So say there's a filter with an impulse response of $h(t) = (0.8)^t u(t)$. I'd like to pass white noise through this and figure out the autocorrelation and power spectral density of the output.

I'm having a bit of trouble figuring out exactly how to do this. I know if you pass a signal through a filter, you simply take the Fourier transforms of the signal and the impulse response, multiply them together, and that's the output.

However, not being any mathematical expression for the noise, I'm unsure how to proceed. Any tips would be appreciated.


White noise is modelled as a process having spectral density $G_n(f) = k$, a constant.

After the filter, the resulting process $y$ will have spectral density $G_y(f) = G_n(f) |H(f)|^2$. And the autocorrelation will be the inverse Fourier transform of the spectral density.

This is a general result for any stable filter and finite power, wide-sense stationary input process (in this case we are pushing the boundaries because white noise has infinite power as defined above).

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    $\begingroup$ if $h(t) = (0.8)^t u(t) = e^{\ln(0.8) t} u(t)$, then $$H(f) = \frac{1}{-\ln(0.8) + j 2 \pi f} $$ and $$|H(f)|^2 = \frac{1}{(\ln(0.8))^2 + (2 \pi f)^2} $$. so that's what you multiply $G_n(f)$ with. and i think it integrates to a nice finite number, so you know what the power of the filtered noise is. (white noise is really a conceptual vehicle because it has infinite power.) $\endgroup$ Apr 3 '16 at 2:52
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    $\begingroup$ @robertbristow-johnson Since $G_n(f)=k$ is a constant, the power in the output process, which is $$\int_{-\infty}^\infty G_y(f)\,df= k\int_{-\infty}^\infty |H(f)|^2\,df = k \int_{-\infty}^\infty |h(t)|^2\,dt$$ (the last equality via Parseval's theorem) is finite if the filter is BIBO-stable (as this one is). That is, there is no reason to worry about whether the PSD of the output integrates to a finite (positive) number or not: it always does for a BIBO filter. See also Appendix B of this Lecture Note of mine. $\endgroup$ Apr 3 '16 at 15:22
  • $\begingroup$ Oops; that is Appendix A, not Appendix B $\endgroup$ Apr 3 '16 at 15:29

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