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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.

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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.

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    $\begingroup$ 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 '13 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 '13 at 10:20

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