Suppose we have a LTI CT signal y(t)=2x(t)
Unit Impulse Response h(t) and System Function H(s)
i) $ y(t)=2x(t)=> h(t)=2\delta(t) $
ii) $ H(s)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau $
$ =\int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $
$ =\int_{-\infty}^\infty 2\delta(\tau)e^{-s\tau}d\tau $
$ =2e^{-s*0} $
$ =2 $
Response of the Signal and Fourier Series Coefficients
$ x(t)=3cos(3t) $
$ =\frac{3}{2}[(e^{j3t})+(e^{-j3t})] $
$ =\frac{3}{2}(e^{j3t})+\frac{3}{2}(e^{-j3t}) $
since we have $ e^{st} $ has a response of $ H(s)e^{st} $
so we can get
$ y(t)=\frac{3}{2}[(e^{j3t})+(e^{-j3t})]H(s) $
$ =\frac{3}{2}(e^{j3t})H(j3)+\frac{3}{2}(e^{-j3t})H(-3j) $
$ =3e^{j3t}+3e^{-j3t} = 6cos(3t) $
Coefficients:
we observe that the output is just double the input, so the coefficients should be doubleed,too
From Q1 we can get that when k=1, $ a_1=3 $, and when k=-1,$ a_-1=3 $
others are all zero
