讯号与系统/第三章 连续时间系统之时域分析/3.4 零状态响应－LTI系统之重叠积分

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LTI系统之重叠积分(Superposition Integral)

数学推导 :

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

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

    (2)經由LTI系統作用後的輸出$\mathbf {y} (t)$ 為 

$\mathbf {T} [x(t)]=T[\int _{-\infty }^{\infty }x(\tau )\delta (t-\tau )\,d\tau ]$

    (3)假設〝積分〞和LTI系統〝T〞的作用順序可對調，則

$\mathbf {y} (t)=\int _{-\infty }^{\infty }x(\tau )T[\delta (t-\tau )]\,d\tau$

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

$\mathbf {y} (t)=\int _{-\infty }^{\infty }x(\tau )h(t-\tau )\,d\tau$

以系统物理特性推导 :

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

$\mathbf {\hat {x}} (t)=\sum _{n=-\infty }^{\infty }x(n\triangle \lambda )rect({t-n\triangle \lambda \over \triangle \lambda })$

$=\mathbf {\sum } _{n=-\infty }^{\infty }x(n\triangle \lambda ){1 \over \triangle \lambda }rect({t-n\triangle \lambda \over \triangle \lambda })\triangle \lambda$

其中 $\mathbf {r} ect({t-a \over \tau })=u[t-(a-{\tau \over 2})]-u[t-(a+{\tau \over 2})]$

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

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

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

注意 : 当$\mathbf {\Delta } \lambda \to 0$时，$\mathbf {n} \Delta \lambda$用$\mathbf {\lambda }$替代

(4) 假設$\mathbf {\hat {h}} (t)=T[{rect(t/\Delta \lambda ) \over \Delta \lambda }]$當$\mathbf {\Delta } \lambda \to 0$

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

$\mathbf {=} T[\lim _{\Delta \lambda \to 0}{rect(t/\Delta \lambda ) \over \Delta \lambda }]$

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

$\mathbf {=} h(t)$

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

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

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

(线性系统满足重叠定理)

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

(非时变特性)

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

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

(6)取$\mathbf {\Delta } \lambda \to 0$

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

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

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

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

$\mathbf {=} y(t)$

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范例3.9

系统输入为$\mathbf {x} (t)=(t+10)e^{-0.4(t+10)}u(t+10)$，系统之单位脉冲响应为$\mathbf {h} (t)=e^{-t}u(t)$，求系统输出$\mathbf {y} (t)$。

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

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范例3.10

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

【解】

$y(t)\mathbf {=} \int _{-\infty }^{\infty }u(t-\tau )h(\tau )\,d\tau$

$\mathbf {=} \int _{-\infty }^{t}h(\tau )\,d\tau$

$\mathbf {=} \int _{-\infty }^{t}{1 \over RC}e^{-\tau /RC}u(\tau )\,d\tau$

$\mathbf {\int } _{0}^{t}{1 \over RC}e^{-\tau /RC}\,d\tau$

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

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

且 $\mathbf {y} (t)=\int _{-\infty }^{t}h(\tau )\,d\tau$

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线性时变系统之重叠积分

●非時變系統之單位脈衝響應：

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

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

●時變系統之單位脈衝響應：

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

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

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

(假设在$\mathbf {\delta } (\lambda )$输入前，系统是rest)

●任意輸入訊號$\mathbf {x} (t)$，則系統的響應(輸出)為：

$\mathbf {y} (t)=\int _{-\infty }^{\infty }x(\lambda )h(t,\lambda )\,d\lambda$

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