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SJTU 1
Chapter 8
Second-Order Circuit
SJTU 2
A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.
What is second-order circuit?
Typical examples of second-order circuits: a) series RLC circuit, b) parallel RLC circuit, c) RL circuit, d) RC circuit
SJTU 3
1. The Series RLC Circuit
2. The Parallel RLC Circuit
3. Second-Order Circuit Complete Response
SJTU 4
1. The Series RLC CircuitFORMULATING SERIES RLC CIRCUIT EQUATIONS
Eq.(7-33)
SJTU 5
The initial conditions
To solve second-order equation, there must be two initial values.
SJTU 6
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
Eq.(7-39)
characteristic equation
SJTU 7
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 , there are two complex conjugate
roots
SJTU 8
A source-free series RLC circuit
Special case: Vc(0)=V0, IL(0)=0
V(t)V0
I(t)
tM
SJTU 9
tM>t>0 t > tM
What happens when R=0?
SJTU 10
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
SJTU 11
SJTU 12
SJTU 13
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
SJTU 14
SJTU 15
SJTU 16
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
SJTU 17
SJTU 18
SJTU 19
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
SJTU 20
SJTU 21
SJTU 22
EXAMPLE 7-14A 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.
*
SJTU 23
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.
*
*
SJTU 24
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.
Eq.(7-48b)tt
C teKeKtV 21)(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.
)()( 21 dtSinKdtCosKetV tC Eq.(7-48c)
SJTU 25
EXAMPLE 7-15The 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 corres
ponding zero-input solution takes the form in Eq.(7-48a).
SJTU 26
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
SJTU 27
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
•(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):
SJTU 28
•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):
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
)()( 21 dtSinKdtCosKetV tC
SJTU 29
Fig. 7-32
SJTU 30
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 , there are two complex conjugate
roots
Overdamped situation
Ciritically damped situation
Underdamped situation
SJTU 31
2. The Parallel RLC Circuit
FORMULATING PARALLEL RLC CIRCUIT EQUATIONS
Eq. 7-55
SJTU 32
Equation(7-55) is second-order linear differential equation of the same form as the series RLC circuit equation in Eq.(7-33). 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
SJTU 33
set iN=0 in Eq.(7-55) and obtain a homogeneous equation in the i
nductor current:
A trial solution of the form IL=Kest leads to the characteristic eq
uation
Eq. 7-56
SJTU 34
There are three distinct cases:
Case A: If (GNL)2-4LC>0, there are two unequal real roots a
nd 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 r
oots and the zero-input response is the underdamped form
0)()(
,
21
21
ttSinKtCosKeti
jsst
L
SJTU 35
EXAMPLE 7-16
In a parallel RLC circuit RT=1/GN=500ohm, C=1uF, L=0.2H. T
he initial conditions are Io=50 mA and Vo=0. Find the zero-input
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
SJTU 36
Evaluating this expression at t=0 yields
SJTU 37
SJTU 38
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 t0 (c) Find the capacitor voltage and current through the switch for t 0
Fig. 7-34
SOLUTION:
(a) For t<0 the circuit is in the dc steady state
SJTU 39
(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
SJTU 40
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:
SJTU 41
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 re
sistor plus the current through the 250 ohm resistor
SJTU 42
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
SJTU 43
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.
Fig. 7-35
SOLUTION:
SJTU 44
By inspection, the forced response is vCF=10V. In standard for
mat the homogeneous equation is
the natural response is underdamped (case C)
SJTU 45
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
SJTU 46
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
SJTU 47
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 analysis tools.
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