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!



  • $\begingroup$ 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, 2016 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, 2016 at 14:30

1 Answer 1


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 $H(j\omega)$.

  • $\begingroup$ Just remember 10 log10 for power and 20 log10 for amplitude $\endgroup$
    – user28715
    Jun 14, 2017 at 22:18

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.