{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EE301

# EE301 - 00(g Causal because Mn = 0 for n< 0 Stable...

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

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

View Full Document
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

{[ snackBarMessage ]}

### Page1 / 2

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

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

View Full Document
Ask a homework question - tutors are online