Sampling of Bandpass Signal
Sampling of Bandpass Signal
For a signal m(t) whose highest-frequency spectral component is fM ,
the sampling frequency fs, must be no less than fs = 2fM only if the lowest-frequency spectral component of m(t) is fL =0.
In the more general case, where fL ≠ 0 , it may be that the sampling frequency need be no larger than fs = 20(Fm – fL).
For example, if the spectral range of a signal extends from 10.0 to 10.1 MHz,
the signal may be recovered from samples taken at a frequency fs = 2(10.1 – 10.0) = 0.2 MHz.
To establish the sampling theorem for such bandpass signals, let us select a sampling frequency fs = 2(fM-fL) and let us initially assume that it happens that the frequency fL turns out to be an integral multiple of fs, that is. fL=nfs, with n an integer.
Such a situation is represented in Fig.
In part a is shown the two-sided spectral pattern of a signal m(t) with Fourier transform M(jω).
Here it has been arranged that n =2, that is, fL coincides with the second harmonic of the sampling frequency,
while the sampling frequency is exactly fs = 2(fM-fL).
In part b is shown the spectral pattern of the sampled signal S(t)m(t).
The product of m(t) and the de term of s(t) duplicates in part b the form of the spectral pattern in part a and leaves it in the same frequency range from fL, to fM.
The product of m(t) and the spectral component in S(t) of frequency fs (= l/Ts) gives rise in part b to a spectral pattern derived from part a by shifting the pattern in part a to the right and also to the left by amount fs.
Similarly, the higher harmonics of fs, in S(t) give rise to corresponding shifts, right and left, of the spectral pattern in part a.
We now note that if the sampled signal S(t)m(t) is passed through a bandpass filter with arbitrarily sharp cutoffs and with passband from fL , to fM the signal m(t) will be recovered exactly.
In Fig. the spectrum of m(t) extends over the first half of the frequency interval between harmonics of the sampling frequency, that is, from 2.0fs, to 2.5fs.
As a result, there is no spectrum overlap, and signal recovery is possible.
It may also be seen from the figure that if the spectral range of m(t) extended over the second half of the interval from 2.5fs, to 3.0fs, there would similarly be no overlap.
Suppose, however, that the spectrum of m(t) were confined neither to the first half nor to the second half of the interval between sampling-frequency harmonics.
In such a case, there would be overlap between the spectrum patterns,
and signal recovery would not be possible.
Hence the minimum sampling frequency allowable is fs = 2 (fM – fL) provided that either fM or fL is a harmonic of fs.
If neither fM nor fL is a harmonic of fs, a more general analysis is required.
In Fig. we have reproduced the spectral pattern of Fig.
The positive-frequency part and the negative-frequency part of the spectrum are called PS and NS respectively.
Let us, for simplicity, consider separately PS and NS and the manner in which they are shifted due to the sampling and let us consider initially what constraints must be imposed so that we cause no overlap over, say, PS.
The product of m(t) and the de component of the sampling waveform leaves PS unmoved and it is this part of the spectrum which we propose to selectively draw out to reproduce the original signal.
If we select the minimum value of fs, to be fs = 2(fH – fL)=2B then the shifted PS patterns will not overlap PS.
The NS will also generate a series of shifted patterns to the left and to the right.
The left shiftings cannot cause an overlap of PS.
However, the right shiftings of NS might cause an overlap and these right shiftings of NS are the only possible source of such overlap over PS.
Shown in Fig. are the right shifted patterns of NS due to the (N-1)st and Nth harmonics of the sampling waveform.
It is clear that to avoid overlap it is necessary that
(N – 1)fs – fL<= fL. ……(1)
Nfs – fM >=fM …….(2)
So that , with B ≡ fM -fL , we have
(N-1) fs <= 2(fM – B) ……..(3)
And Nfs >= 2fM ….(4)
If we let k ≡ fM/B, Eqs (3) and (4) become
fs <= 2B(k-1)/(N-1) ……(5)
And fs >= 2B (k/N)…..(6)
in which k>= N since fs>=2B.
Equations (5) and (6) establish the constraint which must be observed to avoid an overlap on PS.
It is clear from the symmetry of the initial spectrum and the symmetry of the shiftings required that this same constraint assures that there will be no overlap on NS.
Equations (5) and (6) have been plotted in Fig. (C) for several values of N.
The shaded regions are the regions where the constraints are satisfied,
while in the unshaded regions the constraints are not satisfied and overlap will occur.
As an example of the use of these plots, consider a case in which a baseband signal has a spectrum which extends from fL = 2.5 kHz to fM = 3.5 kHz.
Here B= 1 kHz and k= fM/B = 3.5.
On the plot of Fig. (C), we have accordingly erected a dashed vertical line at k=3.5.
We observe that for this value of k, the selection of a sampling frequency fs = 2B =2 kHz brings us to a point in an overlap region.
As fs is increased there is a small range of fs, corresponding to N = 3, where there is no overlap.
Further increase in fs , again takes us to an overlap region,
while still further increase in f, provides a nonoverlap range, corresponding to N = 2 (from fs = 3.5B to fs = 5B).
Increasing fs further we again enter an overlap region while at fs = 7B we enter the nonoverlap region for N= 1.
When , fs>= 7B we do not again enter an overlap region.
(This is the region where fs>= 2 fM : that is, we assume we have a lowpass rather than a bandpass signal.)
From this discussion, we can write bandpass sampling theorem as follows – A bandpass signal with highest frequency fH and bandwidth B,
can be recovered from its samples through bandpass filtering by sampling it with frequency fs = 2fH/k,
where k is the largest integer not exceeding fH/B .
All frequencies higher than fs, but below 2f (lower limit from low pass sampling theorem)
may or may not be useful for bandpass sampling depending on overlap of shifted spectrums. (Sampling of Bandpass Signal)