# Wiener Filter in Frequency Domain: What it does to a specific Frequency?

As I understand Wiener filter in time domain tries to estimate a signal as close as possible to its (original) non-degraded signal using the degraded signal by white noise.

$$H(\omega)=\frac{\Phi_{ss}(\omega)}{\Phi_{ss}(\omega)+\Phi_{nn}(\omega)}$$

How does this function work in the frequency domain? Say for an example at 30 Hz there is noise and how the transfer function looks like?

## 1 Answer

You already know the formula, so that's how the frequency response looks like. In frequency regions where the signal is stronger than the noise, the frequency response is close to $1$, so the portion of the input signal in that frequency range passes the filter with almost no attenuation, whereas in frequency regions where the noise is stronger than the signal, the filter's frequency response becomes small, and the corresponding part of the input signal is attenuated.

If you define a frequency dependent SNR

$$\textrm{SNR}(\omega)=\frac{\Phi_{ss}(\omega)}{\Phi_{nn}(\omega)}\tag{1}$$

you can rewrite the frequency response of the Wiener filter as

$$H(\omega)=\frac{\textrm{SNR}(\omega)}{1+\textrm{SNR}(\omega)}\tag{2}$$

Eq. $(2)$ clearly shows that if $\textrm{SNR}(\omega)\gg 1$ you have $H(\omega)\approx 1$ (i.e., no attenuation), whereas if $\textrm{SNR}(\omega)\ll 1$ you get $H(\omega)\approx \textrm{SNR}(\omega)\ll 1$ (i.e., strong attenuation).