NATURAL AND STEP RESPONSES OF NATURAL AND STEP RESPONSES OF RLC CIRCUITSRLC CIRCUITS
NATURAL RESPONSE OF A PARALLEL RALC CIRCUIT
V
R LVd I C
dV
dt
R
dV
dt
V
LC
d V
dt
d V
dt RC
dV
dt
V
LC
t
1
0
10
10
00
2
2
2
2
Ordinary, second-order differential equation with constant coefficients. Therefore, this circuit is called a second-order circuit.
CL R V
++V0
I0iciL iR
GENERAL SOLUTION OF SECOND-ORDER DIFFERENTIAL EQUATIONS
Assume that the solution of the differential equation is of exponential form V=Aest where A and s are unknown constants.
As eAs
RCe
Ae
LC
Ae ss
RC LC
st stst
st
2
2
0
10
( )
This equation can be satisfied for all values of t only if A=0 or the term in parentheses is zero.
A=0 cannot be used as a general solution because to do so implies that the voltage is zero for all time-a physical impossibility if energy is stored in either inductor or capacitor.
THE CHARACTERISTIC EQUATION
ss
RC LC2 1
0 is called the characteristic equation
The two roots of the characteristic equation are
sRC RC LC
sRC RC LC
1
2
2
2
1
2
1
2
1
1
2
1
2
1
If either s1 or s2 is substituted into Aest, the assumed solution satisfies the differential equation, regardless of the value of A
V Ae V A es t s t1 1 2 2
1 2 ,
These two solutions as well as their summation V=V1+V2 satisfy the differential equation
V Ae A es t s t 1 21 2
FREQUENCIES
The behavior of V(t) depends on the values of s1 and s2. Writing the roots in a notation as widely used in the literature
s s
LC
12
02
22
02
02 1
,
=1
2RC
The exponent of e must be dimensionless, so both s1 and s2 (and hence and 0) must have the dimension of the reciprocal of time (frequency). s1 and s2 are referred to as complex frequencies, is called the neper frequency, and 0 is called the resonant radian frequency. They have the unit radians per second (rad/s)
The nature of the roots s1 and s2 depends on the values of and 0. There are three possible cases:
1) If 02<2, both roots will be real and distinct. The voltage
response is said to be overdamped.
2) If 02>2, both roots will be complex and, in addition, will be
conjugates of each other. The voltage response is said to be underdamped.
3) If 02=2, roots will be real and equal. The voltage response
is said to be critically damped.
EXAMPLE
CL R V
+ Find the roots of the characteristic equation if R=200 , L=50 mH, and C=0.2 F.
1
2
10
400 0 2125 10
1 10 10
50 0 210
125 10 15625 10 10 5000
125 10 15625 10 10 20000
64
3 68 2
14 8 8
24 8 8
RC
LCs
s
s
( )( . ).
( )
( )( . )/
. .
. .
rad / s
rad
rad / s
rad / s
02 2
Overdamped
+V0
I0
Repeat the problem for R=312.5
1
2
10
625)(0 28000
1 10 10
50 0 210
8000 0 64 10 10 8000 6000
8000 0 64 10 10 8000 6000
6
3 68 2
18 8
28 8
RC
LCs
s j
s j
( . )
( )
( )( . )/
.
.
rad / s
rad
rad / s
rad / s
02 2
Underdamped
Find the value of R for a critically damped circuit.
For critical damping, 2=02
1
2
110
110
10
2 10 0 2250
28
46
4
RC LC
RCR
( )( . )
THE OVERDAMPED RESPONSE
When the roots of the characteristic equation are real and distinct, the response is said to be overdamped in the form
VtAeAestst
() 1212
The constant A1 and A2 are to be determined by the initial conditions V(0+) and dV(0+)/dt.
V A AdV
dts A s A
V VdV
dt
i
C
iV
RI
c
c
( ) ,( )
( )( ) ( )
( )
00
00 0
0
1 2 1 1 2 2
0
00
EXAMPLE
CL R V
+ Fin V(t), if R=200 , L=50 mH, and C=0.2 F. V0=12V, I0=30 mA.
+V0
I0iciL iR
i I mA
iV
R
V
RmA
i i i mA
dV
dt
i
CkV s
L
Rc
c L R
c
( )
( )( )
( ) ( ) ( )
( ) ( )
./
0 30
00 12
20060
0 0 0 90
0 0 90 10
0 2 10450
0
0
3
6
From the previous example, we determined s1=-5000 rad/s and s2=-20000 rad/s. Then
V t Ae A e
V A A
dV t
dtAe A e
dV
dtA A
A V A V
V t e e V t
t t
t t
t t
( )
( )
( )
( )
( ) ( )
15000
220000
1 2
15000
220000
1 23
1 2
5000 20000
0 12
5000 20000
05000 20000 450 10
14 26
14 26 0
0 0.5 1 1.5 2 2.5
x 10-4
-6
-4
-2
0
2
4
6
8
10
12
i tV t
e e mA t
i t CdV t
dte e
e e mA t
i t i t i t
e e mA t
Rt t
ct t
t t
L R c
t t
( )( )
( )
( )( )
. ( )
( )
( ) ( ) ( )
( )
20070 130 0
0 2 10 70000 520000
14 104 0
56 26 0
5000 20000
6 5000 20000
5000 20000
5000 20000
THE UNDERDAMPED RESPONSE
When 02>2, the roots of the characteristic equation are complex,
and the response is underdamped.
s j
s j
d
d d
1 02 2
2 02 2
( )
d is called the damped radian frequency.
V t A e A e
A e e A e e
e A t jA t A t jA t
e A A t j A A t
j t j t
t j t t j t
td d d d
td d
d d
d d
( )
( cos sin cos sin )
[( )cos ( )sin ]
( ) ( )
1 2
1 2
1 1 2 2
1 2 1 2
A1 and A2 are complex conjugates. Therefore, their sum is a real number and their difference is imaginary. Then j(A1-A2) is also a real number. Denoting B1=A1+A2, and B2=j(A1-A2)
V t e B t B ttd d( ) ( cos sin ) 1 2
DAMPING FACTOR
The trigonometric functions indicate that the response is oscillatory; that is, the voltage alternates between positive and negative values. The rate at which the voltage oscillates is fixed with d. The rate at which the amplitude decreases is determined by . Because determines how quickly the amplitude decreases, it is called as the damping factor. If there is no damping, =0 and the frequency of oscillations is 0. When there is a dissipative element, R, in the circuit, is not zero and the frequency of oscillations is, d, less than 0. Thus, when is not zero, the frequency of oscillation is said to be damped.
EXAMPLE
CL R V
++V0
I0iciL iR Find V(t) if R=20 k, L=8H, C=0.125F,
V0=0, and I0=-12.25 mA
1
2
10
2 20 10 0125)200
1 10
8 0125)10
979 8
200 979 8
200 979 8
6
3
63
202 2
RC
LC
j j
j j
d
d
d
( ) ( .
( .
.
.
.
rad / s
rad / s
rad / s
s rad / s
s rad / s
0
02
1
2
V(0+)=V0=0, then iR(0+)=V(0+)/R=0.
ic(0+)=-iL(0+)=12.25 mA
dV
dt
B B V
V t e t V t
d
t
( ) .
.
,
( ) sin .
0 12 25 10
0125 1098000
098000
100
100 979 8 0
3
6
1 2
200
V / s
THE CRITICALLY DAMPED RESPONSE
The second-order circuit is critically damped when 2=02. The
two roots of the characteristic equation are equal
s sRC1 2
1
2
For a critically damped circuit, the solution takes the following form
V t D te D et t( ) 1 2
D1 and D2 are constant which must be determined using the initial conditions V(0+) and dV(0+)/dt
EXAMPLE
CL R V
++V0
I0iciL iR Determine the value of R for a critically
damped response when, L=8H, C=0.125F, V0=0, and I0=-12.25 mA
From the previous example 02=106. Then
101
240003
RCR
Again, from the previous example V(0+)=0 and dV(0+)/dt=98000 V/s Then D1=0 and D2=98000 V/s.
V t te V tt( ) 98000 01000
THE STEP RESPONSE OF A PARALLEL RLC CIRCUIT
IC
L R V+t=0 ic iL iR
i i i I
iV
RC
dV
dtI
L R c
L
V Ldi
dt
dV
dtL
d i
dt
iL
R
di
dtLC
d i
dtI
d i
dt RC
di
dt
i
LC
I
LC
L L
LL L
L L L
2
2
2
2
2
2
1
The equation describing the step response of a second-order circuit is a second-order differential equation with constant coefficients and with a constant forcing function. The solution of this differential equation equals the forced response which is in the same form of the forcing function (constant for a step input) plus a response function identical in form to the natural response. Thus, the solution for the inductor current is in the form
i IL f
function of the same form
as the natural response
EXAMPLE
IC
L R V+t=0 ic iL iR C=25nF, L=25mH, R=400
The initial energy in the circuit is zero. I=24 mA. Find iL(t)
Since there is no initial energy in the circuit, iL(0+)=0 and V(0+)=0V(0+)=L[diL(0+)/dt]=0, then diL(0+)/dt=0
02
128
94
802
1 10
25)(25)16 10
1
2
10
2 400 25)5 10
25 10
LC
RC
(
( )( rad / s
2
Overdamped circuit
s
s
14 4
24 4
5 10 3 10 20000
5 10 3 10 80000
rad / s
rad / s
As t , circuit reaches dc steady state where inductor is short and capacitor is open. All of the input current flows through the inductor. Then If=24 mA.
i t Ae A e A
i A A
di
dtAe A e
di
dtA A
Lt t
L
L t t
L
( )
( )
( )
24 10
0 24 10
20000 80000
020000 80000 0
31
200002
80000
31 2
120000
280000
1 2
A mA A mA
i t e e mA tLt t
1 2
20000 80000
32 8
24 32 8 0
( ) ( )
EXAMPLE: If the resistor in the circuit is increased to 625, find iL(t) in the circuit.
Since L and C remain fixed, resonant frequency has the same value. But neper frequency decreases to 3.2x104 rad/s. With these values circuit is underdamped with complex conjugate roots.
s j
s j
i t I B e t B e t
I A
i B
di
dte B t B t
e B t B t
di
L ft
dt
d
f
L
L td d
td d
L
14 4
24 4
1 2
3
31
1 2
1 2
32 10 2 4 10
32 10 2 4 10
24 10
0 24 10 0
0
. .
. .
( ) cos sin
( )
( cos sin )
( sin cos )
(
rad / s
rad / s
+ d
)
dtB Bd 1 2 0
B mA B mA
i t e t t mA tLt
1 2
32000
24 32
24 24 24000 32 24000 0
( ) ( ( cos sin ))
EXAMPLE: Find iL(t) if R=500
The resonant frequency remains the same, but neper frequency becomes 4x104 rad/s. These values correspond to critical damping. Roots of the characteristic equation are real and equal at s=-40000
i t I D te D e AL ft t( )
140000
240000
I A
i D
di
dte D t D D e
di
dtD D
D mA s D mA
i t te e mA t
f
L
L t t
L
Lt t
24 10
0 24 10 0
00
960000 24
24 960000 24 0
3
32
1 2 1
2 1
1 2
40000 40000
( )
( )
( )
/
( ) ( )
0 0.5 1 1.5 2 2.5
x 10-4
0
0.005
0.01
0.015
0.02
0.025
Underdamped Critically damped
Overdamped
EXAMPLE
IC
L R V+t=0 ic iL iR
C=25nF, L=25mH, R=500 The initial current in the inductor is 29 mA, the initial voltage across the capacitor is 50 V. I=24 mA. Find iL(t) and V(t)
From the previous example, we know that this circuit is critically damped with s1=s2=-40000 rad/s
i i mA
V V V
L L
c c
( ) ( )
( ) ( )
0 0 29
0 0 50
+
Vc
V V Ldi
dt
di
dtA s
i t I D te D e A
I A
i D D A
di
dtD D D A s
i t te
L cL L
L ft t
f
L
L
Lt
( ) ( )( ) ( )
/
( )
( )
( )/
( ) ( .
0 00 0 50
25 102000
24 10
0 24 10 29 10 5 10
02000 2200
24 2 2 10 5
3
140000
240000
3
32
32
3
1 2 1
6 40000
e mA tt 40000 0)
V t Ldi
dt
te
e e
te e V
L
t
t t
t t
( )
( )[( . )( )
. ( ) ]
.
25 10 2 2 10 40000
2 2 10 5 40000 10
2 2 10 50
3 6 40000
6 40000 40000 3
6 40000 40000
THE NATURAL AND STEP RESPONSE OF A SERIES RLC CIRCUIT
R L
C +
V0
I0
i
Ri Ldi
dt Cid V
Rdi
dtL
d i
dt
i
C
t
1
0
0
0 0
2
2
d i
dt
R
L
di
dt
i
LC
sR
Ls
LC
2
2
2
0
10
sR
L
R
L LC
R
L LC
1 2
22
02
0
2 2
1
2
1
,
rad / s rad / s
These equations are in the same form that of the equations for the parallel RLC circuit. Therefore, the solution will be overdamped, critically damped, or underdamped depending on relative magnitudes of the resonant frequency and neper frequency.
i t Ae A e
i t B e t B e t
i t D te D e
s t s t
td
td
t t
( )
( ) cos sin
( )
1 2
1 2
1 2
1 2 (overdamped)
(underdamped)
(critically damped)
R L
C +
Vci
+ VR + VL
+V
t=0V Ri L
di
dtV
i CdV
dt
di
dtC
d V
dt
c
c c
2
2
d V
dt
R
L
dV
dt
V
LC
V
LC
V t V Ae A e
V t V B e t B e t
V t V D te D e
c c c
c fs t s t
c ft
dt
d
c ft t
2
2
1 2
1 2
1 2
1 2
( )
( ) cos sin
( )
(overdamped)
(underdamped)
(critically damped)
t=0
+
100V
100mH
+
Vc 560i
The capacitor is initially charged to 100V. At t=0, switch closes. Find i(t) and Vc(t) for t0
02
3 68
3
602
02 2
1 10 10
100 0110
2
560
2 10010 2800
7 84 10
9600
LC
R
L
( )( )
( . )
( )
.
rad / s
Underdamped
rad / s
2
d
i t B e t B e t
i B
V Ri Ldi
dtdi
dt
V
LA s
di
dtB e t B e t
di
dtB B A
td
td
c
c
t t
( ) cos sin
( )
( ) ( )( )
( ) ( )/
sin cos
( )
1 2
1
3
22800
22800
2 2
0 0
0 00
0
0 0 100
10010 1000
2800 9600 9600 9600
09600 1000
1000
9600
i t e t t
V t iR Ldi
dt
V t t t e V t
t
c
ct
( ).
sin
( )
( ) ( cos . sin )
1
9 69600 0
100 9600 2917 9600 0
2800
2800
A
EXAMPLE
0.4F
+
Vc
+48V
t=0
280 0.1H There is no initial stored energy in the circuit at t=0. Find Vc(t) for t0.
s
j
s j
V t V B e t B e t tc cft t
1
2 6
2
11400
21400
280
2 01
280
0 2
10
01 0 4
1400 4800
1400 4800
4800 4800 0
( . ) . ( . )( . )
( ) cos sin ,
rad / s
rad / s
As t , circuit reaches dc steady state (inductor short, capacitor open). Thus, Vcf=48V
VdV
dt
i
C
i
C
V B B V
dV
dte B t B t
e B t B t
cc c L
c
c t
t
( ) ,( ) ( ) ( )
( )
( cos sin )
( cos )
0 00 0 0
0
0 48 0 48
1400 4800 4800
4800 4800 4800
1 1
14001 2
14002 1
dV
dtB B
B V
V t e t e t V t
c
ct t
( )
( ) ( cos sin ) ,
04800 1400 0
14
48 48 4800 14 4800 0
2 1
2
1400 1400
A CIRCUIT WITH TWO INTEGRATING AMPLIFIERS
++Vo1
R1
++Vo
R2
C2
C1
+Vg
Assuming ideal opamps, find the relation between Vo and Vg
00 0
1
11 1
1
1 1
V
RC
d
dtV
dV
dt R CV
g
oo
g( )
00 0
1
1
1 1
1
22
2 21
20
22 2
1
20
21 1 2 2
V
RC
d
dtV
dV
dt R CV
d V
dt R C
dV
dt
d V
dt R C R CV
oo
oo
o
g
( )
EXAMPLE
++Vo1
250k
++Vo
0.1F
+Vg
500k
1F
-5V-9V
9V5V
No energy is stored in the circuit when the input voltage Vg jumps instantaneously from 0 to 25 mV.
Derive the expression for Vo(t) for 0 ttsat.
How long is it before the circuit saturates?
1 1000
250 0140
1 1000
500 12
40 2 25 10 2
2 2
2
1 1
2 2
2
23
0
2
0
R C
R C
d V
dtdV
dtdx t
V xdx t
o
ot
o
t
( )( . )
( )( )
( )( )
0 t tsat
The second integrating amplifier saturates when Vo reaches 9V or t=3s. But it is possible that the first opamp saturates before t=3s. To explore this possibility use the following equation
dV
dt R CV
V t
og
o
1
1 1
3
1
140 25 10 1
( )
Thus, at t=3s, Vo1=-3V. The first opamp does not reach saturation at t=3s. The circuit reaches saturation when the second amplifier saturates.
TWO INTEGRATING AMPLIFIERS WITH FEEDBACK RESISTORS
++Vo1
Ra
++Vo
Rb
C2
C1
+Vg
R1 R2
The reason that the op amp saturates in the integrating amplifier is the feedback capacitor’s accumulation of charge. To overcome this problem, a resistor is placed in parallel with each feedback capacitor.
0 00 0
1
0 00 0
1
11 1
1
1 11
1
1 1 11 1
1 1
1
22
2
1
22 2 2
V
R
V
RC
d
dtV
dV
dt R CV
V
R C
R CdV
dt
V V
R C
V
R
V
RC
d
dtV
dV
dt
V V
R CR C
g
a
oo
oo
g
a
o o g
a
o
b
oo
o o o
b
( )
( )
Let
,
d V
dt
dV
dt R C
dV
dt
dV
dt
V V
R C
V R CdV
dt
R CV
d V
dt
dV
dtV
V
R C R C
o
b
o
o o g
a
o bo b
o
o oo
g
a b
20
22 2
1
1 1
1 1
1 22
2
2
21 2 1 2 1 2
1 1
1 1 1
The characteristic equation is
s s2
1 2 1 2
1 1 10
The roots of the characteristic equation are real
s s11
22
1 1
EXAMPLE
The parameters of the circuit are Ra=100 k, Rb=25 k, R1=500 k, R2=100 k, C1=0.1F, and C2=1F. The power supply of each op amp is 6V. The input voltage jumps from 0 to 250 mV at t=0. No energy is stored in the feedback capacitors at the instant the input is applied. Find Vo(t) for t0.
1 1 1 21 2 2
1 2
2
2
0 05 01
30 200 1000
R C s R C s
V
R C R C
d V
dt
dV
dtV
g
a b
o oo
. .
= 1000 V / s
+
2
The characteristic equation is s2+30s+200=0. Thens1=-20 rad/s and s2=-10 rad/s. The final values of the output is the input voltage times the gain of each stage, because capacitors behave as open circuits as t
V V
V t Ae A e
VdV
dtA V A V
V t e e V t
o
ot t
oo
ot t
( ) ( )
( )
( )( )
( ) ( )
250 10500
100
100
255
5
0 00
0
10 5
5 10 5 0
3
110
220
1 2
10 20
and