarly: m3 (t ) : (? 1 ? ? 2 ) = ( ? 1 ? ? 2 ) ± ? (? 3 ? ? 2 ) = ( ? 3 ? ? 2 ) ± ? m4 (t ) : (? 1 ? ? 3 ) = ( ? 1 ? ?3 ) ± ? (? 2 ? ? 3 ) = ( ? ? ? 3 ) ± ? ?1 = ? 1 ± ? (? 2 ? ?1 ) = ( ? 2 ? ?1 ) ± ? (? 3 ? ?1 ) = ( ? 3 ? ?1 ) ± ? (b) Given that the maximum bandwidth of mi(t) is W, then the separation between fa and fb must be | fa- fb|>2W in order to account for the modulated components corresponding to fa- fb. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 3. 25 b) The charging time constant is (rf + Rs )C = 1? s The period of the carrier wave is 1/fc = 50 ? s. The period of the modulating wave is 1/fm = 0. 025 s. ?The time constant is much shorter than the modulating wave and therefore should track the message signal very well.

The discharge time constant is: Rl C = 100? s . This is twice the period of the carrier wave, and should provide some smoothing capability. From a maximum voltage of V0, the voltage Vc across the capacitor after time t = Ts is: T Vc = V0 exp(? s ) Rl C Using a Taylor series expansion and retaining only the linear terms, will result in the T linear approximation of VC = V0 (1 ? s ) . Using this approximation, the voltage will Rl C decay by a factor of 0. 94 from its initial value after a period of Ts seconds. From the code, it can be seen that the voltage decay is close to this figure.

However, it is somewhat slower than what was calculated using the linear approximation. In a real circuit, it would also be expected that the decay would be slower, as the voltage does not simply turn off, but rather decreases over time. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 3. 25 c) The output of a high-pass RC circuit can be described according to: V0 (t ) = I (t ) R Qc (t ) = C (Vin (t ) ? V0 (t )) dQc dt ? dV (t ) dV (t ) ? V0 (t ) = RC ? in ? 0 ? dt ? ? dt Using first order differences to approximate the derivatives results in the following difference equation: RC RC V0 (t ) = V0 (t ? ) + (Vin (t ) ? Vin (t ? 1)) RC + Ts RC + Ts I (t ) = The high-pass filter applied to the envelope detector eliminates the DC component. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 3. 25. MATLAB code function [y,t,Vc,Vo]=AM_wave(fc,fm,mi) %Problem 3. 25 %Inputs: fc % fm % mi Carrier Frequency Modulation Frequency modulation index %Problem 3. 25 (a) fs=160000; %sampling rate deltaT=1/fs; %sampling period t=linspace(0,. 1,. 1/deltaT); %Create the list of time periods y=(1+mi*cos(2*pi*fm*t)). *cos(2*pi*fc*t); %Create the AM wave %Problem 3. 5 (b) %%%%Create the envelope detector%%%% Vc=zeros(1,length(y)); Vc(1)=0; %inital voltage for k=2:length(y) if (y(k)>(Vc(k-1))) Vc(k)=y(k); else Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Vc(k)=Vc(k-1)-0. 023*Vc(k-1); end end %Problem 3. 25 (c) %%%Implement the high pass filter%%% %%This implements bias removal Vo=zeros(1,length(y)); Vo(1)=0; RC=. 001; beta=RC/(RC+deltaT); for k=2:length(y) Vo(k)=beta*Vo(k-1)+beta*(Vc(k)-Vc(k-1)); end Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Chapter 4 Problems Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

From a table of Bode plots, the following gain response can be obtained: | X 1 ( f ) |= 1 ? f ? fB ? 1+ ? ? ? B ? 2 Where fB is the frequency of the resonant peak, and B is the bandwidth. For the slope circuit, B is the filter’s bandwidth or cutoff frequency. For convenience, we can shift the filter to the origin (with X 1 ( f ) as the shifted version). | X 1 ( f ) |= 1 ? f ? 1+ ? ? ? B? =? f = kB 2 d | X1 ( f ) | df k B(1 + k ) 3 2 2 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Because the filters are symmetric about the central frequency, the contribution of the second filter is identical.

Adding the filter responses results in the slope at the central frequency being: d | X(f )| df =? f = kB 2k B(1 + k ) 3 2 2 In the original definition of the slope filter, the responses are multiplied by -1, so do this here. This results in a total slope of: 2k B(1 + k ) 3 2 2 As can be seen from the following plot, the linear approximation is very accurate between the two resonant peaks. For this plot B = 500, f1=-750, and f2=750. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

All Rights Reserved. Problem 4. 26. a) Beta 1 2 5 10 # of side frequencies 1 2 8 14 b)By experimentation, a modulation index of 2. 408, will force the amplitude of the carrier to be about zero. This corresponds to the first root of J0(? ), as predicted by the theory. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 4. 27. a)Using the original MATLAB script, the rms phase error is 6. 15 % b)Using the plot provided, the rms phase error is 19. 83% Problem 4. 28 a)The output of the detected signal is multiplied by -1. This results from the fact that m(t)=cos(t) is integrated twice.

Once to form the transmitted signal and once by the envelope detector. In addition, the signal also has a DC offset, which results from the action of the envelope detector. The change in amplitude is the result of the modulation process and filters used in detection. f ? ? b)If s (t ) = sin(2? f mt ) + 0. 5cos ? 2? m t ? , then some form of clipping is observed. 3 ? ? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. The above signal has been multiplied by a constant gain factor in order to highlight the differences with the original message signal. c)The earliest signs of distortion start to appear above about fm =4. kHz. As the message frequency may no longer lie wholly within the bandwidth of either the differentiator or the low-pass filter. This results in the potential loss of high-frequency message components. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 4. 29. By tracing the individual steps of the MATLAB algorithm, it can be seen that the resulting sequence is the same as for the 2nd order PLL. e(t ) is the phase error ? e (t ) in the theoretical model. The theoretical model of the VCO is: ?2 (t ) = 2? kv ? v(t )dt 0 t and the discrete-time model is: VCOState = VCOState + 2? kv (t ? )Ts which approximates the integrator of the theoretical model. The loop filter is a PI-controller, and has the transfer function: a H ( f ) = 1+ jf This is simply a combination of a sum plus an integrator, which is also present in the MATLAB code: Filterstate = Filterstate + e(t ) Integrator v(t ) = Filterstate + e(t ) Integrator +input b)For smaller kv, the lock-in time is longer, but the output amplitude is greater. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. c)The phase error increases, and tracks the message signal. d)For a single sinusoid, the track is lost if f m ?

K 0 where K 0 = k f kv Ac Av For this question, K0=100 kHz, but tracking degrades noticeably around 60-70 kHz. e)No useful signal can be extracted. By multiplying s(t) and r(t), we get: Ac Av ? sin(k f ? ? VCOState) + sin(4? f c t + k f ? + VCOState) ? ? 2 ? This is substantially different from the original error signal, and cannot be seen as an adequate approximation. Of particular interest is the fact that this equation is substantially more sensitive to changes in ? than the previous one owing to the presence of the gain factor kv Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc.

All Rights Reserved. Chapter 5 Problems 5. 1. ( x ? ? x )2 exp(? ) (a) Given f ( x) = 2 2 2? x 2?? x 1 exp(?? f 2 ) , then by applying the time-shifting and scaling properties: 2 2 2?? x exp(?? ( 2?? x ) 2 ? f 2 ) exp( j 2? f ? x ) and exp(?? t 2 ) F( f ) = 1 2?? 2 x 2 = exp(?? 2 2? x f 2 + j ? x 2? f ) 1 2 = exp( j?? x ? ? 2? x ) 2 and let ? = 2? f (b)The value of ? x does not affect the moment, as its influence is removed. Use the Taylor series approximation of ? x(x), given ? x = 0. ?x (? ) = exp(? ? 2? x2 ) x2 exp( x) = ? n =0 n ! E[ X n ] = d n? x (? ) d? n v =0 k ? 1 2 2k 2k ? ? 1? ? ? ? ? x (? ) = ? ? ? ? x k! 2? k =0 ?

All Rights Reserved. Problem 5. 28 c)For a given filter, H ( f ) , let ? = ln H ( f ) ? and the Paley-Wiener criterion for causality is: ?? ? 1 + (2? f ) df < ? 2 ?( f ) For the filter of part (b) 1 ? ( f ) = [ ln(2) + ln( S x ( f ) ? ln( N 0 )] 2 The first and the last terms have no impact on the absolute integrability of the previous expression, and so do not matter as far as evaluating the above criterion. This leaves the only condition: ? ln S x ( f ) ? 1 + (2? f )2 df < ? ?? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 29 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 30

Distance 0? 1? 2? 3? 4? Relative Error 0. 94% 2. 6 % 4. 8 % 47. 4% 60. 7% The error increases further out from the centre. It is also important to note that the random numbers generated by this MATLAB procedure can never be greater than 5. This is very different from the Gaussian distribution, for which there is a non-zero probability for any real number. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 5. 34 Code Listing %Problem 5. 34 %Set the number of samples to be 20,000 N=20000 M=100; Z=zeros(1,20000); for i=1:N for j=1:5 Z(i)=Z(i)+2*(rand(1)-0. ); end end sigma=sqrt(var(Z-mean(Z))); %Calculate a histogram of Z [X,C]=hist(Z,M); l=linspace(C(1),C(M),M); %Create a gaussian function with the same variance as Z G=1/(sqrt(2*pi*sigma^2))*exp(-(l. ^2)/(2*sigma^2)); delta2=abs(l(1)-l(2)); X=X/(20000*delta2); Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 5. 35 (a) For the generated sequence: ? ? y = ? 0. 0343 + j 0. 0493 ? 2 ? y = 5. 597 The theoretical values are: ? y = 0 (by inspection). 2 The theoretical value of ? y =5. 56. See 5. 35 (c) for the calculation. 5. 35 (b) From the plots, it can be seen that both the real and imaginary components are approximately Gaussian.

In addition, from statistics, the sum of tow zero-mean Gaussian signals is also Gaussian distributed. As a result, the filter output must also be Gaussian. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 5. 35 (c) y (n) = ay (n ? 1) + w(n) Y ( z ) = aY ( z ) z ? 1 ? H ( z) = 1 1 ? az ? 1 -1 h( n) = a u ( n) n Rh(z) = H(z)H(z ) = 1 (1 ? az )(1 ? az ) ?1 = But, Ry(z) = Rh(z)Rw(z) Taking the inverse z-transform: ry (n) = 2 ? w a z ? 1 1 1 + 2 ? 1 2 1 ? a 1 ? az 1 ? a 1 ? az 1 ? a2 an ?? < n < ? From the plots, the measured and observed autocorrelations are almost identical. Copyright © 2009 John Wiley & Sons, Inc.

The following Matlab script simulates the generation and detection of an AM-modulated signal in noise. %—————————————-% Matlab code for Problem 6. 16 %—————————————function Prob6_16() Fs = 143; % sample rate (kHz) t = [0: 1/Fs : 100]; % observation period (ms) Fc = 20; % carrier frequency (kHz) Fm = 0. 1; % modulation frequency (kHz) Ka = 0. 5; % modulation index SNRc = 25; % Channel SNR (dB) Ac = 1; tau = 0. 25/4; %————————————% Modulated signal %————————————m = cos(2*pi*fm*t); C = Ac*cos(2*pi*fc*t); s = (1 + ka*m). c; subplot(4,1,1), plot(t,s), grid on P = std(m)^2; %—————————————————% Add narrowband noise % Create bandpass noise by low-pass % filtering AWGN noise and converting to % bandpass %———————————————————–P_AM = Ac^2*(1+ka^2*P)/2; N = P_AM/10. ^(SNRc/10); sigma = sqrt(N); %— Create bandpass noise by low-pass filtering complex noise –noise = randn(size(s)) + j*randn(size(s)); LPFnoise = LPF(Fs, noise, tau); BPnoise = real(LPFnoise . * exp(j*2*pi*fc/Fs*[1:length(s)])); scale = 2*sigma / std(BPnoise); s_n = s scale * BPnoise; subplot(4,1,2), plot(t,s_n), grid on %— Envelope detection of both noisy and noise-free signals –ED = EnvDetector(t,s); ED_n = EnvDetector(t,s_n); %— Remove transient and dc –ED = ED(400:end); ED_n = ED_n(400:end); t = t(400:end); ED = ED – mean(ED); ED_n = ED_n – mean(ED_n); %— Low pass filter —BBsig = LPF(Fs,ED,tau); BBsig_n = LPF(Fs,ED_n,tau); Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. %— plot results —–subplot(4,1,3), plot(t,BBsig); subplot(4,1,4), plot(t,BBsig_n) %————————————–% Envelope Detector from Problem 3. 5 %————————————–function Vc = EnvDetector(t,s); Vc(1) = 0; % initial capacitor voltage for i = 2:length(s) if s(i) > Vc(i-1) % diode on Vc(i) = s(i); else % diode off Vc(i) = Vc(i-1) – 0. 023*Vc(i-1); end end % plot(t, Vc), grid on return; %————————————–% Low pass filter %————————————–function y = LPF(Fs, x, tau); % tau = 1; % time constant of RC filter (ms) t1 = [0: 1/Fs : 5*tau]; h = exp(-t1/tau) * 1/Fs; y = filter(h, 1, x); return; The Matlab script produces the following plot: Copyright © 2009 John Wiley & Sons, Inc.

All Rights Reserved. 2 0 -2 2 0 -2 0. 05 0 -0. 05 0. 05 0 -0. 05 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 Time (ms) 70 80 90 100 Figure 6. 16 Plot from Matlab script (a) AM modulated carrier (b) AM modulated carrier plus noise (c) AM demodulated signal in absence of noise (d) AM demodulated signal in noise Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 17 The following Matlab script simulates the generation and detection of an FM modulated signal in noise. —————————————-% Problem 6. 17 %—————————————-function b = Prob6_17; %— Parameters ——————fc = 100; % Carrier frequency (kHz) Fs = 1024; % Sampling rate (kHz) fm = 0. 5; % Modulating frequency (kHz) Ts = 1/Fs; % Sample period (ms) t = [0:Ts:10]; % Observation period (ms) C_N = 20 % channel SNR (dB) Ac = 1; Bt = 20 % (kHz) W = 5; % (kHz) SNRc = C_N+10*log10(Bt/W); %— Message signal ————–m = cos(2*pi*fm*t); % modulating signal kf = 2. ; % modulator sensitivity index (~Bt/2) (kHz/V) %— FM modulate —————FMsig = FMmod(fc,t,kf,m,Ts); %— Add narrowband noise ——–%— Create bandpass noise by low-pass filtering complex noise –P = 0. 5; N = P/10. ^(SNRc/10); sigma = sqrt(N); noise = randn(size(FMsig)) + j*randn(size(FMsig)); LPFnoise = LPF(Fs, noise, 0. 05); % 0. 01 => Bt ` 50 kHz eq. Noise BW BPnoise = real(LPFnoise . * exp(j*2*pi*fc/Fs*[1:length(FMsig)])); scale = sigma / std(BPnoise); FMsign = FMsig + scale * BPnoise; subplot(4,1,1), plot(t,FMsig), grid on subplot(4,1,2), plot(t,FMsign), grid on — FM receiver —Rx_c = FMdiscriminator(fc,FMsig,Ts); Rx_n = FMdiscriminator(fc,FMsign,Ts); t = t(round(1/Ts):end); % remove transient subplot(4,1,3), plot(t,Rx_c); grid on subplot(4,1,4), plot(t,Rx_n); grid on %— Plot result ——–% FFTsize = 4096; %S = spectrum(FMsig,FFTsize); % % Freq = [0:Fs/FFTsize:Fs/2]; Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. % subplot(2,1,1), plot(t,s), xlabel(‘Time (ms)’), ylabel(‘Amplitude’); % axis([0 0. 5 -1. 5 1. 5]), grid on % subplot(2,1,2), stem(Freq,sqrt(S/682)), xlabel(‘Frequency (kHz)’), ylabel(‘Amplitude Spectrum’); % axis([95 105 0 1]), grid on ———————————————–% FM modulator %———————————————–function s = FMmod(fc,t,kf,m,Ts); theta = 2*pi*fc*t+ 2*pi*kf * cumsum(m)*Ts; % integrate signal s = cos(theta); %———————————————–% FM discriminator %———————————————–function D3 = FMdiscriminator(fc,S, Ts) t = [0:Ts:10*Ts]; % for filter %— FIR differentiator (Fs = 1024 kHz, BT/2 = 10 kKhz) –FIRdiff = [ 1. 0385 0. 0 0. 0 0. 0 -0. 0 0. 0 0. 0 -0. 0 -0. 0 -0. 0 -1. 60385]; BP_diff = real(FIRdiff . * exp(j*2*pi*fc*t)); %— Lowpass filter – Fs = 1024 kHz, f3dB = 5 kHz —-LPF_B = 1E-4 *[ 0. 0706 0. 2117 0. 2117 0. 0706]; LPF_A = [1. 0000 -2. 9223 2. 8476 -0. 252]; D1 = filter(BP_diff, 1, S); % Bandpass discriminator D2 = EnvDetect(D1); % Envelope detection D2 = D2 – mean(D2); % remove dc D3 = filter(LPF_B,LPF_A, D2); % Low-pass filtering D3 = D3(round(1/Ts):end); % remove transient (approx 1s) %————————————–% Envelope Detector %————————————–function Vc = EnvDetect(s); Vc(1) = 0; % initial capacitor voltage for i = 2:length(s) if s(i) > Vc(i-1) % diode on Vc(i) = s(i); else % diode off Vc(i) = Vc(i-1) – 0. 05*Vc(i-1); end end return; %————————————–% Low pass filter %————————————–function y = LPF(Fs, x, tau); % tau = 1; % time constant of RC filter (ms) t1 = [0: 1/Fs : 5*tau]; h = exp(-t1/tau) * 1/Fs; y = filter(h, 1, x); return; Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

### Cite this Solution for Communication Systems Essay

Show less • Use multiple resourses when assembling your essay
• Get help form professional writers when not sure you can do it yourself
• Use Plagiarism Checker to double check your essay