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I am currently looking over filters and understand the frequency response of a system, e.g. a low-pass filter. However I am confused by the 'Power/Amplitude response' of a system as detailed in my notes below:

Square of the amplitude response, or power response:

$$|H(\omega)|^2=H(\omega)H^*(\omega)=H(\omega)H(-\omega) $$

I created a basic matlab scenario for the a low-pass filter and plotted the frequency response and the power response:

enter image description here

From the top graph I understand that as the frequency increases along the x-axis you get an attenuation for difference frequency components in db. However I do not understand what the significance or intuitive value is gained by squaring the attenuation (or gain). I can see that the gain becomes more negative as frequency increases and the amplitude response increases. So for example my 3dB cut-off point is at 1rad/s, 10^(0). This has a corresponding amplitude response value of 9. Does this mean the amplitude has decreased by a ninth? Because surely if my gain were positive my amplitude response value would also be a positive number and so the opposite cannot be true?

Any help would be greatly appreciated!

Regards

Pat

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  • $\begingroup$ Isn't the plot for $|H(j\omega)|^2$ wrong? If $|H(j\omega)|<1$, then its square is even smaller, right? $\endgroup$ – MBaz May 20 '16 at 0:36
  • $\begingroup$ Plot them both in dB and you will see that they are the same. It's defined as $20\cdot log()$ for amplitude/fields/linear quantities and $10 \cdot log()$ for energy/power quantities $\endgroup$ – Hilmar Aug 18 '16 at 14:30
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Your second plot is the square of the magnitude of the frequency response in dB, which indeed doesn't make much sense. You must use $|H(j\omega)|^2$, and not $(20\log_{10}|H(j\omega)|)^2$, as you did. Of course, $|H(j\omega)|^2$ will not give you any new information compared to $H(j\omega)$.

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  • $\begingroup$ Just remember 10 log10 for power and 20 log10 for amplitude $\endgroup$ – Stanley Pawlukiewicz Jun 14 '17 at 22:18

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