Let $x_c(t)=\cos(\omega_0t)$. This signal is sampled with $\omega_s$, which is greater than the Nyquist rate. It is then interpolated with a zero-order interpolator. The signal obtained is $y_c(t)$. What should be the relationship between $\omega_0$ and $\omega_s$ so that $y_c(t)$ is periodic? Also, find the relationship between the amplitudes of the first two non-zero Fourier coefficients of $y_c(t)$.

I guess that, for $y_c(t)$ being periodic, $Y(j\omega )$ should consist of just deltas. The problem is that I don't see how this could happen. Mathematically,

$$X_p(j\omega)=\frac{\omega_s}{2}\sum\limits_{k=-\infty}^\infty [\delta(\omega -k\omega_s- \omega_0)+\delta(\omega -k\omega_s+ \omega_0)]$$ $$H_0(j\omega )=\frac{2\pi}{\omega_s}\text{sinc}\left(\frac{\omega}{\omega_s}\right)e^{-j\frac{\pi\omega}{\omega_s}} $$

With $x_p(t)$ being the sampled signal and $h_0(t)$ the impulse response of the interpolator. So

$$Y_c(j\omega)=X_p(j\omega)\cdot H_0(j\omega )$$


2 Answers 2


I'd like to clear up some confusion in your question. First you state that for $y_c(t)$ to be periodic, its Fourier transform should consist only of Dirac delta impulses. That's not true, at least it's not sufficient. As an example, take a signal with Fourier transform

$$Y(j\omega)=\delta(\omega -1)+\delta(\omega - \sqrt{2})\tag{1}$$

The corresponding signal is not periodic even though its spectrum only consists of Dirac delta impulses. The reason is that the discrete frequency contributions of a periodic signal must be at integer multiples of the fundamental frequency, which is not the case for the spectrum given in $(2)$. So it's important to realize that a discrete spectrum is only necessary but not sufficient for periodicity.

The second confusion is that you don't see how $Y_c(j\omega)$ can only consist of Dirac impulses. Note that for any function $H(j\omega)$ (which is continuous at $\omega=\omega_0$), we have


So no matter which form $H_0(j\omega)$ takes, $Y_c(j\omega)$ will always be an infinite sum of Dirac impulses, where the weight of each impulse is determined by the value of $H_0(j\omega)$ at the respective frequency. So the question is not whether $Y_c(j\omega)$ is a sum of weighted Dirac impulses (it always is!), but whether this sum of Dirac impulses corresponds to a periodic signal.

So it is only about the relation of the frequencies of the Dirac impulses, not about their weights, which are determined by $H_0(j\omega)$. So for the determination of periodicity, $H_0(j\omega)$ is irrelevant, as already pointed out in Marcus Müller's answer.

Since you already know that the ratio $\omega_s/\omega_0$ must be rational for the sampled signal to be periodic, let me show you how to obtain the ratio of the amplitudes of the first two non-zero Fourier coefficients of $y_c(t)$. Since $\omega_s>2\omega_0$, the first frequency component of $y_c(t)$ is at $\omega_0$, and the second one is at $\omega_s-\omega_0$. As mentioned before, the weights of the Dirac impulses are given by $H_0(j\omega)$ evaluated at the respective frequencies. So the amplitude ratio of the first and second Fourier coefficient is


because $\sin(\pi-x)=\sin(x)$.

  • $\begingroup$ The periodicity condition is not met in $(1)$ since there is a irrational number and definitely $(1)$ is not periodic. In paragraph two you gave a reference to spectrum $(2)$ and I am bit confused here about what it refers. $\endgroup$
    – jomegaA
    Feb 10, 2020 at 11:25

You're thinking too complicated.

For a digital signal to be periodic, it simply has to repeat every $P\in\mathbb N$ samples.

Now, you have a continuous signal which is $\frac{2\pi}{\omega_0}$-periodic.

Obviously, if a period happens to be exactly "worth" a natural number $N\in\mathbb N$ sample times, the resulting sample from sampling it will repeat with a period of $N$.

More generally, if $M\in\mathbb N$ periods of the continuous signal are as long as $N$ sampling times, the digital signal will also be periodic.

So, as a general rule, one period of the input signal must be worth $\frac{N}{M}$ sampling times, or:

$$\begin{align*} \frac{2\pi}{\omega_0} &\overset != \frac NM \frac {1}{f_\text{sampl}}&\quad N,M\in \mathbb N\\ &= R\frac {1}{f_\text{sampl}} &\quad R\in \mathbb Q\text,\\ \text{which means that}\\ \frac{f_\text{sample}}{\omega_0} &= \frac {R}{2\pi}\text;\\ \text{with}\\ f_{sample}&=\frac{\omega_s}{2\pi}\\ \text{it follows that}\\ \frac{\omega_s}{\omega_0} &= R\text, \end{align*}$$

i.e. the relationship between signal frequency and sampling rate needs to be rational.

The fact that you have a zero-order interpolator afterwards doesn't change this consideration -- the interpolator's output will be periodic iff its input is.

  • $\begingroup$ "... the digital signal will also be continuous." I guess that should be "periodic" ... Otherwise, you're of course right. $\endgroup$
    – Matt L.
    Feb 8, 2016 at 20:43
  • $\begingroup$ @MattL. d'oh! yes. :) corrected that! $\endgroup$ Feb 8, 2016 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.