University of Illinois at Urbana-Champaign ECE 310: Digital Signal Processing

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1 University of Illinois at Urbana-Champaign ECE : Digital Signal Processing Chandra Radhakrishnan PROBLEM SET : SOLUTIONS Peter Kairouz Problem Solution:. ( 5 ) + (5 6 ) + ( ) cos(5 ) + 5cos( 6 ) + cos( ) + j(sin(5 ) + 5sin( 6 ) + sin( )) ( ) ( ) ( ) ( ( ) ( ) ( )) j ( + ) + j.98e j.88. ( +j) 5 +j ( ) 5 e ( jπ )5 e jπ e j5π e jπ e j6π e jπ j. 5 6 j + e jπ j 5/ + j5 / j ( j j + j ) ( ) + j ) ( 5 5 ( ) + j 5j j + j 5.5e j.76. ( ) n +j j + +j j ( ) n ( + j)( + j) + ( + j)( j) ( ) n e jπn ( j)( + j) Problem Sketch the following functions (u[n] is the step function in the discrete time domain):. The function n(u[n] u[n 5]) is shown in Fig.. The function u[n + ]u[n ] is sketched in Fig.. the function u[n ] + nu[n ] is shown in Fig. Problem Assume z re jθ

2 - Figure : Problem. Figure for Problem (). Figure : Problem. Figure for Problem (). 5 Figure : Problem. Figure for Problem (c).. Consider z +. This can be written as, r e jθ + r e jθ r e jθ e π+kπ θ π + kπ

3 The values of θ can be found as follows, After this the values of θ repeat. Hence we have,. The polynomial G(z) can be written as, G(z) k, θ π k, θ π k, θ 5π π z ej π, e jπ, e j 5π ( z + ) ( z ) ( ej π z ) ej 5π. G(z) can be written as, ( z + ) ( z z + ) Problem (t + 5t + 6)δ(t)dt 6 (t + 5t + 6)δ(t)dt (t + 5t + 6)δ(t)dt 6 (t + 5t + 6)δ(t )dt (t + 5t + 6)δ(t )dt.59 (t + 5t + 6)δ(t /)dt ( ) [e t u(t)] δ(t ), where is convolution e t u(t) δ(t ) e(t /) u(t /) Problem 5 The facts used here are: δ(t)e jωt

4 and, e jωt πδ(ω). δ(t ) Using Shifting property of Fourier Transform, (t t ) F (Ω)e jωt δ(t ) e jω/. e αt u(t). u(t) u(t T ) F ( e αu(t)) F (u(t) u(t T )) e αt u(t)e jωt dt e αt e jωt dt e t( α+jω) dt α + jω α + jω T [e t(α+jω)] (u(t) u(t T ))e jωt dt e jωt dt jω ( e jωt ) T e jωt sinc ( ΩT ). sin(ω t + φ) [ e j(ω t+φ) e j(ωt+φ) ] F (sin(ω t + φ)) F j [ e jφ F [e jωt ] e jφ F [e jωt ] ] j jπ [ e jφ δ(ω Ω ) e jφ δ(ω + Ω ) ] jπ [ e jφ δ(ω + Ω ) e jφ δ(ω Ω ) ] Problem 6 Determine whether or not each of the following signals is periodic. If a signal is periodic, determine its fundamental (smallest) period.

5 . x[n] cos ( ) πn Hence periodic with period N 6 π (n + N) π n + πk π N πk N k 6. x[n] cos ( πn + ) π(n + N) Hence periodic with period N. x[n] sin ( ) πn Hence periodic with period N. x[n] e ( jπn ) + πn + + πk πn N π(n + N) πn N πk k πn + πk πk k Hence periodic with period N 8 5. x[n] sin ( ) n Non-periodic since it has no integer solution 6. x[n] sin (πn) π(n + N) πn N 8 (n + N) πn + πk πk k n + πk π(n + N) πn + πk πn πk Hence periodic with period N 5

6 7. x[n] e jπn π(n + N) πn + πk πn πk Hence periodic with period N Problem 7 We know that, Consider now, F [ x(t) e t (u(t) u(t )) ], Now consider the given signals,. x(t)e j5t F [ e at] a + jω F [ e t (u(t) u(t )) ] F [ e t u(t) e t u(t ) ] F [ e t u(t) ] F [e t u(t) e (t ) u(t )e 8] + jω e jω + jω e 8, using time shift X(Ω) e 8 jω + jω F [ x(t)e j5t] X(Ω 5) (Frequency Shift) e 8 j(ω 5) + j(ω 5). x(t) + x(t + ) F [x(t) + x(t + )] X(Ω) + X(Ω)e jω ( + e jω )X(Ω) e jω cosωx(ω). x(t)cos(t) [ ( e jt + e jt) ] F [x(t)cos(t)] F x(t). (X(ω ) + X(Ω + )) [ e 8 j(ω ) + ] e 8 j(ω+) + j(ω ) + j(ω + ). tx(t) F [tx(t)] j dx(ω) dω Ωje 8 jω ( + jω) Problem 8 6

7 G( ) G( ) - Figure : Problem 8. Magnitude and Phase of G(ω). The magnitude and phase is shown in Fig.. Assuming that X(ω) is only defined between ω X(ω) G(ω) ) x(t) sinc ( t However, since we have a rectangular function multiplied with the triangle function, the function value outside of the region [-, ] can be arbitrary. But they will still be suppressed to zero. If this argument is shown, credit will still be given.. Determine g(t) dt. Problem 9 The difference equation for the system is g(t) dt π π π ( G(ω) dω ( + ω) dω + y[n] y[n ] y[n ] ) ( ω) dω 7

8 Problem. x[n] u[n + ] u[n 5]. X d (ω) 5 n e jωn ejω ( e j8ω) e jω ejω e jω5 e jω ( e jω e jω e jω) e jω/ (e jω/ e jω/ ) jω/ sin (ω) e sin (ω/). x[n] δ[n + ] + δ[n ] X d (ω) e jω + e jω cos(ω). x[n] ( ) n u[n ] X d (ω).e jω e jω. x[n] ( n) n u[n] This does not converge. Hence Fourier Transform is undefined. 5. The magnitude and phase for Part () are shown in Fig 5 and 6 respectively. The magnitude and phase for Part () are shown in Fig 7 and 8 respectively Figure 5: Problem. Magnitude for Part (). Note: the zero crossing points are ( π, π/, π/, π/, π/, π/, π/, π) Problem. X d (ω) + e jω + e jω e 5jω Using linearity and time-shift properties of the DTFT: X d (ω) + e jω + e jω e jω δ[n] + δ[n ] + δ[n ] δ[n 5] 8

9 Figure 6: Problem. Phase for Part (). Note: the phase changes at ( π, π/, π/, π/, π/, π/, π/, π) Figure 7: Problem. Magnitude for Part (). Note: the zero crossing points are ( 5π/6, π/, π/6, π/6, π/, 5π/6) Figure 8: Problem. Phase for Part (). Note:The phase is shown in the interval (, π). The zero crossing points are (π/6, π/, 5π/6). The phase is a odd function.. X d (ω) {, ω ω, ω ω π x[n] π pi π X d ωdω π 9 ω π e jωn dω + π e jωn dω π ω

10 For n, For n x[] π (π ω ) + π (π ω ) π ω π x[n] jnπ (e jωn e jπn + jnπ (ejπn e jωn ) sin(nω ) nπ Thus, where, x[n] δ[n] ω π sinc(nω ) sinc(n) { sin(n) n, n, n. π x[n] X d ωe jωn dω π π 9π/ e jωn dω + π π 8π/ 9π/ ( 9π/ (e jωn + e jωn )dω + π 8π/ π πn πn 9π 8π/ cos(ωn)dω + π 9π/ e jωn dω + π [( ( ) ( )) 9π 8πn sin n sin [ ( ) ( )] 9π 8π sin n + sin n 9π/ 9π/ 8π/ e jωn dω + (e jωn + e jωn ) cos(ωn)dω, n ( sin (πn) sin π 9π/ ) e jωn dω ( ))] 9πn, n for n Hence, x[n] π X d (ω)dω π π π ( ) 6π x[n] δ[n] 9 ( ) 9π sinc n 8 ( ) 8π sinc n. One way is to follow the procedure in part of () and solve the inverse DTFT integral. But note that convolving two rectangular functions results in a triangle. Hence the given function X d (ω) can be seen as the convolution of Y (ω) with itself, where in the interval [ π, π], X (ω) is given by {, π/ ω π/ Y (ω), otherwise We need to solve for x[n] F [X d (ω)]. We know the following, If y [n] y [n] y[n], F (y [n]y [n]) π [Y (ω) Y (ω)] F (y [n]) [Y (ω) Y (ω)] π

11 Multiply by and take inverse Fourier transform, y [n] F [ π [Y (ω) Y (ω)] ] or the required time domain signal, x[n] is where, and, x[n] y [n] y[n].5sinc(.5n) x[n].5sinc (.5n) Problem. Evaluate X d (ω) ω X d ω ω n n n x[n] Re{x[n]} + j n Im{x[n]}. Evaluate X d (ω) ωπ X d ω ωπ x[n]( ) n n ( j) ( + j) + ( j) ( + j) + ( j) ( j) j π. Evaluate X d (ω)dω π π π ( π ) π X d (ω)e jω dω πx[] π jπ π π. Determine and sketch the signal whose DTFT is X d ( ω) X d ( ω) is the DTFT of x [n]. The signal x [n] is shown Fig.9, Problem. x[n] δ[n ], n X[k] δ[n ]e j π kn n e j πk X[k] {, j,, j}

12 e xn [ ] xn m [ ] Figure 9: Problem. Magnitude and Phase of X(k) in Part (). x[n] { n n 5 X[k] X[k] X[k] { { n e jπkn/, k else e jπk e jπk/,, k else e jπk/(sin(πk/)) e jπk/6 sin(πk/6),. x[n] cos ( ) πn, n 7 for k, 7 for k, 7 cos( nπ ) ejnπ/ + e jnπ/ X[k] 7 (e jnπ/ + e jnπ/ )e jπn/8 n X[k] X[k] [ ] e jπ(k ) e jπ(k ) + e jπ(k )/ e jπ(k )/

13 Hence, X[] [8 + ] X[7] [ + 8] X[k] (δ(k ) + δ(k 7)). x[n] { n even, n 6 n odd, n 6 X[k] 5. Sketch the magnitude and phase for parts () and (). X[k] N m N m { N+ e j π N km πk j (e 7 ) m, k e j π 7 k N+ e j 6πk 7, else e j π 7 k e j π 7 The magnitude and phase of () and () are shown in Figs. and respectively. X( k) k Phase Xk [ ] k Figure : Problem. Magnitude and Phase of X(k) in Part ()

14 X( k) k. Phase Xk [ ] ( rad/s).. k Figure : Problem. Magnitude and Phase of X(k) in Part ()

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