0
\$\begingroup\$

We know that any continuous time signal can be expressed as follows: $$x(t)=\int_{-\infty}^\infty x(τ)δ(t-τ)dτ$$ I came across a certain relation regarding linear time invariant systems . Using $$x(t)*δ(t)=x(t)$$ we get , since it's an LTI system : $$x(t)*δ(t-t_0)=x(t-t_0)$$ How did we get here? Is there a property I'm not aware of? I try to imagine the convolution graphically and since we have the x(t) how can the multiplication with the dirac function give x(t-t0) for every point in the convolution integral?

\$\endgroup\$
9
  • 2
    \$\begingroup\$ Just substitute t with t-t0 in the integral equation and you will get it. Even without solving the integral... \$\endgroup\$
    – Eugene Sh.
    Commented Jun 29, 2017 at 19:24
  • 1
    \$\begingroup\$ The dirac delta is also called 'the selection function'. You don't get x(t-t0) for every point in the convolution integral, just at t-t0, and zero everywhere else, so the whole convolution comes out to be x(t-t0). It's one of those mathematical things that I regard as a cheat, the dirac delta definition gives it this property, which is consistent when you flog through the convolution. Remember the dirac delta is not something you can draw, it's infinitely high at (0), but with a unity area under it, not very physical, very contrived. \$\endgroup\$
    – Neil_UK
    Commented Jun 29, 2017 at 19:31
  • \$\begingroup\$ @Neil_UK So you should consider i and even negatives as cheats. I call them abstractions. And very useful ones. \$\endgroup\$
    – Eugene Sh.
    Commented Jun 29, 2017 at 19:58
  • \$\begingroup\$ @EugeneSh. No, i, quarternions, and octonians are all useful, but the relation \$e^{i \pi}=-1\$ says more about the construction of the number system than it says about the symbols themselves, so that's a cheat as well in my terms. My complaint with the dirac delta is that it's designed to do a specific job, and it can't be drawn, only described. It's like the difference between infinities, it gets a bit, silly really. I'm happier with integration Lebegue style. \$\endgroup\$
    – Neil_UK
    Commented Jun 29, 2017 at 20:08
  • \$\begingroup\$ @Neil_UK It's a bit primitive to consider anything you can't draw as a non-existent or "cheat" in my opinion, sorry. Difference between infinities does exist. Quantum mechanics is working. Even though it is described using mostly these "cheats" and cannot be imagined or understood by most (if not all). The whole Math is a tool designed for a specific job. And the fact you can draw some stuff in it is completely accidental. \$\endgroup\$
    – Eugene Sh.
    Commented Jun 29, 2017 at 20:17

2 Answers 2

Reset to default
1
\$\begingroup\$

How can the multiplication with the dirac function give x(t-t0) of every point in the convolution integral?

It can't, that is what the integral does.

-- I try to imagine the convolution graphically.

This is very helpful, but inside the integral, it is unnecessary. Inside the integral it is just multiplication. The integral operation handles the infinite summation for all values of tau.

So I agree, the multiplication cannot get every point, that is why the integral is needed.

If you find a good video/gif of convolution some here, you can think of each frame as the multiplication for a particular tau. The combined effect is the integration.

Hope that helps connect the visual to the equation.

\$\endgroup\$
0
0
\$\begingroup\$

A convolution is simply reversing one of the signals w.r.t the vertical (y(t)) axis and then shifting it towards the right, left continously till it reaches infinity at both the ends.in that process the signals comes across the other unshifted signal once and crossesthrough it.while doing so, at each instant it will have some common area,the plot of this common area is the convolution signal output.Now, If we take the impulse function in the same way,the common area will be simply get the other stationery signal.but, if we take a shifted version of the impulse in the same way,again we get the same unshifted signal as the common area but,at a different instant of time compared to earlier output,the reason is that,initially we have shifted by some t0 the impulse and started the sliding ..so the difference in time of the outputs is equal to the difference of intial shift of the impulse.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.