# Unit Impulse Function

Discussion in 'Homework Help' started by dpizano, Sep 5, 2013.

1. ### dpizano Thread Starter New Member

Sep 5, 2013
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Hi Guys, I do not understand the solution to this question

∫t ζ(t-3)dt=-3

Limits -5,-2 ; -5 being the lower limit

I know

ζ(n)= 0 for every n ≠ 0

2. ### dpizano Thread Starter New Member

Sep 5, 2013
3
0
I think I understand. Since -3 is between -5 and -2, it makes the impulse function 1 and 1*t which is -3 = -3

One more question

What is the difference when the integrals limits are from -infinity to t compared to -infinity to infinity???

3. ### WBahn Moderator

Mar 31, 2012
20,247
5,758
But the unit impulse function doesn't fire until t=3, so for it is zero for all values of t between the limits of your integral.

Are you sure it isn't ∫t ζ(t+3)dt=-3 ?

Let's assume that it is. The way to think about it is that you have three regions of integration. -5 to -3-, 3- to -3+, and -3+ to -2 where -3- means an infinitesimal amount less than -3 and -3+ means an infinitesimal amount more than -3.

For the first and third regions, the impulse function is identically zero, so they go away.

For the second, the value of t isn't changing (except by an infinitesimal amount that we can make zero in the limit) and so it is effectively a constant that can be brought outside the integral. The integral itself evaluates to 1. So the result is just the value of the function at the point that the impulse fired.

4. ### dpizano Thread Starter New Member

Sep 5, 2013
3
0
Thanks Wbahn

The limits are definitely -5 and -2; I think the impulse function fires at -3 and 1*(t); where t=-3 is the reason the answer for the problem is -3.

I am still semi confused about the difference between -infinity to t and -infinity to infinity

5. ### WBahn Moderator

Mar 31, 2012
20,247
5,758

ζ(t-3)

At t = -3, this is ζ(-3-3) = ζ(-6)

This most certainly does NOT fire at t = -3.

If your limits are from -∞ to +∞ (or any other pair of constant limits), then the integral is simply a number (unless it has variable other than the variable of integration).

But if your integration limits contain a variable, then the result is a function in that variable.

Consider the following analogy.

You get paid an amount of money each day that is equal to $1 for each day of the year up to that day. So on Jan 1st you get$1, on Jan 10th you get $10, and on Dec 31st you get$365 (ignoring leap years here). You get nothing before the beginning of this year and nothing after the end of this year.

So the function that determines how you much you get pay on any particular day is

p(d) = (\$1/day)*d

where d is the number of days that have elapsed.

Now, this is a discrete function and so we have to talk about sums and not integrals, but the ideas are comparable so we will use the integral terminology. If you integrate from -∞ to +∞, you will get a number which will be the total amount paid to you over all time. But if you integrate from -∞ to d, you will get the total amount paid to you up through day d and it will be as a function of the day.

What should be done, both to be precisely correct and to help avoid confusion, is that the integrand should use a dummy variable that is different that the variable in the limits. Usually, for functions involving time, the integrand variable is substituted out for tau.

So you might have

$
f(t) \ = \ \int_{-\infty}^{t} g(\tau)\zeta (\tau-t_o) d \tau \ = \ g(t_o)U(t-t_o)
$