Convolution of shifted signal [closed]

If $$y(t) = x(t)*h(t)$$, then what is the expression for $$y(t+a)$$?

Is it $$x(t+a)*h(t+a)$$ or $$x(t+a)*h(t)$$?

closed as unclear what you're asking by MBaz, Laurent Duval, Matt L., lennon310, Dan BoschenOct 13 '18 at 21:38

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• Please show what work you have done in your own attempts to prove this – Dan Boschen Oct 12 '18 at 12:55
• It is x(t+a)*h(t) or x(t)*h(t+a) hint: x(t+a)*h(t+b)=y(t+a+b). – Ch.Siva Ram Kishore Oct 12 '18 at 23:43
• The convolution of $y(t+a)$ by what? – Laurent Duval Oct 13 '18 at 0:14
• If $y(t)$ is the output of an LTI system with impulse response $h(t)$ whose input is $x(t)$, then $y(t+a)$ is just $y(t)$ advanced in time by $a$ and so, apply time invariance property to figure out what the input signal needs to be in order to produce $y(t+a)$ as the output. – Dilip Sarwate Oct 13 '18 at 3:59

From your confusion of $$x(t+a) \star h(t+a)$$ vs $$h(t) \star x(t+a)$$ I guess that a little help on the argument manipulations on functions and convolutions could be appropriate here, moving in simple examples:

First let us express the usual simplistic case. Consider the relation: $$y(t) = h(t) \cdot x(t) + g(t) \tag{1}$$

then manipulations on the argument $$t$$ is applied to all functions on both sides

$$y(t+a) = h(t+a) \cdot x(t+a) + g(t+a)$$

or an arbitrary transform on $$t$$ would similary be: $$y(\phi(t)) = h(\phi(t)) \cdot x(\phi(t)) + g(\phi(t))$$

Now consider the case where two functions convolved to produce the third: $$y(t) = \int_{-\infty}^{\infty} h(\tau) x(t-\tau) d\tau$$

which is abbreviated as $$y(t) = h(t) \star x(t) \tag{2}$$

Now be careful to interpret the case-2. The variable $$t$$ shows in all functions as an argument but you may not apply the transform on $$t$$ as you did with the case-1, so assume you have an arbitrary transform on $$t$$ as $$\phi(t)$$ then

$$y(\phi(t)) \neq h(\phi(t)) \star x(\phi(t)) \tag{3}$$

For example, as in your case, if $$\phi(t) = t+a$$ then you get

$$y(t+a) \neq h(t+a) \star x(t+a)$$ but $$y(t+a) = h(t) \star x(t+a) = h(t+a) \star x(t)$$

The justification of this can (only) be seen when you consider the integral definition of the convolution operator:

\begin{align} y(t) &= h(t) \star x(t) \\ & = \int_{-\infty}^{\infty} h(\tau) x(t-\tau)d\tau \\ y(t+a) & = \int_{-\infty}^{\infty} h(\tau) x((t+a)-\tau)d\tau \\ & = h(t) \star x(t+a) \\ \end{align}

Note that since the live variable inside the integral only happens in one function ($$x(t-\tau)$$ in this case) then a change in $$t$$ will only affect one of them and you get:

$$y(t+a) = h(t) \star x(t+a)$$ or from commutativity of convolution you get $$y(t+a) = h(t+a) \star x(t)$$

So this provides the answer you were looking for. However, it's not over. Because the following case represents an exception:

$$y(-t) \neq h(t) \star x(-t)$$ but $$y(-t) = h(-t) \star x(-t) \tag{4}$$

So how to see this case-4. Again, using the integral definition :

Assuming that $$y(t) = h(t) \star x(t)$$, then compute the convolution between two new signals $$g(t)=h(-t)$$ and $$z(t)=x(-t)$$ as: \begin{align} w(t) &= g(t) \star z(t) \\ & = \int_{-\infty}^{\infty} g(\tau) z(t-\tau)d\tau &g(\tau)=h(-\tau),z(t-\tau)=x(-(t-\tau)) \\ & = \int_{-\infty}^{\infty} h(-\tau) x(-(t-\tau))d\tau &\text{ let } \tau'=-\tau \\ & = -\int_{\infty}^{-\infty} h(\tau') x(-(t+\tau'))d\tau' &\text{ replace } \tau' \text{ with } \tau \\ & = \int_{-\infty}^{\infty} h(\tau) x(-t-\tau) d\tau \\ & = y(-t) \\ \end{align}

hence we conclude that $$h(-t) \star x(-t) = w(t) = y(-t)$$. As stated before, you must always consult to the (explicit) integral definition to decide on the correct functions used in the convolution operator.