訊號與系統/第三章 連續時間系統之時域分析/3.4 零狀態響應－LTI系統之重疊積分

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LTI系統之重疊積分(Superposition Integral)

數學推導 :

    (1)利用單位脈衝函數的篩選特性(sifting property) ，任意輸入

訊號$𝐱(t)$可表示成 :${\mathbf{\int }}_{-\mathrm{\infty }}^{\mathrm{\infty }}x(\tau )\delta (t-\tau )d\tau$

    (2)經由LTI系統作用後的輸出$𝐲(t)$ 為 

$𝐓[x(t)]=T[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\tau )\delta (t-\tau )d\tau ]$

    (3)假設〝積分〞和LTI系統〝T〞的作用順序可對調，則

$𝐲(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\tau )T[\delta (t-\tau )]d\tau$

    (4)由於非時變的特性可知，$𝐓[\delta (t-\tau )]=h(t-\tau )$

$𝐲(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\tau )h(t-\tau )d\tau$

以系統物理特性推導 :

(1)由下圖可知，LTI系統的任意輸入訊號$𝐱(t)$可用單位脈波函數$𝐫ect(t)$所形成之階梯函數(stairstep function)來近似

$\widehat{𝐱}(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n\mathrm{△}\lambda )rect(\frac{t-n\mathrm{△}\lambda }{\mathrm{△}\lambda })$

$={\mathbf{\sum }}_{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n\mathrm{△}\lambda )\frac{1}{\mathrm{△}\lambda }rect(\frac{t-n\mathrm{△}\lambda }{\mathrm{△}\lambda })\mathrm{△}\lambda$

其中 $𝐫ect(\frac{t-a}{\tau })=u[t-(a-\frac{\tau }{2})]-u[t-(a+\frac{\tau }{2})]$

© Rodger E. Ziemer, William H. Tranter, D. Ronald Fannin, Signals & Systems: Continuous and Discrete, 4th ed., Prentice Hall International, 1998.

(2)明顯地，$\widehat{𝐱}(t)$會趨近$𝐱(t)$當$\mathrm{\Delta }\lambda$趨近為 0

(3)$\underset{\mathrm{\Delta }\lambda \to 0}{\mathbf{lim}}\frac{rect[(t-n\mathrm{\Delta }\lambda )/\mathrm{\Delta }\lambda ]}{\mathrm{\Delta }\lambda }=\delta (t-\lambda )$

注意 : 當$\mathrm{\Delta }\lambda \to 0$時，$𝐧\mathrm{\Delta }\lambda$用$\mathit{\lambda }$替代

(4) 假設$\widehat{𝐡}(t)=T[\frac{rect(t/\mathrm{\Delta }\lambda )}{\mathrm{\Delta }\lambda }]$當$\mathrm{\Delta }\lambda \to 0$

$\underset{\mathrm{\Delta }\lambda \to 0}{\mathbf{lim}}\hat{h}(t)=\underset{\mathrm{\Delta }\lambda \to 0}{\lim }T[\frac{rect(t/\mathrm{\Delta }\lambda )}{\mathrm{\Delta }\lambda }]$

$\mathbf{=}T[\underset{\mathrm{\Delta }\lambda \to 0}{\lim }\frac{rect(t/\mathrm{\Delta }\lambda )}{\mathrm{\Delta }\lambda }]$

$\mathbf{=}T[\delta (t)]$

$\mathbf{=}h(t)$

(5)考慮LTI系統對$\widehat{𝐱}(t)$的響應 : 假設$\widehat{𝐡}(t)=T[\frac{rect(t/\mathrm{\Delta }\lambda )}{\mathrm{\Delta }\lambda }]$

$\widehat{𝐲}(t)=T[\hat{x}(t)]$

$\mathbf{=}T[\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n\mathrm{\Delta }\lambda )\frac{1}{\mathrm{\Delta }\lambda }rect(\frac{t-n\mathrm{\Delta }\lambda }{\mathrm{\Delta }\lambda })\mathrm{\Delta }\lambda ]$

(線性系統滿足重疊定理)

$\mathbf{=}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n\mathrm{\Delta }\lambda )T[\frac{1}{\mathrm{\Delta }\lambda }rect(\frac{t-n\mathrm{\Delta }\lambda }{\mathrm{\Delta }\lambda })]\mathrm{\Delta }\lambda$

(非時變特性)

${\mathbf{\sum }}_{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(n\mathrm{\Delta }\lambda )\hat{h}(t-n\mathrm{\Delta }\lambda )\mathrm{\Delta }\lambda$

© Rodger E. Ziemer, William H. Tranter, D. Ronald Fannin, Signals & Systems: Continuous and Discrete, 4th ed., Prentice Hall International, 1998.

(6)取$\mathrm{\Delta }\lambda \to 0$

$\mathbf{\Rightarrow }n\mathrm{\Delta }\lambda \to \lambda ,\hat{x}(t)\to x(t),\hat{h}\to h(t)$

$\underset{\mathrm{\Delta }\lambda \to 0}{\mathbf{lim}}\hat{y}(t)=\underset{\mathrm{\Delta }\lambda \to 0}{\lim }T[\hat{x}(t)]$

$\mathbf{=}T[\underset{\mathrm{\Delta }\lambda \to 0}{\lim }\hat{x}(t)]$

$\mathbf{=}T[x(t)]$

$\mathbf{=}y(t)$

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範例3.9

系統輸入為$𝐱(t)=(t+10)e^{-0.4(t+10)}u(t+10)$，系統之單位脈衝響應為$𝐡(t)=e^{-t}u(t)$，求系統輸出$𝐲(t)$。

© Rodger E. Ziemer, William H. Tranter, D. Ronald Fannin, Signals & Systems: Continuous and Discrete, 4th ed., Prentice Hall International, 1998.

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範例3.10

考慮一系統的單位脈衝響應為$𝐡(t)=\frac{1}{RC}{e}^{-t/RC}$，令輸入訊號為單位步階函數$𝐱(t)=u(t)$，試求系統的輸出$𝐲(t)$。

【解】

$y(t)\mathbf{=}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}u(t-\tau )h(\tau )d\tau$

$\mathbf{=}\underset{-\mathrm{\infty }}{\overset{t}{\int }}h(\tau )d\tau$

$\mathbf{=}\underset{-\mathrm{\infty }}{\overset{t}{\int }}\frac{1}{RC}{e}^{-\tau /RC}u(\tau )d\tau$

${\mathbf{\int }}_0^t\frac{1}{RC}{e}^{-\tau /RC}d\tau$

$\mathbf{=}1-e^{-t/RC}$

注意：此一輸出也稱為系統的單位步階響應(unit step response)

且 $𝐲(t)=\underset{-\mathrm{\infty }}{\overset{t}{\int }}h(\tau )d\tau$

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線性時變系統之重疊積分

●非時變系統之單位脈衝響應：

$\mathit{\delta }(t)\to T\to h(t)$

$\mathit{\delta }(t-\lambda )\to T\to h(t-\lambda )$

●時變系統之單位脈衝響應：

$\mathit{\delta }(t)\to T\to h(t,0)$

$\mathit{\delta }(t-\lambda )\to T\to h(t,\lambda )$

可知： (1)$𝐡(t,0)=h(t)$(2)$𝐡(t,\lambda )=0$ $𝐭<\lambda$

(假設在$\mathit{\delta }(\lambda )$輸入前，系統是rest)

●任意輸入訊號$𝐱(t)$，則系統的響應(輸出)為：

$𝐲(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(\lambda )h(t,\lambda )d\lambda$

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