what is the bandwidth of a periodic signal, if fourier transformed?

Question

my question is if a periodic signal is fourier transformed, will the bandwidth be equal to its frequency in time domain?

For example, if a sine wave has the frequency w_c, then in frequency domain of fourier transform, it will have two impulses at -w_c and +w_c, right ? So the bandwidth or the highest frequency in frequency domain is supposed to be w_c, same as the frequency of the original signal in time domain.

But is this true for any periodic signal ? And another thing, how can I generate fourier transform of a periodic signal? Doesn't it violate the first condition of fourier transform ?

I need it because Nyquist theorem states that, If a continuous time signal contains no frequency components higher than B Hz, then the sampling frequency should be F_S>2B. In exercise, I am given the sum of a bunch of sine sine functions and I need to calculate the minimum sampling frequency. So I need to know the highest frequency component?

Answer 1

First of all, the bandwidth of a periodic signal is usually defined as the difference between it's highest and lowest frequency component. It is the frequency range in which the signal's spectral density is nonzero. If you have a periodic signal with frequency components down to zero, you have a low-pass or baseband signal where the bandwidth is equal to it's highest frequency component. The Nyquist theorem states that perfect reconstruction of a continuous-time signal (analog signal) with finite bandwidth is possible with a discrete sequence of samples (digital signal), if certain constrains are satisfied.

In the case of a baseband signal (frequency components down to zero), the sampling rate has to be bigger than two times the highest frequency component of the signal.

In the case of a bandpass signal (all frequency components higher than zero), the sampling rate has to be higher than two times the bandwidth of the signal e.g . This leads to a range of different possible sampling frequencies in the range:

fc is the center frequency of the bandpass signal, B is the bandwidth and m is an arbitrary, positive integer ensuring that fs is at least two times the bandwidth of the periodic signal.

Answer 2

Generally, as I can recall from a prior career that involved radar image processing, the "bandwidth" is the highest 3db portion of the power spectrum. The power spectrum isn't the same as the FFT output; power = square of the amplitude (FFT outputs complex numbers, so this would be the square of the magnitude of each complex vector).

If you're doing an FFT, remember to swap the halves of the frequency domain output before figuring out the bandwidth, because the output of an FFT starts with the middle frequency and goes to the highest, then continues with the minimum frequency and ends before the middle freq.