EECS 120 Signals & Systems University of California, Berkeley: Fall 2005 Gastpar November 16, Solutions to Exam 2

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1 EECS 0 Signals & Systems University of California, Berkeley: Fall 005 Gastpar November 6, 005 Solutions to Exam Last name First name SID You have hour and 45 minutes to omplete this exam. he exam is losed-book and losed-notes; alulators, omputing and ommuniation devies are not permitted. No form of ollaboration between the students is allowed. If you are aught heating, you may fail the ourse and fae disiplinary onsequenes. However, two handwritten and not photoopied double-sided sheet of notes is allowed. Additionally, you reeive ables 3., 3., 4., 4., 5., 5., 9., 9. from the lass textbook. If we an t read it, we an t grade it. We an only give partial redit if you write out your derivations and reasoning in detail. You may use the bak of the pages of the exam if you need more spae. *** Good Luk *** Problem Points earned out of Problem 9 Problem 8 Problem 3 7 Problem 4 33 otal 7

2 Problem (Short Questions.) 9 Points (a) (4 Pts) For the system in Figure, H(jω) = {, for ω ω0 () Sketh the frequeny response G(jω) of the overall system between x(t) and y(t). x(t) H(j ) y(t) Solution: Figure : y(t) = x(t) x(t) h(t) Y (jω) = X(jω) X(jω)H(jω) = X(jω)( H(jω)) G(jω) = Y (jω) X(jω) = H(jω) Remark: his problem was a hint for the sampling system design in Problem 3.(). G(j ω) ω 0 ω 0 ω (b) (5 Pts) A ausal LI system is desribed by the following differential equation: d dt y(t) + d d y(t) + y(t) = x(t) x(t) () dt dt Is this system stable? Does this system have a ausal and stable inverse system? Solution: We take the Laplae transform of both sides of the differential equation to find the transfer funtion H(s) of the LI system. s Y (s) + sy (s) + Y (s) = s X(s) X(s) H(s) = Y (s) X(s) = s s + s + = (s + )(s ) (s ( + j))(s ( j)) he system H(s) has poles at s = +j and s = j, and zeros at s = and s =. Sine we are given that H(s) is ausal, the region of onvergene (ROC) of H(s) is Re{s} >. hus the ROC of H(s) inludes the jω -axis, whih implies that H(s) is stable.

3 Im j Re j he inverse system H(s) has poles at s = and s =, and zeros at s = + j and s = j. he inverse system (whih has a rational transfer funtion) is ausal iff the ROC of H(s) is the right-half plane. However the inverse system is stable iff the ROC inludes the jω -axis. herefore the inverse system annot be both ausal and stable. Im j Re j 3

4 x(t) os( t) r(t) os( t) H(j ) x(t) x(t) os( t) G(j ) r(t) ~ os( t) H(j ) x(t) y(t) H(j ) y(t) y(t) G(j ) H(j ) y(t) sin( t) sin( t) sin( t) sin( t) Figure : Quadrature modulation. Figure 3: Improved quadrature modulation. () (0 Pts) As you have seen in the homework, quadrature multiplexing is the system shown in Figure, where H(jω) = {, for ω ωm and G(jω) = {, for ω ω (3) Both original signals are assumed to be bandlimited: X(jω) = Y (jω) = 0, for ω > ω M ; and the arrier frequeny is ω > ω M. he interesting feature is that the effetive bandwidth of the signal r(t) is only ω M, the same as for a regular AM system with only the signal x(t). Hene, y(t) an ride along for free. Now, your olleague remembers single-sideband AM and suggests to add the filters G(jω) as shown in Figure 3. he effetive bandwidth of the transmitted signal r(t) is only ω M, half as muh as in the original quadrature multiplexing system Show that the improved system will not work. Hint: Find a pair of example spetra X(jω) and Y (jω) for whih R(jω) is not zero, but R(jω) = 0 for all ω. hen, argue (in a few keywords) why this invalidates the improved quadrature modulation. Solution: he basi fat to remember from the homework problem about quadrature modulation is that the spetra overlap and get added up. Consider the example spetra X(jω) and Y (jω) in the figure below. he spetrum X(jω) is purely real-valued. After multiplying by a os(ω t), it remains a purely real-valued spetrum. he trik is to selet the spetrum Y (jω) as purely imaginary; that way, after multiplying by the sin(ω t), it beomes a purely real-valued spetrum, and hene, there is a hane for it to anel out the spetrum X(jω). o atually make this happen, we still need to pik the right shape. One example that works is given in the figure below. Note that if R(jω) looks as skethed in the figure, then R(jω) will be zero the filter G(jω) removes anything below ω. 4

5 X(j ) Im{Y(j )} R(j ) R(j ) A few remarks: he signal with purely imaginary spetrum as given in Y (jω) is a real-valued signal (think of the spetrum of the sin funtion - it s purely imaginary). his is beause it is onjugate-symmetri ( Y (jω) = Y ( jω) ). Also, there are many other ways to onvine your olleague that the system will not work. A longer approah is to selet example spetra and to show that demodulation as suggested in Figure 3 will not reover the desired data-arrying signals x(t) and y(t). 5

6 Problem (Disrete-time proessing of ontinuous-time signals.) 8 Points x(t) H(j ) Sampling >nterval x[n] y[n] Delay /3 y[n] Conversion of disretetime sequene to inpulse train H r (j ) y(t) Figure 4: For the system in Figure 4, {, for ω π H(jω) = and H r (jω) = {, for ω π (4) (a) (0 Pts) Give the formula for the overall system response G(jω), relating x(t) and y(t). Also give a sketh of the magnitude G(jω), paying partiular attention to the labeling of the frequeny axis. No derivation is neessary to get full redit. Solution: he differene equation and frequeny response of the disrete-time blok an be found as: y[n] = x[n] + y[n ] 3 Y ( e jω) = X ( e jω) + 3 e jω Y ( e jω) G d ( e jω ) = 3 e jω here is no aliasing beause of the filter H(jω), so the ontinuous-time system response an be easily found using Equation 7.5 in OWN. G(jω) = { 3 e jω ω π 0 ω > π he magnitude G(jω) for ω < π an be found as follows: 6

7 G(jω) = = = = = 3 e jω 3 (os( ω ) + j sin( ω )) ( 3 os(ω )) + j ( 3 sin(ω )) 3 os(ω ) + 9 os (ω ) + 9 sin (ω ) os(ω ) G(j ω) 3/ 3/4 π/ π/ ω (b) (8 Pts) For x(t) = e jπt/( ), determine the orresponding output signal y(t). Your answer should not ontain an integral, but apart from that, there is no need to simplify it down. Solution: After writing x(t) = e j(π/( ))t we see that: ( y(t) = G j π ) e j(π/( ))t = ejπt/( ) = 3 e jπ/ + j e jπt/( ) 3 7

8 Problem 3 (Sampling System Design.) he signal x(t) has the Fourier transform shown in Figure 5. 7 Points X(j ) H(j ) $"/# $"/# "/# "/# $"/# "/# Figure 5: (a) (5 Pts) As a funtion of (as in Figure 5), determine the smallest sampling frequeny ω s = π/ s (where s is the sampling interval) for whih perfet reonstrution an be guaranteed for the signal x(t). A graphial justifiation (sketh with labels on the frequeny axis) is suffiient. Solution: By the sampling theorem, the bandlimited signal x(t), with X(jω) = 0 for ω > π, an be perfetly reonstruted if we sample at frequeny ω s ( ) π = 4π. Graphially, we see that the shifted replia of X(jω) that is entered at ω s does not overlap with the replia at 0 if ω s π π. X p (j ω) 3 ω s π/ π/τ ω s π/τ ω s ω (b) (0 Pts) Consider the signal y(t) = h(t) x(t), where h(t) is the impulse response of the filter H(jω) in Figure 5. Sketh the spetra of the two disrete-time signals x[n] = x(n ) (5) y[n] = y(n ), (6) where is the same as in Figure 5. Whih effet explains the differene between x[n] and y[n]? Solution: Now we sample the signals x(t) and y(t) with frequeny π, and then onvert the resulting impulse trains x p (t) and y p (t) to disrete-time signals x[n] and y[n]. In the frequeny domain, X p (jω) = k X(j(ω πk/ )) and X(ejΩ ) = X p (j Ω ) (and similarly for Y (ejω ) ). Note that sampling sales the vertial axis by, and onverting to disrete-time sales the frequeny axis by. 8

9 4/ X(e jω ) 3/ 4π π π 4π Ω Y(e jω ) 3/ / 4π π π 4π Ω Sine we are sampling x(t) at a frequeny less than ω s = 4π, the shifted replias of X(jω) overlap, produing the aliasing effet in X(e jω ). In ontrast, beause y(t) is bandlimited by π, the shifted replias of Y (jω) do not overlap, and there is no aliasing effet in the spetrum of Y (e jω ). 9

10 () ( Pts) he goal is now to implement a sampler with sampling interval 0 = /, where is as in Figure 5. Unfortunately, suh a fast sampler is not available in the urrent tehnology. Instead, you have aess to the following devies: samplers with sampling interval, where is the same as in Figure 5 (any number) anti-aliasing filters with the frequeny response given in Figure 5 (any number) ontinuous-time signal adders/subtrators (any number) any disrete-time proessing devies (ideal filters inluded). Draw the blok diagram of a system that takes as an input the signal x(t) (with spetrum as shown in Figure 5) as outputs the disrete-time signal x 0 [n] = x(n 0 ). Hint: o maximize your hane of partial redit, give spetral plots of intermediate signals in your system. Solution: he key insight is that we have to sample different parts of the signal separately. Clearly, we an low-pass filter the signal x(t) to obtain the signal y(t) in the figure below. Sine y(t) is now bandlimited to π/, we an sample it with our sampler (interval ) with no aliasing. Separately, we want to sample the high-pass part of the signal x(t). Here, we an use two standard triks: first, to get the high-pass part, we an just take x(t) and subtrat out the low-pass part, leading to the signal z(t) in the figure below. Seond, in this ase, the resulting signal an be sampled diretly with sampling interval, with no aliasing. his is just like in the homework problem about band-pass sampling. Alternatively, if we had aess to a ontinuous-time mixer (multipliation by a osine funtion), we ould modulate it down before sampling (but as we said, for our example ase, this is not neessary). he resulting signals y[n] and z[n] now have to be ombined. he main insight at this point is that we need twie the sampling rate in the end, and so, we should upsample both y[n] and z[n]. o see how to implement the ombining, one really needs the spetral plots of the upsampled signals, U(e jω ) and V (e jω ) : It is immediately lear that we want to low-pass filter U(e jω ) and high-pass filter V (e jω ). Let s define: F (e jω ) = {, for Ω π and G(e jω ) = {, for Ω > π Reall that these are really π -periodi; we are merely speifying one period. he spetral plots in the figure on the next page shows that the system below works. Grading: Most of the points were given for the two key ideas, namely, that we need two different signal paths, and that we need upsampling. x(t) H(j ) y(t) z(t) Sampler Sampler y[n] z[n] u[n] v[n] j " F(e ) j " G(e ) a[n] b[n] / x[n] 0

11 Y(e j ) Z(e j ) 3/ / / #" " #" " U(e j ) 6/ 4/ V(e j ) 4/ #" " #" " A(e j ) 6/ 4/ B(e j ) 4/ #" " #" "

12 Problem 4 (PAM.) wo pulses are suggested for a PAM system: q (t) = {, t /4 0, otherwise /4 t < 0 and q (t) =, 0 t /4 0, otherwise 33 Points he PAM signal is then x m (t) = n= s[n]q m (t n ), for m =,. (7) hroughout this problem, we assume that the data signal is merely s[n] =, for all n. (a) (6 Pts) Find the powers P and P of the two PAM signals x (t) and x (t). Solution: Beause x (t) and x (t) are periodi, we an alulate their power by looking at one period. P = / / x (t) dt = /4 /4 dt = P = / / x (t) dt = /4 /4 dt = (b) (0 Pts) Give the formula for the Fourier series oeffiients of the signal x (t), and expliitly evaluate the oeffiients a 0, a and a. hen, do the same for the signal x (t). Solution: he oeffiients of x (t) are in able 4., 6th entry. For us, = /4 and ω 0 = π/, hene a k = sin(kπ/) kπ where the oeffiient at k = 0 an be found as usual as the limit of the above expression as k 0, or by noting that the above formula an be written as a k = sin(k/), where we remember that sin(0) =. hus, a 0 = /, a = π and a = π. Alternatively, you an evaluate by hand: / (8) a 0 = / x (t)dt = a k = / / /4 x (t)e jk(π/ )t dt = e jk(π/ )t dt /4 = (e jkπ/ e jkπ/) jkπ/ = kπ sin(kπ/)

13 For the signal x (t), let s first onsider a symmetri box of width /4, entered at zero, and repeated at intervals of. Call this signal v(t). Hene, again from the table, with = /8, we find the FS oeffiients of v(t) : k = sin(kπ/4) kπ = sin(k/4). (9) 4 Clearly, x (t) = v(t /8) v(t + /8), and hene, using the seond property (time shifting) in able 3., we find the FS oeffiients of the signal x (t) as hus, b 0 = 0, b = j π and b = j π. Alternatively, you an evaluate by hand: b k = k e jkπ/4 k e jkπ/4 (0) = sin(k/4)( j) sin(kπ/4) () 4 = j sin(k/4) sin(kπ/4) () b 0 = / / x (t)dt = 0 b k = / / 0 x (t)e jk(π/ )t dt = e jk(π/ )t dt + /4 = ( e jkπ/) jk(π/ ) = ( e jkπ/ e jkπ/) jkπ = ( os(kπ/)) jkπ /4 0 e jk(π/ )t dt jk(π/ ) ( ) e jkπ/ Exerise: Show that the above two formulas for b k are, in fat, equal. 3

14 () (7 Pts) o atually transmit our PAM signal, we first low-pass filter it: x m (t) = h(t) x m (t), where H(jω) = {, for ω 0π (3) hen, we transmit the signals y (t) = x (t) os( 40π t) and y (t) = x (t) os( 40π t). Sketh the Fourier transforms of these two signals in the plots provided below, arefully labeling the frequeny axis. In the magnitude plots (i.e., Y (jω) and Y (jω), respetively), the amplitudes need not be exat. Remark: he urrent labels on the frequeny axis in the plots are for your onveniene only. If you prefer, you an ross them out and start from srath. Solution: Sine x (t) is periodi, its spetrum X (jω) is a line spetrum, i.e., omposed of delta funtions. From able 4., first entry, the delta funtions are spaed π/ apart (the fundamental frequeny of x (t) ), and their amplitudes are πa k, where a k are the Fourier series oeffiients of x (t) : X (jω) = π k= a k δ(ω k π ). (4) he low-pass filtering throws away anything at frequenies higher than 0π/, i.e., only the enter and 5 delta funtions on eah side of the origin survive, but otherwise, the spetrum X (jω) looks exatly like the spetrum of X (ω) : X (jω) = π 5 k= 5 a k δ(ω k π ). (5) Multiplying by os( 40π t) simply plaes two opies of the spetrum of X (jω), one entered at 40π/, the other at 40π/. o get the amplitudes right (but no points were taken off for minor errors in the amplitudes), remember that the spetrum of the osine onsists of two delta funtions of amplitude π eah, and the multipliation property has a fator of /(π), hene, Y (jω) = π π π ( 5 k= 5 a k δ(ω k π 40π ) + 5 k= 5 For the signal y (t), the argument is exatly the same, leading to Y (jω) = π π π ( 5 k= 5 b k δ(ω k π 40π ) + 5 k= 5 ) a k δ(ω k π + 40π ). (6) ) b k δ(ω k π + 40π ). (7) 4

15 Y (j # ) "/ # Z (j #) Z (j #) [ "/$ ] arg{y (j # )} " # [ "/$ ] Y (j # ) # Z (j #) Z (j #) [ "/$ ] arg{y (j # )} "/ "/ 0 30 # [ "/$ ] Figure 6: 5

16 (d) (0 Pts) he ommuniation hannel s effet on the signal an be desribed by the following band-pass filter: sin (ω/4 π) for 36π/ < ω < 38π/, for 38π/ ω 4π/ H hannel (jω) = sin (8) (ω/4 π) for 4π/ < ω < 44π/ he hannel output signal is then z (t) = y (t) h hannel (t) and z (t) = y (t) h hannel (t), respetively. Assuming that s[n] =, for all n, find the power of z (t) and z (t). hese are the reeived powers. Whih pulse is more effiient for transmission aross this hannel? Solution: he key insight is that z (t) is still a periodi signal. Hene, its power is alulated as P r = z (t) dt. (9) From Parseval, this is the same as P r = n= k, (0) where k are the Fourier series oeffiients of the signal z (t). From Part (), we know the Fourier ransform Z (jω) : Passing Y (jω) through the bandpass filter, only 6 delta pulses survive, namely the three entered around 40π/ and the three entered around 40π/. o find the Fourier series oeffiients, we have to divide the amplitude of the delta funtion by π (see able 4., first entry), and so we an read diretly our of the figure: 0 = 0 = /4, 9 = 9 = = = /(π), () and all other Fourier series oeffiients are zero, from whih we find P r = n= k = /6 + 4 /(π) = 8 + π. () By the same token, we an read out the Fourier series oeffiients for the signal z (t), and find P r = n= d k = 4 /(π) = π. (3) Hene, the reeived power for the pulse x (t) is higher, and we onlude that this pulse is more effiient. Note: here is also a subtlety that we swept under the arpet. Really, it is not lear whether both pulses have the same transmitted power We only alulated the powers of x (t) and x (t), respetively, but the transmitted signals, really, are y (t) and y (t). hese powers are most easily found along the lines of the above alulation, but we would need some more of the Fourier series oeffiients. 6