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Solution Manual for Signals and Systems using MATLAB , 3rd Edition

Solution Manual for Signals and Systems using MATLAB, 3rd Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 1 preview imageSolution Manual forSIGNALS AND SYSTEMSUSING MATLABLuis F. Chaparro and Aydin Akan
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 2 preview image
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 3 preview imageChapter 0From the Ground Up1
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 4 preview imageChaparro-Akan — Signals and Systems using MATLAB0.20.1Basic Problems0.1Letz= 8 +j3andv= 9−j2,(a) Find(i)Re(z) +Im(v),(ii)jz+vj,(iii)jzvj,(iv)∠z+∠v,(v)jv/zj,(vi)∠(v/z)(b) Find the trigonometric and polar forms of(i)z+v,(ii)zv,(iii)z∗(iv)zz∗,(v)z−vAnswers:(a)Re(z) +Im(v) = 6;jv/zj=p85/p73; (b)zz∗=jzj2= 73.Solution(a)i.Re(z) +Im(v) = 8−2 = 6ii.jz+vj=j17 +j1j=p172+ 1iii.jzvj=j72−j16 +j27 + 6j=j78 +j11j=p782+ 112iv.∠z+∠v= tan−1(3/8)−tan−1(2/9)v.jv/zj=jvj/jzj=p85/p73vi.∠(v/z) =−tan−1(2/9)−tan−1(3/8)(b)i.z+v= 17 +j=p172+ 1ejtan−1(1/17)ii.zv= 78 +j11 =p782+ 112ejtan−1(11/78)iii.z∗= 8−j3 =p64 + 9(e−jtan−1(3/8))∗=p73ejtan−1(3/8)iv.zz∗=jzj2= 73v.z−v=−1 +j5 =p1 + 25e−jtan−1(5)
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 5 preview imageChaparro-Akan — Signals and Systems using MATLAB0.30.2Use Euler’s identity to(a) show that(i) cos(θ−π/2) = sin(θ),(ii)−sin(θ−π/2) = cos(θ),(iii) cos(θ) = sin(θ+π/2).(b) to find(i)∫10cos(2πt) sin(2πt)dt,(ii)∫10cos2(2πt)dt.Answers:(b)0and1/2.Solution(a) We havei.cos(θ−π/2) = 0.5(ej(θ−π/2)+e−j(θ−π/2)) =−j0.5(ejθ−e−jθ) = sin(θ)ii.−sin(θ−π/2) = 0.5j(ej(θ−π/2)−e−j(θ−π/2)) = 0.5j(−j)(ejθ+e−jθ) = cos(θ)iii.sin(θ+π/2) = (jejθ+je−jθ)/(2j) = cos(θ)(b)i.cos(2πt) sin(2πt) = (1/4j)(ej4πt−e−j4πt)so that∫10cos(2πt) sin(2πt)dt=14jej4πt4πjj10+ 14je−j4πt4πjj10= 0 + 0 = 0ii. We havecos2(2πt) = 14 (ej4πt+ 2 +e−j4πt) = 12 (1 + cos(4πt))so that its integral is1/2since the integral ofcos(4πt)is over two of its periods and itis zero.
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 6 preview imageChaparro-Akan — Signals and Systems using MATLAB0.40.3Use Euler’s identity to(a) show the identities(i)cos(α+β) = cos(α) cos(β)−sin(α) sin(β)(ii)sin(α+β) = sin(α) cos(β) + cos(α) sin(β),(b) find an expression forcos(α) cos(β), and forsin(α) sin(β).Answers:ejαejβ= cos(α+β) +jsin(α+β) = [cos(α) cos(β)−sin(α) sin(β)] +j[sin(α) cos(β) +cos(α) sin(β)].Solution(a) Using Euler’s identity the productejαejβ=(cos(α) +jsin(α))(cos(β) +jsin(β))=[cos(α) cos(β)−sin(α) sin(β)] +j[sin(α) cos(β) + cos(α) sin(β)]whileej(α+β)= cos(α+β) +jsin(α+β)so that equating the real and imaginary parts of the above two equations we get the desiredtrigonometric identities.(b) We havecos(α) cos(β)=0.5(ejα+e−jα) 0.5(ejβ+e−jβ)=0.25(ej(α+β)+e−j(α+β)) + 0.25(ej(α−β)+e−j(α−β))=0.5 cos(α+β) + 0.5 cos(α−β)Now,sin(α) sin(β)=cos(α−π/2) cos(β−π/2)=0.5 cos(α−π/2 +β−π/2) + 0.5 cos(α−π/2−β+π/2)=0.5 cos(α+β−π) + 0.5 cos(α−β)=−0.5 cos(α+β) + 0.5 cos(α−β)
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 7 preview imageChaparro-Akan — Signals and Systems using MATLAB0.50.4Consider the calculation of roots of an equationzN=αwhereN≥1is an integer andα=jαjejφa nonzero complex number.(a) First verify that there are exactlyNroots for this equation and that they are given byzk=rejθkwherer=jαj1/Nandθk= (φ+ 2πk)/Nfork= 0,1,· · ·, N−1.(b) Use the above result to find the roots of the following equations(i)z2= 1,(ii)z2=−1,(iii)z3= 1,(iv)z3=−1.and plot them in a polar plane (i..e., indicating their magnitude and phase). Explain howthe roots are distributed in the polar plane.Answers:Roots ofz3=−1 = 1ejπarezk= 1ej(π+2πk)/3,k= 0,1,2, equally spaced aroundcircle of radiusr.Solution(a) Replacingzk=jαj1/Nej(φ+2πk)/NinzNwe getzNk=jαjej(φ+2πk)=jαjej(φ)=αfor anyvalue ofk= 0,· · ·, N−1.(b) Applying the above result we have:•Forz2= 1 = 1ej2πthe roots arezk= 1ej(2π+2πk)/2,k= 0,1. Whenk= 0,z0=ejπ=−1andz1=ej2π= 1.•Whenz2=−1 = 1ejπthe roots arezk= 1ej(π+2πk)/2,k= 0,1. Whenk= 0,z0=ejπ/2=j,andz1=ej3π/2=−j.•Forz3= 1 = 1ej2πthe roots arezk= 1ej(2π+2πk)/3,k= 0,1,2. Whenk= 0,z0=ej2π/3;fork= 1,z1=ej4π/3=e−j2π/3=z∗0; and fork= 2,z2= 1ej(2π)= 1.•Whenz3=−1 = 1ejπthe roots arezk= 1ej(π+2πk)/3,k= 0,1,2. Whenk= 0,z0=ejπ/3;fork= 1,z1=ejπ=−1; and fork= 2,z2= 1ej(5π)/3= 1ej(−π)/3=z∗0(c) Notice that the roots are equally spaced around a circle of radiusrand that the complexroots appear as pairs of complex conjugate roots.
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 8 preview imageChaparro-Akan — Signals and Systems using MATLAB0.60.5Consider a function ofz= 1 +j1,w=ez(a) Find (i)log(w), (ii)Re(w), (iii)Im(w)(b) What isw+w∗, wherew∗is the complex conjugate ofw?(c) Determinejwj,∠wandjlog(w)j2?(d) Expresscos(1)in terms ofwusing Euler’s identity.Answers:log(w) =z;w+w∗= 2Re[w] = 2ecos(1).Solution(a) Ifw=ezthenlog(w) =z= 1 +j1given that thelogandefunctions are the inverse of each other.The real and imaginary ofwarew=ez=e1ej1=ecos(1)︸︷︷︸real part+jesin(1)︸︷︷︸imaginary part(b) The imaginary parts are cancelled and the real parts added twice inw+w∗= 2Re[w] = 2ecos(1)(c) Replacingzw=ez=e1ej1so thatjwj=eand∠w= 1.Using the result in (a)jlog(w)j2=jzj2= 2(d) According to Euler’s equationcos(1) = 0.5(ej+e−j) = 0.5(we+w∗e)which can be verified usingw+w∗obtained above.
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 9 preview imageChaparro-Akan — Signals and Systems using MATLAB0.70.6A phasor can be thought of as a vector, representing a complex number, rotating around thepolar plane at a certain frequency in radians/second.The projection of such a vector ontothe real axis gives a cosine with a certain amplitude and phase. This problem will show thealgebra of phasors which would help you with some of the trigonometric identities that arehard to remember.(a) When you ploty(t) =Asin(Ω0t)you notice that it is a cosinex(t) =Acos(Ω0t)shifted intime, i.e.,y(t) =Asin(Ω0t) =Acos(Ω0(t−∆t)) =x(t−∆t)how much is this shift∆t?Better yet, what is∆θ= Ω0∆tor the shift in phase?Onethus only need to consider cosine functions with different phase shifts instead of sinesand cosines.(b) From above, the phasor that generatesx(t) =Acos(Ω0t)isAej0so thatx(t) =Re[Aej0ejΩ0t].The phasor corresponding to the siney(t)should then beAe−jπ/2. Obtain an expressionfory(t)similar to the one forx(t)in terms of this phasor.(c) From the above results, give the phasors corresponding to−x(t)=−Acos(Ω0t)and−y(t) =−sin(Ω0t).Plot the phasors that generatecos,sin,−cosand−sinfor a givenfrequency. Do you see now how these functions are connected? How many radians doyou need to shift in positive or negative direction to get a sine from a cosine, etc.(d) Suppose then you have the sum of two sinusoids, for instancez(t)=x(t) +y(t)=Acos(Ω0t) +Asin(Ω0t), adding the corresponding phasors forx(t)andy(t)at some time,e.g.,t= 0, which is just a sum of two vectors, you should get a vector and the correspond-ing phasor. Forx(t),y(t), obtain their corresponding phasors and then obtain from themthe phasor corresponding toz(t) =x(t) +y(t).(e) Find the phasors corresponding to(i) 4 cos(2t+π/3),(ii)−4 sin(2t+π/3),(iii) 4 cos(2t+π/3)−4 sin(2t+π/3)Answers:sin(Ω0t) = cos(Ω0(t−T0/4)) = cos(Ω0t−π/2)sinceΩ0= 2π/T0;z(t) =p2Acos(Ω0t−π/4); (e) (i)4ejπ/3; (iii)4p2ej7π/12.Solution(a) Shifting to the right a cosine by a fourth of its period we get a sinusoid, thussin(Ω0t) = cos(Ω0(t−T0/4)) = cos(Ω0t−Ω0T0/4) = cos(Ω0t−π/2)sinceΩ0= 2π/T0orΩ0T0= 2π.(b) The phasor that generates a sine isAe−jπ/2sincey(t) =Re[Ae−jπ/2ejΩ0t] =Re[Aej(Ω0t−π/2)] =Acos(Ω0t−π/2)which equalsAsin(Ω0t).(c) The phasors corresponding to−x(t) =−Acos(Ω0t) =Acos(Ω0t+π)isAejπ. For−y(t) =−Asin(Ω0t) =−Acos(Ω0t−π/2) =Acos(Ω0t−π/2 +π) =Acos(Ω0t+π/2)the phasor isAejπ/2. Thus, relating any sinusoid to the corresponding cosine, the magni-tude and angle of this cosine gives the magnitude and phase of the phasor that generatesthe given sinusoid.
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 10 preview imageChaparro-Akan — Signals and Systems using MATLAB0.8(d) Ifz(t) =x(t) +y(t) =Acos(Ω0t) +Asin(Ω0t), the phasor corresponding toz(t)is the sumof the phasorsAej0, corresponding toAcos(Ω0t), with the phasorAe−jπ/2, correspondingtoAsin(Ω0t), which givesp2Ae−jπ/4(equivalently the sum of a vector with length A andangle0with another vector of length A and angle−π/2). We have thatz(t) =Re[p2Ae−jπ/4ejΩ0t]=p2Acos(Ω0t−π/4)(e)i. Phasor4ejπ/3ii.−4 sin(2t+π/3) = 4 cos(2t+π/3 +π/2)with phasor4ej5π/6iii. We have4 cos(2t+π/3)−4 sin(2t+π/3)=Re[(4ejπ/3+ 4ej(π/2+π/3))ej2t]=Re[4ejπ/3(1 +ejπ/2)︸︷︷︸√2ejπ/4ej2t]=Re[4p2ej7π/12ej2t]so that the phasor is4p2ej7π/12
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 11 preview imageChaparro-Akan — Signals and Systems using MATLAB0.90.7To get an idea of the number of bits generated and processed by a digital system consider thefollowing applications:(a) A compact disc (CD) is capable of storing 75 minutes of “CD quality” stereo (left and rightchannels are recorded) music. Calculate the number of bits that are stored in the CD asraw data.Hint: find out what ’CD quality’ means in the binary representation of each sample.(b) Find out what the vocoder in your cell phone is used for. To attaining “telephone quality”voice you use a sampling rate of10,000samples/sec, and that each sample is representedby8bits. Calculate the number of bits that your cell-phone has to process every secondthat you talk. Why would you then need a vocoder?(c) Find out whether text messaging is cheaper or more expensive than voice. Explain howthe text messaging works.(d) Find out how an audio CD and an audio DVD compare. Find out why it is said that a vinyllong-play record reproduces sounds much better. Are we going backwards with digitaltechnology in music recording? Explain.(e) To understand why video streaming in the internet is many times of low quality, considerthe amount of data that needs to be processed by a video compressor every second. As-sume the size of a video frame, in pixels, is352×240, and that an acceptable quality forthe image is obtained by allocating8bits/pixel and to avoid jerking effects we use60frames/second.•How many pixels need to be processed every second?•How many bits would be available for transmission every second?•The above is raw data, compression changes the whole picture (literally), find outwhat some of the compression methods are.Answers:(a) About6.4Gbs; vocoder (short for voice encoder) reduces number of transmittedbits while keeping voice recognizable.Solution(a) Assuming a maximum frequency of22.05kHz for the acoustic signal, the numbers of bytes(8bits per byte) for two channels (stereo) and a75minutes recording is greater or equal to:2×22,050samples/channel/second×2bytes/sample×2channels×75minutes×60sec-onds/minute= 7.938×108bytes. Multiplying by8we get the number of bits. CD qualitymeans that the signal is sampled at44.1kHz and each sample is represented by16bits or2bytes.(b) The raw data would consist of8(bits/sample)×10,000(samples/sec)=80,000bits/sec. Thevocoder is part of a larger unit called a digital signal processor chip set. It uses various proce-dures to reduce the number of bits that are transmitted while still keeping your voice recogniz-able. When there is silence it does not transmit, letting another signal use the channel duringpauses.(c) Texting between cell phones is possible by sending short messages (160characters) using theshort message services (SMS). Whenever your cell-phone communicates with the cell phonetower there is an exchange of messages over the control channel for localization, and call setup.This channel provides a pathway for SMS messages by sending packets of data. Except for thecost of storing messages, the procedure is rather inexpensive and convenient to users.
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 12 preview imageChaparro-Akan — Signals and Systems using MATLAB0.10(d) For CD audio the sampling rate is44.1kHz with16bits/sample. For DVD audio the sam-pling rate is192kHz with24bits/sample. The sampling process requires getting rid of highfrequencies in the signal, also each sample is only approximated by the binary representation,so analog recording could sound better in some cases.(e) The number of pixels processed every second is:352×240pixels/frame×60frames/sec.The number of bits available for transmission every second is obtained by multiplying theabove answer by8bits/pixel. There many compression methods JPEG, MPEG, etc.
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 13 preview imageChaparro-Akan — Signals and Systems using MATLAB0.110.8The geometric seriesS=N−1∑n=0αnwill be used quite frequently in the next chapters so let us look at some of its properties:(a) Supposeα= 1what isSequal to?(b) Supposeα6= 1show thatS= 1−αN1−αVerify that(1−α)S= (1−αN). Why do you need the constraint thatα6= 1? Would thissum exist ifα >1? Explain.(c) Suppose now thatN=1, under what conditions willSexist? if it does, what wouldSbe equal to? Explain.(d) Suppose again thatN=1in the definition ofS. The derivative ofSwith respect toαisS1=dSdα=∞∑n=0nαn−1obtain a rational expression to findS1.Answers:S=Nwhenα= 1,S= (1−αN)/(1−α)whenα6= 1.Solution(a) Ifα= 1thenS=N−1∑n=01 = 1 + 1 +· · ·+ 1︸︷︷︸Ntimes=N(b) The expressionS(1−α)=S−αS=(1 +α+· · ·+αN−1)−(α+α2+· · ·+αN−1+αN)=1−αNas the intermediate terms cancel. So thatS= 1−αN1−α ,α6= 1Since we do not want the denominator1−αto be zero, the above requires thatα6= 1. Ifα= 1the sum was found in (a). As a finite sum, it exists for any finite values ofα.Putting (a) and (b) together we haveS={(1−αN)/(1−α)α6= 1Nα= 1(c) IfNis infinite, the sum is of infinite length and we need to impose the condition thatjαj<1so thatαndecays asn! 1. In that case, the termαN!0asN! 1, and the sum isS=11−αjαj<1
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 14 preview imageChaparro-Akan — Signals and Systems using MATLAB0.12Ifjαj ≥1this sum does not exist, i.e., it becomes infinite.(d) The derivative becomesS1=dSdα=∞∑n=0nαn−1=1(1−α)2.
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 15 preview imageChaparro-Akan — Signals and Systems using MATLAB0.130.2Problems using MATLAB0.9 Derivative and finite difference —Lety(t) =dx(t)/dt, wherex(t) = 4 cos(2πt),−1< t <1.Findy(t)analytically and determine a value ofTsfor which∆[x(nTs)]/Ts=y(nTs)(consideras possible valuesTs= 0.01andTs= 0.1). Use the MATLAB functiondiffor create your ownto compute the finite difference. Plot the finite difference in the range[0,1]and compare it withthe actual derivativey(t)in that range. Explain your results for the given values ofTs.Answers:y(t) =−8πsin(2πt)has same sampling period asx(t),Ts≤0.5;Ts= 0.01givesbetter results.SolutionThe derivative isy(t) =dx(t)dt=−8πsin(2πt)which has the same frequency asx(t), thus the sampling period should be like in the previousproblem,Ts≤0.5.%Pr.0.9clearall%actualderivativeTss=0.0001;t1=0:Tss:3;y=-8*pi*sin(2*pi*t1);figure(2)%forwarddifferenceTs=0.01;t=[0:Ts:3];N=length(t);subplot(211)xa=4*cos(2*pi*t);%sampledsignalder1_x=forwardiff(xa,Ts,t,y,t1);clearder1_x%forwarddifferenceTs=0.1;t=[0:Ts:3];N=length(t);subplot(212)xa=4*cos(2*pi*t);%sampledsignalder1_x=forwardiff(xa,Ts,t,y,t1);The functionforwardiffcomputes and plots the forward difference and the actual derivative.functionder=forwardiff(xa,Ts,t,y,t1)%%forwarddifference%%xa:sampledsignalusingTs%%y:actualderivativedefinedintN=length(t);n=0:N-2;der=diff(xa)/Ts;stem(n*Ts,der,’filled’);grid;xlabel(’t,nT_s’)holdonplot(t1,y,’r’);legend(’forwarddifference’,’derivative’)holdoffForTs= 0.1the finite difference looks like the actual derivative but shifted, while forTs= 0.01it doesnot.
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Solution Manual for Signals and Systems using MATLAB , 3rd Edition - Page 16 preview imageChaparro-Akan — Signals and Systems using MATLAB0.1400.511.522.53−40−2002040t, nTs00.511.522.53−40−2002040t, nTsforward differencederivativeforward differencederivativeFigure 1: Problem 9:Ts= 0.01sec (top) andTs= 0.1sec (bottom)
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