Homework_8_solution

Homework_8_solution - 1 EE113 Homework 8 Problem 12.5...

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Unformatted text preview: 1 EE113 Homework 8 Problem 12.5 Determine the unilateral z -transforms of each of the following sequences and indicate their regions of convergence: (a) x(n) = (n − 1)u(n + 1). (b) x(n) = (1 + n2 αn−2 )u(n + 31). (c) x(n) = α|n| with α > 0. (d) The impulse response sequence of the relaxed causal system y (n) − 3 y (n − 1) + 1 y (n − 2) = x(n − 1). 4 8 Solution: Note that the unilateral z -transform of x(n) is the bilateral z -transform of x(n)u(n). (a) The equivalent sequence is x′ (n) = x(n)u(n) = (n − 1)u(n) = nu(n) − u(n). The unilateral z -transform of x(n) then is X + (z ) = z z − , 2 (z − 1) z−1 |z | > 1 (b) The equivalent sequence is x′ (n) = x(n)u(n) = (1 + n2 αn−2 )u(n) = u(n) + α−2 n2 αn u(n). The unilateral z -transform of x(n) then is X + (z ) = z z+α − , z − 1 α(z − α)3 {|z | > 1} ∩ {|z | > |α|} (c) The equivalent sequence is x′ (n) = x(n)u(n) = α|n| u(n) = αn u(n) with α > 0. The unilateral z -transform of x(n) then is X + (z ) = z , z−α |z | > α (d) Because this system is causal, its impulse response sequence h(n) = h(n)u(n) is a casual sequence. The unilateral z -transform of h(n) then is the same as its transfer function H (z ): H (z ) = November 21, 2011 Y (z ) 8z =2 , X (z ) 8z − 6z + 1 |z | > 1 2 DRAFT 2 Problem 12.21 Consider a causal system that is described by the difference equation y (n) = 5 1 y (n − 1) − y (n − 2) + x(n − 2), 6 6 y (−2) = 0, y (−1) = 1 Use the unilateral z -transform to determine its complete response to the sequence 1 4 x(n) = (n − 1) n−2 u(n − 1) Solution: Because x(n) = (n − 1) 1 4 n−2 u(n − 1) = (n − 2) n−2 1 4 u(n − 2) if we introduce an equivalent input sequence xo (n) as 1 4 xo (n) = n n u(n) and modify the difference equation as y (n) = 5 1 y (n − 1) − y (n − 2) + x(n − 4), 6 6 y (−2) = 0, y (−1) = 1 then we should have the same output sequence. Let us find out the unilateral z -transform of the equivalent input sequence xo (n): + Xo (z ) = z 1 4z 2, −1 4 |z | > 1 4 Take the unilateral z -transform on both sides of the equivalent difference equation. We can get Y + (z ) = 5 −1 + 1 −2 + + z Y (z ) + y (−1) − z Y (z ) + y (−1)z −1 + y (−2) + z −4 Xo (z ) 6 6 Substituting the initial conditions leads to Y + (z ) = 5 −1 + 1 −2 + + z Y (z ) + 1 − z Y (z ) + z −1 + z −4 Xo (z ) 6 6 Solving for Y + (z ) yields Y + (z ) = 1− 5 1 −1 6 − 6z 5 −1 + 1 z −2 6z 6 + 1− 1 z −4 4z 1 −2 · 1 + 6z z−4 5 −1 6z 2, |z | > 1 2 Resorting to the partial fractions, we can recover the sequence y (n) as y (n) = 3 · November 21, 2011 1 2 n−1 −2· 1 3 n−1 u(n) + 96 1 2 n−1 −9· 1 3 n−1 + 2 · (n + 3) · 1 4 n−1 u(n − 4) DRAFT 3 Problem 12.25 Consider the constant-coefficient difference equation 1 1 y (n) − y (n − 1) − y (n − 2) = x(n − 1) 6 6 with initial conditions y (−2) = 0 and y (−1) = 6. Use the z -transform technique to find the answers to parts (a)-(c): (a) The zero-input response. (b) The zero-state response to x(n) = u(n). (c) The complete response using the unilateral z -transform. (d) Check that your answer to part (c) is the sum of the answers to parts (a) and (b). (e) Now determine the answers to parts (a)-(c) by using the time-domain techniques you learned earlier for solving constant-coefficient difference equations by working with the modes of the system. Compare your answers to those obtained by using the z -transform technique. (f) Find the response of the system to x(n) = ( 1 )n u(n − 2) in three different ways. 3 Solution: (a) The zero-input response has to be solved by the general solutions. The general solution for the homogeneous equation is n 1 2 yh (n) = C1 + C2 − 1 3 n Substituting the initial conditions y (−2) = 0 and y (−1) = 6 leads to C1 = 9 , 5 C2 = − 4 5 Hence, the zero-input response for n ≥ 0 is yzi = 9 5 n 1 2 − 4 5 − n 1 3 u(n) (b) The zero-state response can be solved by using the z -transform technique. The transfer function of the system is H (z ) = 1− z −1 z =2 1 , 1 z − 6z − 1 − 6 z −1 6 |z | > 1 −1 6z 1 2 The z -transform of the input sequence x(n) = u(n) is X (z ) = z , z−1 |z | > 1 The z -transform of the zero-state response can be expressed as Yzs (z ) = z z2 − 1 z − 6 1 6 · z z2 = z−1 (z − 1) z − 1 2 z+ 1 3 = 3 2 z−1 + 3 1 −5 + 10 1 , z−1 z+ 3 2 |z | > 1 which leads to yzs (n) = November 21, 2011 33 − 25 1 2 n−1 + 1 10 − 1 3 n−1 u(n − 1) DRAFT 4 (c) Taking the unilateral z -transform on both sides of the difference equation yields Y + (z ) − 1 −1 + 1 −2 + z Y (z ) + y (−1) − z Y (z ) + y (−1)z −1 + y (−2) = z −1 X + (z ) 6 6 Substituting the initial conditions yields Y + (z ) − 1 −1 + 1 −2 + z Y (z ) + 6 − z Y (z ) + 6z −1 = z −1 X + (z ) 6 6 Solving for Y + (z ) yields Y + (z ) = = = 1 + z −1 z −1 X + (z ) 1 −2 + 1 −1 1− − 6z 1 − 6 z − 1 z −2 6 1 −1 6z 2 z2 z +z − 1z − 6 9 5z 1 2 z− 1+ 6 4 −5z z+1 3 + z − 1z 6 3 2 z2 + z−1 · z z−1 − 1 6 + 1 −3 5 10 , 1+ z−2 z+1 3 |z | > 1 which leads to y (n) = 9 5 1 2 n − 4 5 − n 1 3 33 − 25 u(n) + 1 2 n−1 + 1 10 − 1 3 n−1 u(n − 1) (d) Compare the two results straightforwardly. (e) We only need to find out the zero-state response in the time domain. The impulse response sequence can be solved from the general solution by plugging the corresponding initial conditions h(0) = 0 and h(1) = 1, which results in n 1 2 6 5 h(n) = 6 5 − − n 1 3 u(n − 1) Hence, the zero-state response can be found by 6 5 1 2 n 6 5 yzs (n) = = k=1 = = 1 2 6 5 n − 33 − 25 − k 1 2 −1 2 1− 6 5 − 6 5 n+1 − 1 2 1 2 1 3 n u(n − 1) ⋆ u(n) − 1 3 k u(n − 1) 1 1 −3 − −3 1+ 1 3 n−1 1 10 − 33 − 25 1 2 + 1 3 n+1 u(n − 1) n−1 u(n − 1) Therefore, by part (a), the complete response is y (n) = 9 5 1 2 n − 4 5 − 1 3 n u(n) + n−1 + 1 10 − 1 3 n−1 u(n − 1) which coincides with the result obtained in part (c). (f) One can use the z -transform, or use the time-domain convolution, or use the unilateral z -transform to find 1 out the zero-state or complete response to x(n) = ( 3 )n u(n − 2). Here we only give the first method. The November 21, 2011 DRAFT 5 z -transform of the input sequence is X (z ) = z −2 z · 1, 9 z−3 1 3 |z | > The z -transform of the zero-state response can be expressed as Yzs (z ) = z z2 − 1 6z − 1 6 · z −2 z · 9 z− 1 3 = 1 9 z− z+ 1 2 1 3 1 3 n−1 − 1 2 n−1 = z− 1 3 4 5 z− 1 2 + 1 5 z+ 1 3 + −1 , z−1 3 |z | > 1 2 which leads to 4 5 yzs (n) = 1 2 1 3 n−1 n + 1 5 1 3 − n−1 u(n − 1) Then, the complete response is y (n) = November 21, 2011 9 5 1 2 n − 4 5 − u(n) + 4 5 + 1 5 − 1 3 n−1 − 1 3 n−1 u(n − 1) DRAFT 6 Problem 13.9 Find the DTFTs of the following sequences: (a) x(n) = (b) x(n) = (c) x(n) = 1 n−1 u(n + 1). 2 2n−1 1 u(n − 1). 3 1 n−2 −4 u(−n − 1). In each case, find expressions for the magnitude and phase of the DTFT. Solution: (a) Notice that 1 2 x(n) = 4 · n+1 u(n + 1) Thus we can get X (ejω ) = 4ejω = 1 − 1 e−jω 2 4 · exp(jω ) 5 4 − cos(ω ) · exp j arctan sin(ω ) 2−cos(ω ) where the amplitude and the phase are 4 |X (ejω )| = 5 4 ∠X (ejω ) = ω − arctan , − cos(ω ) sin(ω ) 2 − cos(ω ) (b) Notice that x(n) = 1 · 3 1 9 n−1 u(n − 1) Thus we can get X (ejω ) = 1 3 1 3 · e−jω = 1 − 1 e−jω 9 82 81 − 2 9 · exp(−jω ) · cos(ω ) · exp j arctan sin(ω ) 9−cos(ω ) where the amplitude and the phase are 1 3 |X (ejω )| = 82 81 − 2 9 , ∠X (ejω ) = −ω − arctan · cos(ω ) sin(ω ) 9 − cos(ω ) (c) Notice that x(n) = −64 · 4−n−1 u(−n − 1) Thus we can get X (ejω ) = −64 · ejω = 1 − 4 ejω −64 · exp(jω ) 17 − 8 · cos(ω ) · exp −j arctan sin(ω ) 1 4 −cos(ω ) where the amplitude and the phase are |X (ejω )| = November 21, 2011 64 17 − 8 · cos(ω ) , ∠X (ejω ) = π + ω + arctan 1 4 sin(ω ) − cos(ω ) DRAFT 7 Problem 13.13 Find the DTFTs of the following sequences: (a) x(n) = (b) x(n) = 1 n−3 2 1 n−1 4 (c) x(n) = cos 1n 3 u(n) + π 3n u(n) + cos π 4n u(n − 1). . + δ (n − 1) + δ (n + 1). In each case, find expressions for the magnitude and phase of the DTFT. Solution: (a) Notice that n 1 2 x(n) = 8 · u(n) + 1 · 3 n−1 1 3 u(n − 1) Thus we can get X (ejω ) = 1 −jω e 8 + 3 1 −jω = 1 −jω 1 − 2e 1 − 3e 1 3 + 10 9 − 2 3 8 5 4 − cos(ω ) · exp j arctan sin(ω ) 2−cos(ω ) · exp(−jω ) cos(ω ) · exp j arctan sin(ω ) 3−cos(ω ) Let a1 − arctan θ1 8 5 4 , 1 3 a2 10 9 − cos(ω ) sin(ω ) 2 − cos(ω ) , θ2 − 2 3 cos(ω ) −ω − arctan sin(ω ) 3 − cos(ω ) Then, X (ejω ) can be rewritten as X (ejω ) = a1 ejθ1 + a2 ejθ2 = [a1 cos(θ1 ) + a2 cos(θ2 )] + j [a1 sin(θ1 ) + a2 sin(θ2 )] a2 + 2a1 a2 cos(θ1 − θ2 ) + a2 · e 1 2 = j arctan a1 sin(θ1 )+a2 sin(θ2 ) a1 cos(θ1 )+a2 cos(θ2 ) Hence, the amplitude and the phase are |X (ejω )| = a2 + 2a1 a2 cos(θ1 − θ2 ) + a2 1 2 ∠X (ejω ) = arctan a1 sin(θ1 ) + a2 sin(θ2 ) a1 cos(θ1 ) + a2 cos(θ2 ) (b) Notice that x(n) = 4 · 1 4 n u(n) + cos π n 3 Thus we can get X (ejω ) = = November 21, 2011 4 1− 1 −jω 4e +π δ ω− π π +δ ω+ 3 3 4 17 16 − 1 2 · cos(ω ) · exp j arctan sin(ω ) 4−cos(ω ) +π δ ω− π π +δ ω+ 3 3 DRAFT 8 Let a= 4 17 16 − 1 2 , θ = − arctan · cos(ω ) sin(ω ) 4 − cos(ω ) The amplitude is |X (ejω )| = a, and the phase is jω ∠X (e ) = (c) We can get X (ejω ) = π δ ω − a2 + 2πa cos(θ) + π 2 , otherwise arctan a sin(θ ) a cos(θ )+π θ, π π +δ ω+ 4 4 ω = ±π 3 , ω = ±π 3 otherwise + e−jω + ejω = π δ ω − π π +δ ω+ 4 4 + 2 cos (ω ) The amplitude is |X (ejω )| = 2 cos (ω ) + π δ ω − π π +δ ω+ 4 4 and the phase is ∠X (ejω ) = 0 November 21, 2011 DRAFT 9 Problem 13.21 Find the DTFTs of the following sequences: (a) x(n) = (b) x(n) = sin( π n) 3 . n π sin( 3 (n−1)) . n−1 Solution: See example 13.7 Low-pass DTFT on the textbook. (a) We can rewrite x(n) as x(n) = and identify ωc = π 3. π sin π n 3 · π 3 n 3 Then, we can get X (ejω ) = π · rect where we define the rectangular function rect(·) as 1 , ω rect ωc 0 , ω π 3 |ω | < ω c ω c ≤ |ω | ≤ π (b) Notice that this is a time-shifted version of part (a). We can immediately get X (ejω ) = π · rect November 21, 2011 ω π 3 · e−jω DRAFT 10 Problem 13.23 Invert the following DTFTs: (a) X (ejω ) = cos( π ω ). 2 (b) X (ejω ) = (c) X (ejω ) = 1 . 1− 1 e−jω 2 π cos( 3 ω ) · sin( π ω ). 3 (d) X (ejω ) = cos2 ( π ω ). 6 (e) X (ejω ) = sin( π ω ) · cos2 ( π ω ). 3 4 Solution: (a) Recall the Euler formula. Then, mathematically, we can get x(n) = 1 π π δ n+ +δ n− 2 2 2 However, from the signal processing point of view, where we interpret the delay apparently, the non-integer number π 2 π 2 as multiples of samples, is not a legal one. In other words, in general we shall not see this kind of DTFT in signal processing area. (b) We can get x(n) = 1 2 n u(n) (c) We notice that X (ejω ) = 2π ω 3 1 sin 2 which leads to x(n) = j 2π δ n+ 4 3 −δ n− 2π 3 Again, this is not a legal signal. (d) We notice that X (ejω ) = cos π 3ω +1 2 which leads to x(n) = 1 π π δ n+ + 2δ (n) + δ n − 4 3 3 Again, this is not a legal signal. (e) We notice that X (ejω ) = sin cos π ω· 3 π 2ω 2 +1 = 1 π 1 sin ω+ sin 2 3 4 5π π ω − sin ω 6 6 which leads to x(n) = j π π δ n+ −δ n− 2 3 3 + j 5π δ n+ 4 6 −δ n 5π 6 −δ n+ π π +δ n− 6 6 Again, this is not a legal signal. November 21, 2011 DRAFT 11 Problem 13.35 The DTFTs of two sequences {x(n), y (n)} are shown in Fig. 13.18. Determine the sequences. Find also their energies. Solution: Notice that X (ejω ) can be expressed as ω X (ejω ) = 2 · rect π 4 ω − rect π 8 Therefore, the sequence x(n) is x(n) = 1 π 1 π sinc n − sinc n 2 4 8 8 where the sinc(·) function is defined as sin(x) x sinc(x) = The energy of the signal is Ex = 1 2π π |X (ejω )|2 dω = −π 5 8 jω Similarly, we can express Y (e ) as Y (ejω ) = 2 · rect ω − 2 · tri π 4 ω π 4 where we define the triangular function as 1− |ω | ωc The sequence corresponding to the triangular function tri ω ωc tri ω ωc x(n) = November 21, 2011 rect ω ωc is ωc ωc · sinc2 n 2π 2 DRAFT 12 Therefore, the sequence y (n) is y (n) = 1 π 1 π sinc n − sinc2 n 2 4 8 8 The energy of the signal is Ey = November 21, 2011 1 2π π |Y (ejω )|2 dω = −π 1 3 DRAFT ...
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This note was uploaded on 01/24/2012 for the course EE 113 taught by Professor Walker during the Fall '08 term at UCLA.

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