Definition of convolution question

[transgalactic]

<br /> (f*g)=\int_{-\infty}^{+\infty}f(\tau)g(t-\tau)<br />

i was told that because of the definition of the function
we can substitute the itervals to
<br /> (f*g)=\int_{0}^{t}f(\tau)g(t-\tau)<br />


why??

 

[Cyosis]

What is the definition of the function you were provided?

 

[transgalactic]

a convolution function is 0 at t<0
and f(t)>0 at t>=0

i am not sure if this is a correct definition of a convolution function
??

 

[Mark44]

The two integrals are equal if f(\tau) = 0 for \tau <= 0 and for \tau > t.

I don't believe there is such a thing as a "convolution function." Convolution is an operation involving two functions, and defined as you show in your first integral. You can get to the second integral if function f is nonzero on the interval above, and zero everywhere else.

 

[transgalactic]

i am using the type of function you talked about.
i was told that if we put a value in tau which is bigger then t
then we get a negative time for which it defined as 0.

so what the problem in that??

why we can't put values bigger then "t" into tau??

 

[Mark44]

i am using the type of function you talked about.
i was told that if we put a value in tau which is bigger then t
then we get a negative time for which it defined as 0.

so what the problem in that??

why we can't put values bigger then "t" into tau??

Let's cut to the chase here. What is the function f you are using? Be sure to include the domain for this function.

Instead of saying stuff like this, and reporting things that some mysterious person whispered to you

i was told that if we put a value in tau which is bigger then t
then we get a negative time for which it defined as 0

just tell us what the function is.

 

[transgalactic]

i am not having any function
i am using a function of type you were talking about

 

[conway]

<br /> (f*g)=\int_{-\infty}^{+\infty}f(\tau)g(t-\tau)<br />

i was told that because of the definition of the function
we can substitute the itervals to
<br /> (f*g)=\int_{0}^{t}f(\tau)g(t-\tau)<br />


why??

If you look at the physical motivation it makes sense. Think of f(t) as your driving function e.g. an electric voltage, and g(t) as your impulse response e.g. the current through a capacitor. Then if f(t) begins at t=0, then the response only needs to be tracked on the interval from[0,t]. The current at time t can be calculated by adding up the impulse responses on the interval [0,t].

 

[transgalactic]

i am not having any function
i am using a function of type you were talking about

 

[transgalactic]

but what happens after time of t>tau
?

 

[Mark44]

f(t) = 0 for those values, so
\int_t^{\infty} f(\tau) d\tau = 0
If a function is identically zero over some interval, its integral is zero over that interval as well. It's really not very complicated.

 

[transgalactic]

ye but let's take a step function or shock delta function
they are defined at t<0 as 0
but after they don't have any 0 value
like in the step function (which is 1 till infinity)

 

[transgalactic]

the solution is simpler than that.
its just when t>tau the function gets a negative time
which is by the original definition gives us 0.

 

[Mark44]

the solution is simpler than that.
its just when t>tau the function gets a negative time
which is by the original definition gives us 0.

That doesn't make any sense. Time can't increase through positive values, and then all of a sudden become negative.