BackEMF said:
Another way to look at this is that the time domain signal is finite in duration, and hence, by the same idea as is used to derive Hesinberg's Uncertainty principle, the frequency domain signal must be infinite in duration i.e. time and frequency are Fourier conjugates.
The bandwidth is infinite as you say but practically there is a bandwidth where most energy lies and hence the rules of thumb like Carson's Rule which I think is adequate to answer the question.
The infinite bandwidth can be shown by representing the FM signal as the product of two exponentials:
V
fm = cos([w
c+km(t)]t)=cos[w
ct+ka(t)]
where wc=carrier frequency, m(t)=message, and a(t)=∫m(t)dt (changed message to phase)
then
Vfm=Re{e
jwcte
jka(t)}
e
jka(t) can be expanded in a power series in a so you have a modulating carrier at w
c multiplied by a power series in a, and each term of the power series is thus modulated to w
c in the frequency domain. a(t) has the same bandwidth as m(t) so the power series terms a, a
2, a
3, ... have bandwidth B, 2B, 3B, ... and the result is the sum so the bandwidth is infinite.
Carson's Rule comes from sampling m(t) at the the minimum frequency and representing m(t) using a staircase. Then the FM signal consists of rectangular bursts of sinusoids of duration equal to the sample period of m(t) and frequency equal to the carrier plus the constant amplitude of the sampled km(t) in that interval. In the frequency domain this is a sinc function (from the time domain rectangle) modulated to the frequency during the time period. The main lobe of the sinc is assumed to contain most of the signal energy and the summation of all these sincs leads to the bandwidth estimate.The question is further complicated by saying they would only sample the signal during the interval (t
o,t
f). This is the same as multiplying the signal by a rectangle of width (t
f-t
o) in the time domain and in the frequency domain this means convolving a sinc with the spectrum of the signal. This also makes the bandwidth infinite, if it wasn't already, but practically we can estimate the bandwidth of the sinc function as the width of the main lobe. So the sinc adds its main lobe width to the bandwidth of the function found with Carson's Rule. The longer the sample period (t
f-t
o), the skinnier the sinc lobe and the less it adds to the bandwidth of the resulting signal. *** Note this assumes we want to try to reproduce the signal outside the sample interval and if we only try to reconstruct the signal in the interval (t
f-t
o) then there is no multiplication by a rect in the time domain. Reproduction of a signal outside the sampling interval depends on finite bandwidth of the signal and is therefore not possible to achieve perfectly. So the question is arithmetically simple to answer since the equations are very simple. But the theory behind it is more sophisticated than it seems. That's why I was wondering what sort of text this question came from. The question's hint is actually wrong (and was in fact the same mistake early pioneers in FM made!) so maybe the question comes from an elementary signals text and the 'answer' they are looking for really is twice the highest instantaneous frequency, which is approximately correct for loud FM signals (Δf >> bandwidth of m).Also notice we've been talking about the bandwidth occupied around the central carrier frequency which is k/2. If there won't be any demodulation or bandpass sampling, then the highest frequency component will be k/2+B/2 and sampling must occur at twice that rate. (The other half of B is to the left of k/2)
