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saltydog
saltydog is offline
#6
Oct8-05, 05:31 PM
Sci Advisor
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Quote Quote by mathwurkz
But doesn't that integral have to be evaluated to get the answer? That's what I can't evaluate. Hence, I went looking to try partial fractions which I can't get either.
So we got:

[tex]i(t)=\frac{1}{L}\int_0^t E(\beta)e^{-R/L(t-\beta)}d\beta[/tex]

with E(t) being a square-wave.

For now, let's just let L and R both be 1:

What is i(t) in the interval [0,1]? Wouldn't that just be:

[tex]i(t)=\int_0^t e^{-(t-\beta)}d\beta\quad\text{for}\quad t\in[0,1][/tex]

What about in the interval [1,2]? So that would be:

[tex]i(t)=\int_0^1 E(\beta)e^{-(t-\beta)}d\beta+\int_1^t E(\beta)e^{-(t-\beta)}d\beta[/tex]

but the second integral is zero because E(t) is zero in that region so:

[tex]i(t)=\int_0^1 e^{-(t-\beta)}d\beta\quad\text{for}\quad t\in[1,2][/tex]

Can you figure out what i(t) would be for the next interval?