Question

Convolution of $x(t+5)$ with impulse function $\delta(t-7)$ is equal to

• A. $x(t-12)$
• B. $x(t+12)$
• C. $x(t-2)$
• D. $x(t+2)$

Hint:

Convolution with $\delta(t-t_0)$ simply shifts the signal by $t_0$.

Answer 1

The Correct Option is C

Step 1: Use convolution property of delta function.
\[ x(t)*\delta(t-t_0)=x(t-t_0) \]
Step 2: Apply to the given signal.
\[ x(t+5)*\delta(t-7)=x\big((t-7)+5\big) \]
Step 3: Simplify.
\[ x(t-2) \]
Step 4: Final conclusion.
The result of the convolution is $x(t-2)$.

Answer 2

Convolving any signal with a shifted impulse simply relocates the signal by that shift amount. Here the task is to convolve \( x(t+5) \), which is \( x(t) \) advanced (shifted left) by \( 5 \), with \( \delta(t-7) \), an impulse placed at \( t = 7 \).

1. Option \( x(t-12) \): This would result from shifting \( x(t+5) \) by \( -17 \) overall (since \( t+5-17=t-12 \)), which does not correspond to convolving with an impulse at \( t=7 \); it does not match the correct net shift.
2. Option \( x(t+12) \): This would follow only if the two shifts (the internal \( +5 \) and the impulse's \( -7 \), i.e., a shift of \( +7 \)) combined by simple addition of magnitudes in the wrong direction; but shifting by an impulse at \( t_0 \) replaces \( t \) with \( t - t_0 \) inside the argument, so this does not follow the correct substitution rule.
3. Option \( x(t-2) \): Using the identity \( g(t) * \delta(t-t_0) = g(t-t_0) \) with \( g(t) = x(t+5) \) and \( t_0 = 7 \), the result is \( g(t-7) = x\big((t-7)+5\big) = x(t-2) \). This matches exactly, correctly combining the internal shift of \( +5 \) with the impulse shift of \( -7 \) to give a net shift of \( -2 \).
4. Option \( x(t+2) \): This would arise from an error in the sign of the impulse shift, treating \( \delta(t-7) \) as though it shifted the signal by \( +7 \) instead of by \( -7\) relative to its argument, which contradicts the standard convolution-with-impulse identity.

Applying the impulse's shifting property correctly to the already-shifted signal \( x(t+5) \) gives one specific net shift.

So the correct answer is \( x(t-2) \).