# What is the relationship between the PSD of a continuous signal and the PSD of its periodically sampled one?

In Oppenheim's "discrete time signal processing", it says,

How is the equation (10.65) derived?

PS: you may try to derive it with Fourier transform of the original and the sampled signals. But i think it is not appropriate. Because when you are considering about the PSD of a signal, the assumption will be the fourier transform of the signal doesn't exist...

the definition of the PSD of a signal actually comes with a truncated signal:

the same is for the discrete version:

I guess there may exist something like... "sampling a random process" theory?

Regards.

If you sample a finite-power continuous-time random process $x(t)$ you get a discrete-time random process $y_k$. If $x(t)$ is wide-sense stationary (WSS) you get for the autocorrelation function of $y_k$

$$R_y(k,l)=E\{X(kT)X^*(lT)\}=R_x((k-l)T)=R_x(mT)$$

where $T$ is the sampling period. Obviously, $y_k$ is also WSS (it only depends on the difference $m=k-l$), and its autocorrelation function is a sampled version of the autocorrelation $R_x(\tau)$ of the original continuous-time process. Since the power spectrum is the Fourier transform of the autocorrelation function, the relation between the two power spectra is the same as the relation between the spectrum of a continuous signal and its sampled version:

$$S_y(e^{j\omega T})=\frac{1}{T}\sum_{k=-\infty}^{\infty}S_x(\omega-2\pi k/T)\tag{1}$$

Since the anti-aliasing filter has a cut-off frequency of $\pi/T$ it is sufficient to consider the interval $-\pi/T\le\omega<\pi/T$ (i.e. just consider the term $k=0$ in (1)):

$$S_y(e^{j\omega T})=\frac{1}{T}S_x(\omega)\tag{2}$$

This equation is equivalent to Eq. (10.65) in your question. Just note that I've used $\omega$ in a different way (analog frequency in radians), whereas Oppenheim uses $\omega$ as the normalized angular frequency. So with $\omega'=\omega T$ you finally get

$$S_y(e^{j\omega'})=\frac{1}{T}S_x\left(\frac{\omega'}{T}\right)$$

• wow!! cool!! thank you very much @Matt. It is very clear!! But I think there should be a 2 pi before the right hand of the equation(1),because the unit for frequency is radians... Or is there something wrong with my reasoning? ref:link – alexyangfox Jul 15 '14 at 9:10
• @alexyangfox: You're welcome! As for the factor $2\pi$, I'm sure that Eq. (1) is correct, but think I understand why you think it isn't. You're right that the Fourier transform of an impulse train $\sum_k\delta(t-kT)$ is $2\pi/T\sum_k\delta(\omega-2\pi k/T)$ but the factor $2\pi$ is canceled afterwards because multiplication in the time domain corresponds to convolution in the frequency domain times $1/(2\pi)$ if we use $\omega$ as the frequency variable. – Matt L. Jul 15 '14 at 9:31
• Fantastic!! That's just What I have missed!!! Thank you very very much... :).. Good luck.. – alexyangfox Jul 15 '14 at 9:58