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I know normalised bandwidth can be found from FFT plot of the signal. But I'm confused that at what stage I have to take the FFT.

I mean first binary data is input to a convolutional encoder, then to NRZ encoder, Then GMSK modulation (sampling frequency till now is 100KHz), then upsampling to 30MHz sampling frequency and then the quadrature modulation.

At what stage I'm supposed to take the FFT to find normalised bandwidth?

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You can take the FFT on any signal after modulation, but would be simplest (fewer samples required) to do it immediately after modulation. With proper upsampling the FFT of your signal prior to upsampling would be identical as long as you used proportionally more points to keep the time length identical. In your example of 100 KHz and 30 MHz, you could take a N point FFT at 100 KHz or a $300 N$ point FFT at 30 MHz and the result should look identical (with just more points interpolated in the FFT due to the upsampling). Given the 100 KHz sampling rate and room for your interpolation filter, I would assume that your GMSK modulated signal BW you are trying to view is less than 70 KHz (as a complex signal)?

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  • $\begingroup$ BW of my signal is 25KHz. Can you please explain the bold part here: In your example of 100 KHz and 30 MHz, you could take a N point FFT at 100 KHz or a 300N point FFT at 30 MHz and the result should look identical (with just more points interpolated in the FFT due to the upsampling). How can I know the number of points to be interpolated to make them identical? $\endgroup$ – anonymous Jun 16 '17 at 6:15
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    $\begingroup$ I meant the say the result would be identical if you used the same duration for the signal length. The frequency resolution in the FFT is 1/T where T is the length of your buffer in time containing all the samples to take the FFT. For example if you had a buffer of 1 ms you would have a 1 KHz resoluiton in your FFT; each FFT bin would be spaced by 1 KHz. At 100 KHz sampling the buffer is 100 samples and your FFT goes from 0 to 99 KHz. With 30 MHz sampling the buffer is 30,000 samples, and your FFT goes from 0 to 30.999 MHz. Zoom in on your signal and it will be identical... $\endgroup$ – Dan Boschen Jun 16 '17 at 10:33
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    $\begingroup$ ...if you did your interpolation properly when going from 100 KHz sampling to 30 MHz sampling. Since they are the same, the point is to do your FFT at the lower rate (less processing)! $\endgroup$ – Dan Boschen Jun 16 '17 at 10:33
  • $\begingroup$ Got that.Thanks! If I find fs right after modulation by using B_analog=25KHz and it comes out to be 100KHz then what analog bandwidth should I use to calculate fs after upsampling to 30MHz? $\endgroup$ – anonymous Jun 16 '17 at 10:51
  • $\begingroup$ Upsampling does not change the signal bandwidth. For example, imagine a sine wave that is 1 Hz sampled at 10 Hz (it has 10 samples per cycle and takes 1 second to complete). Sample that same sine wave at 10 MHz, and there will be 10E6 samples in the one cycle, but it still takes 1 second to complete. We resampled (interpolated) but we did not change the signal frequency or any of its characteristics. $\endgroup$ – Dan Boschen Jun 16 '17 at 10:54

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