I'm studying for test tomorrow and i feel that i know how to do this but the numbers are not "nice" number so i wanna check it see if they are right. it 3 problems and I'm very sure the first one is right. from these i can check to see if the others are right. I'm going to put the question, circuit and my work for the answer (the circuits and graphs are attached.) thank you for you help.
(i'm sorry if i uploaded a bunch of stuff)
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1) Sketch the given function and find it's Laplace transform.
this is what i got:
(graph is in attachment not sure how it shows up)
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2)the switch in the circuit has been closed for a long time. at t=0 the switch is opened. find v0(t) for t >= 0 using the Laplace transform.
(pic is 2nd in attachment )
first i found initial conditions across the capacitor and inductor.
v = 0
i = .06A
then did the transformation to s domain using a voltage source from the inductor. (3rd in attachment)
used voltage divider to find
then partial fraction decomposition (PFD) yields the numerator to be
in polar
making the answer:
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last one
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3)there is no energy stored in the capacitors at the time the switch is closed. find I1(s), V1(s),V2(s), i1(t), v1(t), and v2(t)
(circuit is 4th in attachment)
first i added the parallel capacitors giving one of 40uF
then i converted to s domain (5th in attachment)
then i found I1(s) to be
and i1(t)
then used I1(s) to find V1 and V2
then PFD on V1 and V2
giving v1(t) and v2(t)
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thank you
(i'm sorry if i uploaded a bunch of stuff)
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1) Sketch the given function and find it's Laplace transform.
Rich (BB code):
f(t) = (8t-8)[u(t-1)-u(t-2)] + (24-8t)[u(t-2)-u(t-4)] + (8t-40)[u(t-4)-u(t-5)](graph is in attachment not sure how it shows up)
Rich (BB code):
L{(6t-8)u(t-1) - (8t-8)u(t-2) + (24-8)u(t-2) - (24-8t)u(t-4) + (8t-40)u(t-4) - (8t-40)u(t-5)}
L{8(t-1)u(t-1) + (24-8t-8t+8)u(t-2) + (8t-40-24+8t)u(t-4) - 8(t-5)u(t-5)}
(8e^-s)/s^2 + L{(-16t + 32)u(t-2) + (16t - 64)u(t-4)} - (8e^-5s)/s^2
(8e^-s - 16e^-2s + 16e^-4s - 8e^-5s)/s^22)the switch in the circuit has been closed for a long time. at t=0 the switch is opened. find v0(t) for t >= 0 using the Laplace transform.
(pic is 2nd in attachment )
first i found initial conditions across the capacitor and inductor.
v = 0
i = .06A
then did the transformation to s domain using a voltage source from the inductor. (3rd in attachment)
used voltage divider to find
Rich (BB code):
(12s + 48.72)/(s^2+4.06s +8.33333)
Rich (BB code):
6.23 <-94.024°making the answer:
Rich (BB code):
12.46*e^(-2.03t)*cos(28.796t-94.024)u(t)last one
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3)there is no energy stored in the capacitors at the time the switch is closed. find I1(s), V1(s),V2(s), i1(t), v1(t), and v2(t)
(circuit is 4th in attachment)
first i added the parallel capacitors giving one of 40uF
then i converted to s domain (5th in attachment)
then i found I1(s) to be
Rich (BB code):
.02/(s+20000)
Rich (BB code):
.02e^-20000t
Rich (BB code):
V1(s) = .02/(s^2+s/2000)
V2(s) = .02/(s^2 + s/500)
Rich (BB code):
V1(s) = 40/s + 80000/s+.0005
V2(s) = 10/s + 5000/x+.002
Rich (BB code):
v1(t) = 40t + 80000e^-.0050t
v2(t) = 10t + 5000e^-.002tthank you
