GATE 2018 IN Q.9: solution and answer

Official answer: DSource: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Time reversal keeps the period unchanged. Shifting $t$ by $T$ gives $x(-(t+T))=x(-t-T)$, which equals $x(-t)$ for every $t$ exactly when $T$ is a period of $x$. So $x(-t)$ has the same fundamental period as $x(t)$: $T_1=3\text{s}$.

2. A time shift keeps the period unchanged; time scaling by $a$ divides it by $|a|$. Shifting $t$ by $T$ gives $y(2(t+T)+1)=y(2t+1+2T)$, which equals $y(2t+1)$ for every $t$ exactly when $2T$ is a period of $y$. The smallest such $T$ satisfies $2T=3$, so $y(2t+1)$ has fundamental period $T_2=\frac{3}{2}=1.5\text{s}$. The $+1$ only shifts the signal and does not affect the period.

3. Periodicity of the sum. The ratio $\frac{T_1}{T_2}=\frac{3}{1.5}=2$ is rational, so $z(t)$ is periodic. Since $3=1\times T_1=2\times T_2$, both terms repeat after 3 s, so $3\text{s}$ is a period of $z(t)$. It is the LCM: $\text{LCM}(3,1.5)=3\text{s}$.

4. Rule out 1.5 s and 2 s. Every period of $z$ is an integer multiple of its fundamental period $T_0$, so $3=n{T}_0$ for some integer $n$. If $n$ were even, $1.5=\frac{n}{2}{T}_0$ would also be a period of $z$. But the $y(2t+1)$ term repeats every 1.5 s, so $$z(t+1.5)-z(t)=x(-t-1.5)-x(-t),$$ and this is not zero for all $t$, because $1.5\text{s}$ is shorter than 3 s, the fundamental period of $x$, so it is not a period of $x$. Hence $n$ is odd, and 1.5 s (B) is ruled out. 2 s (C) is also impossible, because $2\ne \frac{3}{n}$ for any integer $n$.

5. Rule out 1 s: it needs an exact cancellation. The only remaining option shorter than 3 s is $1\text{s}$ ($n=3$). It would require $x(-t-1)-x(-t)=-[y(2t+3)-y(2t+1)]$ for every $t$. That means the change in one term exactly cancels the change in the other, and the question gives no such relation between $x$ and $y$. For ordinary signals it fails. For example, $x(t)=y(t)=\cos \frac{2\pi t}{3}$ gives $z(t)=\cos \frac{2\pi t}{3}+\cos \frac{2\pi (2t+1)}{3}$. The two terms have different frequencies, so they cannot cancel, and the $\frac{1}{3}\text{Hz}$ term does not repeat after 1 s. (A pair can be built to cancel on purpose. $x(t)=\cos \frac{4\pi t}{3}+\cos 2\pi t$ and $y(t)=-\cos \frac{2\pi (t-1)}{3}$ both have fundamental period 3 s, yet they give $z(t)=\cos 2\pi t$, which has period 1 s. This is why the question can only be asking for the period that holds for any $x$ and $y$.)

6. Answer. For $x(t)$ and $y(t)$ with fundamental period 3 s and no special relation between them, the fundamental period of $z(t)$ is the LCM, $T_0=3\text{s}$. Option (D).