113_1_midterm_08SP_A

# 113_1_midterm_08SP_A - EE113: Digital Signal Processing...

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Unformatted text preview: EE113: Digital Signal Processing Solution to Midterm Exam (A) Spring 2008 1. Consider the following two systems y1 (n) = 1 4 n x(k ), k=n−3 y2 (n) = x 0 n 3 n = 3k, k integer n = 3k, k integer. (a) Determine whether the systems are linear, time-invariant, relaxed, BIBO stable, and causal. Justify your answer to receive full credit. Solution: Properties Relaxed Linear Time-Invariant BIBO Stable Causal y1 (n) Yes Yes Yes Yes Yes y2 (n) No Yes No Yes No • System y1 (n): n 1 1 · y1 (n) = 4 k=n−3 x(k ) = 4 [x(n − 3) + x(n − 2) + x(n − 1) + x(n)] can be considered as a relaxed constant coeﬃcient diﬀerence equation. It is relaxed, linear, time-invariant, BIBO stable, and causal. • System y2 (n): · Not relaxed: Counter example: if the ﬁrst non-zero sample of x(n) is at n = −1, then y2 (−3) will be non-zero, which is before −1. · Linear: Let two outputs be a y2 (n) = xa 0 n 3 n = 3k, k integer , n = 3k, k integer. and b y2 (n) = xb 0 n 3 n = 3k, k integer , n = 3k, k integer. for an input xa (n) and xb (n), respectively. The output y (n) for an input Axa (n)+ Bxb (n) with some constants A and B is given by i) if n = 3k , k integer, y (n) = Axa n n + Bxb 3 3 a b = Ay2 (n) + By2 (n). ii) if n = 3k , k integer, y (n) = 0. Based on i) and ii), y (n) for an input Axa (n) + Bxb (n) is a b y (n) = Ay2 (n) + By2 (n), which implies that system y2 (n) is linear. · BIBO Stable: For a bounded sequence |x(n)| ≤ M < ∞ with a ﬁnite positive number M , |y2 (n)| = |x n | ≤ M < ∞ n = 3k, k integer 3 , 0<∞ n = 3k, k integer. implying that |y2 (n)| is bounded output. · Not Time-Invariant: Consider inputs δ (n) and δ (n − 1). The outputs are δ (n) and δ (n − 3). This shows that y2 (n) is not time-invariant. · Not causal: For example, y2 (−3) depends on x(−1), which is on the future of y2 (−3). (b) Given x(n) = nu(n), compute and plot y1 (n) and y2 (n) for 0 ≤ n ≤ 5. Solution: n y1 (n) y2 (n) 0 0 0 1 1/4 0 2 3/4 0 3 6/4 = 3/2 1 4 10/4 = 5/2 0 5 14/4 = 7/2 0 The corresponding plots are shown in Figure 1. (c) Find the z -transform of y1 (n) and y2 (n) in terms of the z -transform of x(n). Solution: z -transform for y1 (n) = 1 4 n k=n−3 1 x(k ) = 4 [x(n − 3) + x(n − 2) + x(n − 1) + x(n)] gives Y1 (z ) = 1 −3 [z X (z ) + z −2 X (z ) + z −1 X (z ) + X (z )] 4 z −3 + z −2 + z −1 + 1 X (z ). = 4 2 y (n) 1 y (n) 2 3.5 1 0.9 3 0.8 2.5 0.7 0.6 0.5 1.5 0.4 0.3 0.2 0.5 0.1 0 0 2 n 4 6 0 0 2 n 4 6 2 1 Figure 1: Plot for y1 (n) and y2 (n). z -transform for y2 (n) is given by ∞ Y2 (z ) = n=−∞ ∞ y2 (n)z −n x n=−∞ ∞ = = k=−∞ ∞ n −n z 3 x(k )z −3k x(k )(z 3 )−k k=−∞ 3 = = X (z ). 3 2. A causal system is described by the following block diagram. x(n) −1 6 z −1 y (n) 1 1 3 z −1 (a) Determine the constant coeﬃcient diﬀerence equation that describes the system, and ﬁnd its impulse response. Solution: Interchanging the order of the two systems in cascade, we get 1 1 y (n) = − y (n − 1) + y (n − 2) + x(n) + x(n − 1). 6 3 The impulse response can be found by computing the zero-state response to x(n) = δ (n). We ﬁrst ﬁnd the homogeneous solution: 1 1 λ2 + λ − = 0 6 3 and hence, yh (n) = C1 1 2 n ⇒ λ1 = 2 3 1 , 2 n λ2 = − 2 3 + C2 − , for all n. Letting x(n) = δ (n) and assuming that the system is relaxed, we get y (0) = 1 and y (1) = − 1 y (0) + 1 = 5 . From the homogeneous solution we get 6 6 y (0) = C1 + C2 = 1 1 2 5 y (1) = C1 − C2 = 2 3 6 and solving for C1 and C2 yields C1 = h(n) = 9 7 and C2 = − 2 , and consequently, 7 1 2 n 9 7 − 2 7 − 2 3 n u(n). (b) Given x(n) = n2n u(−n), ﬁnd the output of the system using the z -transform. Solution: The output y (n) when x(n) = n2n u(−n) can be found by computing the inverse z -transform of Y (z ) = H (z )X (z ). The z -transform of h(n) is readily found from the impulse response: H (z ) = 1 + z −1 , 1+ − 1 z −2 3 1 −1 6z ROC: |z | > 2 . 3 4 The z -transform of x(n) is given by X (z ) = Z{x(n)} = −z = −z d dz ∞ d d Z{2n u(−n)} = −z dz dz n 0 2 n z −n n=−∞ n=0 z 2 = −z d 1 2z , =− 1 dz 1 − 2 z (2 − z )2 ROC: |z | < 2. Consequently, Y (z ) is given by Y (z ) = H (z )X (z ) = −2z 3 − 2z 2 , (z − 1 )(z + 2 )(z − 2)2 2 3 ROC: 2 < |z | < 2. 3 Using partial fraction expansion, we get A Y (z ) B = 1+ z z−2 z+ where A= B= C= z− z+ 1 2 2 3 Y (z ) z Y (z ) z =− 1 z= 2 2 3 + D C + z − 2 (z − 2)2 4 7 3 56 =− z =− 2 3 d Y (z ) 5 (z − 2)2 = dz z 8 z =2 2 Y (z ) = −3. D = (z − 2) z z=2 Thus, Y (z ) = − 4z 7z− − 2 3 1 2 3z 56 z + 2 3 + 5z 3 2z − 8 z − 2 2 (z − 2)2 n and since the ROC of Y (z ) is given by z z− < |z | < 2, we get ↔ ↔ ↔ ↔ 1 2 − 2 3 u(n) n 1 2 z 2 z+3 z z−2 2z (z − 2)2 and consequently, y (n) = − 4 7 1 2 n u(n) −2n u(−n − 1) −n2n u(−n − 1) − 3 56 − 2 3 n 53 u(n) + − + n 2n u(−n − 1). 82 Alternatively, partial fraction expansion on Y (z ) yields y (n) = − 2 7 1 2 n−1 + 1 28 − 2 3 n−1 5 u(n − 1) + − + 3n 2n−1 u(−n). 4 5 3. The diﬀerence equation of a relaxed system is: y (n) + 0.5y (n − 1) − 0.14y (n − 2) = x(n) (a) Find a closed form for the impulse response h(n) (i.e., the zero state system output when the input is an impulse). Solution: By setting x(n) = δ (n), we have h(n) + 0.5h(n − 1) − 0.14h(n − 2) = δ (n) = 1, n = 0 0, n = 0. Hence, for n ≥ 1, h(n) can be found by solving homogeneous diﬀerence equation h(n) + 0.5h(n − 1) − 0.14h(n − 2) = 0. The characteristic polynomial for (1) can be solved by λ+ thus, the modes are λ1 = − Hence, h(n) = C1 − 7 10 7 10 λ− 1 5 = 0, (1) 7 , 10 n λ2 = 1 5 1 . 5 n + C2 , n ≥ 0. Since the system is relaxed, y (−1) = h(−1) = 0 and h(0) = δ (0) = 1, which gives C1 = Therefore, h(n) = 7 9 − 7 10 7 , 9 n C2 = 2 . 9 1 5 n + 2 9 u(n). (b) If the input to the system is x(n) = 2δ (n) + δ (n − 2), what is the output? Solution: The output y (n) of the system can be expressed as the convolution of x(n) and h(n), i.e., y (n) = x(n) ∗ h(n) = [2δ (n) + δ (n − 2)] ∗ h(n) = 2h(n) + h(n − 2). 6 Hence, using h(n) in part (a) gives us i) n ≥ 2, y (n) = 2h(n) + h(n − 2) =2 7 9 − 7 10 n + 7 10 +6 2 9 −2 1 5 n u(n) + − 7 10 n 7 9 − 7 10 n−2 + 1 5 2 9 −2 1 5 1 5 n−2 u(n − 2) n = 2· = ii) 0 ≤ n ≤ 1, 22 7 77 +· 99 − 7 10 n + 2· 22 +· 99 1 5 n . y (n) = 2h(n) =2 = Therefore, y (n) = 0, 14 9 22 7 7 9 14 9 7 21 u(n) + 10 95 n n 7 41 − + . 10 95 − n n n < 0, 7n − 10 + 4 9 7n − 10 + 6 1n , 5 1n , 5 n = 0, 1, n ≥ 2. (c) For what values of α is g (n) = αn h(n) a ﬁnite energy sequence? Solution: The energy Eg of g (n) is given by ∞ Eg = n=−∞ ∞ |g (n)|2 |αn h(n)|2 n=−∞ ∞ = = n=0 ∞ αn 7 9 7 9 − − 7 10 n n + + 2 9 ∞ 2 9 1 α 5 1 5 n 2 = n=0 ∞ 7 α 10 n2 = 49 81 n=0 49 2 α 100 n + 4 81 n=0 12 α 25 n + 28 81 n=0 ∞ − 72 α 50 n . Thus, for the energy to be ﬁnite, 49 2 α < 1, 100 12 72 α < 1, and α < 1, 25 50 7 or equivalently, |α| < Hence, α should satisfy |α| < 10 7. 10 , 7 |α| < 5, and |α| < 50 . 7 8 ...
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