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This assignment was given by Hema Bachan at University of Kalyani for Electrical Circuital Analysis course. It includes: Energy, Current, Transfer, Function, Steady, State, Response, Frequency, Phase, Pass, Filter
Typology: Exercises
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Question for Quiz 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?
Solution:
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