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<math>Y(jw)=\frac{\frac{1}{4}}{jw+1}+\frac{\frac{1}{2}}{(jw+1)^2}-\frac{\frac{1}{4}}{jw+4}</math> | <math>Y(jw)=\frac{\frac{1}{4}}{jw+1}+\frac{\frac{1}{2}}{(jw+1)^2}-\frac{\frac{1}{4}}{jw+4}</math> | ||
| + | <math?y(t)=[\frac{1}{4}e^{-t}+\frac{1}{2}te^{-t}-\frac{1}{4}e^{-3t}]u(t)</math> | ||
Revision as of 16:52, 24 October 2008
System Characterized By Linear Constant-Coefficient Differential Equations
$ \sum_{k=0}^{N}a_k\frac {d^ky(t)}{dt^k} = \sum_{k=0}^{M}b_k\frac {d^kx(t)}{dt^k} $
$ Y(jw)=H(jw)X(jw), H(jw)=\frac{Y(jw)}{X(jw)} $
Example
Consider a LTI system that is chracterized by the differential equation
$ \frac{d^2y(t)}{dt^2}+4\frac{dy(t)}{dt}+3y(t) = \frac{dx(t)}{dt}+2x(t) $
$ H(jw)=\frac{(jw)+2}{(jw)^2+4(jw)+3} $
$ H(jw)=\frac{jw+2}{(jw+1)(jw+3)} $
$ H(jw)=\frac{\frac{1}{2}}{jw+1} + \frac{\frac{1}{2}}{jw+3} $
$ h(t)=\frac{1}{2}e^{-t}u(t)+\frac{1}{2}e^{-3t}u(t) $
Consider the system above, and suppose that the input is
$ x(t)=e^{-t}u(t) $
$ Y(jw)=H(jw)X(jw)=[\frac{jw+2}{(jw+1)(jw+3)}][\frac{1}{jw+1}] $
$ =\frac{jw+2}{(jw+1)^2(jw+3)} $
$ Y(jw)=\frac{A_{11}}{jw+1}+\frac{A_{12}}{(jw+1)^2}+\frac{A_{21}}{jw+3} $
$ A_{11}=\frac{1}{4}, A_{12}=\frac{1}{2},A_{21}=-\frac{1}{4} $
$ Y(jw)=\frac{\frac{1}{4}}{jw+1}+\frac{\frac{1}{2}}{(jw+1)^2}-\frac{\frac{1}{4}}{jw+4} $ <math?y(t)=[\frac{1}{4}e^{-t}+\frac{1}{2}te^{-t}-\frac{1}{4}e^{-3t}]u(t)</math>
