EE102B Signal Processing and Linear Systems II. Solutions to Problem Set Nine Spring Quarter

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1 EE02B Signal Processing and Linear Systems II Solutions to Problem Set Nine Spring Quarter Problem 9. (25 points) (a) 0.5( + 4z + 6z 2 + 4z 3 + z 4 ) + 0.2z 0.4z z 3 x[n] 0.5 y[n] -0.2 Z Z Z Z (b) ( + z 2 )( 0.8z 2 ) ( 0.64z 2 )( + z + 0.8z 2 ) x[n] y[n] Z - Z Z - Z Z - Z

2 Problem 9.2 (25 points) (a) y[n] = y[n 5] + x[n] y[n] + y[n 5] = x[n] (b) Y (z) + z 5 Y (z) = X(z) + z 5 Poles are values of z such that + z 5 = 0, or z 5 = e jπ. Therefore the poles are z k = e jπ(2k+)/5 for k = 0,, 2, 3, 4. Also note e j9π/5 = e jπ/5 and e j7π/5 = e j3π/5, so the poles can be simplified to z = e ±jπ/5, e ±j3π/5, e jπ. We can therefore express H(z) as: ( e jπ/5 z )( e jπ/5 z )( e j3π/5 z )( e j3π/5 z )( e jπ z, ROC: z > ) Since all of these poles are on the unit circle, the two possible regions of convergence are the areas inside and ouside the unit circle. Since the system is causal, the area outside the unit circle is the correct ROC. There are 5 zeros at z = Imaginary Part Real Part 5 The white region is the ROC and the gray region is outside the ROC. (c) The system is not stable since the ROC does not contain the unit circle. 2

3 (d) This can be solved using the difference equation: y[n] = y[n 5] + x[n] By repeatedly using the formula above on x[n] we see that: y[n] = ( ) k (δ[n 5k] + δ[n 5k]) k=0 Alternatively, we can solve this in the z-domain using the formula for a geometric series: + z 5 = ( ) k z 5k z > h[n] = y[n] = h[n] x[n] = k=0 ( ) k δ[n 5k] k=0 ( ) k (δ[n 5k] + δ[n 5k]) k=0 n = 0:20; yy = filter(,[,0,0,0,0,],[, zeros(,20)]); stem(n,yy); y[n] n (e) From the above formula, the period is 0 samples. 3

4 Problem 9.3 (25 points) (a) h[n] = r n cos(ˆω 0 n + θ)u[n] h[n] = 2 rn (e j ˆω 0n+jθ + e j ˆω 0n jθ )u[n] h[n] = 2 ejθ r n e j ˆω 0n u[n] + 2 e jθ r n e j ˆω 0n u[n] (b) 2 ejθ re j ˆω 0 z + 2 e jθ re j ˆω 0 z ROC: z > r 2 ejθ ( re j ˆω 0 z ) + 2 e jθ ( re j ˆω 0 z ) ( re j ˆω 0 z )( re j ˆω 0 z ) cos(θ) r cos(ˆω 0 θ)z ( re j ˆω 0 z )( re j ˆω 0 z ) ROC: z > r ROC: z > r The poles are at z = re ±j ˆω 0 and the zeroes are at z = r cos(ˆω 0 θ) cos(θ), z = 0. r = 0.8, ˆω 0 = 2π/3, θ = 0 in the figure below: Imaginary Part Real Part The white region is the ROC and the gray region is outside the ROC. 4

5 (c) The DTFT exists if the time-domain sequence is absolute summable: sufficient condition is when r <. (d) n= x[n] <. A Since the system is causal, for the system to be stable the poles must be within the unit circle, so the sufficient and necessary condition is r <. (e) cos(θ) r cos(ˆω 0 θ)z 2r cos(ˆω 0 )z + r 2 ROC: z > r z 2 y[n] = 2r cos(ˆω 0 )y[n ] r 2 y[n 2] + cos(θ)x[n] r cos(ˆω 0 θ)x[n ] y[n] = 0.8y[n ] 0.64y[n 2] 0.5x[n] + 0.4x[n ] 5

6 Problem 9.4 (25 points) PZ- looks approximately like: h[n] = z ( re j ˆω 0 z )( re j ˆω 0 z ) r(e j ˆω 0 e j ˆω 0) (r n e jnˆω 0 u[n] r n e jnˆω 0 u[n]) h[n] = rn sin(nˆω 0 ) u[n] sin(ˆω 0 ) h[n] will oscillate and die out slowly. The oscillation frequency is approximately ˆω 0 = π/3, therefore the period of the oscillation should be around 6. Thus PZ- matches IR-N PZ-2 looks approximately like: + z rz h[n] = r δ[n] + ( + r )rn u[n] Since there is only one pole, h[n] decays but does not oscillate. The zero at z = means h[n] is not just an exponential decay, so PZ-3 does not match IR-L. Thus PZ-2 matches IR-M PZ-3 looks approximately like: z ( r)z h[n] = r δ[n] + ( + r )( r)n u[n] Since there is only one pole, h[n] decays but does not oscillate. However, it does alternate in sign because the pole is at negative z. Thus PZ-3 matches IR-J PZ-4 looks approximately like: rz h[n] = r n u[n] This decays but does not oscillate, so PZ-4 matches IR-L 6

7 Problem 9.5 (25 points) PZ- is a highpass filter, so PZ- matches FR-D PZ-2 has a pole at around z = 0.5, which gives a slight peak at ˆω = 0 and then falls off slowly. Thus PZ-2 matches FR-B PZ-3 has a pole at approximately z = 0.9 and a zero at z =. This should give a lowpass frequency response with H(e jπ ) = 0, so PZ-3 matches FR-A PZ-4 is a bandpass filter. Since the poles have angles less than π/4, PZ-4 matches FR-E (and not FR-C) 7

8 Problem 9.6 (0 points) (a) We use approach #2: Step #: Find X(z) Step #2: Find H(z) Step #3: Find Y (z) = X(z)H(z) Step #4: Find y[n] Y (z) = Y (z) = X(z) = ROC: z > z + z 2 ROC: z > z + z 2 ( 0.8z )( z ) + z 2 (.8z + 0.8z 2 ) ROC: z > ROC: z > To have a lower order of numerator than denominator in z, we do polynomial long division: 0.8z 2.8z z 2 +0z + z z z 0.25 Thus: Y (z) = z ( 0.8z )( z ) ROC: z > Y (z) = ROC: z > 0.8z z y[n] =.25δ[n] n u[n] + 0u[n] 8

9 (b) We use a special approach using an identity that we have: Step #: FindH(e j ˆω ) Step #2: Use identity y[n] = H(e j ˆω 0 ) cos(ˆω 0 n + θ + H(e j ˆω 0 )) when x[n] = cos(ˆω 0 n + θ),0 ˆω 0 < π + e 2j0.25π 0.8e j0.25π H(e j ˆω ) = + e 2j ˆω 0.8e j ˆω + e 2j0.5π 0.8e j0.5π = 0 + e 2j0.5π =.98, = 0.542π 0.8e j0.5π y[n] = cos(0.25πn π 0.542π) (c) Since x[n] is so short and y[n] only depends on one previous y[n], we use approach #: Step #: Compute successive y[n] by hand until we see a pattern. Step #2: Write a formula for the remaining y[n]. y[n] = 0.8y[n ] + x[n] + x[n 2] By inspection: y[n] = 0 n < 5 y[n] = n = 5 y[n] = n = 6 y[n] = n = 7 y[n] = 0.8 (n 7) 6.4 n > 7 (d) We use approach #, which has O(N) complexity (where N is the length of the input signal). Step #: Compute y[n] from x[n] using MATLAB s filter() function. yy=filter([,0,],[,-0.8],xx); 9