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

Question

A periodic signal $x(t)$ of period $T_0$ is given by $ x(t) = \begin{cases} 1, & |t|<T_1 \\ 0, & T_1<|t|<\frac{T_0}{2} \end{cases} $. The dc component of $x(t)$ is

Answer 1

Step 1: Let $ g(t) = x(t+5) $, which is the signal $ x(t) $ shifted so it occurs $ 5 $ units earlier in time.
Step 2: Recall the sifting/shifting identity for convolution with an impulse: $ g(t) * \delta(t-t_0) = g(t-t_0) $; here $ t_0 = 7 $, so the result is $ g(t-7) $.
Step 3: Substitute back $ g(t-7) = x\big((t-7)+5\big) $, and simplify the argument. \[ \boxed{x(t+5)*\delta(t-7) = x(t-2)} \]

Answer 2

Another way to view this is to track a single reference point through the convolution rather than applying the identity abstractly.

Option $ x(t-12) $: Track where the "marker" at $ x(t) $'s origin (where its argument is zero) ends up: for $ g(t) = x(t+5) $, this marker sits at $ t=-5 $; convolving with an impulse at $ 7 $ moves every point by $ +7 $, so the marker moves to $ -5+7=2 $, meaning the result peaks at $ t=2 $, i.e., it is $ x(t-2) $, not $ x(t-12) $, which would place the marker at $ t=12 $.
Option $ x(t+12) $: This would place the marker at $ t=-12 $, which does not match the tracked location of $ t=2 $ found by shifting the reference point by $ +7 $.
Option $ x(t-2) $: Tracking the reference point of $ g(t)=x(t+5) $, which sits at $ t=-5 $ (since $ g(-5) = x(0) $), and shifting it by the impulse location $ +7 $, lands it exactly at $ t = -5+7 = 2 $; a signal whose defining reference point sits at $ t=2 $ is precisely $ x(t-2) $, since $ x(t-2) $ reaches its own reference point when $ t-2=0 $, i.e. $ t=2 $.
Option $ x(t+2) $: This would place the reference point at $ t=-2 $, which does not agree with the tracked value of $ t=2 $.

Following where the signal's own reference point physically moves to after the shift confirms one particular option.

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

Answer 3

Another way to view this is to track a single reference point through the convolution rather than applying the identity abstractly.

Option $ x(t-12) $: Track where the "marker" at $ x(t) $'s origin (where its argument is zero) ends up: for $ g(t) = x(t+5) $, this marker sits at $ t=-5 $; convolving with an impulse at $ 7 $ moves every point by $ +7 $, so the marker moves to $ -5+7=2 $, meaning the result peaks at $ t=2 $, i.e., it is $ x(t-2) $, not $ x(t-12) $, which would place the marker at $ t=12 $.
Option $ x(t+12) $: This would place the marker at $ t=-12 $, which does not match the tracked location of $ t=2 $ found by shifting the reference point by $ +7 $.
Option $ x(t-2) $: Tracking the reference point of $ g(t)=x(t+5) $, which sits at $ t=-5 $ (since $ g(-5) = x(0) $), and shifting it by the impulse location $ +7 $, lands it exactly at $ t = -5+7 = 2 $; a signal whose defining reference point sits at $ t=2 $ is precisely $ x(t-2) $, since $ x(t-2) $ reaches its own reference point when $ t-2=0 $, i.e. $ t=2 $.
Option $ x(t+2) $: This would place the reference point at $ t=-2 $, which does not agree with the tracked value of $ t=2 $.

Following where the signal's own reference point physically moves to after the shift confirms one particular option.

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

LIST-I	LIST-II
A. \( e^{-at}u(t), a>0 \)	I. \( \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)] \)
B. \( \cos \omega_0 t \)	II. \( \frac{1}{j\omega + a} \)
C. \( \sin \omega_0 t \)	III. \( \frac{1}{(j\omega + a)^2} \)
D. \( te^{-at}u(t), a>0 \)	IV. \( -j\pi[\delta(\omega - \omega_0) - \delta(\omega + \omega_0)] \)

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

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The Correct Option is C

Approach Solution - 1

Approach Solution -2

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