# 訊號與系統/由系統微分方程式求系統脈衝響應

## 線性非時變系統

●連續時間線性非時變系統一般可以線性常微分方程式表示 :

${\displaystyle \sum _{n=0}^{N}a_{n}{d^{n}y(t) \over dt^{n}}=\sum _{m=0}^{M}bm{d^{m}x(t) \over dt^{m}}}$

●此一系統對於任意輸入訊號${\displaystyle \mathbf {x} (t)}$ 的零狀態響應可用旋積運算求得 :

${\displaystyle \mathbf {y} (t)=x(t)*h(t)}$

●本附錄將介紹如何由描述系統的線性常微分方程式找出系統的單位脈衝響應。

## 尋找系統單位脈衝響應的第一種方法

${\displaystyle \mathbf {h} (t)=\sum _{m=0}^{M}bm{d^{m}x(t) \over dt^{m}}}$

## 範例3.23

【解】

${\displaystyle \Rightarrow \tau _{0}[h_{a}(0^{+})-h_{a}(0^{-})]+0=1}$

${\displaystyle \Rightarrow h_{a}(0^{+})={1 \over \tau _{0}}}$

${\displaystyle \tau _{0}{dh_{a}(t) \over dt}+h_{a}(t)=0}$  ${\displaystyle t>0}$

${\displaystyle \Rightarrow h_{a}(t)=Ae^{-t/\delta _{0}}}$  ${\displaystyle t>0}$

${\displaystyle h_{a}(0^{+})={1 \over \tau _{0}}}$

${\displaystyle \Rightarrow h_{a}(t)={1 \over \tau _{0}}e^{-t/\tau _{0}}}$

${\displaystyle h(t)={d \over dt}h_{a}(t)}$

${\displaystyle ={d \over dt}[{1 \over \tau _{0}}e^{-t/\tau _{0}}u(t)]+{1 \over \tau _{0}}e^{-t/\tau _{0}}u(t)}$

${\displaystyle =-{1 \over \tau _{0}^{2}}e^{-t/\tau _{0}}u(t)+{1 \over \tau _{0}}e^{-t/\tau _{0}}\delta (t)+{1 \over \tau _{0}}e^{-t/\tau _{0}}u(t)}$

${\displaystyle =({1 \over \tau _{0}}-{1 \over \tau _{0}^{2}})e^{(}-t/\tau _{0})u(t)+{1 \over \tau _{0}}\delta (t)}$

## 尋找系統單位脈衝響應的第二種方法

●已知系統的單位脈衝響應等於系統的單位步階響應的微分。
●當系統的微分方程式等號右邊沒有輸入訊號${\displaystyle x(t)}$ 的微分時，可先求出系統的單位步階響應再將其微分而得系統的單位脈衝響應${\displaystyle h(t)}$

## 範例3.24

【解】先求單位步階響應${\displaystyle s(t)}$

${\displaystyle {d^{2}s(t) \over dt^{2}}+3{ds(t) \over dt}+2s(t)=u(t)}$

(1)對於${\displaystyle \mathbf {t} >0}$

${\displaystyle \mathbf {d^{2}s(t) \over dt^{2}} +3{ds(t) \over dt^{2}}+2s(t)=1}$

${\displaystyle \mathbf {(} D^{2}+3D+2)s(t)=1}$

(2)令${\displaystyle \mathbf {\lambda } ^{2}+3\lambda +2=0}$   ${\displaystyle \mathbf {\Rightarrow } \lambda =-2,-1}$

(3)利用未定係數法求特解${\displaystyle \mathbf {s} _{p}(t)=k}$ 代入方程式

${\displaystyle {d^{2}s_{p}(t) \over dt^{2}}+3{ds_{p}(t) \over dt}+2s_{p}(t)=1}$

${\displaystyle \mathbf {0} +0+2k=1}$

${\displaystyle \Rightarrow k={1 \over 2}}$

${\displaystyle s_{p}(t)={1 \over 2}}$

(4)由(2)(3)知

${\displaystyle \mathbf {s} (t)=s_{h}(t)=s_{h}(t)+s_{p}(t)}$

${\displaystyle \mathbf {=} Ae^{-2t}+Be^{-t}+{1 \over 2}}$

(5)因為${\displaystyle \mathbf {d^{2}s(t) \over dt^{2}} +3{ds(t) \over dt}+2s(t)=u(t)}$ ，故${\displaystyle \mathbf {s} (t)}$ ${\displaystyle \mathbf {ds(t) \over dt} }$ ${\displaystyle \mathbf {t} =0}$ 處均 必須為連續。

${\displaystyle \mathbf {s} (0^{+})=s(0^{-})=0}$

${\displaystyle s\prime (0^{+})=s\prime (0^{-})=0}$

${\displaystyle \Rightarrow {\begin{cases}A+B+{1 \over 2}=0\\-2A-B=0&\end{cases}}}$

${\displaystyle \Rightarrow {\begin{cases}A={1 \over 2}\\B=-1\end{cases}}}$

(6)系統的單位脈衝響應 :

${\displaystyle h(t)={ds(t) \over dt}=[-e^{-2t}+e^{-t}u(t)]+[{1 \over 2}e^{-2t}+{1 \over 2}]\delta (t)}$

${\displaystyle \mathbf {=} [-e^{-2t}+e{-t}]u(t)}$

## 參考資料

●B. P. Lathi, Signal Processing and Linear Systems, Berkeley-Cambridge Press, 1998.

●G. E. Carlson, Signal and Linear System Analysis, 2nd ed., John Wiley & Sons, 1998.

●余兆棠、李志鵬，信號與系統， 滄海書局，2007。

●Edward W. Kamen and Bonnie S. Heck, Fundamentals of Signals and System Using the Web and Matlab, 2nd ed.,Prentice Hall International, 2000.

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

●Charls L. Phillips, John M. Parr, Eve A. Riskin, Signals, Systems, and Transforms, 3rd ed., Pearson Education, 2003.

●Rodger E. Ziemer and William H. Tranter, Principles of Communications, John Wiley & Sons, 2002.

●Simon Haykin, Communication Systems, 4th ed., John Wiley & Sons, 2001.

●John G. Proakis and M. Salehi, Communication Systems Engineering, 2nd ed., Prentice Hall International, 2002.

●Benoit Boulet, Fundamentals of Signals and Systems, Da Vinci Engineering Press, 2005.