# why i has that result..

Status
Not open for further replies.

#### transgalactic

##### Banned
why
$$\int_{0^+}^{t} (x+1)\delta (x)dx=0$$

delta is defined that in 0 it is infinity
and on x differs 0 its value is 0

there heard of some formula
i am not sure if its the right form
$$\int f(x)\delta (x)dx=f(0)$$

why i get 0
??

#### MrAl

##### Well-Known Member
why
$$\int_{0^+}^{t} (x+1)\delta (x)dx=0$$

delta is defined that in 0 it is infinity
and on x differs 0 its value is 0

there heard of some formula
i am not sure if its the right form
$$\int f(x)\delta (x)dx=f(0)$$

why i get 0
??

Hi there,

It's appears that you are investigating the properties of the impulse
function a little, and that's a good idea.

the following response has to be qualified by:
"If i understand you correctly".

Also, in the following discussion
"Intg[a to b]"
means:
"the integral from a to b of"
and the body of the integral follows.
Thus,
Intg[-inf to +inf] f(x) dx
simply means:
"the integral from minus infinity to plus infinity of f(x) dx".

First, is your first equation formed properly? Is it really supposed to
be equal to 1 instead of 0 ?

Second, your second equation looks like you are trying to take the
integral of the function of x times delta(x), and seeing if it is equal
to f(0).

To start with, there is a property of delta(x) that is of great interest here.
That is, that mathematically:
Intg[-inf to +inf] delta(x) dx = 1

and physically you already know that the function exists only at
t=0, where it is an impulse, and for other times t=tau the function
is equal to zero. This physical interpretation also leads us to find
that since f(x) is multiplied by delta(x), that multiplication leads to
a result that only involves f(0), because delta(x) only exists at t=0
and that multiplication result is zero otherwise. The constant f(0)
is singled out of all the other f(x) because of the temporal quality
of the impulse function.

Taking both of these interpretations at once, we can write:
Intg[-inf to +inf] f(x)*delta(x) dx is equal to:
Intg[-inf to +inf] f(0)*delta(x) dx, and now since f(0) is a constant
we can take that out of the integral and this becomes:
f(0)*Intg[-inf to +inf] delta(x) dx, and since this is just f(0) times
the integral we just looked at above we can now write:
f(0)*1
which is of course simply f(0), and so the following is true:
Intg[-inf to +inf] f(x)*delta(x) dx = f(0).

Now while all this is nice and neat, it doesnt help as much as the next
interpretation of the impulse i have been saving. The mathematical
and physical interpretation is interesting, but this next property is
even more useful because it says something about how the impulse
is of use in network analysis...

The impulse is the derivative of the step, or:
delta(t)=u'(t)

But even more interesting, the impulse response of a system is
equal to the derivative of the step response of a system!
This is interesting, because the delta(t) response of a system can be
found by first finding the step response, then taking the first
derivative. Petty nice huh?

Another property is also interesting, and that is that the unforced
response (also called natural response, or ZIR) is equal to the
impulse response.
Thus, when we convolve the impulse response of the network with
the excitation, we are simply finding the forced response of the system.

Last edited:
Status
Not open for further replies.

Replies
6
Views
1K
Replies
14
Views
2K
Replies
0
Views
430
Replies
4
Views
11K
Replies
14
Views
7K