MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROBLEM SET 11 SOLUTIONS Problem 1(O&W 1029(d)) In this problem we are asked to sketch the magnitude of the Fourier transform associated with the pole-zero diagram, Figure P10.29(d). In order to do so, we need to make some assumptions (a) The ROC includes the unit circle to ensure the existence of the Fourier transform (b)The gain factor is one so that the system we are dealing with has the following form H(2)= (z-z1)(x+1) where z1 is a positive real number whose magnitude is less than 1 To obtain the frequency response of a DT system, we need to evaluate the magnitude and phase of H(a) along the unit circle in z-plane, i.e., 2=eu for 0<w<2. First,we look at the magnitude plot. In general, for a fixed w we can think of H(e3u)l as tot zeros length of a vector connecting i th zero to eju) IH(eju) i=I( length of a vector connecting j th pole to eju If of zeros or poles is zero, then we define the product above to be 1. In are two poles and no zeros. Thus, the above expression can be s implied lo. our case. there HH(e)l where v and U2 are vectors shown in the figure below:
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 11 Solutions Problem 1 (O&W 10.29 (d)) In this problem we are asked to sketch the magnitude of the Fourier transform associated with the pole-zero diagram, Figure P10.29 (d). In order to do so, we need to make some assumptions: (a) The ROC includes the unit circle to ensure the existence of the Fourier transform. (b) The gain factor is one so that the system we are dealing with has the following form: H(z) = 1 (z − z1)(z + z1) , where z1 is a positive real number whose magnitude is less than 1. To obtain the frequency response of a DT system, we need to evaluate the magnitude and phase of H(z) along the unit circle in z− plane, i.e., z = e jω for 0 ≤ ω < 2π. First, we look at the magnitude plot. In general, for a fixed ω we can think of |H(e jω )| as |H(e jω )| = Π # of zeros i=1 ( length of a vector connecting i th zero to e jω ) Π # of poles j=1 ( length of a vector connecting j th pole to e jω ) If # of zeros or poles is zero, then we define the product above to be 1. In our case, there are two poles and no zeros. Thus, the above expression can be s implied to: |H(e jω )| = 1 |v1||v2| , where v1 and v2 are vectors shown in the figure below: 1
∠t From the plot, we see that when w=2, the product of lun and lu2l becomes maximum Thus, we expect to have the minimum magnitude at the frequency. Also, when w=0 or T, the product of the lengths or the two vectors becomes minimum; thus the magnitude of H(eJu) becomes maximum. The magnitude plot of H(eu) for 0 <w<T is shown below t Although you are not asked to sketch the phase, here is a briefly outline on how to sketch
<e =m 1 1 −1 −1 ω × ∠v~1 v~1 × ∠v~2 v~2 −z1 z1 e jω From the plot, we see that when ω = π 2 , the product of |v1| and |v2| becomes maximum. Thus, we expect to have the minimum magnitude at the frequency. Also, when ω = 0 or π, the product of the lengths or the two vectors becomes minimum; thus the magnitude of H(e jω ) becomes maximum. The magnitude plot of H(e jω ) for 0 ≤ ω < π is shown below: |H(e jω )| ω π 2 π Although you are not asked to sketch the phase, here is a briefly outline on how to sketch the phase. 2
The phase ZH(e) can be described as: ∠H(e) angle of vector connecting i th zero to e of poles angle of vec ng j th pole to e) Again for our specific case, using the vectors u and u2 defined above we have ∠H(e-)=-(∠v1+∠v2), The phase starts off at 0 when w=0 and decreases to when w=3 symmetric poles, the phase keeps decreasing to -2T at w = T. The phase plot is bele ∠H(e) Problem 2(O&W 10.34) ym]=yn-1]+yn-2]+r{n-1] Taking the z-transform of this equation Y(z)=z-1Y()+x-2Y(2)+x-1Xx(x2) H(z) X 1+√5 1-√5 H(z) has a zero at z=0 and poles at z1 and Since the system is causal, the ROC of H(a)will be outside the circle containing its outermost pole:[=>|a1 The pzmap and ROC are depicted below
The phase ∠H(e jω ) can be described as: ∠H(e jω ) = # X of zeros i=1 ( angle of vector connecting i th zero to e jω ) − # X of poles j=1 ( angle of vector connecting j th pole to e jω ). Again for our specific case, using the vectors v1 and v2 defined above we have ∠H(e jω ) = −(∠v1 + ∠v2), The phase starts off at 0 when ω = 0 and decreases to −π when ω = π 2 . Because of the symmetric poles, the phase keeps decreasing to −2π at ω = π. The phase plot is shown below: ∠H(e jω ) ω π 2 π −π −2π Problem 2 (O&W 10.34) (a) y[n] = y[n − 1] + y[n − 2] + x[n − 1] Taking the z-transform of this equation: Y (z) = z −1Y (z) + z −2Y (z) + z −1X(z) H(z) = Y (z) X(z) = z −1 1 − z −1 − z −2 = z z 2 − z − 1 = z z − 1 + √ 5 2 ! z − 1 − √ 5 2 ! H(z) has a zero at z = 0 and poles at z1 = 1+√ 5 2 and z2 = 1− √ 5 2 . Since the system is causal, the ROC of H(z) will be outside the circle containing its outermost pole: |z| > |z1|. The pzmap and ROC are depicted below: 3
H(z) 1+√5 z-1+ B 1+√5 (2) 21=-4、l B H(2) 25 H
0 × 1+√ 5 2 × 1− √ 5 2 1 + √ 5 2 4
Taking the inverse z-transform: h/n= √5(1+√3 an (c)The system is unstable, as its ROC does not contain the unit circle. The instability is 山m() term will grow indefinitely as In order to make the system stable, the rOC must contain the unit circle. For stabilit the ROC should be: ==v5<|z<=tv5 The inverse z-transform of H(a) with this ROC is 4间=(125)(-2)时n Problem 3(O&W 10.42 Because we are dealing with a system which has initial conditions, we may want to the unilateral z-transform. From the properties of the unilateral z-transform, we get the following relationships (x)+y{-1 In each part of this problem, the first step is to take the unilateral z-transform of both sides of the difference equation. To find the zero-input response(ZIR), set the input to 0 and solve for Y(z). To find the zero-state response(ZSR), set the initial conditions to zero and solve for Y(a) ym]+3n-1=xm,y-1=1 In/=/1 uny Taking the unilateral z-transform of both sides of the difference equation Solving for Y(a) gives X 1+ ZIR
Taking the inverse z-transform: h[n] = − 1 √ 5 − 2 1 + √ 5 n u[n] + 1 √ 5 − 2 1 − √ 5 n u[n] (c) The system is unstable, as its ROC does not contain the unit circle. The instability is also apparent in h[n], as the − 2 1− √ 5 n term will grow indefinitely as n → ∞. In order to make the system stable, the ROC must contain the unit circle. For stability, the ROC should be: 2 −1− √ 5 < |z| < 2 −1+√ 5 . The inverse z-transform of H(z) with this ROC is: h[n] = − 1 √ 5 − 2 1 + √ 5 n u[n] − 1 √ 5 − 2 1 − √ 5 n u[−n − 1] Problem 3 (O&W 10.42) Because we are dealing with a system which has initial conditions, we may want to use the unilateral z-transform. From the properties of the unilateral z-transform, we get the following relationships: y[n] ←→ Y (z) y[n − 1] ←→ z −1Y (z) + y[−1] In each part of this problem, the first step is to take the unilateral z-transform of both sides of the difference equation. To find the zero-input response (ZIR), set the input to 0 and solve for Y (z). To find the zero-state response (ZSR), set the initial conditions to zero and solve for Y (z). (a) y[n] + 3y[n − 1] = x[n], y[−1] = 1, x[n] = 1 2 n u[n] Taking the unilateral z-transform of both sides of the difference equation, Y (z) + 3z −1Y (z) + 3y[−1] = X(z). Solving for Y (z) gives, Y (z) = X(z) 1 + 3z −1 | {z } ZSR + −3y[−1] 1 + 3z −1 | {z } ZIR . 5
To find the ZIR, set X (a)=0 and use the fact that y[-1=1 回m]=-3(-3)uln]=(-3)+um] Co find the ZSR, set y[-1]=0 and use x[n]=(5)(n] as given in the problem, X(z (1+32-1)(1 6 2=(-37+(2) m]-n-1],y-1=0,xn]=ul Taking the unilateral z-transform of both sides of the difference equation Y(z)-32-Y(x)-5y-1]=X(z) X(a-o[ Solving for Y(z) gives To find the ZIR, set X(a)=0(note that this means x[-1=0), and use the fact that y-1]=0, 0 To find the ZSR, set yl-1=0 and use n]=un as given in the problem, X(a) Usain= unl
To find the ZIR, set X(z) = 0 and use the fact that y[−1] = 1, YZIR (z) = −3 1 + 3z −1 yZIR [n] = −3(−3)nu[n] = (−3)n+1u[n]. To find the ZSR, set y[−1] = 0 and use x[n] = 1 2 n u[n] as given in the problem, X(z) = 1 1 − 1 2 z −1 YZSR (z) = 1 (1 + 3z −1 ) 1 − 1 2 z −1 = 6 7 1 + 3z −1 + 1 7 1 − 1 2 z −1 yZSR [n] = 6 7 (−3)nu[n] + 1 7 1 2 n u[n]. (b) y[n] − 1 2 y[n − 1] = x[n] − 1 2 x[n − 1], y[−1] = 0, x[n] = u[n] Taking the unilateral z-transform of both sides of the difference equation, Y (z) − 1 2 z −1Y (z) − 1 2 y[−1] = X(z) − 1 2 z −1X(z) − 1 2 x[−1]. Solving for Y (z) gives, Y (z) = (1 − 1 2 z −1 )X(z) 1 − 1 2 z −1 + − 1 2 x[−1] + 1 2 y[−1] 1 − 1 2 z −1 = X(z) | {z } ZSR + − 1 2 x[−1] + 1 2 y[−1] 1 − 1 2 z −1 | {z } ZIR . To find the ZIR, set X(z) = 0 (note that this means x[−1] = 0), and use the fact that y[−1] = 0, YZIR (z) = 0 =⇒ yZIR [n] = 0. To find the ZSR, set y[−1] = 0 and use x[n] = u[n] as given in the problem, X(z) = 1 1 − z −1 YZSR (z) = 1 1 − z −1 yZSR [n] = u[n]. 6
1 n]-亓yn-1 cIn-1, yl-1=l, xn]=un Since this is the same difference equation as given in part(b), we can use the equation for Y(a) derived above. To find the ZIR, set X()=0(note that this means x[-1=0) and use the fact that yl-1=1 1/1 unl To find the ZSR, set yl-1=0 and use xn= un as given in the problem. This is exactly what we did in part(b) Problem 4(O&W 10.47) (a) Recall that complex exponentials are the eigenfunction of LTI systems. Thus, the re- sponse of Lti systems to complex exponentials is of the form zn→→H(2)2n Since the output of the system to an input n=(2) mis 0, we conclude that the ROC of H (a) contains 2=-2 and H(a) has a zero at z=-2 The second input-output relation gives us H(a) X(a) (1+a-2-1)(1 (1+a) assuming that 1 +a+. H(a)has two poles; at z=0 and z=1. In order to include 2 in its ROC, we know that the system is causal and its ROC is outside of the circle of radius. Since H(z) has a zero at 2 =-2, 1+a-21=2=0→a=-3
(c) y[n] − 1 2 y[n − 1] = x[n] − 1 2 x[n − 1], y[−1] = 1, x[n] = u[n] Since this is the same difference equation as given in part (b), we can use the equation for Y (z) derived above. To find the ZIR, set X(z) = 0 (note that this means x[−1] = 0), and use the fact that y[−1] = 1, YZIR (z) = 1 2 1 − 1 2 z −1 yZIR [n] = 1 2 1 2 n u[n] = 1 2 n+1 u[n]. To find the ZSR, set y[−1] = 0 and use x[n] = u[n] as given in the problem. This is exactly what we did in part (b) above, so yZSR [n] = u[n]. Problem 4 (O&W 10.47) (a) Recall that complex exponentials are the eigenfunction of LTI systems. Thus, the response of LTI systems to complex exponentials is of the form: z n −→ H(z)z n . Since the output of the system to an input x[n] = (−2)n is 0, we conclude that the ROC of H(z) contains z = −2 and H(z) has a zero at z = −2. The second input-output relation gives us: H(z) = Y (x) X(z) = 1 + a 1− 1 4 z−1 1 1− 1 2 z−1 , |z| > 1 2 = (1 + a − 1 4 z −1 )(1 − 1 2 z −1 ) 1 − 1 4 z −1 = (1 + a) (z − 1 4(1+a) )(z − 1 2 ) z(z − 1 4 ) , assuming that 1 + a 6= 0. H(z) has two poles; at z = 0 and z = 1 4 . In order to include z = −2 in its ROC, we know that the system is causal and its ROC is outside of the circle of radius 1 4 . Since H(z) has a zero at z = −2, 1 + a − 1 4 z −1 |z=−2 = 0 ⇒ a = − 9 8 . 7
(b) Now we know the expression for H(2) H 1 The input in this case is a complex exponential an]=1n The output will therefore be of the form yn]=H(1) Problem 5(O&W 10.50) 0) From the pole-zero pattern, the system function takes the following form Showing that H(eju)l is constant is equivalent to showing that H(e3u)I2 is constant H(ej)l2= H(eu)H"(eu) e e]u-a e-Ju-a 2 1- 2a cos(w)+a 1-2a cos(w)+a Thus, H(e)l= fal which is constant and is completely determined by the location of (b) Using the law of cosines Ju112=|112+la)2-2 1al cos(w) 1+a-2a cos(w) (c) Following the same procedure as in(b) l2=12+|-21|1(c(0
(b) Now we know the expression for H(z): H(z) = (− 1 8 − 1 4 z −1 )(1 − 1 2 z −1 ) 1 − 1 4 z −1 . The input in this case is a complex exponential x[n] = 1 n . The output will therefore be of the form: y[n] = H(1) · 1 n = − 3 8 1 2 3 4 = − 1 4 . Problem 5 (O&W 10.50) (a) From the pole-zero pattern, the system function takes the following form: H(z) = z − 1 a z − a . Showing that |H(e jω )| is constant is equivalent to showing that |H(e jω )| 2 is constant. |H(e jω )| 2 = H(e jω )H ∗ (e jω ) = e jω − 1 a e jω − a · e −jω − 1 a e −jω − a = 1 − 2 a cos(ω) + 1 a 2 1 − 2a cos(ω) + a 2 = 1 a 2 · a 2 − 2a cos(ω) + 1 1 − 2a cos(ω) + a 2 = 1 a 2 . Thus, |H(e jω )| = 1 |a| which is constant and is completely determined by the location of the pole and zero. (b) Using the law of cosines, |v1| 2 = |1| 2 + |a| 2 − 2|1||a| cos(ω) = 1 + a 2 − 2a cos(ω). (c) Following the same procedure as in (b), |v2| 2 = |1| 2 + 1 a 2 − 2|1| 1 a cos(ω) = 1 a 2 a 2 + 1 − 2a cos(ω) = 1 a 2 |v1| 2 . ∴ |v2| = v1 a . 8
It clearly shows that the length of u2 is proportional in length to i independently of w Problem 6(O&W 11.25(a)) The system given is G(z)H(x)=- First, we would like to find the poles and zeros of the system. It is easy to see that G(a)H(a) can be expressed as G(z)H(z) (2+)(2-) (1) The system has poles at+, and a zero at 1. The pole-zero map as is as shown below. Note that dashed circle in all the figures in this problem denotes a unit circle Recall that the derivation of the angle criteria does not concern if the system is ct or DT. Thus, by evoking the angle criteria we have ZG(a)H(a)= odd integer multiple of if K>0 ZG(z)H(z)= even integer multiple of T if K <0
It clearly shows that the length of v2 is proportional in length to v1 independently of ω. Problem 6 (O&W 11.25 (a)) The system given is: G(z)H(z) = z − 1 z 2 − 1 4 . First, we would like to find the poles and zeros of the system. It is easy to see that G(z)H(z) can be expressed as: G(z)H(z) = z − 1 z + 1 2 z − 1 2 . (1) The system has poles at ± 1 2 and a zero at 1. The pole-zero map as is as shown below. Note that dashed circle in all the figures in this problem denotes a unit circle. − 1 2 × 1 2 × 0 1 0, ∠G(z)H(z) = even integer multiple of π if K < 0. 9
for all z on the locus Let's look at the case where K>0. Using the angle criteria, we can identify that the two real line segments belong to the locus. One is(00,-5)and the other(, 1). In this case, these two segments completely specify the locus. From this, we can see that a pole at will approach to the zero at1asK→∞. The other pole, one at- will move to-∞ Thus, as K-o0, one of the poles will become unstable(Note that in feedback systems, we only consider causal systems ). Thus, we would like to know up to which K, the closed loop system remains stable, i.e., we would like to find the value of K when the closed loop system has a pole at -1. From Eqn(1) G(x)H(2)-=-1=-K z-1 1 (2+)(2-号) F 之=-1 +)( Thus the root locus for K >0 is shown above
for all z on the locus. Let’s look at the case where K > 0. Using the angle criteria, we can identify that the two real line segments belong to the locus. One is (−∞, − 1 2 ) and the other ( 1 2 , 1). In this case, these two segments completely specify the locus. From this, we can see that a pole at 1 2 will approach to the zero at 1 as K → ∞. The other pole, one at − 1 2 will move to −∞. Thus, as K → ∞, one of the poles will become unstable (Note that in feedback systems, we only consider causal systems). Thus, we would like to know up to which K, the closed loop system remains stable, i.e., we would like to find the value of K when the closed loop system has a pole at −1. From Eqn (1), G(z)H(z)| z=−1 = − 1 K z − 1 z + 1 2 z − 1 2 z=−1 = − 1 K −1 − 1 −1 + 1 2 −1 − 1 2 = − 1 K ∴ K = 3 8 . − 1 2 × 1 2 × 0 1 0 is shown above. 10