第3次作业讲è¯

第3次作业讲è¯

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Unformatted text preview: s ª º3 ( -‚ 2.1 ( º -‚ x(t) s h(t) (‚ º -ª * x(t)*h(t) s (1) x(t ) = e − at u (t ) h(t ) = u (t ) t ∞ ∞ −∞ −∞ x(t ) * h(t ) = ∫ x(τ )h(t − τ )dτ = ∫ e − aτ u (τ )u (t − τ )dτ t = u (t ) ∫ e 0 − aτ 1 − e − at u (t ) dτ = a (2) x(t ) = δ (t ) h(t ) = cos(ω 0t ) + sin(ω 0t ) x(t ) * h(t ) = δ (t ) * h(t ) = h(t ) = cos(ω 0t ) + sin(ω 0t ) shi9876@hzcnc.com 1 s ª ¹3 Ø ØÅ (4) x(t ) = sin( 2t )u (t ) h(t ) = u (t ) t ∞ ∞ −∞ −∞ x(t ) * h(t ) = ∫ x(τ )h(t − τ )dτ = ∫ sin( 2τ )u (τ )u (t − τ )dτ t = u (t ) ∫ sin( 2τ )dτ = 0 2.2 Ø ¹ ØÅ 1 − cos(2t ) u (t ) 2 x[n] s h[n] ØÅ ¹ª * x[n]*h[n] s (1) x[n] = nu[n] h[n] = δ [n − 2] x[n] * h[n] = x[n] * δ [n − 2] = x[n − 2] = (n − 2)u[n − 2] shi9876@hzcnc.com 2 s ª º3 ˆ -• (2) x[n] = 2 n u[n] h[n] = u[n] s x[n] * h[n] = ∞ ∑ x[k ]h[n − k ] = k =− ∞ ∞ 2 k u[k ]u[n − k ] ∑ k =− ∞ n = u[n]∑ 2 k = (2 n +1 − 1)u[n] k =0 (4) x[n] = α n u[n] h[n] = β n u[n] α ≠ β s x[n] * h[n] = ∞ ∑ x[k ]h[n − k ] = k =− ∞ ∞ α k u[k ]β n − k u[n − k ] ∑ k =− ∞ β n +1 − α n +1 = u[n]∑ α k β n − k = u[n] β −α k =0 n shi9876@hzcnc.com 3 s ª ¹3 À Ø 2.5 s x(t) s h(t) x ¨ x(t)*h(t) s (1) t ∞ x(t ) * h(t ) = ∫ x(τ )h(t − τ )dτ −∞ shi9876@hzcnc.com 4 s ª ¹3 x ØÄ (3) t ≤ 0t x(t ) * h(t ) = 1 t > 0s shi9876@hzcnc.com x(t ) * h(t ) = 2 − e − t 5 s ª º3 h -‚ 1 x(t ) * h(t ) = 2 − e −t t≤0 t >0 ( 4) shi9876@hzcnc.com 6 s ª ¹3 X ØÇ t ≤ π t t ≥ 5π t x(t ) * h(t ) = 0 π < t ≤ 3π s x(t ) * h(t ) = ∫ t −π 0 sin(τ )dτ = 1 + cos(t ) 3π < t ≤ 5π t x(t ) * h(t ) = ∫ 2π t −3π sin(τ )dτ = −1 − cos(t ) shi9876@hzcnc.com 7 s ª ¹3 h ØË 1 + cos(t ) π < t ≤ 3π y (t ) = − 1 − cos(t ) 3π < t < 5π 0 other n 1 LTI h[n] = u[n] 2 h[n] − Ah[n − 1] = δ [n] 2 .7 Ø η ˹ª (1) A (2)1 ì (3)t Ë hعª x[n]t h1[n] x[n] * h[n] = 2 n ( u[n] − u[n − 4]) shi9876@hzcnc.com 8 s ª ¹3 ˜ Ø t (1)h[n] − Ah[n − 1] = 0.5n u[n] − A ⋅ 0.5n −1 u[n − 1] = δ [ n] 0.5n − A ⋅ 0.5n −1 = 0 n ≥1 t A = 0.5 (2)h[n] − 0.5h[n − 1] = δ [n] h[n] − 0.5h[n − 1] = h[n] * ( δ [n] − 0.5δ [n − 1]) t h1[n] = δ [n] − 0.5δ [n − 1] n 1 (3) x[n] = x[n] * h[n] * h1[n] = 2 n ( u[n] − u[n − 4]) * ( δ [n] − 0.5δ [n − 1]) = 2 n ( u[n] − u[n − 4]) − 2 n − 2 ( u[n − 1] − u[n − 5]) shi9876@hzcnc.com 9 s ª ¹3 ¸ ØÄ 2.9 x[n] = δ [n] + 0.5δ [n − 1] h1[n] = α nu[n] + β nu[n] h2 [n] = ( − 0.5) u[n] n y[n] = x[n] * h1[n] * h2 [n] x[n] * h2 [n] = ( δ [n] + 0.5δ [n − 1]) * ( − 0.5) u[n] n = ( − 0.5) u[n] + 0.5 ⋅ ( − 0.5) n n −1 u[n − 1] = ( − 0.5) u[n] − ( − 0.5) u[n − 1] n n = ( − 0.5) ( u[n] − u[n − 1]) = ( − 0.5) δ [n] = δ [n] n n y[n] = x[n] * h2 [n] * h1[n] = δ [ n] * h1[n] = α n [ n] + β nu[n] shi9876@hzcnc.com 10 s ª 3½c þ 2.12 LTI (ª 1½c þ ) t y (t ) = ∫ e −(t −τ ) x(τ − 2)dτ èÈ −∞ (2) x(t ) = u (t + 1) − u (t − 2) h(t ) t (1)h(t ) = ∫ e −∞ 0 δ (τ − 2)dτ = −( t − 2 ) e − ( t −τ ) y (t ) t<2 t>2 = e − (t − 2 )u (t − 2) (2) x(t ) * h(t ) = (1 − e − t +1 )u (t − 1) − (1 − e − t + 4 )u (t − 4) x(t ) * h(t ) = y (t ) x(t − t1 ) * h(t − t 2 ) = y (t − t1 − t 2 ) shi9876@hzcnc.com 11 s ª º3 ¸ -• 2.14 LTI h(t ) (1)h(t ) = e −4t u (t − 2) ª º - ¸• (3)h(t ) = e −2t u (t + 50) (5)h(t ) = e 2 .1 5 • 1 * ª º - ¸• −6 t * ª º- ÷ • LTI h[n] • 1 n 4 (2)h[n] = u[n + 2] 5 * ª º - ¸• n 1 (3)h[n] = u[−n] 2 * ª º- ÷ • shi9876@hzcnc.com 12 s ª ¹3 ˜ ØÈ n 1 (5)h[n] = − u[n] + (1.01) n u[n − 1] 2 2.18 * LTIÈ ª ¹Ø x1ª (tØ ) È ¹ x2 (t ) = sin(π t )[ u (t ) − u (t − 1)] y1 (t ) = x1 (t ) * h(t ) * ª ¹ Ø ˜È y1 (t ) y2 (t ) dy1 (t ) dx1 (t ) = * h(t ) dt dt shi9876@hzcnc.com 13 s ª ¹3 Ø ØÀ h(t ) = u (t ) − u (t − 1) x2 (t ) = sin(π t )[ u (t ) − u (t − 1)] 1 − cos(π t ) [ u (t ) − u (t − 2)] y2 (t ) = x2 (t ) * h(t ) = π shi9876@hzcnc.com 14 ...
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This note was uploaded on 08/22/2011 for the course EE 201 taught by Professor Yhm during the Spring '05 term at Zhejiang University.

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