This is a question I feel too stupid asking my professor about. I'm having a mental block remembering how this works even though I think I understood it at one point:

I know the following properties:

$$x(t) {\longrightarrow}\boxed{\textrm{LTI System}}{\longrightarrow} y(t) = x(t) \star h(t) \longleftrightarrow X(j\omega)H(j\omega)$$

$$x(t) = e^{(j\omega_0 t)} \overset{\mathcal F}{\longleftrightarrow}X(j\omega) = 2\pi\delta(\omega-\omega_0)$$

So why is this true:

$$x(t) = e^{(j\omega_0t)}{\longrightarrow}\boxed{\textrm{LTI}}{\longrightarrow} y(t) = e^{(j\omega_0t)}H(j\omega)$$

instead of this:

$$e^{(j\omega_0t)}{\longrightarrow}\boxed{\textrm{LTI}}{\longrightarrow} y(t) = 2\pi\delta(\omega-\omega_0)H(j\omega)$$

I know I'm missing something here or have some fundamental misunderstanding, but I can't seem to catch what it is. I'm taking a DSP course, but it's been quite a while since I took basic signals and systems. If anyone could help me out I'd really appreciate it.


1 Answer 1


You're confusing time and frequency domain. Why should the (time domain) output signal $y(t)$ be a function of $\omega$? If $x(t)=e^{j\omega_0t}$ is the input signal to an LTI system with frequency response $H(j\omega)$, then the output signal is given by


Note that in $(1)$, $H(j\omega_0)$ is a constant, not a function of $\omega$.

Equation $(1)$ can be easily derived from the convolution integral


with $x(t)=e^{j\omega_0t}$ and the Fourier transform relation

$$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt\tag{3}$$

I'm sure you can fill in the details yourself.

Your last equation would be true for the Fourier transform of the output signal:


And with $X(j\omega)=2\pi\delta(\omega-\omega_0)$ you obtain


which is of course just the Fourier transform of Eq. $(1)$.


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