I'm studying a book about DSP and trying to make exercises. Here is one I'm interested in:

A scientist acquires 65,536 samples from an experiment at a sampling rate of 1 MHz. He knows that the signal contains a sinusoid at 100 kHz. He needs to determine is if there is a second sinusoid at 103 kHz. As a start, he takes a 65,536 point DFT of the signal. To his surprise, all he can see in the spectrum is noise. He estimates that the signal he is looking for is 4 times lower in amplitude than the random noise (i.e., SNR = 0.25). He also estimates that the SNR will need to be at least 3.0 for the signal to be detected, if present. To improve the SNR, he decides to break the signal into segments, and average their spectra. Arrange your answers to the following questions in a table.

a. If he uses a segment length of 16,384 samples, what will be the frequency resolution (i.e., the spacing between data points) in the averaged spectrum? Give your answer in hertz. How many segments will be averaged? What is the SNR of the averaged spectrum? Does this have the required frequency resolution? Does this have the required SNR?

b. Repeat for segment lengths of: 8192, 4096, 2048, 1024, 512, 256, and 128.

Here is my solutions:


16384 samples

4 segments will be averaged

freq. resolution would be (25*10^4)/8193Hz ~ 30Hz

(it is told in the book that random noise reduces in proportion to the square-root of the number of segments) snr = 0.25 *$\sqrt{4}$ =0.5 (not enough)

freq. resolution is enough to resolve between 100kHz and 103kHz sinusoid


2048 samples

32 segments will be averaged

$10^6$/32 = 31250

freq. resolution would be 31250/1025Hz ~ 30Hz

snr = 0.25 *$\sqrt{32}$ = 1.4 (not enoght)

freq. resolution is ok


128 samples

512 segments will be averaged

freq. resolution would be 1953/65Hz ~ 30Hz

snr = 0.25 *$\sqrt{512}$ = 5.6 (is ok)

freq. resolution is ok

Am I right or wrong? Where I can be wrong?

Thank you so much in advance!


1 Answer 1

  1. Your calculation of the SNR is right and the DFT Bin Resolution is also OK.
  2. One thing you're missing is the effective resolution due to "Windowing" effect.
    The DFT of finite number of samples is interpolated by a Dirichlet Kernel (Like Sinc). It means you resolution is also limited by the main lobe width of the Sinc which is proportional to the inverse of the window length.
    Namely, using less samples means wider lobe which reduces the resolution and the ability to differ between two harmonic signals.
  • $\begingroup$ Thank you! But how I can corretcly calculate frequency resolution in this excercise based on given data? $\endgroup$
    – Sharov
    Commented Aug 1, 2014 at 13:45
  • 1
    $\begingroup$ Rule of thumb says the resolution is the width of the main lobe of the Sinc function. $\endgroup$
    – Royi
    Commented Aug 1, 2014 at 13:46

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