Signals and Systems Profs. Byron Yu and Pulkit Grover Fall Midterm 2 Solutions

Transcription

1 8-90 Signals and Systems Profs. Byron Yu and Pulkit Grover Fall 08 Midterm Solutions Name: Andrew ID: Problem Score Max Total 00

2 Midterm Solutions. (8 points) Indicate whether the following statements or equations are true or false by circling the answer. No justification is needed. (a) ( points) The following graph describes a high-pass filter: h[n] n True False (b) ( points) When the input to an LTI system is a complex exponential, the output is also a complex exponential. True False (c) ( points) The step response of an LTI system is the convolution of the impulse response with a unit step function. True False (d) ( points) Differentiation in the time domain corresponds to integration in the frequency domain. True False (a) True; it s a HPF. H(e jω ) = cos(ω) + j sin(ω) H(e jω ) = ( cos(ω)). Notice that H(e j0 ) = 0, H(e jπ ) = 4. Hence this is a high pass filter. (See HW6, Problem 7(a) for more details.) (b) True (c) True (d) False; differentiation and integration have separate duals.

3 Midterm Solutions 3. (5 points) Consider the interconnected LTI system below, where h (t), h (t) and h 3 (t) are impulse responses of three LTI systems. Let H (jω), H (jω) and H 3 (jω) be the corresponding frequency responses. x(t) h (t) h (t) + y(t) h(t) h 3 (t) Find H(jω), the frequency response of the interconnected system. h(t) = h (t) (h (t) + h 3 (t)) H(jω) = H (jω)(h (jω) + H 3 (jω))

4 4 Midterm Solutions 3. (6 points) Consider the following periodic, discrete time signals: ( π ) x [n] = cos 4 n, ( π ) ( π ) x [n] = cos n + cos 4 n Let x[n] = x [n] x [n]. Compute the DTFS, X[k], of x[n]. First, the fundamental periods of x [n] and x [n] are N = π π/4 = 8, and N = π π/4 = 8, respectively. Then, the fundamental frequency of X[k], Ω 0 = π N = π 4. Therefore, So, ( π ) cos n + cos ( π ) cos 4 n DTFS; Ω0 = π 4 δ[k ] + δ[k 7], periodic with N = 8 ( π ) 4 n DTFS; Ω0 = π 4 δ[k ] + δ[k ] + δ[k 6] + δ[k 7], x [n] x [n] DTFS; Ω 0= π 4 periodic with N = 8 DTFS; Ω 0 = π 4 8 NX [k]x [k] ( 4 δ[k ] + δ[k 7] 4 DTFS; Ω 0 = π 4 δ[k ] + δ[k 7], periodic with N = 8 Any equivalent solution due to the periodic nature of the signal is acceptable, e.g. δ[k ] + δ[k + ], periodic with N = 8. )

5 Midterm Solutions 5 4. (7 points) A system has input x[n], output y[n], and impulse response h[n] = u[n] + ( ) n u[n ] (a) (5 points) Find H(e jω ), the frequency response of the system. Simplify your answer to not include any complex exponentials. (b) ( points) Is the system stable? Justify your answer.

6 6 Midterm Solutions (a) See piazza post as to why this problem has some complicated intricacies. Everybody was given full credit for this problem. (b) The system is not stable. It can be shown that n= h[n] is not finite. n= h[n] = = n= u[n] + ( ) n u[n ] + ( ) n u[n ] n=0 = + + ( ) n n= Note that { + ( ) n } n= = {0,, 0,, 0,,...}

7 Midterm Solutions 7 5. (8 points) Consider the Fourier Transform X(jω) of a continuous-time signal x(t): X(jω) = { cos (ω), ω π 0, otherwise. (a) ( points) Plot X(jω). Label the axes appropriately. 0 (b) (6 points) Find x(t). Simplify your answer to be in terms of sinc functions, where the sinc function is defined as sinc(u) = sin(πu). πu

8 8 Midterm Solutions (a) X(jω) is one period of a cosine wave in the frequency domain, i.e. a cosine wave multiplied by a zero-centered, box pulse of length π. X(jω) 0 π π π π ω (b) We utilize knowledge that multiplication in one domain corresponds to convolution in the other domain. Note that X(jω) = cos(ω) (u(ω + π) u(ω π)) From the table we have = ejω (u(ω + π) u(ω π)) + e jω (u(ω + π) u(ω π)) x 0 (t) = sin(πt) πt = sinc(t) X 0 (jω) = u(ω + π) u(ω π) Thus X(jω) = ejω X 0 (jω)+ e jω X 0 (jω). Using the linearity and time shift properties x(t) = x 0(t + ) + x 0(t ) = sinc(t + ) + sinc(t )

9 Midterm Solutions 9 6. ( points) Determine whether an LTI system with each of the following impulse responses is i) memoryless, ii) causal, and iii) stable. Justify your answers. (a) (6 points) h(t) = 3 m=0 δ(t 3m) (b) (6 points) h[n] = sin ( π 5 n) u[n + ] (a) Not memoryless, causal, stable i h(t) = δ(t) + δ(t 3) + δ(t 6) + δ(t 9) cδ(t) = not memoryless ii h(t) = 0 for t < 0 = causal iii h(t) dt = 4 = stable (b) Not memoryless, not causal, not stable i. h[n] cδ[n] = not memoryless ii. h[ ] = sin( π ) 0 = not causal 5 iii. h[n] = sin( π n) = = not stable 5 n= n=

10 0 Midterm Solutions 7. (6 points) Consider an LTI system with impulse response ( ) n h[n] = u[n]. 5 (a) (6 points) Determine the system s frequency response H(e jω ) using the defining equation of the DTFT. (b) (4 points) Determine the system s step response.

11 Midterm Solutions (c) (6 points) Find the system s output y[n] for the input x[n] = 6 cos (πn) without 5 using convolution. (a) (b) (c) H(e jω ) = = = = n= h[n]e jωn ( ) n e jωn 5 ( ) n 5 e jω n=0 n=0 5 e jω (Valid because s[n] = n k= = u[n] = ( 5 ( ) k u[k] 5 n ( ) k 5 k=0 ) n+ 5 = 5 ( 5 4 ) n u[n] u[n] 5 e jω < ) x[n] = 3 5 ejπn e jπn y[n] = 3 5 ejπn H(e jπ ) e jπn H(e jπ ) = 3 5 ejπn 5 e jπ e jπn 5 ejπ = 3 5 ejπn + 5 = ejπn + e jπn = cos(πn) e jπn + 5

12 Midterm Solutions 8. (6 points) Consider an LTI system that yields the output y[n] when the input is x[n], where ( ) n x[n] = u[n] 3 ( ) n y[n] = u[n 3] 3 (a) (6 points) Let X(e jω ) and Y (e jω ) be the DTFT of x[n] and y[n] respectively. Express Y (e jω ) in terms of X(e jω ). (b) (6 points) Based on part (a), find the frequency response H(e jω ) of the system. Plot the magnitude and phase of the frequency response. Label the axes appropriately. 0

13 Midterm Solutions 3 0 (c) (4 points) Find and plot the impulse response, h[n], of the system. Label the axes appropriately. 0

14 4 Midterm Solutions (a) Note that y[n] = ( ) n u[n 3] = 3 7 ( ) n 3 u[n 3] 3 If X(e jω ) is the DTFT of x[n], applying the time shift and linearity properties we have Y (e jω ) = 7 e j3ω X(e jω ) (b) Since y[n] = h[n] x[n], we have h H(e jω ) = Y (ejω ) X(e jω ) = 7 e j3ω = H(e jω ) = 7 and H(ejΩ ) = 3Ω H(e jω ) H(e jω ) -3π 7 3π π π π π 3π Ω 3π π π π π 3π Ω 7-3π (c) From the table we have z[n] = δ[n] Z(e jω ) = So applying the time shift and linearity properties again we have h[n] = δ[n 3] 7

15 Midterm Solutions 5 h[n] n 7

16 6 Midterm Solutions 9. (0 points) Consider the following discrete-time signal x[n]. Find and plot the DTFT of x[n]. Label the axes appropriately. x[n] = ( ) π sin n n 0 x[n] = ( sin( n π ) ) ( sin( n = π ) n πn ) has a DTFT Y (e jω ) whose frequency domain representa- From the table, y[n] = sin( n ) πn tion is shown below

17 Midterm Solutions 7 Y (e jω ) 0 π π - π π Ω Since we are multiplying two of these y[n] s in the time domain we need to circularly convolve them in the frequency domain to get X(e jω ). Note that we have a π coefficient and a scaling factor that multiply together to yield a coefficient of. π X(e jω ) 0 π π - π π Ω

18 8 Midterm Solutions 0. ( points) Consider the following continuous-time signal x(t): x(t) = where x(t) is periodic with period 3. { e ( j 4π 3 t u ( ( t + ) u t )),.5 t <.5 0, otherwise, (a) (9 points) Find X[k], the Fourier series of x(t). (b) (3 points) Let y(t) = d x(t). Find Y [k], the Fourier series of y(t). dt x(t) has a period of T = 3. Then, ω 0 = π T = π 3. (a) First, from the table, e j 4π 3 t F S,ω 0= π 3 δ[k ] Second, from the table, ( u t + ) ( u t ) F S,ω 0 = π 3 sin(k π 3 kπ F S,ω 0 = π 3 sin(k π) 3 kπ F S,ω 0 = π 3 3 sinc ( k 3 ) )

19 Midterm Solutions 9 Therefore, since multiplication in the time domain is convolution ( ) in the frequency domain, ( e j4 π 3 (u t t + ) ( u t )) F S,ω 0 = π 3 δ[k ] ( ) k 3 sinc 3 F S,ω 0 = π 3 ( ) k 3 sinc 3 (b) We have from the table that, d dt x(t) F S,ω 0 jkω 0 X[k] Therefore, d dt x(t) F π S,ω 0= 3 jk π 3 3 sinc ( ) k 3 ) F S,ω 0 = π 3 jk π 9 sinc ( k 3