Download First Order And Second Order Response Of RL And

First Order And Second Order
Response Of RL And RC Circuit
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
voltan
  
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
Ex.
The switch in the circuit has been closed
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.
Ans.
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 saat
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
volt
 v(t )  Vs 
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
Step Response: RC circuit
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 
Current
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
Ex.
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
computed 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 switching cannot
be t = 0.
Example…
The first switch moves from a to b at t=0.
The second switch closed at t=1ms. Find the
current, i for t ≥ 0.
•
•
Step 1: the current at t=0- is determined
by assuming that the first switch at point
a and the second switch opened for along
time. Therefore, the current, i(0-)=10A.
At t=0, the time constant of RL circuit is
obtained 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
RLC circuit: consists of resistor,
inductor and capacitor
• Second order response : response
from RLC circuit
• Type of RLC circuit:
•
1.
2.
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:
1

2 RC
0 
1
LC
• Summary
Parameter
Terminology
s1, s2
Characteristic
equation
α
0
Neper frequency
Resonant radian
frequency
Value in natural
response
s1     2  0
2
s2     2  0
1

2 RC
0 
1
LC
2
•
•
1.
2.
3.
The two roots, s1 and s2 depend on
the value of α and ωo.
3 possible conditions are:
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 expression:
v  A1 e  A2 e
s1t
s2t
•
The constants of A1 and A2 can be
obtained from these two equations:

v(0 )  A1  A2  (1)

dv(0 )
 s1 A1  s2 A2  (2)
dt
•
Whereby, vo+ is obtained from the circuit at
the instant when the switching occurred.
•
The value of v(0+) = V0 and the 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.
The values of A1 and A2 is obtained by
solving the equation below:

v(0 )  A1  A2


dv(0 ) iC (0 )

 s1 A1  s2 A2
dt
C
4.
Substitute the value for s1, s2, A1 dan A2
to determine the expression for v(t) for t
≥ 0.
• The
response of overdamped
voltage for v(0) = 1V and i(0) = 0
Underdamped voltage response
ωo2 > α2, the roots of the
characteristic equation are
complex and the response is
underdamped.
• When
• 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
number.
The
two simultaneous equation that
determine B1 and B2 are:

v(0 )  V0  B1


dv(0 ) iC (0 )

 1B1  d B2
dt
C
Response of underdamped voltage
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:
1
s1  s2    
2 RC
•
The solution for the voltage is
v(t )  D1t e
t
 D2 e
t
•The two simultaneous equation that
determine D1 and D2 are,

v(0 )  V0  D2


dv(0 ) iC (0 )

 D1  D2
dt
C
Response of the critically damped
voltage 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
are two approaches to solve
the equation:
– direct
approach
– 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
determined from:
iL (0)
and
diL (0)
dt
• The
solution for a second-order
differential equation consists of
the forced response and the
natural response.
•
If If and Vf is the final value of the
response function, the solution for the
step function can be written as:
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 similar as the natural
response of a parallel RLC circuit as
both circuits are described by the
similar form of differential equations.
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
RLC,
frequency (α) for series
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 for
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 are 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
2
into KVL equation,
d vC R dvC vC
V



2
dt
L dt
LC LC
• Three
possible solutions 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
untuk
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 1400t cos 4800 t 
V
vC (t )  
  14 e 1400t sin 4800 t



for
t0