CENG3313, Spring 2021 HW 6 Not Graded !!SOLUTIONS – Students: DO NOT SHARE WITH ANYONE!!
Due Date: This HW & solutions are to guide you for your next test since the material is included in that test. So you
have to go over these questions/problems’ solutions and understand them by then.
Question 1
Design a series RLC circuit with ζ= 0.5 and 0 = 100,000 rads/s. Select C = X nF, where X is the last digit of your student
ID (if the last digit is 0, use X=1). Then calculate R and L values. Assume the initial conditions for vC(t=0) = 5V and
iL(t=0)=0. Express the natural response of vC(t) for your design and sketch vC(t) by marking vC(0) and vC(∞) values. Sketch
the circuit. Is the vC(t) response under-, over- or critically- damped, why?
Solution:
The characteristic equation of the series RLC circuit are:
and
.
Here in this solution, C is selected to be 0.001 F (or X=1). So, the students’ R,L,C values will differ based ID last digit.
Here s =  + j where  = -50000 and  = 86603.
The two I.C. equations: vC(0) = 5 = K1 + K2.0 = K1. And v’(0) = 0 = -50000 K1 + 86603 K2, using the derivative of a
product property: ( f(t)g(t) )’ = f’(t)g(t) + f(t)g’(t) , where f’ denotes the df(t)/dt.
From these two equations, we obtain: K1 = 5, and K2 = 250000/86603 = 2.887.
IC=5V
vC(t) response is underdamped since it has cosine & sine terms in it (or,
since it has complex roots, or since it has oscillations in the response,
not just simple exponential decays; or since the damping coefficient
<1).
Question 2
The switch in the RLC circuit, which includes R,L,C elements in
parallel, has been open for a long time and is closed at t = 0. Given
R1 = R2 = 2k, L = 1.25 H, C = 50 nF, and VA(0)=20V.
a. What are the initial conditions before the switch closes, i.e. vC(0)=?,
iL(0)=?
b. Find vC(t) = ? and iL(t) = ? for t≥0.
c. Is the circuit overdamped, underdamped or critically damped? Why?
Solution: For t≥0, you can do a source transformation (of the VA that is
in series with R1, to convert to a current source IA that is in parallel with
R1) to convert the circuit into a pure parallel RLC circuit which has a
current source in parallel with the R,L,C elements which are in parallel.
Here, IA= VA/R1 = 20V/2kOhms = 10mA.
Then R1 and R2 are in parallel and they can be replaced with an
equivalent resistor RN = (2k)(2k)/(2k+2k)=1kOhms. So we have the
equivalent circuit on the right for t≥0.
For a parallel RLC circuit, the homogenous equation of the inductor
current and its characteristic eqn. are given as
or
The roots of this polynomial are, if distinct:
or s2 + s/(RC) + 1/(LC) = 0.
S1t
then the solution of the homogenous eqn. is of the form K1e
+K2eS2t.
(a)
(b) After the switch closes,
or
or
The roots of the characteristic equation are approximately s1 = 19165.15
(-1.25/1000 +/- sqrt( (-1.25/1000)^2 -4*1.25*50e-9 ) ) /(2*1.25*50e-9)
.
and s2 = -834.9,
so the natural response has the following form:
The input signal is a constant, so the forced response is also a constant and we have iLF(t) = 0.01 A.
The complete, total response has the following form:
.
Apply the ICs to find K1 and K2:
The inductor current is then:
The capacitor voltage then is:
Note the conversion to mA units here.
(c)
Based on the form of the response, the circuit is overdamped because the roots of the characteristic polynomial are
distinct and real (i.e. they are not repeating nor complex).
Question 3:
The RLC circuit shown is in the zero state when the step function input is applied.
Is it a 1st order or a 2nd order differential equation?
Hint: Write the corresponding differential equation. Then find its characteristic
equation. Then find the roots of the characteristic equation. Find the capacitor
voltage first. Then compute the inductor current and the output (inductor) voltage
Solution:
Question 4:
a.
t 
b.
Solution
Use Euler’s formula: eja = exp(ja) = cos(a) + jsin(a)
a.
(since a = -4 <0, we added  to the angle (37 degrees) to obtain 180+37 = 217 degrees = -143 degrees)
b.
Question 5:
An element in a circuit operating in the sinusoidal steady state has a voltage across it and a current
through it, as given as
Find the impedance of the element.
Realize (sketch a design of) this element using a series combination of the two of the following: resistor,
capacitor, inductor.
Calculate the R, or C, or L of the elements. Do not forget to indicate the units.
Solution:
Step 1: Convert to Phasors:
This can be realized with a resistor of resistance R = 6.25 Ohms in series with an inductor of L = 0.02166 H,
since
jL=j10.83 Ohms, so L = 10.83/ = 10.83/500 = 0.02166 H.
Question 6:
a. Find the equivalent impedance Z in this phasor circuit.
b. If  = 10 krad/s, what two elements (R, L, and/or C) could be
used to replace the phasor circuit (find the values of your new
R, L or C)? So combine the individual impedances and then
determine an equivalent circuit.
In other words, “we could use a ___ Ohms resistor in series with a ____
(F or H) (capacitor or inductor) to replace the circuit when  = 10 krad/s.
Solution
Question 7:
A resistor R is connected in parallel with a capacitor C. Design/select the values of R and C so that the equivalent
impedance of the parallel combination is 40 - j30 Ohms at f = X kHz (note that it is not in rad/s but in kHz), where X is the
last digit of your ID. If the last digit of your ID is 0, then use X=10. Two equations two unknowns, so first find R and then
find C.
Use  = 3.141.
Solution:
Solution for X = 1: So f = 1000 Hz. (The students’ solutions for C will change based on the  (C will be inversely
proportional to ).  = 2f = (6.282)(1000) = 6282 rads/s.
Question 8 (we covered this example together in class also):
k
i(t) = 10 cos(10000t - 45º)
F
currents iC(t), iL(t), and iR(t)
V, IC, IL, and IR
The three currents all have the _____ (same? different?) magnitudes.
The current through the resistor___________ (leads?lags?matches? in terms of phase) the source current by ____ (?)
degrees. The capacitor current __________ the resistor current by ______ degrees. The inductor current ________ the
resistor current by _____ degrees. The voltage across the parallel combination ____________ the source and current by
_____ (?) degrees. KCL holds, because the sum of the three component currents __________ the source current.”
Solution
First convert the i(t) to its phasor I :
Question 9:
Find the Laplace Transform of the following waveforms in (a) and (b) and their pole/zero diagrams. Simplify the resulting
expressions.
(a) [-2e-t -t +2] u(t)
(b) [2+2sin(2t) - 2cos(2t)] u(t)
Solutions
(a) Multiply everything by 2 in the following F(s):
{ There is a zero at s = 1 (i.e. sF(s) = 0 for s = 1). Also, since the difference between the denominator polynomial order
(3rd degree) and the numerator order (1st degree) is 2, we say that there are two “zeros” “at infinity” (i.e. as s approaches
infinity, the F(s) value approaches 0, and even the sF(s) value approaches 0.) }
(b)
Zero at -2, two zeros at infinity. Pole at zero and poles at +/- 2j .
Question 10: Find the Laplace transforms of the following waveforms, simplify F(s) in factored form.
Solutions: Using the LT pairs and properties tables: