Download First Order And Second Order Response Of RL And

Topic 5
First-Order and Second-Order
Response of RL and RC Circuit
 Natural response of RL and RC Circuit
 Step Response of RL and RC Circuit
 General solutions for natural and step response
 Sequential switching
 Introduction to the natural and step response of RLC
circuit
 Natural response of series and parallel RLC circuit
 Step response of series and parallel RLC circuit
Natural response of RL and RC Circuit
 RL- resistor-inductor
 RC-resistor-capacitor
 First-order circuit: RL or RC circuit
because their voltages and currents are
described by first-order differential
equation.
 Natural response: refers to the behavior
(in terms of voltages and currents) of
the circuit, with no external sources of
excitation.
Natural response of RC circuit
Consider the conditions below:
1. At t < 0, switch is in a closed position
for along time.
2. At t=0, the instant when the switch is
opened
3. At t > 0, switch is not close for along
time
For t ≤ 0, v(t) = V0.
For t ≥ 0:
ic  iR  0
dv(t ) v(t )
C

0
dt
R
dv(t ) v(t )

0
dt
RC
dv(t )
v(t )

dt
RC
dv(t )
1

dt
v(t )
RC
du
1

dv
u
RC
v (t ) 1
1
V0 u du   RC

t
0
dv
1
ln v(t )  ln V0  
(t  0)
RC
 v(t ) 
t
voltage
  
ln 
RC
 V0 
v(t )  V0 e
 t RC
 Thus for t > 0,
v(t )  V0 e
 t RC
v(t )
V0
ic (t )  
 e
R
R
t
RC
1
1
2
2
W (t )  C v(t )   C V0 e
2
2
 2t
RC
The graph of the natural response of RC
circuit
v(t )  V0
 V0 e
t0
 t RC
t0
 The time constant, τ = RC and thus,
v(t )  V0 e
t


 The time constant, τ determine how fast
the voltage reach the steady state:
Natural response of RL circuit
Consider the conditions below:
1. At t < 0, switch is in a closed position
for along time.
2. At t=0, the instant when the switch is
opened
3. At t > 0, switch is not close for along
time
 For t ≤ 0, i(t) = I0
For t > 0,
v(t )  R i (t )  0
1
R t
du    dv

i (0) u
0
di (t )
L
L
 R i (t )  0
dt
R
ln i (t )  ln i (0)   (t  0)
di (t )
L
  R i (t )
L
dt
 i (t ) 
R current
di (t )
R
  t
ln 
  dt
i (t )
L
L
 i (0) 
du
R
  dv
u
L
i (t )
i (t )  i (0) e
t R L
 Thus for t > 0,
i(t )  I 0 e
t R L
v(t )  i (t ) R
  RI 0 e
t R L
1
2
w(t )  Li (t ) 
2
1
2  2t R L
 LI 0 e
2
The switch in the circuit has been closed
Example…
for along time before is opened at t=0.
Find
a) IL (t) for t ≥ 0
b) I0 (t) for t ≥ 0+
c) V0 (t) for t ≥ 0+
d) The percentage of the total energy
stored in the 2H inductor that is
dissipated in the 10Ω resistor.
Solution
a) The switch has been closed for along
time prior to t=0, so voltage across the
inductor must be zero at t = 0-.
Therefore the initial current in the
inductor is 20A at t = 0-. Hence iL (0+)
also is 20A, because an instantaneous
change in the current cannot occur in an
inductor.
 The equivalent resistance and time
constant:
Req  2  40 10  10
L
2

  0.2 sec
Req 10
 The expression of inductor current, iL(t)
as,

i L (t )  i(0 ) e
 20 e
5 t
 t
A
t0
b) The current in the 40Ω resistor
can be determine using current
division,
 10 
i0  i L 

 10  40 
 Note that this expression is valid for
t ≥ 0+ because i0 = 0 at t = 0-.
 The inductor behaves as a short circuit
prior to the switch being opened,
producing an instantaneous change in
the current i0. Then,
i0 (t )  4e
5t
A
t0

c) The voltage V0 directly obtain
using Ohm’s law
V0 (t )  40i0
5t
 160e V
t0

d) The power dissipated in the 10Ω
resistor is
2
V0
p10 (t ) 
10
10t
 2560 e W
t0

The total energy dissipated in the 10Ω
resistor is

W10 (t )   2560e
0
 256 J
10t
dt
The initial energy stored in the 2H
inductor is
1 2
W (0)  L i (0)
2
1
 2 400   400 J
2
 Therefore the percentage of energy
dissipated in the 10Ω resistor is,
256
100  64%
400
First-Order and Second-Order
Response of RL and RC Circuit
 Natural response of RL and RC Circuit
 Step Response of RL and RC Circuit
 General solutions for natural and step response
 Sequential switching
 Introduction to the natural and step response of RLC
circuit
 Natural response of series and parallel RLC circuit
 Step response of series and parallel RLC circuit
Step response of RC circuit
 The step response of a circuit is its behavior when
the excitation is the step function, which maybe a
voltage or a current source.
Consider the conditions below:
1. At t < 0, switch is in a closed and
opened position for along time.
2. At t=0, the instant when the switch is
opened and closed
3. At t > 0, switch is not close and
opened for along time
 For t ≤ 0, v(t)=V0
For
t > 0,
Vs  v(t )  Ri (t )
1
du

dv 
RC
u  Vs
dv(t )  t  ln v(t )  V   ln V  V 
s
0
s
Vs  v(t )  RC
RC
dt
 v(t )  Vs  voltage
t
1
dv(t )


 ln 
dt 
RC
V

V
0
s 

RC
Vs  v(t )
 t RC
v(t )  Vs  V0  Vs e
1
dv(t )

dt 
 t
RC
v(t )  Vs
 Vs  V0  Vs e
 Thus for t >0
V  Vs  V0  Vs e
 V f  Vn
Where
V f V s
Vn  V0  Vs e
 t
 t
Vf = force voltage or also known as
steady state response
Vn = transient voltage is the
circuit’s temporary response that
will die out with time
Graf Sambutan Langkah Litar RC
force
total
Natural
 The current for step response of RC circuit
dv
i (t )  C
dt
 1
 t 
 C   (V0  Vs )e 
 

1
 t
  V0  Vs e
R
 Vs V0   t 
   e
R R

i (t )  i(0 )e
 t
Step response of RL circuit
Consider the conditions below:
1. At t < 0, switch is in a opened position
for along time.
2. At t=0, the instant when the switch is
closed
3. At t > 0, switch is not open for along
time
 i(t)=I0 for t ≤ 0.
 For t > 0,
Vs  Ri (t )  v(t )
di (t )
Vs  Ri (t )  L
dt
Vs
L di (t )
 i (t ) 
R
R dt
R
di (t )
dt  V
s
L
R  i (t )
R
di
 dt 
L
i  Vs R
R
du
 dv 
L
u  Vs R
i (t )
R t
du
  dv  
I 0 u  Vs
L 0
R
R
 t  ln i (t )  Vs R   ln I 0  Vs R 
L
 i (t )  Vs R 
R

 t  ln 
Curre
Vs
L
 I0  R 
i(t ) 
Vs
R
 I 0 
Vs
R
e
t R L
Thus,
i(t )  I 0

Vs
R
 I 0 
Vs
R
e
di (t )
v(t )  L
dt
t R L
 Vs  R I 0 e
t 0
t R L
t 0
t0
t0
Question
The switch in the circuit has been
open for along time. The initial charge
on the capacitor is zero. At t = 0, the
switch is closed. Find the expression for
a) i(t) for t ≥ 0
b) v(t) when t ≥ 0+
Answer (a)
 Initial voltage on the capacitor is
zero. The current in the 30kΩ resistor
is
(7.5)( 20)
i (0 ) 
 3mA
50

 The final value of the capacitor current
will be zero because the capacitor
eventually will appear as an open
circuit in terms of dc current. Thus if =
0.
 The time constant, τ is
  (20  30)10 (0.1) 10
 5ms
3
6
 Thus, the expression of the current i(t)
for t ≥ 0 is

i ( t )  i (0 )e
 3e
 3e
t
 t
5103
 200t
mA
t0

Answer (b)
The initial value of voltage is zero
and the final value is
Vf  (7.5)(20)  150V
 The capacitor vC(t) is
v C ( t )  Vf  V0  Vf e
 150  (0  150)e
 150  150e
 200t
 t
 200t
V t0
• Thus, the expression of v(t) is
v( t )  150  150e
 (150  60e
200t
 200t
 (30)(3)e
)V t  0

200t
First-Order and Second-Order
Response of RL and RC Circuit
 Natural response of RL and RC Circuit
 Step Response of RL and RC Circuit
 General solutions for natural and step response
 Sequential switching
 Introduction to the natural and step response of RLC
circuit
 Natural response of series and parallel RLC circuit
 Step response of series and parallel RLC circuit
General solutions for natural and step
response
 There is common pattern for voltages, currents and
energies:
v(t )  V f  V0  V f e
i(t )  I f  I 0  I f e
W (t )  W f  W0  W f e
 t
 t
2t 
The general solution can be compute
as:
x(t )  x f  x0  x f  e
 t
Write out in words:
 the unknown
  the final
  the initial   the final

t





 var iable as a
   value of the    value of the  value of the  e time cons tan t
 


 
 
  var iable

function of time   var iable
  var iable
 

When computing the step and natural responses of
circuits, it may help to follow these steps:
1. Identify the variable of interest for the circuit. For RC
circuits, it is most convenient to choose the capacitive
voltage, for RL circuits, it is best to choose the
inductive current.
2. Determine the initial value of the variable, which is
its value at t0.
3. Calculate the final value of the variable, which is its
value as t→∞.
4. Calculate the time constant of the circuit, τ.
First-Order and Second-Order
Response of RL and RC Circuit
 Natural response of RL and RC Circuit
 Step Response of RL and RC Circuit
 General solutions for natural and step response
 Sequential switching
 Introduction to the natural and step response of RLC
circuit
 Natural response of series and parallel RLC circuit
 Step response of series and parallel RLC circuit
Sequential switching
 Sequential switching is whenever switching occurs
more than once in a circuit.
 The time reference for all switchings cannot be t = 0.
Example…
First switch move form a to b at t=0 and
second switch closed at t=1ms. Find the
current, i for t ≥ 0.
 Step 1: current value at t=0- is determine as assume
that the first switch at point a and second switch
opened for along time. Therefore, the current, i(0)=10A.
 When t=0, an RL circuit is obtain as
L
   1ms .
R
• Thus the current for 0 ≤ t ≤ 1ms is,
i  10e
1000t
A
 At t=t1=1ms,
i (t1 )  10e
1
 3.68 A
When switch is closed at t=1ms, the equivalent
resistance is 1Ω. Then,
L 2
1    2ms
R 1
 Thus i for t ≥ 1ms is
i  i (t1 ) e
 3.68 e


( t t1 )
1
( t t1 )
1
A
The graph of current for t ≥ 0
First-Order and Second-Order
Response of RL and RC Circuit
 Natural response of RL and RC Circuit
 Step Response of RL and RC Circuit
 General solutions for natural and step response
 Sequential switching
 Introduction to the natural and step response of RLC
circuit
 Natural response of series and parallel RLC circuit
 Step response of series and parallel RLC circuit
Second order response for RLC c
 RLC circuit: consist of resistor,


1.
2.
inductor and capacitor
Second order response : response
from RLC circuit
Type of RLC circuit:
Series RLC
Parallel RLC
Natural response of parallel RLC
Summing all the currents away
from node,
V 1 t
dv
  vd  I 0  C
0
0
R L
dt
Differentiating once with respect to
t,
2
1 dv v
d v
 C 2  0
R dt L
dt
2
d v 1 dv v



0
2
dt
RC dt LC
Assume that
v  Ae
st
As st
A st
As e 
e 
e 0
RC
LC
2
st
s
1 
 2
Ae  s 

0
RC
LC



st
characteristic equation
Characteristic equation is zero:
s
1 
 2

s 
0
RC LC 

The two roots:
2
1
1
 1 
s1  
 
 
2 RC
 2 RC  LC
2
1
1
 1 
s2  
 
 
2 RC
 2 RC  LC
The natural response of series RLC:
v  A1 e  A2 e
s1t
s2t
 The two roots:
s1      0
2
s2      0
2
2
2
•where:
0 
1

2 RC
1
LC
Summary
Parameter
Terminology
Value in natural
response
s1, s2
Charateristic
equation
s1     2  0
α
0
Neper frequency
Resonant radian
frequency
2
s2     2  0
1

2 RC
0 
1
LC
2
 The two roots s1 and s2 are depend on

1.
2.
3.
α and ωo value.
3 possible condition is:
If ωo < α2 , the voltage response is
overdamped
If ωo > α2 , the voltage response is
underdamped
If ωo = α2 , the voltage response is
critically damped
Overdamped voltage response
 overdamped voltage solution
v  A1 e  A2 e
s1t
s2t
 The constant of A1 dan A2 can be
obtain from,

v(0 )  A1  A2

dv(0 )
 s1 A1  s2 A2
dt
 The value of v(0+) = V0 and initial
value of dv/dt is


dv(0 ) iC (0 )

dt
C
The process for finding the overdamped
response, v(t) :
1. Find the roots of the characteristic
equation, s1 dan s2, using the value of
R, L and C.
2. Find v(0+) and dv(0+)/dt using circuit
analysis.
3.
Find the values of A1 and A2 by solving equation
below simultaneously:

v(0 )  A1  A2

4.

dv(0 ) iC (0 )

 s1 A1  s2 A2
dt the valueC
Substitute
for s1, s2, A1 dan A2 to
determine the expression for v(t) for t ≥ 0.
Example of overdamped voltage
response for v(0) = 1V and i(0) = 0
Underdamped voltage response
When ωo2 > α2, the roots of the
characteristic equation are complex
and the response is underdamped.
The roots s1 and s2 as,
s1     (0   )
2
   j 0  
2
   jd
s2    jd
 ωd : damped radian frequency
2
2
The underdamped voltage response
of a parallel RLC circuit is
v( t )  B1 e
 t
 B2 e
cos d t
 t
sin d t
The constants B1 dan B2 are real
not complex number.
The
two simultaneous equation that
determine B1 and B2 are:

v(0 )  V0  B1


dv(0 ) iC (0 )

 1B1  d B2
dt
C
Example of underdamped voltage
response for v(0) = 1V and i(0) = 0
Critically Damped voltage response
A circuit is critically damped when
ωo2 = α2 ( ωo = α). The two roots of
the characteristic equation are real
and equal that is,
1
s1  s2    
2 RC
 The solution for the voltage is
t
v(t )  D1t e
 D2 e
t
•The two simultaneous equation needed to
determine D1 and D2 are,

v(0 )  V0  D2


dv(0 ) iC (0 )

 D1  D2
dt
C
Example of the critically damped
voltage response for v(0) = 1V and i(0) =
0
The step response of a
parallel RLC circuit
From the KCL,
iL  iR  iC  I
dv
v
I
iL   C
dt
R
Because
•We get
di
vL
dt
2
dv
d iL
L 2
dt
dt
Thus,
2
L diL
d iL
iL 
 LC 2  I
R dt
dt
2
d iL
1 diL
iL
I



2
dt
RC dt LC LC
There is two approach to solve the
equation that is direct approach
and indirect approach.
Indirect approach
From the KCL:
1 t
v
dv
vd



C

I

0
L
R
dt
Differentiate once with respect to t:
2
v 1 dv
d v

C 2  0
L R dt
dt
2
d v
1 dv
v



0
2
dt
RC dt LC
 The solution for v depends on the roots
of the characteristic equation:
v  A1 e  A2 e
s1t
v  B1 e
 t
 B2 e
v  D1t e
s2t
cos d t
t
t
sin d t
 D2 e
t
Substitute into KCL equation :


iL  I  A1 e  A2 e

s1t
s2t
 t
iL  I  B1 e cos d t
 t
 B2 e sin d t
  t
  t
iL  I  D1 t e  D2 e
Direct approach
It is much easier to find the primed
constants directly in terms of the
initial values of the response
function.






A1 , A2 , B1 , B2 , D1 , D2
The primed constants could be find
from
iL (0) and
diL (0)
dt
The solution for a second-order
differential equation equals the
forced response plus a response
function identical in form to
natural response.
 If If and Vf is the final value of the
response function, the solution for the
step function can be write in the form,
Function of the same form 
i  If  

as the natural response

function of the same form 
v  Vf  

as the natural response

Natural response of a series RLC
 The procedures for finding the natural
response of a series RLC circuit is the
same as those to find the natural
response of a parallel RLC circuit
because both circuits are described by
differential equations that have same
form.
Series RLC circuit
Summing the voltage around the
loop,
di 1 t
Ri  L   i d  V0  0
dt C 0
Differentiate once with respect to t
,
2
di
d i i
R L 2  0
dt
dt
C
2
d i R di
i



0
2
dt
L dt LC
The characteristic equation for the
series RLC circuit is,
R
1
s  s
0
L
LC
2
The roots of the characteristic
equation are,
2
s1, 2
R
1
 R 

   
2L
 2 L  LC
@
s1, 2      0
2
2
Neper frequency (α) for series RLC,
R

rad / s
2L
And the resonant radian frequency,
0 
1
rad / s
LC
The current response will be
overdamped, underdamped or
critically damped according to,
0  
2
0  
2
2
0  
2
2
2
 Thus the three possible solutions fo the
currents are,
i(t )  A1 e  A2 e
s1t
i(t )  B1e
t
 B2e
i(t )  D1t e
cos d t
t
 t
s2t
sin d t
 D2 e
 t
Step response of series RLC
The procedures is the same as the
parallel circuit.
Series RLC circuit
Use KVL,
di
v  Ri  L  vC
dt
The current, i is related to the
capacitor voltage (vC ) by
expression,
dv C
iC
dt
Differentiate once i with respect to
t
2
di
d vC
C 2
dt
dt
Substitute into KVL equation,
2
d vC R dvC vC
V



2
dt
L dt
LC LC
Three possible solution for vC are,

s1t

t

vC  V f  A1 e  A2 e
vC  V f  B1 e

 B2 e
t
s2t
cos d t
sin d t
  t
  t
vC  V f  D1 t e  D2 e
Example 1 (Step response of parallel RLC)
The initial energy stored in the circuit is zero.
At t = 0, a DC current source of 24mA is
applied to the circuit. The value of the resistor
is 400Ω.
1. What is the initial value of iL?
2. What is the initial value of diL/dt?
3. What is the roots of the characteristic equation?
4. What is the numerical expression for iL(t) when t ≥ 0?
Solution
1. No energy is stored in the circuit prior
to the application of the DC source, so
the initial current in the inductor is
zero. The inductor prohibits an
instantaneous change in inductor
current, therefore iL(0)=0 immediately
after the switch has been opened.
2. The initial voltage on the capacitor is
zero before the switch has been
opened, therefore it will be zero
immediately after. Because
di L
vL
dt
thus

di L (0 )
0
dt
3. From the circuit elements,
0
12
2
1
10
8


 16  10
LC (25)(25)
9
1
10


2 RC (2)( 400)( 25)
 5 10 rad / s
4
  25  10
2
8
Thus the roots of the characteristic
equation are real,
s1  5  10  3  10
4
4
 20 000 rad / s
s 2  5  10  3  10
4
 80 000 rad / s
4
4. The inductor current response
will be overdamped.


i L  I f  A1 e  A2 e
s1t
s2t
Two simultaneous equation:


i L (0)  I f  A1  A2  0
di L (0)


 s1 A1  s 2 A2  0
dt

A1  32mA

A2  8mA
 Numerical solution:
 24  32e
iL (t )  
80000t
  8e
for
t0
20000t

mA


Example 2 (step response
of series RLC)
 No energy is stored in the
100mH inductor or 0.4µF
capacitor when switch in the
circuit is closed. Find vC(t) for
t ≥ 0.
Solution
 The roots of the characteristic equation:
2
280
10
 280 
s1  
 
 
0.10.4
0 .2
 0. 2 
  1400  j 4800 rad / s
s 2   1400  j 4800 rad / s
6
The roots are complex, so the
voltage response is underdamped.
Thus:

1400t
vC  48  B1 e
cos 4800t
 1400t
 B2 e
sin 4800t
t0
No energy is stored in the circuit
initially, so both vC(0) and
dvC(0+)/dt are zero. Then:

vC (0)  0  48  B1

dvC (0 )


 0  4800 B2  1400 B1
dt
Solving for B1’and B2’yields,

B1  48V

B2  14V
Thus, the solution for vC(t),
 48  48 e
cos 4800t 
V
vC (t )  
1400t

sin 4800t
  14 e

for
t0
1400t