Ideal sampling of audio signals-Periodic spectrum [duplicate]

Question

When applying ideal sampling on audio singals, why is the spectrum periodic ?

Answer 1

Discritizing a signal in the ideal sense is equivalent to multiplying by a Dirac delta impulse train. The Fourier transform of an impulse train is also an impulse train. Lastly, a multiplication in the time domain is equivalent to a convolution in the frequency domain. So when we discritize a signal, we are convolving that signal’s frequency transform with an impulse train, resulting in a frequency spectrum that repeats at intervals equal to the sample rate, making it periodic.

Answer 2

The answer by @Dan Szabo is correct and just about every DSP book that cover Nyquist sampling will use the relationship that a periodic train of delta function in the time domain has a Fourier Transform that is a periodic train of delta functions is the frequency domain. When taken together with multiplication in time is equivalent to convolution in frequency, one gets the periodic spectrum you asked about.

while mathematically apea;ing to many, it doesn't lend itself to an intuitive physical interpretation.

One thing to note is that if the time domain pulse train multiplying the time series isn't periodic, the Fourier Transform will not be periodic either. remove the periodicity in time, you remove the periodicity in frequency. So the simple answer could be that when you multiply something by a periodic function you get something that has some periodic characteristic.

There are some results for non-periodic sampling but they are a lot harder to show reconstruction which is usually involves a bound error as opposed to perfect reconstruction.

Perfect sampling is an idealization. Actual sampling doesn't involve perfect delta functions.

Answer 3

It's a direct consequence of multiplying by the sampling function:

$$\begin{align} \mathrm {III}_T (t) &= \sum\limits_{n=-\infty}^{\infty} \delta (t-nT) \\ &= \sum\limits_{k=-\infty}^{\infty} \tfrac{1}{T} e^{j 2 \pi (k/T) t } \\ \end{align}$$

the latter summation is the Fourier series of former noting that it is periodic with period $T$. all of the Fourier series coefficients are the constant $\frac1T$. i will choose to scale that with $T$ to get rid of that factor.

so what happens (conceptually) when you sample ideally is that you multiply you analog signal with the sampling function.

$$\begin{align} x_\mathrm{s}(t) &= x(t) \ T\,\mathrm {III}_T (t) \\ &= x(t) \ T\sum\limits_{n=-\infty}^{\infty} \delta(t-nT) \\ &= T\sum\limits_{n=-\infty}^{\infty} x(t) \, \delta(t-nT) \\ &= T\sum\limits_{n=-\infty}^{\infty} x(nT) \, \delta(t-nT) \\ \end{align}$$

now that demonstrates sampling. you have thrown away all of the information about $x(t)$ (by multiplying by zero) except for the values of $x(t)$ at the sampling instances, $t=nT$.

now in the frequency domain, it looks like this:

$$\begin{align} x_\mathrm{s}(t) &= x(t) \ T\,\mathrm {III}_T (t) \\ &= x(t) \ T\sum\limits_{n=-\infty}^{\infty} \delta(t-nT) \\ &= x(t) \ T\sum\limits_{k=-\infty}^{\infty} \tfrac{1}{T} e^{j 2 \pi (k/T) t } \\ &= \sum\limits_{k=-\infty}^{\infty} x(t) \, e^{j 2 \pi (k/T) t } \\ \end{align}$$

now, if you were to apply the continuous Fourier Transform to both sides:

$$\begin{align} \mathscr{F}\Big\{ x_\mathrm{s}(t) \Big\} &= \mathscr{F}\Big\{ x(t) \ T\,\mathrm {III}_T (t) \Big\} \\ &= \sum\limits_{k=-\infty}^{\infty}\mathscr{F}\Big\{ x(t) \, e^{j 2 \pi (k/T) t } \Big\} \\ \end{align}$$

if we define

$$ X(f) \triangleq \mathscr{F}\Big\{ x(t) \Big\} $$ and $$ X_\mathrm{s}(f) \triangleq \mathscr{F}\Big\{ x_\mathrm{s}(t) \Big\} $$

then you will see that

$$\begin{align} X_\mathrm{s}(f) &= \sum\limits_{k=-\infty}^{\infty} \mathscr{F}\Big\{ x(t) \, e^{j 2 \pi (k/T) t } \Big\} \\ &= \sum\limits_{k=-\infty}^{\infty} X(f - \tfrac{k}{T}) \\ \end{align}$$

so you can see that the spectrum $X_\mathrm{s}(f)$ of the sampled signal $x_\mathrm{s}(t)$ is periodic with period $\frac{1}{T}$:

$$\begin{align} X_\mathrm{s}(f + \tfrac{1}{T}) &= \sum\limits_{k=-\infty}^{\infty} X(f + \tfrac{1}{T} - \tfrac{k}{T}) \\ &= \sum\limits_{k=-\infty}^{\infty} X(f - \tfrac{k-1}{T}) \\ &= \sum\limits_{m=-\infty}^{\infty} X(f - \tfrac{m}{T}) \\ &= X_\mathrm{s}(f) \\ \end{align}$$

so the the spectrum is periodic with the sampling frequency $\frac1{T}$ as the period or spacing of these many copies of the original spectrum.

for the DTFT, we just scale that sampling frequency to be $2 \pi$.