EE301

EE301 - 00 (g) Causal because Mn] = 0 for n < 0....

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 00 (g) Causal because Mn] = 0 for n < 0. Stable because 2 = 1 < oo. TL=-00 ) 00 2.29. (a) Causal because h(t) = O for t < 0. Stable because / Ih(t)|dt = 9—8/4 < oo. —00 00 (b) Not causal because h(t) aé 0 for t < 0. Unstable because / |h(t)| z oo. —00 00 (c) Not causal because h(t) 75 O for t < 0. a Stable because / |h(t)|dt = 6100/2 < oo. —oo 00 (d) Not causal because h(t) 54$ 0 for t < 0. Stable because / |h(t)|dt = 62/2 < oo. —oo 00 (e) Not causal because h(t) 79 D for t < 0. Stable because / |h(t)|dt = 1/3 < oo. —oo 00 (f) Causal because h(t) = U for t < 0. Stable because / lh(t)|dt = 1 < 00. “W 00 (g) Causal because h(t) = 0 for t < 0. Unstable because / |h(t)|dt = 00. —00 We need to find the output of the system when the input is :c[n] = 6 Since we are asked to assume initial rest, we may conclude that y[n] = 0 for n < 0. Now, 2.30. yl'n] = wlnl - Zyln - 1]- .Therefore, 3/[01 = 340] - 2yl~1l = 1, ylll = wlll - 22/[0] = -2, W] = Il2l + 2yl2] = —4 and so on. In closed form, ylnl : (—2)"u[n1. This is the impulse response of the system. 2.31. Initial rest implies that y[n] = 0 for n < —2. Now y[n] = $[n] + 2$[n —- 2] —- 2y[n — 1]. Therefore, yl—Zl : 1) y[4] = 56,y[5] = —110, y[n] = —110(—2)“‘5 forn 2 5. 2.32. (a) If yh[n] = A(1/2)", then we need to verify 1 n. 1 1 n—l AG) 74(5) =°' Clearly this is true. 48 ylll Figure 82.58 (c) The figures corresponding to the remaining parts of this problem are shown in Figure 82.59. 2.60. (a) Integrating the given differential equation once and simplifying, we get _al 1: an t 1' ya) = ——— y(r)dT-—/ f mowed? ‘12 —oo 0’2 -oo —00 t 1’ t + b—O/ / :1:(a)dad7' + :z:(T)dT + 22-376). “2 -oo —oo “2 —oo “1 Therefore, A = ~a1/a2, B = —ao/a2, C = (lg/(11, D = b1/a2,E = bo/ag. (b) Realizing that x2(t) = y1(t), we may eliminate these from the two given integral equa- tions. (c) The figures corresponding to the remaining parts of this problem are shown in Figure S2.60. 2.61. (a) (i) From Kirchofl’s voltage law, we know that the input voltage must equal the sum of the voltages across the inductor and capacitor. Therefore, fMfl fiw=LC +ma 72 ...
View Full Document

This note was uploaded on 11/26/2008 for the course EE 301 taught by Professor Enright during the Spring '08 term at USC.

Page1 / 2

EE301 - 00 (g) Causal because Mn] = 0 for n &amp;lt; 0....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online