# Integration of unit-step ,rect & delta functions

Discussion in 'Homework Help' started by mo2015mo, Oct 25, 2013.

1. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
Hi guys ...

I have some questions about Integration of unit-step ,rect & delta functions

I have been to solve it by plotting functions as attached photo..
i need u to confirm from my solution and guide me to correct method if they are wrong

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2. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
It's a bit hard to read your explanation for your first answer, but it looks like it says that the result is zero because there are two impulse functions. Unless you are claiming that the integral of any integrand that contains two impulse functions is zero, then this is not a good enough answer. Need to be specific and defensible.

I don't see a solution for the second questions.

Third question looks fine.

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3. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
thanQ u Mr. WBahn

4. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
You've got it pretty much right. A fine point is that

δ(t-T)*f(t)

is not equal to f(T), but rather it contributes f(T) to the integral that includes T in its limits.

Here's one for you:

What is

I = ∫δ(t-T1)*f(t)*δ(t-T2)dt

If the limits of integration include T1 and T2 and if T1 and T2 happen to be equal?

5. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
I = ∫δ(t-T1)*f(t)*δ(t-T2)dt = ∫δ(t-T1)*f(T1+T2)*δ(t-T2) = 0
if equal I = ∫δ(t-T1)*f(2T)*δ(t-T2)dt =0

6. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
NO!!!!

Think about this. Do you really mean that if T1=10s and T2=20s, that the integral all of a sudden depends on what the function happens to be at t=30s? Does that make sense?

If T1 and T2 are equal, then you have

I = ∫δ(t-T1)*f(t)*δ(t-T1)dt

So the real question comes down to what the product of two delta functions that lie on top of each other and, more to the point, what happens when you integrate across them.

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7. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
as we know the product of two delta functions

∫ δ (a-t) δ (b-t) dt = δ (a-b)

I = ∫δ(t-T1)*f(t)*δ(t-T1)dt = ∫δ(0)*f(t)dt = f(0)

8. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Where does this "knowledge" come from?

Again, ask if it makes any sense whatsoever?

Consider the pieces:

∫ δ (a-t) δ (b-t) dt = δ (a-b)

δ (a-t)

This is a function that is identically zero EVERYWHERE except at t=a.

δ (b-t)

This is a function that is identically zero EVERYWHERE except at t=b.

δ (a-b)

What does this even mean? What is δ(7) ? Not δ(t-7), but δ(7) ?

And why is it δ(a-b) and not δ(b-a) ?

But leaving that aside, let's say that a=6s and b=10s, does it make any sense for the product of two functions that are BOTH identically zero everywhere except at t=6s and t=10s, to result in a δ(-4s)?

Go back and look at the definition of the delta function.

9. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
oohhhhh, All my answers were unsuccessful

10. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
It's very possible you got some of the answers "right", but through faulty reasoning. But, if so, then the discussion has been very good because now you can get the your mind wrapped around the topic properly.

One of the best ways to do this is to look at the definition of a couple of approximations to the delta function and see how they behave in the limit as they become true delta functions.

The simplest of these is a rectangular pulse that is defined to be:

f(t) = (1/To) for -To/2 < |t| < +To/2
and
f(t) = 0 elsewhere

Work with that as your delta function and evaluate the integral. Then take the limit as To goes to 0.

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11. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
f(t) = (1/To)t
g(t)=d/dt(f(t)) = 1/To lim (To approaches to 0) g(t) = δ(t-to)
δ(t-to) = 0 for t ≠ to
and ∞ for t = to
and the area under a unit impulse function is equal to unity.
the integral of impulse function is 1 also.

so the integral of two impulse function at to is equal 1 .

12. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
I'm not following this at all.

Let's do this step by step and not just use a bunch of hand waving along the way.

We want to evaluate

$
I \, = \, \int_{-\infty}^{+\infty} \delta(t-T_1) f(t) \delta(t-T_2) dt
$

We want to use the approximation

$
\delta(t) \, = \, \frac{1}{T_0} \ \ for \ |t| \ < \ \frac{T_0}{2}
$

In the event that T1 = T2, we have

$
I \, = \, \int_{-\infty}^{+\infty} \delta(t-T_1) f(t) \delta(t-T_1) dt
$

Restricting the integral to where it is not forced to zero, we have

$
I \, = \, \int_{T_1-T_0/2}^{T_1+T_0/2} \frac{1}{T_0} f(t) \frac{1}{T_0} dt
\
I \, = \, \int_{T_1-T_0/2}^{T_1+T_0/2} \frac{1}{T_0^2} f(t) dt
$

In the limit that T_0 goes to zero, this becomes

$
I \, = \, \frac{1}{T_0^2} f(T_1) [(T_1+T_0/2)-(T_1-T_0/2)]
$

$
I \, = \, \frac{1}{T_0^2} f(T_1) T_0
$

$
I \, = \, \frac{f(T_1)}{T_0}
$

In the limit as T0 goes to zero, this blows up.

The bottom line is that the product of two delta functions is zero everywhere is they do not coincide and result in a singularity if they do.

One way to think of this is that a single delta function integrates to the value of the rest of the integrand evalated at the point at which the delta function "fires". But if the rest of the integrand contains a delta function that fires at the same point, then the value of the rest of the integrand is infinite at that point.

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13. ### mo2015mo Thread Starter Member

May 9, 2013
157
1
100% understood it
really thanks u Mr. Wbhan

Apr 19, 2012
748
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15. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Basically, as T0 gets smaller, then the value of f(t) looks more an more like a constant.

Consider f(t) = (1e6/s)*t and T1 being 10s. The value of f(T1) would be 10,000,000. So if T0 were 1s, the value of f(t) would vary by 1,000,000 between t=9.5s and t=10.5s.

But what about when T0 is 1ns? Now it varies by only 0.01. Pick whatever amount of variation it the maximum allowed before you would be willing to consider f(t) to be constant with a value of f(T1) within that window and I can tell you how small T0 has to be in order to satisfy you. Once we reach that point, I can evaluate f(t) at t=T1 and take it outside the integral. Then I just have the integral of dt between two limits.

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16. ### anhnha Active Member

Apr 19, 2012
748
44
That makes sense. However, I am still a bit confused. Even if T0 is very very small, f(t) is not a constant although its values almost not change over that range(T1 -T0/2 to T1 + T0/2). And therefore, how that can yield a precise result?

17. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
In the limit that T0 goes to zero, it doesn't change at all.

Even for T0 being small but not zero, remember that it is arbitrarily small. We can drive the error associated with f(t) changing over the interval to be less than any non-zero threshold we choose to pick. Want the fractional error to be less then the equivalent of the mass of one electron compared to the mass of the known universe? Fine. We can compute a value of T0 that will be small enough to make that happen. And then we can choose to make the value of T0 one one-billionth of that.

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