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 \(x[n]\)?
\[ x[n]=x_c(nT_s), \]
where \(n\) is an integer.
Show Hint
- \(x[n]\) 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
The Correct Option is D
Solution and Explanation
We can also settle this by testing one rational case and one irrational case directly, instead of deriving the general condition first.
Rational case: Let $T=3$ and $T_s=2$, so $T/T_s=3/2$, a rational number. We look for an integer $N$ and integer $k$ with $NT_s=kT$, that is $2N=3k$. Taking $N=3,k=2$ works: $2(3)=6=3(2)$. So $x[n+3]=x_c((n+3)(2))=x_c(2n+6)$. Since $6=2\times3=2T$, and $x_c$ has period $T=3$, we get $x_c(2n+6)=x_c(2n)=x[n]$. So the sequence repeats with period $N=3$: periodic, as claimed for the rational case.
Irrational case: Let $T=1$ and $T_s=1/\sqrt2$, so $T/T_s=\sqrt2$, which is irrational. Suppose, for contradiction, some integer $N$ and integer $k$ satisfy $NT_s=kT$, that is $N/\sqrt2=k$, so $\sqrt2=N/k$. This says $\sqrt2$ equals a ratio of two integers, which is impossible since $\sqrt2$ is irrational. So no such $N$ exists, and $x[n]$ never repeats: not periodic, as claimed for the irrational case.
These two worked examples rule out option (A) and (B), since periodicity clearly fails in the irrational case, and rule out option (C), since periodicity clearly holds in the rational case. The only statement consistent with both examples is that $x[n]$ is periodic exactly when $T/T_s$ is rational.
\[ \boxed{x[n]\ \text{is periodic iff}\ T/T_s\ \text{is rational}} \]