23.09.2013
Assist .Prof. Sibel ÇİMEN
Electronics and Communication Engineering
University of Kocaeli
CourseBook:
• Fundamentals of Electric Circuits, by Charles K. Alexander and Matthew N. O. Sadiku, McGraw Hill; 5rd edition (2007) 1
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ReferenceBooks:
1) Electric Circuits, by James W. Nilsson and Susan Riedel, Prentice Hall; 8th edition (2007)
2) Schaum's Outline of Electric Circuits, by Mahmood
Nahvi and Joseph Edminister, McGraw‐Hill; 4th edition (2002)
3) Introduction to Electric Circuits, by Richard C. Dorf and James A. Svoboda, Wiley, 7th edition (2006)
4) Schaum's Outline of Basic Circuit Analysis, by John O'Malley and John O'Malley, McGraw‐Hill; 2nd edition (1992)
CourseOutline
1) Second Order DC Circuits (Fund. of Electric Circuits, CH 8)
2) Sinusoids and Phasors (Fund. of Electric Circuits, CH 9)
3) Sinusoidal Steady‐State Analysis (Fund. of Electric Circuits, CH 10)
4) AC Power Analysis (Fund. of Electric Circuits, CH 11)
5) Frequency Response (Fund. of Electric Circuits, CH 14)
6) Laplace Transform (Fund. of Electric Circuits, CH 15
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Overview
• The previous chapter introduced the concept of first order circuits.
• This chapter will expand on that with second order circuits: those that need a second order differential equation.
• RLC series and parallel circuits will be discussed in this context.
• The step response of these circuits will be covered as well.
• Finally the concept of duality will be discussed. 3
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SecondOrderCircuits
• The previous chapter considered circuits which only required first order differential equations to solve.
• However, when more than one “storage element”, i.e. capacitor or inductor is present, the equations require second order differential equations
• The analysis is similar to what was done with first order circuits
• This time, though we will only consider DC independent sources
1.1.Introduction
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1.1.Introduction
Keep in mind;  Capacitor voltage always continuous…
 Inductor current always continuous…
EXAMPLE1.1.
• The switch in Figure has been closed for a long time. İt is open at t=0.
• Find;
a) 0 , (0
b
(0 )/dt, (0 )/dt, c ∞ , ∞
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EXAMPLE1.1./Solution:
• (a) If the switch is closed a long time before t = 0, it means that the circuit has reached dc steady state at t = 0. At dc steady state, the inductor acts like a short circuit, while the capacitor acts like an open circuit, so we have the circuit in Fig. 8.(a) at t = 0−. Thus,
As the inductor current and the capacitor voltage cannot change abruptly,
EXAMPLE1.1./Solution:
• (b) At t = 0+, the switch is open; the equivalent circuit is as shown in Fig. 8 (b). The same current flows through both the inductor and capacitor.
We now obtain by applying KVL to the loop
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EXAMPLE1.1./Solution:
(c) Fort > 0, the circuit undergoes transience. But as t →∞, the
circuit reaches steady state again. The inductor acts like a short circuit and the capacitor like an open circuit, so that the circuit becomes that shown in Fig. 8(c), from which we have
EXAMPLE1.2.
• In figure calculate;
a
0 , (0 , (0
b
(0 )/dt, (0 )/dt
c
∞ , ∞ ,
∞
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EXAMPLE1.2./Solution:
• (a) For t < 0, 3u(t) = 0. At t = 0−, since the circuit has reached steady state, the inductor can be replaced by a short circuit, while the capacitor is replaced by an open circuit as shown in Fig. (a). From this figure we obtain
EXAMPLE1.2./Solution:
• For t > 0, 3u(t) = 3, so that the circuit is now equivalent to that
in Fig. (b). Since the inductor current and capacitor voltage cannot change abruptly,
Applying KCL at node a in Fig. (b) gives
Applying KVL to the middle mesh in Fig.(b) yields
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EXAMPLE1.2./Solution:
But applying KVL to the right mesh in Fig. (b) gives
EXAMPLE1.2./Solution:
• (c) As t →∞, the circuit reaches steady state. We have the equivalent circuit in Fig.(a) except that the 3‐A current source is now operative.
• By current division principle,
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1.3.TheSource‐FreeSeries
RLCCircuits
(1.a)
(1.b)
1.3.TheSource‐FreeSeries
RLCCircuits
 Applying KVL around the loop;
(2)
 To eliminate the integral, we differentiate with respect to t and rearrange terms:….
(3)
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1.3.TheSource‐FreeSeries
RLCCircuits
 with initial values equation (2)…
(4)
or
Based on the first order solutions, we can expect that the solution will be in exponential form.
 equation (3) becomes…
or
i
0
1.3.TheSource‐FreeSeries
RLCCircuits
(5)
Known as characteristic equation
İt’s roots;
Where;
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1.3.TheSource‐FreeSeries
RLCCircuits
:resonantfrequencyorundamped naturalfrequency rad/s ∝: neper frequency or damping factor (Np/s)
İn terms of ∝ equation (5) gets… (6)
The two values of s in Eq. (5) indicate that there are two possible solutions in Eq. (6); that is,
Natural response of series RLC;
1.3.TheSource‐FreeSeries
RLCCircuits
There are three types of solutions;
Overdamped Case (∝
İn this situation; negative and real From this, we should not expect to see an oscillation
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1.3.TheSource‐FreeSeries
RLCCircuits
Critically Damped Case (∝
1.3.TheSource‐FreeSeries
RLCCircuits
Under Damped Case (∝
• ω0 is often called the undamped natural frequency • ωd is called the damped natural frequency
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EXAMPLE1.3.
R=40 Ω, L=4H and C=1/4 F. Calculate the characteristic roots of the circuit. Is the natural response overdamped, underdamped, or critically damped?
Solution:
Since α > ω0, we conclude that the response is overdamped. This is also
evident from the fact that the roots are real and negative.
EXAMPLE1.4.
Find i(t) in the circuit. Assume that the circuit has reached steady state at t=0 .
Solution:
For t < 0, the switch is closed. The capacitor acts like an open circuit while the inductor acts like a shunted circuit. The equivalent circuit is shown in Fig. (a). Thus, at t = 0,
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EXAMPLE1.4./Solution
• For t > 0, the switch is opened and the voltage source is disconnected. The equivalent circuit is shown in Fig. 8.11(b), which is a source‐free series RLC circuit. Hence, the response is underdamped (α < ω); that is,
EXAMPLE1.4./Solution
• We now obtain and using the initial conditions. At t = 0,
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