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Consider a DSBSC signal s(t) = m(t).cos(2.pi.20k.t) where say m(t) is multi-tone signal with frequencies less than 5 kHz.

The spectrum gives a typical bandpass say triangle centred around 20k and -20k. Now, when I calculate the minimum sampling rate;

Since the highest frequency is not an integer multiple of bandwidth of the bandpass signal; I end up getting the sampling rate as 25 KHz.

But when I draw the spectrum for the sampling rate of 10 kHz, the message signal centred at origin was doubled in amplitude and there was no aliasing as well.

Does that mean the minimum sampling rate is indeed 10 kHz?

I have used the typical bandpass sampling rate procedure many times, this was the first time it failed in giving me the minimum sampling rate!

edit: I doubt that the 10 kHz would be the minimum sampling rate to recover the message signal while the bandpass sampling rate of 25 kHz is the one which recovers the bandpass spectrum!

Also, if the above conclusion is correct, is it independent of the carrier frequency (Fc>>Fm)?

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The problem is that the general criterion for bandpass sampling (which results in $25\text{ kHz}\leq f_s\leq 30\text{ kHz}$) applies for general signals, whose spectrum may not be symmetric around $f_c$. In your example, sampling at 10 kHz would fail because spectrums of different shapes overlap.

The most complete resource I know on bandpass sampling is "The theory of bandpass sampling", by Vaughan et al, IEEE Trans. on Signal Processing, 1991, Vol. 39, Issue 9 (http://dx.doi.org/10.1109%2F78.134430). At the start of page 4, it specifies that in the special case of DSB signals, the minimum sampling interval is $1/B$, where $B$ is the bandwidth (for both sidebands). This corresponds to your case, where the minimum sampling rate is 10 kHz, equal to the signal's bandwidth.

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  • $\begingroup$ I want to know why it turns out to be 1/B for DSB signals and is it carrier frequency independent as well? $\endgroup$ – Ashik Anuvar Jan 11 '16 at 4:41
  • $\begingroup$ also my bad i am not in ieee, can you get me that file $\endgroup$ – Ashik Anuvar Jan 11 '16 at 4:44
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    $\begingroup$ @AshiK_741 Say the DSB spectrum goes from $f_l$ to $f_h$ centered at $f_c$. If you shift this spectrum by $k(f_h-f_l)=kB$ for integer $k$, either to the left or to the right, then the alias won't overlap the original spectrum. As you can see, it is independent of $f_c$. $\endgroup$ – MBaz Jan 11 '16 at 15:35
  • $\begingroup$ @AshiK_741Do you have access to a college/university library? Most are subscribed to IEEE. $\endgroup$ – MBaz Jan 11 '16 at 15:36
  • $\begingroup$ no acces now, am a passout $\endgroup$ – Ashik Anuvar Jan 11 '16 at 17:36
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Your rule does not take advantage of the double sideband symmetry.

The positive peaks of the cosine have a value of 1 and occur at a rate of 20 kHz. Sampling at the positive peaks of the cosine gives you directly samples of $m(t)$. The obtained 20 kHz sampling frequency is sufficient because even 2 × 5 kHz = 10 kHz is enough to sample $m(t)$. Throw away every second sample to arrive at the critical sampling frequency 10 kHz, still not losing information about $m(t)$.

The catch is that you must be able to synchronize to the cosine carrier. Otherwise you might even sample where its value is zero, losing all information. You also need the synchronization if you want to reconstruct the carrier which is lost by the sampling.

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