0
$\begingroup$

What does $|H(j\omega)|^2$ in $20 \log_{10}|H(j\omega)|$ mean?

Is it some ratio of energy or power? And why? How to derive it?

I'm sorry for curtness of my questin.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

I suppose you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $h(t)$:

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

Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $X(j\omega)$ is the Fourier transform of the input signal, and $Y(j\omega)$ is the Fourier transform of the output signal, then $$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$

For a sinusoidal input signal $x(t)=\sin (\omega_0 t)$ the output signal $y(t)$ is given by

$$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$

where $\text{arg}\{H(j\omega_0\})$ is the phase of the complex function $H(j\omega)$ at frequency $\omega_0$. So you can see that $|H(j\omega)|$ represents the amplification of the system at frequency $\omega$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $H(j\omega)$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works.

$\endgroup$
2
  • 1
    $\begingroup$ ok, so $H(j\omega)$ is a ratio of output and input signal. And $|H(j\omega)|^2$ is ratio of... energies of output and input signal? Or power? What does it mean physicaly? $\endgroup$
    – user50222
    May 17, 2013 at 10:14
  • $\begingroup$ For deterministic signal you usually talk about energy. The output signal's energy is given by $\frac{1}{2\pi}\int_{-\infty}^{\infty}|Y(j\omega)|^2d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}|H(j\omega)|^2|X(j\omega)|^2d\omega$. For stochastic signals, the power density spectrum of the output signal is $S_y(\omega)=S_x(\omega)|H(j\omega)|^2$. $\endgroup$
    – Matt L.
    May 17, 2013 at 10:20

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.