Question

Can any body show me how??
[integrating from 0 to t]{f(t)*g(t-@)} d@=[integrating from 0 to infnity] {f(t)*g(t-@)*u(t-@) }d@
Here u(t) is the unit step function

Answer 1

this is hard to read

Answer 2

\[\int\limits_{0}^{t}f(t)g(t-\theta) d \theta=\int\limits_{0}^{\infty}f(t) g(t-\theta) u(t-\theta) d \theta\]

Answer 3

is that right?

Answer 4

yes

Answer 5

i never heard of the unit step function

Answer 6

oh thats the heavside function?

Answer 7

u(t-a)={1 for t>a
0 t<a ,a>=0

Answer 8

yes it is

Answer 9

so this is like partial different equations or differential equations class?

Answer 10

this is not any class I see that step in one of my class and I'm trying to understand how it's done

Answer 11

soory of my book

Answer 12

what book?

Answer 13

oh lol i don't have that book
i'm still thinking ok...

Answer 14

yeah i just seen alot about laplace

Answer 15

lets see if this helps
http://www.physicsforums.com/showthread.php?t=434008
i'm still reading this so it may not help us

Answer 16

This'll not help if you see defination of unit step then you can easily answer his post

Answer 17

you need zarkon
let me see if hes online

Answer 18

If we break into two intervals [integrating from 0 to @] f(t)*g(t-@)*0d@+[integrating from @ to infnity] f(t)*g(t-@)*1d@=[integrating from @ to infnity] f(t)*g(t-@)d@
I'm not getting how can be it equal to [integrating from 0 to t]{f(t)*g(t-@)} d@

Answer 19

\[u(t−\theta)=1\] if \[t-\theta\ge 0\]

and 0 otherwise

\[\int\limits_{0}^{\infty}f(t) g(t-\theta) u(t-\theta) d \theta\]

\[=\int\limits_{0}^{t}f(t) g(t-\theta) u(t-\theta) d \theta+\int\limits_{t}^{\infty}f(t) g(t-\theta) u(t-\theta) d \theta\]

\[=\int\limits_{0}^{t}f(t) g(t-\theta) 1 d \theta+\int\limits_{t}^{\infty}f(t) g(t-\theta) 0d \theta\]

\[=\int\limits_{0}^{t}f(t) g(t-\theta) d \theta\]

Answer 20

how can you suppose for 0 to t interval t is greater than theta

Answer 21

and for t to infnity it is less than theta

Answer 22

on 0 to t

we have \[0<\theta<t\]

thus
\[t-\theta>0\]
and

\[u(t-\theta)=1\]

Answer 23

on t to infinity we have

\[t\le \theta\le \infty\]

thus \[t-\theta<0\]

hence \[u(t-\theta)=0\]

Answer 24

\[\int\limits_{0}^{t}f(t) g(t-\theta) d \theta\]

is a \[d\theta\] integral


thus

\[0\le \theta\le t\]

Answer 25

great man !!! realy great ,thank you so much zarkon and myininaya.

Answer 26

zarkon can you solve one integral, you can see in my questions

Answer 27

gj zarkon