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Now we need to find the LTI system above's response to my signal from section 4.1. The response can be found with the equation | Now we need to find the LTI system above's response to my signal from section 4.1. The response can be found with the equation | ||
| − | <math>y(t) = H(j\omega)x(t)</math> | + | <math>y(t) = H(j\omega)x(t) \!</math> |
Using my signal from section 4.1 | Using my signal from section 4.1 | ||
Latest revision as of 16:09, 26 September 2008
Define a CT, LTI System
One possible CT, LTI system would be
$ y(t) = 3x(t-2) \! $
Unit Impulse Response
The unit impulse response of the system is found by substituting $ \delta(t) $ for $ x(t) $. So, for the system
$ y(t) = 3x(t-2) \! $
$ h(t) = 3\delta(t-2) \! $
System Function
The system function is defined as
$ H(j\omega)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau $
Using the sifting property, we can easily find that our system function is defined as
$ H(j\omega)=e^{-2j\omega} \! $
Signal Response
Now we need to find the LTI system above's response to my signal from section 4.1. The response can be found with the equation
$ y(t) = H(j\omega)x(t) \! $
Using my signal from section 4.1
$ x(t) = 3cos(2t) = \frac{3}{2}e^{j2t}+\frac{3}{2}e^{-j2t} $
we can multiply each term by $ H(j\omega) $ to get
$ y(t) = e^{-j4}(\frac{3}{2}e^{j2t}) + e^{j4}(\frac{3}{2}e^{-j2t}) $
Simplifying this gives us
$ y(t) = \frac{3}{2}e^{j2(t-2)} + \frac{3}{2}e^{-j2(t-2)} \! $
