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Question A continuous-time sinusoid $x_a(t)$ with fundamental period $T_p = \frac{1}{F_0}$ is sampled at a rate $F_s = \frac 1 T$ to produce a discrete-time sinusoid $x(n) = x_a(nT)$.

  1. Show that $x(n)$ is periodic if $\frac{T}{T_p} = \frac K N$ (that is, $\frac{T}{T_p}$ is a rational number)
  2. If $x(n)$ is periodic, what is the fundamental period $T_p$ in seconds?

I feel there is some ambiguity in the question itself. I assume $N$ here refers to the period of the sampled discrete-time signal.

I know how to prove that a discrete-time signal is periodic only if its relative frequency is a rational number. But the concept of periodicity in both continuous-time as well as sampled signal confused me.

So here is what I have tried:

Consider a continuous time sinusoid, $$x_a(t) = \cos(2 \pi Ft) \tag{1}$$ For this signal to be continous with a fundamental period $T_p$, it must follow that $$F=\frac{k_1}{T_p} \tag{2}$$ for some $k_1$ which is an integer.

Now that this signal is sampled with samppling rate $F_s$, the resulting discrete-time signal will be: $$x(n) = \cos(2\pi f n)\tag{3}$$ where, $$f = \frac{F}{F_s} \tag{4}$$ This sampled signal will be periodic with period $N$ iff $$f = \frac{k_2}{N} \tag{5}$$ for some $k_2$ which is also an integer.

Using equations (2), (4) and (5), it follows $$\frac{T}{T_p} = \frac{k_2}{k_1 N} \tag{6}$$ If I redefine $\frac{k_2}{k_1}$ to be $K$, then equation (6) becomes $$\frac{T}{T_p} = \frac{K}{N} \tag{7}$$ which proves the first part.

Now that the fundamental period $T_p$ has been asked, solving for $T_p$ in equation (6) gives $$T_p = \frac{k_1 T N}{k_2} \tag{8}$$

Since both discrete time period and continuous time periods are in the same place, this has made me confused whether my approach to the solution is correct or not. Could you please verify that the approach is correct? Or if anything is wrong, could you point that out to me? Thanks.

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  • $\begingroup$ I think this has already been answered but I can't find it right now... anyway, this is related: dsp.stackexchange.com/q/58452/11256 $\endgroup$ – MBaz Sep 8 at 17:51
  • $\begingroup$ @MBaz I checked the link you provided, but it doesn't answer my question. As I have already pointed out in the question, I understand and can prove that a sampled signal will be periodic iff its relative frequency is a rational number. My problem is with this particular proof however where both continuous as well as discrete time periods are involved. :) $\endgroup$ – bikalpa Sep 8 at 18:09
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You might be making this problem harder than it is by thinking in terms of frequency rather than periods. Your sinusoid is $$x(t) = \cos\left(2\pi \frac{t}{T_p}\right)$$ and its samples are $$x[n] = x(nT) = \cos\left(2\pi \frac{nT}{T_p}\right).$$ The samples form a discrete-time periodic sequence of period $N$ if \begin{align} x[n] &= x[n+N]~\forall\, n\\ &{\Large \Downarrow}\\ \cos\left(2\pi \frac{nT}{T_p}\right) &= \cos\left(2\pi \frac{(n+N)T}{T_p}\right)\\ &{\Large \Downarrow}\\ \cos\left(2\pi \frac{nT}{T_p}\right)&= \cos\left(2\pi \frac{nT}{T_p} + 2\pi \frac{NT}{T_p}\right) \end{align} where the last equality will hold only if $\displaystyle\frac{NT}{T_p}$ equals some integer $k$. This is equivalent to the OP's desired requirement that $$\frac{T}{T_p} = \frac kN$$ for some integers $k$ and $N$.

So, what is the fundamental period of $x[n]$? That is, what is the smallest integer $N_0$ such that $\displaystyle\frac{N_0T}{T_p}$ is an integer? Well, suppose that the ratio $\displaystyle\frac{T}{T_p}$ is the rational number $\frac{k_1}{k_2}$ where $k_1$ and $k_2$ are relatively prime integers. Then, $N_0 = k_2$ and $\displaystyle\frac{N_0T}{T_p} = k_1$. Thus, considering $x[n]$ as a periodic discrete-time sequence of samples spaced $T$ seconds apart -- think of it as the impulse train $\sum_n x(nT)\delta(t-nT)$ -- gives a (continuous-time) period of $k_2T$ seconds for the sequence/impulse train. Note that one period of the discrete-time sequence spans $k_1$ periods of the sinusoid of period $T_p$.

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  • $\begingroup$ Thanks for this simple yet helpful answer. Now for the second part of the question, $T_p = \frac{NT}{k}$ seconds right? Are there any restrictions to what the value of integer $k$ should be, for finding the value of $T_p$ in seconds? $\endgroup$ – bikalpa Sep 10 at 3:44
  • $\begingroup$ Here the continuous time period $N_0 T$ as you implied will just be the period of continuous time signal but may or may not be its fundamental period, right? And also since one time period of discrete signal spans $k_1$ periods of the sinusoid of period $T_p$, the fundamental period of continuous signal will be $1/k_1$ of the period $N_0 T$. Am I correct? $\endgroup$ – bikalpa Sep 11 at 4:30
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As far as I understand, the sampling interval $T=1/F_s$ is given, and, hopefully, the period $N$ of the sampled signal. Now the only thing we can say about the period $T_p$ of the continuous-time signal is that it satisfies

$$T_p=\frac{NT}{k},\qquad k\in\mathbb{Z}^+\tag{1}$$

I.e., the period of the discrete time signal measured in units of time $(NT)$ is an integer multiple of the period $T_p$ of the continuous-time signal.

EDIT: Now that I see the exact formulation of the exercise, my guess is that they might mean the period of the discrete-time signal $N$, or $NT$ (in seconds). If they really ask for the period $T_p$ of the continuous-time signal, then see my answer above.

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  • $\begingroup$ I have edited my question to add the exact formulation of the exercise. Please review it. $\endgroup$ – bikalpa Sep 9 at 3:25
  • $\begingroup$ I'm having difficulty understanding the phrase 'the period of discrete time signal measured in actual time'. Could you elaborate it, please? $\endgroup$ – bikalpa Sep 9 at 3:29
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    $\begingroup$ The period of the discrete-time signal is $N$ samples, and those $N$ samples take $NT$ seconds, that's what I mean by "period in actual time". $\endgroup$ – Matt L. Sep 9 at 5:35

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