FFT in ISM Band

hi, I want to generate a signal between 2.4GHz and 2.5GHz which mean bandwidth of 0.1GHz, and and then want to check its FFT in matlab simulink with frequency resolution of 0.04MHz, so please give me some suggestion, as I have tried it with N= fs/freq resultion but N comes very very large, can I get this resolution with low number of dft points.I am confused in selecting fs (sampling frequency) Thanks

Answer 1

[Wayne King]

Hi, I think you will need over 200,000 samples depending on the sampling frequency, but that is not a long duration signal at those sampling frequencies by any means.

Assuming that 2.5 GHz is not the Nyquist, I'll suggest as an example that 8 GHz is your sampling frequency, then you would need:

8e9/4e4

200,000 samples. If your sampling frequency is 10 GHz, you'll need 250,000 samples. Again, that is not a "long" signal at that sampling frequency (around 25 microseconds)

I don't you mean to say: "this resolution with low number of dft points" I think you just mean samples there.

[worker]

My question is I want to see this signal on FFT with this frequency resolution, so what will be my FFT length in this case???

[worker]

As by this calculation you mean to say FFT length will be 200,000, if I select 200,000 I get result, but it seems too large for practical purpose, as normally I have seen FFT length till 1024 points.

I will have to transmit these signals(2.40400GHz,2.40404 GHz) for 2ms and have to check FFT of the both signals.

I will be waiting for your kind respose, if u got my question

Answer 2

[Wayne King]

You are mistaking FFT length with frequency resolution. Frequency resolution in the DFT defined as the spacing between adjacent DFT bins is determined by the length of the input vector (in the time domain) and the sampling frequency, not by any padding on the DFT.

The goertzel.m function in Signal Processing Toolbox allows you to return the DFT at a specified set of frequenices (provided you have the frequency resolution). Perhaps you can use that function in a MATLAB function block?

That would enable you to not return a 200,000 point DFT.

[worker]

Thanks for response, I am out of lab so I ll check .m function tomorrow, but one thing the formula for freq resolution is (samping frequency/FFT points), actually when you select the input vector(in the time domain)as input to the FFT block in Simulink, that take the FFT points=the input vector, so in this case FFT points equals to 200,000 and this is my problem actually.