While decomposing the impulse train using the Fourier series the number of the samples are infinite hence representing a infinite length of train.

But while studying the Nyquist Criteria in the frequency domain it is shown the expression to have $m$ samples where $-\infty<m<\infty$

While studying the concept of the Nyquist Criteria in the ISI model where for the system to be interference free the value of the sampling instant must be given as following:


For more reference here. I cannot understand that the signal is having infinite duration and infnite bandwidth accordingly is it even possible.


2 Answers 2


Let's consider the continuous time case.

Take a Gaussian $f(t) = \frac{1}{\sqrt{2 \pi}} e^{-t^2/2}$, which is positive for all time.

The Fourier transform of a Gaussian is another Gaussian: $F(\omega)= c e^{- d \omega^2 }$ for some constants $c,d$ that you can work out on your own, so a Gaussian occupies all time and all frequency. That is, you can be both infinite in time and infinite in frequency.

Your trade-off between resolution in time and frequency is given by the uncertainty principle. The Gaussian attains equality in the uncertainty principle.

It seems like you're somewhat confused as to what Nyquist's sampling theorem says -- see this page for details (or your favorite signal processing text). You're assuming the signal is band-limited (i.e. is non-zero in some finite interval of frequencies) in order to be able to sample the signal at some prescribed rate in order to be able to perfectly reconstruct the signal.


To complement Batman's answer: what you can't have is a signal that is both band-limited and time-limited. The sampling theorem requires the sampled signal to be band-limited, and this implies that the signal has infinite duration. This is why the theorem requires an infinite number of samples.

Regarding ISI, again the pulses being transmitted are assumed to be band-limited, which implies that they are of infinite duration (for example, sinc or raised cosine pulses).

In practice, we assume that signals tend to zero as $|t|\rightarrow\infty$ and that their spectrum also tends to zero as $|f|\rightarrow\infty$. This allows us to truncate the signals both in time and in frequency with negligible negative consequences.

  • $\begingroup$ what do we say about the energies of these signals.Thanks that puts some light on it. $\endgroup$
    – MaMba
    Commented May 13, 2015 at 19:01

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