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enter image description here


As the signal $m(t)$ is Baseband signal after modulation it wil be bandpaas signal will look like this

enter image description here

So I need to use the Baseband sampling

Here $f_L=15kHz,f_H=25kHz,B=5kHz,$Now Undersampling formula

$n\leq\frac{f_H}{f_H-f_L}$ ,$n\leq2.5,$ so $n=2$ (maximum),

Now $\frac{2f_H}{n}\leq f_s$,$25\leq f_s$,

so minimum sampling rate will be $25kHz$

But the answer given is $10kHz$,What should be the correct answer?

Also i am confused whether will I get back the signal m(t) after bandpass sampling like this?

enter image description here

What is the mistake I am doing?


PS:Please answer it with some graphical approach if possible i am confused with the theory from this question



EDIT:

Please have a look at this

enter image description here

enter image description here

Also that means modulation will not effect sampling if we want to recover the baseband signal back,so for this minimum sampling can be calcuated by baseband sampling as $2f_m\leq f_s$

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  • $\begingroup$ Your bandwidth is litteraly less than 5 kHz as specified in the statement. Why not use bandpass sampling and sample at 10 kHz? $\endgroup$ – Ben Oct 20 '17 at 13:07
  • $\begingroup$ @Ben But its saying sampling after modulation then that signal will become a Bandpas signal..How can we then apply $2f_m\leq f_s$ formula? $\endgroup$ – Rohit Oct 20 '17 at 13:14
  • $\begingroup$ en.wikipedia.org/wiki/Undersampling $\endgroup$ – Ben Oct 20 '17 at 13:18
  • $\begingroup$ Think, what would happen if you sampled at 10 kHz? Where will your 40 kHz component end up ? Maybe 0 Hz? Where will your 44 kHz component end up? Maybe 4 kHz? $\endgroup$ – Ben Oct 20 '17 at 13:19
  • $\begingroup$ @Ben How 40kHz componet will shift into $0$ Hz,did't get you? if I take $f_m-f_s$ that means 40-10 will shift into 30kHz $\endgroup$ – Rohit Oct 20 '17 at 13:23
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I would not use a formula for the understanding of the so called bandpass sampling (or undersampling) operation. Instead try to analyse the situation by yourself considering the signal spectrum, sampling operation and the definition of aliasing which defines the permitted range of sampling frequencies.

First, we state the fundamental principle of sampling: in order to represent a signal $x(t)$ perfectly with a set of samples $x[n]$ taken uniformly at a sampling rate $f_s$ samples per second, there should be no aliasing (spectral overlap) in the sampled signal $x_s(t)$'s spectrum $X_s(\Omega)$.

Then we define the ideally sampled signal as $x_s(t) = x(t) \sum_{k=-\infty}^{\infty} \delta(t - k T_s)$ and its associated CTFT spectrum as: $$X_s(\Omega) = \frac{2\pi}{T_s} \sum_{k=-\infty}^{\infty} X(\Omega - k \frac{2\pi}{T_s}) $$

Finally we ask, given the consequences of sampling on the spectrum $X_s(\Omega)$ of the sampled signal $x_s(t)$, which set of frequencies $f_s$ can satisfy the fundamental principle of no-aliasing. Then we try to determine the minimum of this set of valid sampling frequencies.

Lets apply this to your problem: Given your real bandpass signal with a spectrum as in the figure-1.

First assume that for some $k=m$ the spectrum is shifted to right by $m f_s$ such that the shifted left-piece stands closest from left to the original right piece in its original place. Then consider the next shift at $k=m+1$ so that the shifted left piece jumps over the original right piece and stands closest (from right) to the original right piece. These two conditions are mathematicaly yielding the following constraints on the sampling frequency $f_s$:

For the case one with $k=m$ we have: $$ -f_L + m f_s \leq f_L $$

For the case $k=m+1$ we have: $$ -f_H + (m+1) f_s \geq f_H $$

Combining these two yields: $$ f_s \geq 2(f_H - f_L) $$

In your case $f_H-f_L$ is $10$ kHz, so we conclude that $$f_{smin} = 20 \text{kHz}$$

Note that this is the minimum sampling rate for real valued samples. If you are allowd to perform complex valued sampling, then you can use only one side of the spectrum (via complex analytic signal generation) and sample it at half the real sampling rate, but then you will be getting twice the number samples per second nevertheless)

Note also that this particular analysis just defines the minimum valid sampling frequency without aliasing. However, unlike the baseband sampling theorem which states that any frequency greater than the minimum Nyquist rate will also be a valid sampling frequency, in the case of bandpass sampling this is not so. For example you can easily verify that for the frequency $f_s = 2 f_L + B/2$ which is can be larger than the minimum we computed above, the resulting spectrum shifts will overlapp; there will be aliasing. So bandpass sampling frequencies are selected with care on the resulting spectral shifts.

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  • $\begingroup$ I am trying hard to understand it but not getting sir..:( $\endgroup$ – Rohit Oct 21 '17 at 9:48
  • $\begingroup$ which line do you not understand? Also have a look at this page $\endgroup$ – Fat32 Oct 21 '17 at 13:03
  • $\begingroup$ sir $f_H-f_L$ should be $25-15=10$,but you took its 5 $\endgroup$ – Rohit Oct 21 '17 at 13:32
  • $\begingroup$ oh sorry I updated the answer. It now says the min $f_s$ is 20 kHz for real samples. $\endgroup$ – Fat32 Oct 21 '17 at 13:47
  • $\begingroup$ Sir I really did't get the term "real sample" first time listioning this.Its also a examination question they gave answer $10kHz$ so what should be the answer 10 or 20 kHz? $\endgroup$ – Rohit Oct 21 '17 at 13:50

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