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  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

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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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