We know that SNR is defined as "power" of the transmitted signal to the "power" of the noise. In practice we transmit time limited signals (for simplicity consider the case that there is only one transmission), hence its power is zero, since $$P=\lim_{T\rightarrow\infty} \frac{\int_{-T}^{T}|x(\tau)|^2d\tau}{2T}$$ and as energy of a time limited signal is limited, as $$T\rightarrow\infty$$ the power will be zero, hence in real case, the SNR is always zero. Where is my fault?
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\$\begingroup\$ Perhaps the fault is assuming real things go to infinity. \$\endgroup\$– SamuelCommented Jun 11, 2015 at 20:21
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\$\begingroup\$ @Samuel but the definition of power is as above, we should take the limit at infinity \$\endgroup\$– CLAUDECommented Jun 11, 2015 at 20:22
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\$\begingroup\$ @AMIR, for the average power over an infinite interval, you should take the limit at T to infinity. For the average power over a shorter interval (like maybe the 1 second or microsecond when you are expecting to receive a message) you should take the integral over that interval. \$\endgroup\$– The PhotonCommented Jun 11, 2015 at 21:32
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\$\begingroup\$ Note, power is a function of time. This calculation is not for power, it is for the average power. If the signal is only non-zero for a finite time and you average it over infinite time then you get zero at the limit which is correct. \$\endgroup\$– JonCommented Jun 11, 2015 at 21:37
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\$\begingroup\$ AMIR you're confusing average power with actual power. \$\endgroup\$– FormagellaCommented Jun 11, 2015 at 21:46
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In accordance with your calculations, the power of the noise would also be zero. And 0/0 is indeterminate. In any case, the SNR is only important for the duration of the signal. That determines the detectability of the signal. SNR has no meaning if there is no signal.Also, by dividing by T, you are calculating average power. Again, the meaningful SNR is the peak signal divided by the noise or, in some cases, the average power over the duration of the signal divided by the average noise power over the duration of the signal, not to infinity.
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\$\begingroup\$ but I think as the signal is band limited, its bandwidth is so large(theoretically infinite) and hence the noise power which is $$N_{0}W$$ will be a very large number, not 0 \$\endgroup\$– CLAUDECommented Jun 11, 2015 at 22:12