I have to compare amplitudes of two signal ("ideal " signal and "estimated" signal) and want to compute an amplitude imbalance

For my calculation I used the following equation: $$ A = \frac{\left | A_{max} - A_{aver}\right |}{A_{aver}} \cdot 100% $$

as I do if I need compute a voltage or a current imbalance.

I need the value in dB-scale, so I didnt multiply by 100%... and $ 20 \cdot \log_{10}A$.

Ma result is -25 dB ( the result should be less than 0,5dB, p 20-21 (65)). I think I should use for amplitude imbalance another math equation.

Could someone explain how imbalance should be computed?


1 Answer 1


Compute the dB of each and then subtract the two to get the difference in dB which should be very small. You can get the same result from converting the ratio of the two to dB directly.

For dB of voltage quantities of AC sources that are arbitrary waveforms you would use the rms value, so the difference between two different sources in dB using both approaches would be:

$$20Log_{10}(A_1) - 20Log_{10}(A_2)$$


Where $A_1$ is the rms value for source 1 and $A_2$ is the rms value for source 2. The rms value can never be negative or complex. Since we are using a ratio, if the two waveforms are similar (both sinusoids for example), then any other consistent metric of amplitude could be used, such as peak amplitude instead of rms).

From the link the OP gave, the intention here is to derive an amplitude and phase imbalance specification and for that the relationship between SNR and imbalance is that the SNR will be the normalized error vector. For example, if the error vector normalized to ideal was 0.1 radians, the SNR would be $20Log_{10}(0.1) = 20$ dB. For small angle the amplitude error and phase error would have the same effect, so if the error vector was 0.1 due to amplitude imbalance, we would also get a 20 dB SNR. (Universally, it is the magnitude of the error vector). An error vector of 0.1 dB is an amplitude imbalance of $20Log_{10}(1\pm 0.1) = +0.8$ or $-0.9$ dB.

  • $\begingroup$ I have an "ideal " amplitude of the signal and "estimated" amplitude of the signal. I can compute only one rms value $\endgroup$
    – FrimHart64
    Commented Apr 20, 2022 at 5:20
  • $\begingroup$ i calculated the subtract of the two of them: amplitude of the "ideal"signal and amplitude of the "estimated" signal (from -0.0506 to 0.0482) , and convert it in dB $20 \cdot \log_{10} (\Delta A)$. In dB i got the complex numbers. $\endgroup$
    – FrimHart64
    Commented Apr 20, 2022 at 7:24
  • $\begingroup$ The rms value cannot be negative or complex. If the "ideal" is a sinusoid, divide the peak value by $\sqrt{2}$ to get rms. You don't do $\Delta A$, but the ratio of the rms values as I described, or take the dB of each one separately and then subtract the dB results. $\endgroup$ Commented Apr 20, 2022 at 11:37
  • $\begingroup$ Assuming you have actual samples of each one, then compute the standard deviation of the samples. If there is no DC offset, the standard deviation is the rms value. Or compute the variance and then use $10Log_{10}()$ given the variance is a power quantity. All equivalent. $\endgroup$ Commented Apr 20, 2022 at 11:40
  • $\begingroup$ $\Delta A = A_{estimated} - A_{ideal}$. I want to determine a deviation. The magnitude of estimated signal is found to be slighly varing around the magnitude of the ideal signal. I need to comput the deviation. I thiught I need to work with substract of the magnitudes $\endgroup$
    – FrimHart64
    Commented Apr 20, 2022 at 12:18

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