4
$\begingroup$

I have a filter where the input is:

$$ x(t) = s(t) + w(t) $$ which is signal plus white Gaussian noise and the output is $y(t)$. If the impulse response of the filter is $h(t)$, then:

$$y(t) = s(t)*h(t) + w(t)*h(t)$$

I have given the following equation to calculate the SNR:

$$SNR = \frac{\int_{-\infty}^\infty|s(t)*h(t)|^2 dt}{\int_{-\infty}^\infty|w(t)*h(t)|^2 dt}$$

And in frequency domain, following the Parseval's theorem:

$$SNR = \frac{\int_{-\infty}^\infty|S(f)H(f)|^2df}{\int_{-\infty}^\infty|W(f)H(t)|^2 df}$$

I wonder how this represents the signal power and noise power at numerator and denominator, respectively? Cause it is more like calculating energy than power.

$\endgroup$

1 Answer 1

2
$\begingroup$

Using

$$\int\limits_{-\infty}^\infty|s(t)*h(t)|^2 \,\mathrm dt$$

requires that this integral exists – and that requires, unless an infinite amount of the energy of $s$ is in $\ker((*h))$ (i.e. gets mapped to 0) that $s$ is an energy signal, not only a power signal (as you're probably used to dealing with). (An energy signal is exactly a signal that is magnitude-integrable.)

For that, defining SNR as quotient of energies makes sense. Also, your $S(f), H(f)$ are then energy spectral densities, instead of power spectral densities.

Problem: WGN is not an energy signal, and it's by definition not in the kernel of any deterministic filter. Therefore,

$$\int\limits_{-\infty}^\infty|w(t)*h(t)|^2 \,\mathrm dt$$

simply doesn't exist and the formula doesn't "work out", as it has a diverging denominator.

I can only guess where the formula came from, something like from the derivation of (average) power of a signal $\xi$ being $\lim\limits_{T\to\infty}\frac1T\int_{T/2}^{T/2} |\xi(t)|^2 \,\mathrm dt$, and then eager cancelation of the $\frac 1T$ from denominator and numerator?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.