Unformatted text preview: EECS 216 SOLUTIONS TO PROBLEM SET #3  Winter 2008 h(t)dt is finite. 1. Causal iff h(t)=0 for t<0. (BIBO) stable iff For each question, Y=yes and N=no. Worth 2 points each. PROBLEM CAUSAL? STABLE? (a) (b) Y N Y Y (d) (f) (g) N N Y Y Y Y 2. The overall impulse response is [h1 (t) + h4 (t) + h2 (t) h3 (t)] h5 (t). Each individual system hi (t) is stable. The overall system is stable. 3. Before you pass out, recall x(t) (t  a) = x(t  a) and x(t) u(t  a) = 3a. y(t) =
t+3 3 e d 0 ta  x( )d . 1 = 1 [1  e9 e3t ]u(t + 3) = 3 [1  e3t ]u(t) advanced by 3. 3 Huh? e3t u(t) u(t) integrates e3t u(t); advance this in time by 3. 4 for t < 1; u(  t + 4)d = 5  t for 1 < t < 5; 3b. y(t) = 0 for t > 5 Huh? u(t+3)u(t1)=0 unless 3 < t < 1. u(4t) is timereversed step with jump at t=4. Flip it for convolution (jump at 4), and area is 0 or 4 or linearly varying between.
1 3 3c. y(t) = 2u(t) + u(t  6) = 3d. y(t)=2 p 1st 2nd = = = = 1 p p=0 ( 2 ) (tp)+ 3 for t > 6; 2 for 0 < t < 6; . 0 for t < 0 1 p p=0 ( 2 ) (t2p) =2(t)+(t1)+6 1 p p=0 ( 2 ) (t2p). 0 2 0 2 1 1 0 1 2 1
3 2 1 2 3
1 4 1 2 3 4 4
1 8 1 4 3 8 ... ... ... ... 4. Use ej0 t H(s) H(jo )ejo t = M ej(o t+) where H(jo ) = M ej and A cos(o t + ) H(s) M A cos(o t + + ) (take the real part) Complex exponential functions are eigenfunctions of LTI systems: Outputinput. 4a. y(t) = H(j4)3ej4t =
30 ej4t j4+4 = 30 j(4t/4) e 4 2 = 5.303ej(4t.785) 4b. y(t) = 5.303 cos(4t  .785) since this is just the real part of (a). 4c. y(t) = 5.303 sin(4t  4 + ) = 5.303 sin(4t + 3 12 ). 10 4d. x(t) = cos2 (2t) = 1 + 1 cos(4t). Now need H(j0) = j0+4 = 2.5. 2 2 1 y(t) = 2.5 2 + 4102 1 cos(4t  ) = 1.25 + 0.884 cos(4t  .7854). 2 4 ...
View
Full Document
 Fall '08
 Yagle
 Digital Signal Processing, Exponential Function, LTI system theory, ... ..., Impulse response

Click to edit the document details