Question:
Let \(x_c(t)\) be any continuous-time periodic signal with period \(T\). It is sampled uniformly with a sampling period \(T_s\) where \(T_s\neq T\), resulting in the discrete sequence
\[ x[n]=x_c(nT_s), \]
where \(n\) is an integer.
Which one of the following statements is correct about \(x[n]\)?
Let \(x_c(t)\) be any continuous-time periodic signal with period \(T\). It is sampled uniformly with a sampling period \(T_s\) where \(T_s\neq T\), resulting in the discrete sequence
\[ x[n]=x_c(nT_s), \]
where \(n\) is an integer.
\[ x[n]=x_c(nT_s), \]
where \(n\) is an integer.
Which one of the following statements is correct about \(x[n]\)?
Show Hint
A sampled version of a periodic continuous-time signal repeats only when the sampling period fits an exact whole number of times into a whole number of source periods.
Updated On: Jul 20, 2026
- \(x[n]\) will always be periodic with period \(T/T_s\) for all values of \(T/T_s\)
- \(x[n]\) will always be periodic with period \(1\) for all values of \(T/T_s\)
- \(x[n]\) will never be periodic
- \(x[n]\) will be periodic if and only if \(T/T_s\) is a rational number
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Recall what it means for a discrete sequence to be periodic.
\(x[n]\) is periodic with period \(N\) (a positive integer) if
\[ x[n+N]=x[n]\quad\text{for all }n \]
Step 2: Substitute the sampling relation.
\[ x[n+N]=x_c\big((n+N)T_s\big)=x_c(nT_s+NT_s) \]
Since \(x_c(t)\) is continuous-time periodic with period \(T\), we have \(x_c(nT_s+NT_s)=x_c(nT_s)\) exactly when \(NT_s\) is an integer multiple of \(T\), that is,
\[ NT_s=kT\quad\text{for some integer }k \]
Step 3: Rearrange the periodicity condition.
\[ N=k\left(\frac{T}{T_s}\right) \]
For an integer \(N\) to exist satisfying this for some integer \(k\), \(T/T_s\) must equal a ratio of two integers, that is, \(T/T_s\) must be a RATIONAL number. If \(T/T_s=p/q\) in lowest terms, choosing \(k=q\) gives \(N=p\), an integer period.
Step 4: Show irrational \(T/T_s\) gives no periodicity.
If \(T/T_s\) is irrational, no pair of integers \(N,k\) (other than the trivial \(N=0\)) can satisfy \(N=k(T/T_s)\), because an irrational number can never equal a ratio of two integers. So \(x[n]\) never exactly repeats and is not periodic in this case.
Step 5: Check the given options.
Option (A) claims \(x[n]\) is always periodic with period \(T/T_s\) for every value of \(T/T_s\), but \(T/T_s\) need not even be an integer, and periodicity fails entirely when \(T/T_s\) is irrational, so (A) is false. Option (B) claims the period is always \(1\), which would mean \(x[n]\) is constant, not true in general. Option (C) is false because Step 3 shows periodicity does occur whenever \(T/T_s\) is rational. Only option (D) matches both Step 3 and Step 4.
Step 6: Final Answer.
\[ \boxed{x[n]\ \text{is periodic if and only if}\ T/T_s\ \text{is rational}} \]
\(x[n]\) is periodic with period \(N\) (a positive integer) if
\[ x[n+N]=x[n]\quad\text{for all }n \]
Step 2: Substitute the sampling relation.
\[ x[n+N]=x_c\big((n+N)T_s\big)=x_c(nT_s+NT_s) \]
Since \(x_c(t)\) is continuous-time periodic with period \(T\), we have \(x_c(nT_s+NT_s)=x_c(nT_s)\) exactly when \(NT_s\) is an integer multiple of \(T\), that is,
\[ NT_s=kT\quad\text{for some integer }k \]
Step 3: Rearrange the periodicity condition.
\[ N=k\left(\frac{T}{T_s}\right) \]
For an integer \(N\) to exist satisfying this for some integer \(k\), \(T/T_s\) must equal a ratio of two integers, that is, \(T/T_s\) must be a RATIONAL number. If \(T/T_s=p/q\) in lowest terms, choosing \(k=q\) gives \(N=p\), an integer period.
Step 4: Show irrational \(T/T_s\) gives no periodicity.
If \(T/T_s\) is irrational, no pair of integers \(N,k\) (other than the trivial \(N=0\)) can satisfy \(N=k(T/T_s)\), because an irrational number can never equal a ratio of two integers. So \(x[n]\) never exactly repeats and is not periodic in this case.
Step 5: Check the given options.
Option (A) claims \(x[n]\) is always periodic with period \(T/T_s\) for every value of \(T/T_s\), but \(T/T_s\) need not even be an integer, and periodicity fails entirely when \(T/T_s\) is irrational, so (A) is false. Option (B) claims the period is always \(1\), which would mean \(x[n]\) is constant, not true in general. Option (C) is false because Step 3 shows periodicity does occur whenever \(T/T_s\) is rational. Only option (D) matches both Step 3 and Step 4.
Step 6: Final Answer.
\[ \boxed{x[n]\ \text{is periodic if and only if}\ T/T_s\ \text{is rational}} \]
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