1. Cosine signal of frequency 498 Hz is sampled 20 times in one period : fs = 9960 Hz Ts = 0.0001s T = 0.002s Ns = 20 Resolution bandwidth = fs / Ns = 9960 / 20 = 498 Hz and because fs / 2 = 4980, we only need 10 frequency bins / RBW’s (k). Using the DFT below in a spreadsheet

                N - 1                   

    Yk(K) = 1 / N ∑ y(n) ( cos (2πnk/N) - j sin (2πnk/N) )

               n = 0                    

    We can plot the results as a blue line in the attached graph.
    The red line is the corresponding graph I get when a build a simple model in Simulink/Matlab and observe the results in a signal analyser.

    The signal analyser in Simulink is setup with a word length of 20 and zero overlap. A rectangular window is used with a duration = simulation time = T = 0.002s. If in my spreadsheet example I increase the no. of bins from 10 to 100 i.e. RBW now = 49.8 Hz, I get a lot more data and the resulting graph is exactly the same as the Simulink graph, even though RBW = 498 in Simulink ? I suspect the reason for the difference in the results is to do with data segmentation ? enter image description here


closed as unclear what you're asking by Laurent Duval, Matt L., MBaz, jojek Mar 17 '17 at 12:38

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  • $\begingroup$ apologies about the detail. I am confused re. the power spectrum graph (red line) which is produced by Simulink (MATLAB). How can simulink produce such a smooth graph with only 20 samples ? In order to obtain such a graph using excel (blue line) i need to use at least 100 - 200samples $\endgroup$ – derick007 Mar 18 '17 at 19:32

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