Question

Suppose signal $y(t)$ is obtained by the time-reversal of signal $x(t)$, i.e., $y(t) = x(-t)$,\(-\inftyLt;tLt;\infty\) Which of the following options is always true for the convolution of $x(t)$ and $y(t)$?

• A. It is an even signal
• B. It is an odd signal
• C. It is a causal signal
• D. It is an anti-causal signal

Hint:

Convolution of a signal with its time-reversed version always produces an even signal.

Answer 1

The Correct Option is A

Step 1: Understanding convolution with time-reversed signals. (i) Time Scaling Property:
The time scaling property states that: $$x(at) * y(at) = \frac{1}{|a|} x(t) * y\left(\frac{t}{a}\right)$$ Substituting $a = -1$: $$x(-t) * y(-t) = x(t) * y(t)$$ (ii) Commutative Property:
The commutative property is expressed as: $$x(t) * y(t) = y(t) * x(t)$$ From equation (1), we have: $$z(t) = y(t) * x(t) = x(-t) * y(-t) \tag{2}$$ Substitute $t = -t$: $$z(-t) = y(-t) * x(-t)$$ From the commutative property, we get: $$z(-t) = x(-t) * y(-t) \tag{3}$$ From equations (2) and (3): $$z(t) = z(-t)$$ Step 2: Applying properties of convolution. - Convolution of $x(t)$ with its time-reversed version $x(-t)$ results in a symmetric output. - The symmetry implies $z(t)$ is an even signal: $$z(t) = z(-t).$$
Hence, the correct option is (1) It is an even signal.