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EECE202 NETWORK ANALYSIS I
Dr. Charles J. Kim
Class Note 25: Second-Order Circuits
A. Preface
1. A second-order circuit is a circuit environment where an inductor and a capacitor are
present simultaneously.
2. The second-order circuit analysis is, in this class, is limited to one loop (series RLC) or
one non-reference node (parallel RLC) case.
3. PSPICE analysis practice is encouraged.
A. Basic Circuit Equation of Second-Order Circuit
1. Let’s first consider a parallel RLC circuit powered by a DC current source.
2. Let’s assume that there is no energy initially stored in the capacitor and inductor.
3. The node voltage equation is:
t
t
v 1
dv
v 1
dv
− I s + + ∫ vdx + C
= 0 → + ∫ vdx + C
= Is
R L to
dt
R L to
dt
4. By derivation with respect to time t, we have:
d 2 v 1 dv v
d 2v
v
1 dv
C 2 +
+ = 0 or
+
+
= 0 --------------(1)
2
R dt L
RC dt LC
dt
dt
5. Let’s now consider a series RLC circuit powered by a DC voltage source.
6. Again, let’s assume that there is no energy initially stored in the capacitor and inductor.
7. The loop KVL equation for the current is:
t
t
1
di
1
di
− Vs + Ri + ∫ i ( x)dx + L = 0 → Ri + ∫ i ( x)dx + L = Vs
C to
dt
C to
dt
8.
By derivation with respect to time t, we have:
d 2i
di i
d 2 i R di
i
L 2 +R + =0→ 2 +
+
= 0 ------(2)
L dt LC
dt C
dt
dt
9. We can see that the equation for the node voltage in the parallel RLC [equation (1) above]
and the equation for the loop current in the series RLC [equation (2) above] are identical: a
second-order differential equation with constant coefficients.
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B. Solution of a Second-Order differential Equation (part 1: solution for forced function)
1. Let’s change equations (1) and (2) to a more general second-order differential equation
form:
d 2 x(t )
dx(t )
+ a1
+ a2 x(t ) = K --------------------------------(3)
2
dt
dt
2. As we did in the first-order analysis, the solution of the equation (3) is:
x(t ) = x p (t ) + xc (t )
d 2 x(t )
dx(t )
+ a1
+ a2 x(t ) = K -------------(3a)
2
dt
dt
d 2 x(t )
dx(t )
and x(t ) = xc (t ) is a solution to
+ a1
+ a2 x(t ) = 0 ----------------------(4)
2
dt
dt
3. Let's observe equation (3a) for a while. Since the right hand side is a constant K, therefore
xp(t) must be a constant (at left hand side). Let say xp(t)=C (C is a constant), then the left
K
hand side is: a 2 C . Therefore, x p (t ) = C = .
a2
4. The, the complete solution of equation (3) is of the form:
K
x(t ) = x p (t ) + xc (t ) =
+ xc (t )
a2
5. The solution of the homogeneous equation for xc(t) [eq. (4)] starts in the next section.
where, x(t ) = x p (t ) is a solution to
C. Solution of a Second-Order differential Equation (part 2: solution for homogeneous eq.)
1. For simplicity (you will see why soon), let’s rewrite the equation (4), by simple
2
substitutions for a1 = 2α and a2 = wo , in the form of:
d 2 x(t )
dx(t )
2
+ 2α
+ wo x(t ) = 0
2
dt
dt
----------------------------------------(5)
*Note: For this revised equation form, x p (t ) =
K
2
since a 2 = wo .
2
wo
2. We assume a solution that: xc (t ) = Ae st
3. The substitution of the assumed solution into equation (5) yields
s 2 Ae st + 2αsAe st + wo2 Ae st = 0
4. Simplification of the above equation yields to: ( s 2 + 2αs + wo ) Ae st = 0
2
Since xc (t ) = Ae st cannot be zero, s 2 + 2αs + wo = 0 -------------------------(6)
5. The equation (6) is called the characteristic equation, where
α is referred to Neper Frequency
wo is referred to Undamped Natural Frequency (or Resonant Frequency)
2
α 
and   is referred to Exponential Damping Ratio.
 wo 
6. If the characteristic equation is satisfied, then, the assumed solution x(t ) = Ae st is correct.
7. Employing the quadratic formula, the roots for the characteristic equation are:
− 2α ± 2 α 2 − wo
2
s=
= −α ± α 2 − wo
2
8. Therefore, the two roots are:
2
2
s1 = −α + α 2 − wo
2
and
s2 = −α − α 2 − wo
2
9. This means that we have two solutions of the homogeneous equation:
x1 (t ) = A1e s1t and x2 (t ) = A2 e s2t
10. Note that the sum of two solutions is also a solution. Therefore, in general the solution of
the homogeneous equation is of the form:
xc (t ) = A1e s1t + A2 e s2t -------------------------------------------(7)
11. Finally, the solution for the original second-order equation of
d 2 x(t )
dx(t )
2
+ 2α
+ wo x(t ) = K
2
dt
dt
is:
K
K
x(t ) = 2 + xc (t ) = 2 + A1e s1t + A2 e s2t
wo
wo
with, s1 = −α + α 2 − wo
2
and
s2 = −α − α 2 − wo .
2
12. From the solution, we can easily see the final value: x(∞) =
K
.
2
wo
D. Examination of the solution of the homogeneous equation: Natural Frequency Analysis
1. Let’s have a closer examination of the roots of the characteristics roots:
s1 = −α + α 2 − wo
and s2 = −α − α 2 − wo
2. The roots s1 and s2 are called the natural frequencies because they determine the natural
(unforced) response of the network.
2
3. We see that the roots are dependent upon the value of (α 2 − wo ) .
2
2
4. If α 2 = wo : the roots are real and equal --> “Critically Damped”
2
If α 2 > wo : the roots are real and unequal --> “Overdamped”
2
If α 2 < wo : the roots are complex numbers --> “Underdamped”
2
3
5. “Critically Damped” case: (real and equal s)
2
(a) Condition: α 2 = wo ⇒ s1 = s2 = −α
(b) Solution Form: x(t ) = x(∞) + D1te − at + D2 e −αt
[Note: x(t ) = x(∞) + A1e −αt + A2 e −αt = ( A1 + A2 )e − at = A3 e −αt . This simple form,
dx(t )
dt t =0
with the single constant A3. After applying an approach for repeated roots, the
solution for critically damped case is of the form: x(t ) = x(∞) + A3 e − at ( A1 + A2 t ) ]
(c) Constraints (equations to find the two coefficients, D1, and D2):
i) x(0) = x(∞) + D2
dx(t )
= D1 − αD2
ii)
t =0
dt
however, in general does not satisfy the two initial conditions, i.e., x(0) and
6. “Overdamped” case: (real and unequal s)
2
(a) Condition: α 2 > wo
(b) Solution: x(t ) = x(∞) + A1e s1t + A2 e s2t
(c) Constraints (or equations to find two coefficients A1 and A2)
i) x(0) = x(∞) + A1 + A2
dx(t )
ii)
t = 0 = s1 A1 + s 2 A2
dt
7. “Underdampled” case: (complex s)
(a) Condition: α 2 < wo . Also define wd = wo − α 2
The roots are rewritten as:
2
2
s1 = −α + j wo − α 2 = −α + jwd
and s 2 = −α − j wo − α 2 = −α − jwd
Then, the solution can be rewritten as:
x(t ) = x(∞) + A1e −(α − wd )t + A2e − (α + jwd )t
2
2
= x(∞) + e −αt [ A1e jwd t + A2e − jwd t ]
= x(∞) + e −αt [ A1{cos wd t + j sin wd t} + A2{cos wd t − j sin wd t}]
= x(∞) + e − at [( A1 + A2 ) cos wd t + j ( A1 − A2 ) sin wd t ]
= x(∞) + e − at [ B1 cos wd t + B2 sin wd t ]
(b) The solution form: x(t ) = x(∞) + e − at [ B1 cos wd t + B2 sin wd t ]
(c) Constraints (Coefficient equations)
i) x(0) = x(∞) + B1
dx(t )
ii)
t = 0 = −α 1 B1 + ω d B2
dt
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Second-Order Equation Summary Table
Second-Order
Differential
Equation
Final Value
Characteristics
Roots
S Damping Types
o
l Condition
u
t Solution Form
I x (t ) =
o
n
Coefficient
s
Determination
Relationship
d 2x
dx
2
+ 2α
+ ω0 x = K
2
dt
dt
K
x (∞ ) = 2
wo
s1 = −α + α 2 − ω 0
2
and s 2 = −α − α 2 − ω 0
2
Overdapmed case
Underdamped case
Critically Damped case
α 2 > ω0 2
α 2 < ω02
α 2 = ω02
x(∞) + A1 e s1t + A2 e s2 t
x(∞) + B1 e −αt cos ω d t
x(∞) + D1te −αt + D 2 e −αt
+ B2 e −αt sin ω d t
x(0) = x(∞) + B1
dx(t )
t =0 = −αB1 + ω d B2
dt
x(0) = x(∞) + D2
dx(t )
t =0 = D1 − αD2
dt
x(0) = x(∞) + A1 + A2
dx(t )
dt
t =0
= s1 A1 + s 2 A2
ωd = ω0 2 − α 2
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E. Parallel RLC Natural Response Example
Consider the parallel RLC circuit shown below. Let’s assume that the initial conditions on the
storage elements are: iL(0)= -1[A] and vc(0)=4 [V]. Find the node voltage v(t) and the current
through the inductor iL(t).
SOLUTION:
F. Series RLC Natural Response Example
The switch in the circuit has been closed for a long time. At t=0, the switch opens. Find i(t).
SOLUTION:
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G. Step Response of Parallel RLC Example
Energy is stored in the circuit before the DC current source is applied, with i L (0) = 0.29 [A] and
vC (0) = 5 [V]. Find iL(t).
SOLUTION
H. Step Response of Series RLC Example
Find vC(t).
SOLUTION:
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I. RLC Response Extra Problems
I.1. The switch in the circuit has been in position 1 for a long time. At t=0, it moves from
position 1 to position 2. Compute i(t) for t>0 and use this current to determine the voltage
v(t).
I.2. The switch in the circuit has been in position 1 for a long time. At t=0, it moves from
position 1 to position 2. Compute v(t) for t>0.
I.3. Find v(t) and i(t)
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