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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 Luis F. Chaparro and Aydin Akan

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    Chapter 0 From the Ground Up 1

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    Chaparro-Akan — Signals and Systems using MATLAB 0.2 0.1 Basic Problems 0.1 Let z = 8 + j 3 and v = 9 − j 2 , (a) Find ( i ) R e ( z ) + I m ( v ) , ( ii ) j z + v j , ( iii ) j zv j , ( iv ) ∠ z + ∠ v, ( v ) j v/z j , ( vi ) ∠ ( v/z ) (b) Find the trigonometric and polar forms of ( i ) z + v, ( ii ) zv, ( iii ) z ∗ ( iv ) zz ∗ , ( v ) z − v Answers: (a) R e ( z ) + I m ( v ) = 6 ; j v/z j = p 85 / p 73 ; (b) zz ∗ = j z j 2 = 73 . Solution (a) i. R e ( z ) + I m ( v ) = 8 − 2 = 6 ii. j z + v j = j 17 + j 1 j = p 17 2 + 1 iii. j zv j = j 72 − j 16 + j 27 + 6 j = j 78 + j 11 j = p 78 2 + 11 2 iv. ∠ z + ∠ v = tan − 1 (3 / 8) − tan − 1 (2 / 9) v. j v/z j = j v j / j z j = p 85 / p 73 vi. ∠ ( v/z ) = − tan − 1 (2 / 9) − tan − 1 (3 / 8) (b) i. z + v = 17 + j = p 17 2 + 1 e j tan − 1 (1 / 17) ii. zv = 78 + j 11 = p 78 2 + 11 2 e j tan − 1 (11 / 78) iii. z ∗ = 8 − j 3 = p 64 + 9( e − j tan − 1 (3 / 8) ) ∗ = p 73 e j tan − 1 (3 / 8) iv. zz ∗ = j z j 2 = 73 v. z − v = − 1 + j 5 = p 1 + 25 e − j tan − 1 (5)

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    Chaparro-Akan — Signals and Systems using MATLAB 0.3 0.2 Use 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 ) ∫ 1 0 cos(2 πt ) sin(2 πt ) dt, ( ii ) ∫ 1 0 cos 2 (2 πt ) dt. Answers: (b) 0 and 1 / 2 . Solution (a) We have i. cos( θ − π/ 2) = 0 . 5( e j ( θ − π/ 2) + e − j ( θ − π/ 2) ) = − j 0 . 5( e jθ − e − jθ ) = sin( θ ) ii. − sin( θ − π/ 2) = 0 . 5 j ( e j ( θ − π/ 2) − e − j ( θ − π/ 2) ) = 0 . 5 j ( − j )( e jθ + e − jθ ) = cos( θ ) iii. sin( θ + π/ 2) = ( je jθ + je − jθ ) / (2 j ) = cos( θ ) (b) i. cos(2 πt ) sin(2 πt ) = (1 / 4 j )( e j 4 πt − e − j 4 πt ) so that ∫ 1 0 cos(2 πt ) sin(2 πt ) dt = 1 4 j e j 4 πt 4 πj j 1 0 + 1 4 j e − j 4 πt 4 πj j 1 0 = 0 + 0 = 0 ii. We have cos 2 (2 πt ) = 1 4 ( e j 4 πt + 2 + e − j 4 πt ) = 1 2 (1 + cos(4 πt )) so that its integral is 1 / 2 since the integral of cos(4 πt ) is over two of its periods and it is zero.

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    Chaparro-Akan — Signals and Systems using MATLAB 0.4 0.3 Use 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 for cos( α ) cos( β ) , and for sin( α ) sin( β ) . Answers: e jα e jβ = cos( α + β ) + j sin( 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 product e jα e jβ = (cos( α ) + j sin( α ))(cos( β ) + j sin( β )) = [cos( α ) cos( β ) − sin( α ) sin( β )] + j [sin( α ) cos( β ) + cos( α ) sin( β )] while e j ( α + β ) = cos( α + β ) + j sin( α + β ) so that equating the real and imaginary parts of the above two equations we get the desired trigonometric identities. (b) We have cos( α ) cos( β ) = 0 . 5( e jα + e − jα ) 0 . 5( e jβ + e − jβ ) = 0 . 25( e j ( α + β ) + e − j ( α + β ) ) + 0 . 25( e j ( 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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    Chaparro-Akan — Signals and Systems using MATLAB 0.5 0.4 Consider the calculation of roots of an equation z N = α where N ≥ 1 is an integer and α = j α j e jφ a nonzero complex number. (a) First verify that there are exactly N roots for this equation and that they are given by z k = re jθ k where r = j α j 1 /N and θ k = ( φ + 2 πk ) /N for k = 0 , 1 , · · · , N − 1 . (b) Use the above result to find the roots of the following equations ( i ) z 2 = 1 , ( ii ) z 2 = − 1 , ( iii ) z 3 = 1 , ( iv ) z 3 = − 1 . and plot them in a polar plane (i..e., indicating their magnitude and phase). Explain how the roots are distributed in the polar plane. Answers: Roots of z 3 = − 1 = 1 e jπ are z k = 1 e j ( π +2 πk ) / 3 , k = 0 , 1 , 2 , equally spaced around circle of radius r . Solution (a) Replacing z k = j α j 1 /N e j ( φ +2 πk ) /N in z N we get z N k = j α j e j ( φ +2 πk ) = j α j e j ( φ ) = α for any value of k = 0 , · · · , N − 1 . (b) Applying the above result we have: • For z 2 = 1 = 1 e j 2 π the roots are z k = 1 e j (2 π +2 πk ) / 2 , k = 0 , 1 . When k = 0 , z 0 = e jπ = − 1 and z 1 = e j 2 π = 1 . • When z 2 = − 1 = 1 e jπ the roots are z k = 1 e j ( π +2 πk ) / 2 , k = 0 , 1 . When k = 0 , z 0 = e jπ/ 2 = j , and z 1 = e j 3 π/ 2 = − j . • For z 3 = 1 = 1 e j 2 π the roots are z k = 1 e j (2 π +2 πk ) / 3 , k = 0 , 1 , 2 . When k = 0 , z 0 = e j 2 π/ 3 ; for k = 1 , z 1 = e j 4 π/ 3 = e − j 2 π/ 3 = z ∗ 0 ; and for k = 2 , z 2 = 1 e j (2 π ) = 1 . • When z 3 = − 1 = 1 e jπ the roots are z k = 1 e j ( π +2 πk ) / 3 , k = 0 , 1 , 2 . When k = 0 , z 0 = e jπ/ 3 ; for k = 1 , z 1 = e jπ = − 1 ; and for k = 2 , z 2 = 1 e j (5 π ) / 3 = 1 e j ( − π ) / 3 = z ∗ 0 (c) Notice that the roots are equally spaced around a circle of radius r and that the complex roots appear as pairs of complex conjugate roots.

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    Chaparro-Akan — Signals and Systems using MATLAB 0.6 0.5 Consider a function of z = 1 + j 1 , w = e z (a) Find (i) log( w ) , (ii) R e ( w ) , (iii) I m ( w ) (b) What is w + w ∗ , where w ∗ is the complex conjugate of w ? (c) Determine j w j , ∠ w and j log( w ) j 2 ? (d) Express cos(1) in terms of w using Euler’s identity. Answers: log( w ) = z ; w + w ∗ = 2 R e [ w ] = 2 e cos(1) . Solution (a) If w = e z then log( w ) = z = 1 + j 1 given that the log and e functions are the inverse of each other. The real and imaginary of w are w = e z = e 1 e j 1 = e cos(1) ︸ ︷︷ ︸ real part + j e sin(1) ︸ ︷︷ ︸ imaginary part (b) The imaginary parts are cancelled and the real parts added twice in w + w ∗ = 2 R e [ w ] = 2 e cos(1) (c) Replacing z w = e z = e 1 e j 1 so that j w j = e and ∠ w = 1 . Using the result in (a) j log( w ) j 2 = j z j 2 = 2 (d) According to Euler’s equation cos(1) = 0 . 5( e j + e − j ) = 0 . 5 ( w e + w ∗ e ) which can be verified using w + w ∗ obtained above.

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    Chaparro-Akan — Signals and Systems using MATLAB 0.7 0.6 A phasor can be thought of as a vector, representing a complex number, rotating around the polar plane at a certain frequency in radians/second. The projection of such a vector onto the real axis gives a cosine with a certain amplitude and phase. This problem will show the algebra of phasors which would help you with some of the trigonometric identities that are hard to remember. (a) When you plot y ( t ) = A sin(Ω 0 t ) you notice that it is a cosine x ( t ) = A cos(Ω 0 t ) shifted in time, i.e., y ( t ) = A sin(Ω 0 t ) = A cos(Ω 0 ( t − ∆ t )) = x ( t − ∆ t ) how much is this shift ∆ t ? Better yet, what is ∆ θ = Ω 0 ∆ t or the shift in phase? One thus only need to consider cosine functions with different phase shifts instead of sines and cosines. (b) From above, the phasor that generates x ( t ) = A cos(Ω 0 t ) is Ae j 0 so that x ( t ) = R e [ Ae j 0 e j Ω 0 t ] . The phasor corresponding to the sine y ( t ) should then be Ae − jπ/ 2 . Obtain an expression for y ( t ) similar to the one for x ( t ) in terms of this phasor. (c) From the above results, give the phasors corresponding to − x ( t ) = − A cos(Ω 0 t ) and − y ( t ) = − sin(Ω 0 t ) . Plot the phasors that generate cos , sin , − cos and − sin for a given frequency. Do you see now how these functions are connected? How many radians do you 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 instance z ( t ) = x ( t ) + y ( t ) = A cos(Ω 0 t ) + A sin(Ω 0 t ) , adding the corresponding phasors for x ( t ) and y ( 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. For x ( t ) , y ( t ) , obtain their corresponding phasors and then obtain from them the phasor corresponding to z ( t ) = x ( t ) + y ( t ) . (e) Find the phasors corresponding to ( i ) 4 cos(2 t + π/ 3) , ( ii ) − 4 sin(2 t + π/ 3) , ( iii ) 4 cos(2 t + π/ 3) − 4 sin(2 t + π/ 3) Answers : sin(Ω 0 t ) = cos(Ω 0 ( t − T 0 / 4)) = cos(Ω 0 t − π/ 2) since Ω 0 = 2 π/T 0 ; z ( t ) = p 2 A cos(Ω 0 t − π/ 4) ; (e) (i) 4 e jπ/ 3 ; (iii) 4 p 2 e j 7 π/ 12 . Solution (a) Shifting to the right a cosine by a fourth of its period we get a sinusoid, thus sin(Ω 0 t ) = cos(Ω 0 ( t − T 0 / 4)) = cos(Ω 0 t − Ω 0 T 0 / 4) = cos(Ω 0 t − π/ 2) since Ω 0 = 2 π/T 0 or Ω 0 T 0 = 2 π . (b) The phasor that generates a sine is Ae − jπ/ 2 since y ( t ) = R e [ Ae − jπ/ 2 e j Ω 0 t ] = R e [ Ae j (Ω 0 t − π/ 2) ] = A cos(Ω 0 t − π/ 2) which equals A sin(Ω 0 t ) . (c) The phasors corresponding to − x ( t ) = − A cos(Ω 0 t ) = A cos(Ω 0 t + π ) is Ae jπ . For − y ( t ) = − A sin(Ω 0 t ) = − A cos(Ω 0 t − π/ 2) = A cos(Ω 0 t − π/ 2 + π ) = A cos(Ω 0 t + π/ 2) the phasor is Ae jπ/ 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 generates the given sinusoid.

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    Chaparro-Akan — Signals and Systems using MATLAB 0.8 (d) If z ( t ) = x ( t ) + y ( t ) = A cos(Ω 0 t ) + A sin(Ω 0 t ) , the phasor corresponding to z ( t ) is the sum of the phasors Ae j 0 , corresponding to A cos(Ω 0 t ) , with the phasor Ae − jπ/ 2 , corresponding to A sin(Ω 0 t ) , which gives p 2 Ae − jπ/ 4 (equivalently the sum of a vector with length A and angle 0 with another vector of length A and angle − π/ 2 ). We have that z ( t ) = R e [ p 2 Ae − jπ/ 4 e j Ω 0 t ] = p 2 A cos(Ω 0 t − π/ 4) (e) i. Phasor 4 e jπ/ 3 ii. − 4 sin(2 t + π/ 3) = 4 cos(2 t + π/ 3 + π/ 2) with phasor 4 e j 5 π/ 6 iii. We have 4 cos(2 t + π/ 3) − 4 sin(2 t + π/ 3) = R e [(4 e jπ/ 3 + 4 e j ( π/ 2+ π/ 3) ) e j 2 t ] = R e [4 e jπ/ 3 (1 + e jπ/ 2 ) ︸ ︷︷ ︸ √ 2 e jπ/ 4 e j 2 t ] = R e [4 p 2 e j 7 π/ 12 e j 2 t ] so that the phasor is 4 p 2 e j 7 π/ 12

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    Chaparro-Akan — Signals and Systems using MATLAB 0.9 0.7 To get an idea of the number of bits generated and processed by a digital system consider the following applications: (a) A compact disc (CD) is capable of storing 75 minutes of “CD quality” stereo (left and right channels are recorded) music. Calculate the number of bits that are stored in the CD as raw 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 of 10 , 000 samples/sec, and that each sample is represented by 8 bits. Calculate the number of bits that your cell-phone has to process every second that you talk. Why would you then need a vocoder? (c) Find out whether text messaging is cheaper or more expensive than voice. Explain how the text messaging works. (d) Find out how an audio CD and an audio DVD compare. Find out why it is said that a vinyl long-play record reproduces sounds much better. Are we going backwards with digital technology in music recording? Explain. (e) To understand why video streaming in the internet is many times of low quality, consider the amount of data that needs to be processed by a video compressor every second. As- sume the size of a video frame, in pixels, is 352 × 240 , and that an acceptable quality for the image is obtained by allocating 8 bits/pixel and to avoid jerking effects we use 60 frames/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 out what some of the compression methods are. Answers: (a) About 6 . 4 Gbs; vocoder (short for voice encoder) reduces number of transmitted bits while keeping voice recognizable. Solution (a) Assuming a maximum frequency of 22 . 05 kHz for the acoustic signal, the numbers of bytes ( 8 bits per byte) for two channels (stereo) and a 75 minutes recording is greater or equal to: 2 × 22 , 050 samples/channel/second × 2 bytes/sample × 2 channels × 75 minutes × 60 sec- onds/minute = 7 . 938 × 10 8 bytes. Multiplying by 8 we get the number of bits. CD quality means that the signal is sampled at 44 . 1 kHz and each sample is represented by 16 bits or 2 bytes. (b) The raw data would consist of 8 (bits/sample) × 10 , 000 (samples/sec)= 80 , 000 bits/sec. The vocoder 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 during pauses. (c) Texting between cell phones is possible by sending short messages ( 160 characters) using the short message services (SMS). Whenever your cell-phone communicates with the cell phone tower 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 the cost of storing messages, the procedure is rather inexpensive and convenient to users.

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    Chaparro-Akan — Signals and Systems using MATLAB 0.10 (d) For CD audio the sampling rate is 44 . 1 kHz with 16 bits/sample. For DVD audio the sam- pling rate is 192 kHz with 24 bits/sample. The sampling process requires getting rid of high frequencies 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 × 240 pixels/frame × 60 frames/sec. The number of bits available for transmission every second is obtained by multiplying the above answer by 8 bits/pixel. There many compression methods JPEG, MPEG, etc.

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    Chaparro-Akan — Signals and Systems using MATLAB 0.11 0.8 The geometric series S = N − 1 ∑ n =0 α n will be used quite frequently in the next chapters so let us look at some of its properties: (a) Suppose α = 1 what is S equal to? (b) Suppose α 6 = 1 show that S = 1 − α N 1 − α Verify that (1 − α ) S = (1 − α N ) . Why do you need the constraint that α 6 = 1 ? Would this sum exist if α > 1 ? Explain. (c) Suppose now that N = 1 , under what conditions will S exist? if it does, what would S be equal to? Explain. (d) Suppose again that N = 1 in the definition of S . The derivative of S with respect to α is S 1 = dS dα = ∞ ∑ n =0 nα n − 1 obtain a rational expression to find S 1 . Answers: S = N when α = 1 , S = (1 − α N ) / (1 − α ) when α 6 = 1 . Solution (a) If α = 1 then S = N − 1 ∑ n =0 1 = 1 + 1 + · · · + 1 ︸ ︷︷ ︸ N times = N (b) The expression S (1 − α ) = S − αS = (1 + α + · · · + α N − 1 ) − ( α + α 2 + · · · + α N − 1 + α N ) = 1 − α N as the intermediate terms cancel. So that S = 1 − α N 1 − α , α 6 = 1 Since we do not want the denominator 1 − α to be zero, the above requires that α 6 = 1 . If α = 1 the sum was found in (a). As a finite sum, it exists for any finite values of α . Putting (a) and (b) together we have S = { (1 − α N ) / (1 − α ) α 6 = 1 N α = 1 (c) If N is infinite, the sum is of infinite length and we need to impose the condition that j α j < 1 so that α n decays as n ! 1 . In that case, the term α N ! 0 as N ! 1 , and the sum is S = 1 1 − α j α j < 1

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    Chaparro-Akan — Signals and Systems using MATLAB 0.12 If j α j ≥ 1 this sum does not exist, i.e., it becomes infinite. (d) The derivative becomes S 1 = dS dα = ∞ ∑ n =0 nα n − 1 = 1 (1 − α ) 2 .

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    Chaparro-Akan — Signals and Systems using MATLAB 0.13 0.2 Problems using MATLAB 0.9 Derivative and finite difference — Let y ( t ) = dx ( t ) /dt , where x ( t ) = 4 cos(2 πt ) , −1 < t < 1 . Find y ( t ) analytically and determine a value of T s for which ∆[ x ( nT s )] /T s = y ( nT s ) (consider as possible values T s = 0 . 01 and T s = 0 . 1 ). Use the MATLAB function diff or create your own to compute the finite difference. Plot the finite difference in the range [0 , 1] and compare it with the actual derivative y ( t ) in that range. Explain your results for the given values of T s . Answers : y ( t ) = − 8 π sin(2 πt ) has same sampling period as x ( t ) , T s ≤ 0 . 5 ; T s = 0 . 01 gives better results. Solution The derivative is y ( t ) = dx ( t ) dt = − 8 π sin(2 πt ) which has the same frequency as x ( t ) , thus the sampling period should be like in the previous problem, T s ≤ 0 . 5 . % Pr. 0.9 clear all % actual derivative Tss=0.0001;t1=0:Tss:3; y=-8*pi*sin(2*pi*t1); figure(2) % forward difference Ts=0.01;t=[0:Ts:3];N=length(t); subplot(211) xa=4*cos(2*pi*t); % sampled signal der1_x=forwardiff(xa,Ts,t,y,t1); clear der1_x % forward difference Ts=0.1;t=[0:Ts:3];N=length(t); subplot(212) xa=4*cos(2*pi*t); % sampled signal der1_x=forwardiff(xa,Ts,t,y,t1); The function forwardiff computes and plots the forward difference and the actual derivative. function der=forwardiff(xa,Ts,t,y,t1) % % forward difference % % xa: sampled signal using Ts % % y: actual derivative defined in t N=length(t);n=0:N-2; der=diff(xa)/Ts; stem(n*Ts,der,’filled’);grid;xlabel(’t, nT_s’) hold on plot(t1,y,’r’); legend(’forward difference’,’derivative’) hold off For T s = 0 . 1 the finite difference looks like the actual derivative but shifted, while for T s = 0 . 01 it does not.

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    Chaparro-Akan — Signals and Systems using MATLAB 0.14 0 0.5 1 1.5 2 2.5 3 −40 −20 0 20 40 t, nT s 0 0.5 1 1.5 2 2.5 3 −40 −20 0 20 40 t, nT s forward difference derivative forward difference derivative Figure 1: Problem 9: T s = 0 . 01 sec (top) and T s = 0 . 1 sec (bottom)

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