- #1
bingie
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** disclaimer - not homework ** =)
Is the following correct?
I have a simple series RL circuit with a square wave input (periodic) defined by:
Here the period = 2a thus a = 2
E(s) = 1/(1-exp(-4s) * int (0->2) exp(-st) (10) dt
E(s) = 1/(1-exp(-4s) * 10 * exp(-st)/-s | 0 -> 2
E(s) = 10/s * (1-exp(-2s)/(1-exp(-4s)
Since this is an RL circuit, the equation is
E(s) = I(s) (R + sL)
thus,
I(s) = E(s) / (R+sL)
= 10/s * (1-exp(-2s)/(1-exp(-4s)(R+sL)
= 10 / (s * (R+sL) * (1+exp(-2s))
Now, I don't have a transform where the exponent is on the bottom - how do I get the inverse laplace transform?
Is the following correct?
I have a simple series RL circuit with a square wave input (periodic) defined by:
Code:
e(t) = {10 0 <= t < 2
{ 0 2 <= t < 4
With a period of 4
E(s) = 1/(1-exp(-2as) * int (0 -> 2) exp(-st) f(t) dt
Here the period = 2a thus a = 2
E(s) = 1/(1-exp(-4s) * int (0->2) exp(-st) (10) dt
E(s) = 1/(1-exp(-4s) * 10 * exp(-st)/-s | 0 -> 2
E(s) = 10/s * (1-exp(-2s)/(1-exp(-4s)
Since this is an RL circuit, the equation is
E(s) = I(s) (R + sL)
thus,
I(s) = E(s) / (R+sL)
= 10/s * (1-exp(-2s)/(1-exp(-4s)(R+sL)
= 10 / (s * (R+sL) * (1+exp(-2s))
Now, I don't have a transform where the exponent is on the bottom - how do I get the inverse laplace transform?