I will take two sine waves i) 0.1 AMPLITUDE with no phase shift ii)0.4 amplitude with some phase shift 20 degree (Considering Both same frequency).

In time domain - If i divide signal 1 by signal 2 by taking RMS value,it was nearly 0.25
In frequency domain - I took fft of both the signals and divided signal 1 by signal 2, i got the result in real and imaginary part.

If i take the RMS value of frequency domain output(real and imaginary), i am not able to get the same result as time domain.

So what should i do to get the same result in both the case.


  • $\begingroup$ When you said " RMS value of frequency domain output(real and imaginary)" does that mean you did separate RMS calculations and simply added them? $\endgroup$ Mar 27, 2018 at 12:40
  • 1
    $\begingroup$ Why dont you share your code with us, this way the question and the problem become easier to understand. $\endgroup$ Mar 27, 2018 at 12:42

2 Answers 2


Here is one way to do it properly:

import numpy as np

def main():

        t = np.arange( 0, 200 )  
        N = len( t )

        signal1 = 0.1 * np.sin( .1 * t )
        signal2 = 0.4 * np.sin( .1 * t + .35 )

        rms_time1 = np.sqrt( np.sum( signal1 * signal1 ) / N )
        rms_time2 = np.sqrt( np.sum( signal2 * signal2 ) / N )
        q_time = rms_time1 / rms_time2

        print q_time, rms_time1, rms_time2

        dft1 = np.fft.fft( signal1 )
        dft2 = np.fft.fft( signal2 )

        dft1_real = np.real( dft1 )
        dft1_imag = np.imag( dft1 )
        dft2_real = np.real( dft2 )
        dft2_imag = np.imag( dft2 )

        rms_freq_real1 = np.sqrt( np.sum( dft1_real * dft1_real ) / N )
        rms_freq_imag1 = np.sqrt( np.sum( dft1_imag * dft1_imag ) / N )
        rms_freq_real2 = np.sqrt( np.sum( dft2_real * dft2_real ) / N )
        rms_freq_imag2 = np.sqrt( np.sum( dft2_imag * dft2_imag ) / N )

        rms_freq1 = np.sqrt( rms_freq_real1 * rms_freq_real1 \
                           + rms_freq_imag1 * rms_freq_imag1 )

        rms_freq2 = np.sqrt( rms_freq_real2 * rms_freq_real2 \
                           + rms_freq_imag2 * rms_freq_imag2 )

        q_freq = rms_freq1 / rms_freq2

        print q_freq, rms_freq1, rms_freq2

        print rms_freq1 / rms_time1
        print rms_freq2 / rms_time2
        print np.sqrt( 200 )


The output is:

0.246138603222 0.0699023232816 0.283995774603
0.246138603222 0.988568136262 4.016306761


This is a more straightforward way to calculate the RMS of the DFT bins. I coded the sample above to more closely align with how I thought the OP was doing it.

        sumsquares1 = np.real( np.sum( dft1 * dft1.conjugate() ) )
        sumsquares2 = np.real( np.sum( dft2 * dft2.conjugate() ) )

        rms_freq1 = np.sqrt( sumsquares1 / N )
        rms_freq2 = np.sqrt( sumsquares2 / N )

        q_freq = rms_freq1 / rms_freq2

        print q_freq, rms_freq1, rms_freq2
  • $\begingroup$ And one more thing i need to clarify, After dividing dft1 and dft2 we will get real and imaginary parts, By using these real and imaginary parts can i get the same value as in time domain. $\endgroup$
    – sk gowda
    Mar 27, 2018 at 16:24
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    $\begingroup$ @sk gowda, The key here is Parseval's Theorem which says the sum of the squares will be preserved for a properly normalized DFT, i.e. 1/sqrt(N) which makes the transform unitary. Since your signal is real valued, the magnitude squared of each sample is the value of the sample squared. For your bin values (a+bi) the magnitude squared is a^2+b^2. The RMS calculation sums the squares first. In my sample code I simply summed all the a^2 values and b^2 values and then added them using Pythagorean theorem. The final three prints are to demonstrate the rescaling factor for proper normalization. $\endgroup$ Mar 27, 2018 at 16:34
  • $\begingroup$ @sk gowda, I added a followup. $\endgroup$ Mar 27, 2018 at 17:20

Can you please make more clear what you mean by "divide signal 1 by signal 2". Is it $z[n] = x[n] / y[n] = x[n] \times y^{-1}[n] $ what you mean?

It doesn't make much sense to me such a non-linear operation. What is the purpose of this? For example, if $ y[n] = 0 $ for some $n$, what do you expect as your result? Maybe I can help you if you make this more clear.

Besides, applying a division in the time domain does not match division in the frequency domain. I don't see why you expect to obtain the same result.

  • $\begingroup$ Firstly thank you......As per RLC circuit , Z =sqrt(R^2 + X^2). so i considered one signal 0.1 amplitude as voltage with some phase shift 20 degree and another signal 0.4 as current. As per my knowledge , Zrs=Vrms/Irms . and Zrs= sqrt(R^2 + x^2) , Here i considered real part as Resistance and X as reactance. so both Zrs may be same. correct me if i am wrong. $\endgroup$
    – sk gowda
    Mar 27, 2018 at 16:12
  • $\begingroup$ My bad, I didn't read carefully your question. Of course, if your refer to RMS values, my comment before is out of place. And the answer of @Cedron Dawg is the right one. $\endgroup$ Mar 28, 2018 at 13:27

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