# Relationship between $y(t)$ and $Y(f)$ in LTI system

If I'm given an LTI system. And I'm supposed to find the input $x(t)$, given $y(t)$ and $h(t)$. Am I able to use this relationship:

$$X(F) = \frac{Y(f)}{H(f)}$$

Another question I have is if I'm given $$y(t) = x_1(t) + x_2(t)$$ Does the following hold in the frequency domain:

$$Y(f) = H(f) + X(f)$$

NOTE: $y(t) = x_1(t) + x_2(t)$ Could just be a mistake in the assignment. I think a convolved sign should be there and not addition. I'm not too sure.

• Just a quick comment: The first relationship holds in general, but there are some cases where it doesn't, just picture a case where H(f) has elements with the value of 0. Keep also in mind that the time domain signals have to be zero padded properly for aliasing not to occur when multiplying the transforms. The second one holds by the additivity of fourier transform, which only applies to rectangular coordinates. – Dole Sep 29 '16 at 1:12

## 1 Answer

The Fourier transform is linear, so a linear combination of multiple signals in the time domain also results into a linear combination in the frequency domain. So the Fourier transform of,

$$y(t) = a_1\, x_1(t) + a_2\, x_2(t),$$

yields the following relation in the frequency domain,

$$Y(f) = a_1\,X_1(f) + a_2\,X_2(f),$$

with $G(f)=\mathcal{F}\left\{g(t)\right\}(f)$.

However if instead you mean convolution, instead of addition, then you can use the fact that convolution in the time domain translates to multiplication in the frequency domain (the opposite is also true). So your example then becomes,

$$y(t) = x_1(t)\, \large{*} \, \normalsize{x_2(t)} \equiv \int_{-\infty}^{\infty} x_1(\tau)\, x_2(t-\tau)\,d\tau,$$

$$Y(f) = X_1(f)\,X_2(f).$$