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```Chapter 8
Second-Order Circuit
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What is second-order circuit?
A second-order circuit is characterized by a secondorder differential equation. It consists of resistors and
the equivalent of two energy storage elements.
Typical examples of second-order circuits: a) series RLC circuit, b)
parallel RLC circuit, c) RL circuit, d) RC circuit
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1. The Series RLC Circuit
2. The Parallel RLC Circuit
3. Second-Order Circuit Complete Response
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1. The Series RLC Circuit
FORMULATING SERIES RLC
CIRCUIT EQUATIONS
Eq.(7-33)
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The initial conditions
To solve second-order equation, there must be two
initial values.
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ZERO-INPUT RESPONSE OF THE SERIES
RLC CIRCUIT
With VT=0(zero-input) Eq.(7-33) becomes
Eq.(3-37)
try a solution of the form
then
characteristic equation
Eq.(7-39)
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In general, a quadratic characteristic equation has two roots:
Eq.(7-40)
three distinct possibilities:
Case A: If
, there are two real, unequal
roots
Case B: If
, there are two real, equal
roots
Case C: If
roots
, there are two complex conjugate
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A source-free series RLC circuit
Special case: Vc(0)=V0, IL(0)=0
V0
V(t)
tM
I(t)
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t > tM
tM>t>0
What happens when R=0?
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Second Order Circuit with no Forcing Function
vc(0) = Vo , iL(0) = Io.
I. OVER DAMPED:
R=2 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A
iL(t) = -0.7 e -0.354t +2.7 e -5.646t A
vc(t) = 1.318 e -0.354t -0.318 e -5.646t V
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II. CRITICALLY DAMPED:
R=0.943 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A
iL(t) = 2e -1.414t -5.83t e -1.414t A
vc(t) = e -1.414t+ 2.75 t e -1.414t V
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III. UNDER DAMPED:
R=0.5 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A
iL(t) =4.25 e -0.75t cos(1.2t + 1.081) A
vc(t) = 2 e -0.75t cos(1.2t - 1.047) V
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IV. UNDAMPED:
R=0 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A
iL(t) =2.915 cos(1.414t + 0.815) A
vc(t) =1.374 cos(1.414t - 0.756) V
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EXAMPLE 7-14
A series RLC circuit has C=0.25uF and L=1H. Find the roots of
the characteristic equation for RT=8.5kohm, 4kohm and 1kohm
*
SOLUTION:
For RT=8.5kohm, the characteristic equation
is
whose roots are
These roots illustrate case A. The quantity under the
radical is positive, and there are two real, unequal roots
at S1=-500 and S2=-8000.
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*
For RT=4kohm, the characteristic equation is
whose roots are
This is an example of case B. The quantity under the
radical is zero, and there are two real, equal roots at
S1=S2=-2000.
* For RT=1kohm the characteristic equation is
whose roots are
The quantity under the radical is negative, illustrating case
C.
In case C the two roots are complex conjugates.
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In case A the two roots are real and
unequal
and the zero-input response
is the sum of two exponentials of the form
Eq.(7-48a)
In case B the two roots are real and equal
and
the zero-input response is the sum of an exponential and a
damped ramp.
VC (t )  K1e t  K 2tet
Eq.(7-48b)
In case C the two roots are complex
conjugates
and the zero-input
response is the sum of a damped cosine and a damped sine.
t
VC (t )  e ( K1Cosdt  K 2 Sindt )
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Eq.(7-48c)
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EXAMPLE 7-15
The circuit of Figure 7-31 has C=0.25uF and L=1H. The
switch has been open for a long time and is closed at t=0. Find
the capacitor voltage for t  0 for (a) R=8.5k ohm, (b) R=4k
ohm, and (c) R=1k ohm. The initial conditions are Io=0 and
Vo=15V.
SOLUTION:
Fig. 7-31
•(a) In Example 7-14 the value
R=8.5kohm yields case A with
roots S1=-500 and S2=-8000. The
corresponding zero-input solution
takes the form in Eq.(7-48a).
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The initial conditions yield two equations in the constants
K1 and K2:
Solving these equations yields K1=16 and K2 =-1, so that the
zero-input response is
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•(b) In Example 7-14 the value R=4kohm yields case B with
roots S1=S2=-2000. The corresponding zero-input response
takes the form in Eq.(7-48b):
The initial conditions yield two equations in the constants
K1 and K2:
Solving these equations yields K1=15 and K2= 2000 x 15,
so the zero-input response is
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•c) In Example 7-14 the value R=1k ohm yields case C with
roots
. The corresponding zero-input
response takes the form in Eq.(7-48c):
VC (t )  e t ( K1Cosdt  K 2 Sindt )
The initial conditions yield two equations in the constants K1
and K2:
Solving these equations yields K1=15 and K2=( ) , so the
zero-input response is
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Fig. 7-32
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In general, a quadratic characteristic equation has two roots:
Eq.(7-40)
three distinct possibilities:
Case A: If
roots
Case B: If
roots
Case C: If
roots
, there are two real, unequal
Overdamped situation
, there are two real, equal
Ciritically damped situation
, there are two complex conjugate
Underdamped situation
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2. The Parallel RLC Circuit
FORMULATING PARALLEL RLC
CIRCUIT EQUATIONS
Eq. 7-55
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Equation(7-55) is second-order linear differential equation
of the same form as the series RLC circuit equation in Eq.(733). In fact, if we interchange the following quantities:
we change one equation into the other. The two circuits are
duals, which means that the results developed for the series
case apply to the parallel circuit with the preceding duality
interchanges.
The initial conditions
iL(0)=Io and
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set iN=0 in Eq.(7-55) and obtain a homogeneous equation in the
inductor current:
A trial solution of the form IL=Kest leads to the characteristic
equation
Eq. 7-56
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There are three distinct cases:
Case A: If (GNL)2-4LC>0, there are two unequal real roots
and the zero-input response is the overdamped form
Case B: (GNL)2-4LC=0, there are two real equal roots and
the zero-input response is the critically damped form
Case C:(GNL)2-4LC<0, there are two complex, conjugate
roots and the zero-input response is the underdamped form
s1 , s2    j
 t
iL (t )  e ( K1Cos t  K 2 Sin t ) t  0
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EXAMPLE 7-16
In a parallel RLC circuit RT=1/GN=500ohm, C=1uF, L=0.2H.
The initial conditions are Io=50 mA and Vo=0. Find the zeroinput response of inductor current, resistor current, and
capacitor voltage
SOLUTION:
From Eq.(7-56) the circuit characteristic equation is
The roots of the characteristic equation are
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Evaluating this expression at t=0 yields
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EXAMPLE 7-17
The switch in Figure 7-34 has been open for a long time and is
closed at t=0
(a) Find the initial conditions at t=0
(b) Find the inductor current for t0
(c) Find the capacitor voltage and current through the switch for t
0
SOLUTION:
(a) For t<0 the circuit is in the
Fig. 7-34
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(b) For t  0 the circuit is a zero-input parallel RLC circuit with
initial conditions found in (a). The circuit characteristic equation
is
The roots of this equation are
The circuit is overdamped (case A), The general form of the
inductor current zero-input response is
using the initial conditions
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The initial capacitor voltage establishes an initial condition on
the derivative of the inductor current since
The derivative of the inductor response at t=0 is
The initial conditions on inductor current and capacitor voltage
produce two equations in the unknown constants K1 and K2:
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Solving these equations yields K1=30.3 mA and K2=-0.309 ma
The zero-input response of the inductor current is
(c) Given the inductor current in (b), the capacitor voltage is
For t 0 the current isw(t) is the current through the 50 ohm
resistor plus the current through the 250 ohm resistor
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3. Second-order Circuit Complete
Response
The general second-order linear differential equation with a step
function input has the form
Eq. 7-60
The complete response can be found by partitioning y(t) into
forced and natural components:
Eq. 7-61
yN(t) --- general solution of the homogeneous equation (input set
to zero), yF(t) is a particular solution of the equation
∴ yF=A/ao
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Combining the forced and natural responses
Eq. 7-67
EXAMPLE 7-18
The series RLC circuit in Figure 7-35 is driven by a step
function and is in the zero state at t=0. Find the capacitor
voltage for t  0.
SOLUTION:
Fig. 7-35
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By inspection, the forced response is vCF=10V. In standard
format the homogeneous equation is
the natural response is underdamped (case C)
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The constants K1 and K2 are determined by the initial conditions.
These equations yield K1= -10 and K2= -2.58. The complete
response of the capacitor voltage step response is
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General second-order circuit
Steps:
1. Set a second-order differential equation
2. Find the natural response yN(t) from the homogeneous
equation (input set to zero)
3. Find a particular solution yF(t) of the equation
4. Determine K1 and K2 by the initial conditions
5. Yield the required response
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Summary
•Circuits containing linear resistors and the equivalent of two
energy storage elements are described by second-order
differential equations in which the dependent variable is one of
the state variables. The initial conditions are the values of the
two state variables at t=0.
•The zero-input response of a second-order circuit takes
different forms depending on the roots of the characteristic
equation. Unequal real roots produce the overdamped response,
equal real roots produce the critically damped response, and
complex conjugate roots produce underdamped responses.
•Computer-aided circuit analysis programs can generate
numerical solutions for circuit transient responses. Some
knowledge of analytical methods and an estimate of the general
form of the expected response are necessary to use these
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analysis tools.
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