# Exercise related to frequency resolution and SNR

I'm studying a book aboud dsp and trying to make excercises. 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:

1)

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

2)

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

3)

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!