I have two questions that are bugging me:

  1. Does the duration of an audio signal affect the amplitude of frequency components of that same signal? For example, I am recording the sound of a fan using a microphone. At first, I record only for 10 sec and convert the audio signal into frequency spectrum. Then, I record the same sound for 20 sec and then convert the audio signal into frequency spectrum. In both the cases, the sound of the fan is same, but does the duration of the signal affect the amplitude of frequency components in the spectrum plot?
  2. For example, I have 2 audio signals. For the first one, I have that same fan sound recording for 10 sec and the sampling frequency is 5KHz, and for the second recording, I have that same audio signal but now the sampling frequency is changed to 15KHz. I used MATLAB to check the power for both the signals and the power for both the signals was same, however I want to know why. Formula that I used was Power=rms(signal)^2. According to me the second signal should have more power because now there are more number of samples compared to the first recording and since those extra samples would also have a random amplitude, the average shouldn't be the same as for the first one. Am I thinking it right? Can anyone provide their thoughts? Thank You!
  • $\begingroup$ Recall the definition of rms. There is a $1/N$ in the square root. $\endgroup$
    – ZR Han
    May 10 at 4:55

Power is energy per unit time. If you increase the duration, you increase the energy, but due to the normalization with time the power would be the same.

The DFT as given by

$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j 2\pi n k/N}$$

will scale the frequency component by $N$ as given by the summation over $N$ samples. This can be normalized by multiplying the result by $1/N$.

The frequency components of the signal levels will be the same in a normalized DFT (normalized by dividing by the total number of samples) for signal components that occupy one bin (pure tones), but the noise floor as observed may be lower by the change in sampling rate: if the noise floor is limited by quantization noise, the total quantization noise (well approximated as white noise, meaning constant across all frequencies) will be spread over a wider frequency range, so increasing the sampling rate will cause the contribution of quantization noise on a per Hz basis (noise density) to be lower. Further the duration will effect the frequency resolution in each bin in the frequency domain; for an unwindowed DFT, the equivalent noise bandwidth per bin is $f_s/N$ where $f_s$ is the sampling rate and $N$ is the number of bins. At a given sampling rate, increasing the number of bins will increase the total duration and thus reduce the noise bandwidth per bin; causing the overall DFT noise floor as observed to decrease. (Same noise power just less measured in each bin). Windowing if done in the time domain will reduce the signal level by the coherent gain of the window, but increase the equivalent noise bandwidth such that pure tones will be reduced more than noise that is spread over multiple bins. This detail is explained further in this post.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.