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Assignment No.
Question No.
The switch in the circuit shown in has been in position ’a’ for a long time. At time t=0, the switch moves instantaneously to position b.
a) Construct an s-domain circuit for t>0. b) Find Vo(s) c) Find vo(t)
Question No.
The switch in the circuit has been in position a for ’a’ long time. At time t=0 the switch moves instantaneously to position ‘b’.
a) Construct the s-domain circuit for t>0. b) Find Vo(s) c) Find IL(s) d) Find vo(t) for t> e) Finf iL(t) for t>
Question No.
The switch in the circuit has been closed for a long time before opening at time t=0.
a) Construct the s-domain equivalent circuit for t> b) Find Vo(s) c) Find vo(t)
Question No.
Find vo(t) in the circuit shown if ig = 5u(t) mA. There is no energy stored in the circuit at time t=0.
Question No. 7
There is no energy stored in the circuit at the time the sources are energized.
a) Find I1(s) and I2(s) b) Find i1(t) and i2(t)
Question No.
There is no energy stored in the circuit at time t=0. The value of the current source is ig(t) = 100cos(10^4 t) mA. The output variable is current io.
a) Find the transfer function of Io(s)/Ig(s). b) Find Io(s). c) Find io(t). d) Determine the steady state response of the system using the transfer function. e) Determine the magnitude of the steady state response if the frequency of the current source is increased from 10^4 to 10^5, 10^5 and 10^6. What do you observe?
f) Determine the magnitude of the steady state response if the frequency of the current source is decreased from 10^4 to 10^3, 10^2 and 10^1. What do you observe? g) Determine the phase of the steady state response if the frequency of the current source is increased from 10^4 to 10^5, 10^5 and 10^6. What do you observe? h) Determine the phase of the steady state response if the frequency of the current source is decreased from 10^4 to 10^3, 10^2 and 10^1. What do you observe?
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Since H(s) =
Hence H(jw) =
Put w = 10^4
we get H(jw) = 1.7241 + 0.6897j
hence H(jw) = 1.8570 < 21.
Now we know that the steady state response is given as
Io(t) (^) steady-state = A|H(jw)|cos(wt + φ + θ(jw))
Where |H(jw)| is the magnitude of H(jw) and θ(jw) is the angle of H(jw)
So the answer is
Io(t) (^) steady-state = 1001.857cos(10^4 t + 0 + 21.80)
Io(t) (^) steady-state = 185.7cos(10^4 t + 21.80)
Now if the frequency increases from 10^4 to 10^5 then put w=10^5
H(jw) = 1.0046 + 0.0202j
Hence
H(jw) = 1.0048 < 1.
So magnitude of output signal is 100*1.0048 = 100.
So phase of output signal is 1.15^0
Now if the frequency increases from 10^5 to 10^6 then put w=10^6
H(jw) = 1 + 0.002j
Hence
H(jw) = 1 <0.
So the magnitude is 100
And the phase is 0.
Now if the frequency decreases from 10^4 to 10^3 then put w=10^3
H(jw) = 0.02 <177.
So the magnitude is 100*0.02 = 2
