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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
First-Order and Second-Order Response of RL
and RC 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 opened3. At t > 0, switch is not close for
along time
•For t ≤ 0, v(t) = V0.
dtRCtv
tdvRC
tv
dt
tdvRC
tv
dt
tdvR
tv
dt
tdvC
ii Rc
1
)(
)(
)()(
0)()(
0)()(
0
RCt
eVtv
RC
t
V
tv
tRC
Vtv
dvRC
duu
dvRCu
du
tv
V
t
0
0
0
)(
0
)(
)(ln
)0(1
ln)(ln
11
1
0
voltage
For t ≥ 0:
• Thus for t > 0,
RCt
eVtv 0)(RCt
eR
V
R
tvtic
0)()(
RCt
eVCtvCtW2
20
2
2
1)(
2
1)(
The graph of the natural response of RC circuit
0
0)(
0
0
teV
tVtv
RCt
• The time constant, τ = RC and thus,
t
eVtv 0)(
• 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 opened3. At t > 0, switch is not close for
along time
• For t ≤ 0, i(t) = I0
For t > 0,
dvL
R
u
du
dtL
R
ti
tdi
tiRdt
tdiL
tiRdt
tdiL
tiRtv
)(
)(
)()(
0)()(
0)()(
LRt
tti
i
eiti
L
Rt
i
ti
tL
Riti
dvL
Rduu
)0()(
)0(
)(ln
)0()0(ln)(ln
10
)(
)0(
current
• Thus for t > 0,
LRteIti 0)(
LRteRI
Rtitv
0
)()(
LRteLI
tiLtw
220
2
2
1
)(2
1)(
Example…The switch in the circuit has been
closedfor 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.
Solutiona) 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:
1010402eqR
sec2.010
2
eqR
L
• The expression of inductor current, iL(t) as,
020
)0()(5
tAe
eitit
L
t
b) The current in the 40Ω resistor can be determine using current division,
4010
100 Lii
• 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, 04)( 5
0 tAeti t
c) The voltage V0 directly obtain using Ohm’s law
0160
40)(5
00
tVe
itVt
d) The power dissipated in the 10Ω resistor is
02560
10)(
10
20
10
tWe
Vtp
t
•The total energy dissipated in the 10Ω resistor is
J
dtetW t
256
2560)(0
1010
•The initial energy stored in the 2H inductor is
J
iLW
40040022
1
)0(2
1)0( 2
• Therefore the percentage of energy dissipated in the 10Ω resistor is,
%64100400
256
• 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
First-Order and Second-Order Response of RL
and RC 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,
s
s
s
s
Vtv
tdvdt
RC
tvV
tdvdt
RC
dt
tdvRCtvV
tRitvV
)(
)(1
)(
)(1
)()(
)()(
s
s
ss
s
VV
Vtv
RC
t
VVVtvRC
t
Vu
dudv
RC
0
0
)(ln
ln)(ln
1
t
RCt
eVVV
eVVVtv
ss
ss
0
0)(
voltage
• Thus for t >0
t
t
eVVV
VV
VV
eVVVV
sn
sf
nf
ss
0
0
Where
•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
Natural
total
• The current for step response of RC circuit
t
t
t
eR
V
R
V
eVVR
eVVC
dt
dvCti
s
s
s
0
0
0
1
)(1
)(
t
eiti )0()(
Step response of RL Step response of RL circuitcircuit
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 closed3. At t > 0, switch is not open for
along time
• i(t)=I0 for t ≤ 0. For t > 0,
RV
RV
s
s
s
s
s
i
didt
L
R
ti
tdidt
L
Rdt
tdi
R
Lti
R
Vdt
tdiLtiRV
tvtiRV
)(
)(
)()(
)()(
)()(
RV
RV
RV
RV
ti
IR
V
t
RV
s
s
ss
s
s
I
ti
L
Rt
ItiL
Rt
u
dudv
L
R
u
dudv
L
R
0
0
)(
0
)(ln
ln)(ln
0
LR
ss tR
VR
V eIti 0)(
Current
•Thus,
0
0)(
0
0
teI
tIti
LR
ss tR
VR
V
0
0)(
)(
0
teIRV
tdt
tdiLtv
LRt
s
QuestionThe switch in the circuit has been open for along time. The initial
chargeon the capacitor is zero. At t = 0,
the switch is closed. Find the
expression fora) i(t) for t ≥ 0b) v(t) when t ≥ 0+
Answer (a)
• Initial voltage on the capacitor is zero. The current in the 30kΩ resistor is
mA350
)20)(5.7()0(i
• 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
ms5
10)1.0(10)3020( 63
• Thus, the expression of the current i(t) for t ≥ 0 is
0tmAe3
e3
e)0(i)t(i
t200
105t
3
t
Answer (b)
•The initial value of voltage is zero and the final value is
V150)20)(5.7(Vf
• The capacitor vC(t) is
0tVe150150
e)1500(150
eVVV)t(v
t200
t200
f0fC
t
• Thus, the expression of v(t) is
0tV)e60150(
e)3)(30(e150150)t(vt200
t200t200
• 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
First-Order and Second-Order Response of RL
and RC Circuit
General solutions for natural and step response
• There is common pattern for voltages, currents and energies:
t
eVVVtv ff 0)(
t
eIIIti ff 0)(
t
eWWWtW ff
2
0)(
The general solution can be compute as:
t
exxxtx ff 0)(
Write out in words:
ttanconstime
t
e
iablevar
theofvalue
finalthe
iablevar
theofvalue
initialthe
iablevar
theofvalue
finalthe
timeoffunction
aasiablevar
unknownthe
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.
• 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
First-Order and Second-Order Response of RL
and RC 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 andsecond switch closed at t=1ms. Find thecurrent, 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 .ms1
R
L
• Thus the current for 0 ≤ t ≤ 1ms is,
Ae10i t1000
• At t=t1=1ms,
A
eti
68.3
10)( 11
When switch is closed at t=1ms, the equivalent resistance is 1Ω. Then,
ms21
2
R
L1
• Thus i for t ≥ 1ms is
Ae
etiitt
tt
1
1
1
1
)(
)(
1
68.3
)(
The graph of current for t ≥ 0
• 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
First-Order and Second-Order Response of RL
and RC Circuit
Second order response for RLC c
• RLC circuit: consist of resistor, inductor and capacitor
• Second order response : response from RLC circuit
• Type of RLC circuit:1. Series RLC2. Parallel RLC
Natural response of parallel RLC
•Summing all the currents away from node,
01
0 0 t
dt
dvCIvd
LR
V
•Differentiating once with respect to t,
01
2
2
dt
vdC
L
v
dt
dv
R
01
2
2
LC
v
dt
dv
RCdt
vd
•Assume thatsteAv
02 ststst eLC
Ae
RC
AseAs
0LC
1
RC
sseA
equationsticcharacteri
2st
•Characteristic equation is zero:
012
LCRC
ss
•The two roots:
LCRCRCs
1
2
1
2
12
1
LCRCRCs
1
2
1
2
12
2
•The natural response of series RLC:
tsts eAeAv 2121
• The two roots:
RC2
1
20
21 s
20
22 s
•where:
LC
10
•Summary
20
22
20
21
s
s
RC2
1
0LC
10
Parameter Terminology Value in natural response
s1, s2Charateristic
equation
α Neper frequency
Resonant radian frequency
• The two roots s1 and s2 are depend on α and ωo value.
• 3 possible condition is: 1. If ωo < α2 , the voltage response
is overdamped2. If ωo > α2 , the voltage response
is underdamped3. If ωo = α2 , the voltage response
is critically damped
Overdamped voltage Overdamped voltage responseresponse
• overdamped voltage solution
tsts eAeAv 2121
• The constant of A1 dan A2 can be obtain from,
21)0( AAv
2211
)0(AsAs
dt
dv
• The value of v(0+) = V0 and initial value of dv/dt is
C
i
dt
dv C )0()0(
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:
4. Substitute the value for s1, s2, A1 dan A2 to determine the expression for v(t) for t ≥ 0.
21)0( AAv
2211
)0()0(AsAs
C
i
dt
dv C
•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,
• ωd : damped radian frequency
dj
j
s
220
2201 )(
djs 2
•The underdamped voltage response of a parallel RLC circuit is
tsineB
tcoseB)t(v
dt
2
dt
1
•The constants B1 dan B2 are real not complex number.
10)0( BVv
211
)0()0(BB
C
i
dt
dvd
C
The two simultaneous equation that determine B1 and B2 are:
Example of underdamped voltage response for v(0) = 1V
and i(0) = 0
Critically Damped voltage response
•A circuit is critically damped when ωo
2 = α2 ( ωo = α). The two roots of the characteristic equation are real and equal that is,
RCss
2
121
• The solution for the voltage is tt eDetDtv 21)(
21
20
)0()0(
)0(
DDC
i
dt
dv
DVv
C
•The two simultaneous equation needed to determine D1 and D2 are,
Example of the critically damped voltage response for
v(0) = 1V and i(0) = 0
The step response of a The step response of a parallel RLC circuitparallel RLC circuit
The step response of a The step response of a parallel RLC circuitparallel RLC circuit
•From the KCL,
Idt
dvC
R
vi
Iiii
L
CRL
•Because dt
diLv
2
2
dt
idL
dt
dv L•We get
•Thus,
Idt
idLC
dt
di
R
Li LLL
2
2
LC
I
LC
i
dt
di
RCdt
id LLL 1
2
2
•There is two approach to solve the equation that is direct approach and indirect approach.
Indirect approachIndirect approach
•From the KCL:
Idt
dvC
R
vvd
L
t0
1
•Differentiate once with respect to t:
01
2
2
dt
vdC
dt
dv
RL
v
01
2
2
LC
v
dt
dv
RCdt
vd
• The solution for v depends on the roots of the characteristic equation:
tsts eAeAv 2121
tsineB
tcoseBv
dt
2
dt
1
tt eDetDv 21
•Substitute into KCL equation :
tstsL eAeAIi 21
21
teB
teBIi
dt
dt
L
sin
cos
2
1
ttL eDetDIi 21
Direct approach
•It is much easier to find the primed constants directly in terms of the initial values of the response function.
212121 ,,,B,, DDBAA
•The primed constants could be find from
and dt
diL )0()0(Li
•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,
responsenaturaltheas
formsametheofFunctionIi f
responsenaturaltheas
formsametheoffunctionVv f
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,
01
00 Vdi
Cdt
diLRi
t
•Differentiate once with respect to t ,
02
2
C
i
dt
idL
dt
diR
02
2
LC
i
dt
di
L
R
dt
id
•The characteristic equation for the series RLC circuit is,
012 LC
sL
Rs
•The roots of the characteristic equation are,
LCL
R
L
Rs
1
22
2
2,1
@
20
22,1 s
•Neper frequency (α) for series RLC,
sradL
R/
2
And the resonant radian frequency,
sradLC
/1
0
The current response will be overdamped, underdamped or critically damped according to,
220 22
0
220
• Thus the three possible solutions fo the currents are,
tsts eAeAti 2121)(
teB
teBti
dt
dt
sin
cos)(
2
1
tt eDetDti 21)(
Step response of series RLC
•The procedures is the same as the parallel circuit.
Series RLC circuit
Cvdt
diLiRv
•Use KVL,
•The current, i is related to the capacitor voltage (vC ) by expression,
dt
dvCi C
•Differentiate once i with respect to t
2
2
dt
vdC
dt
di C
•Substitute into KVL equation,
LC
V
LC
v
dt
dv
L
R
dt
vd CCC 2
2
•Three possible solution for vC are,
tstsfC eAeAVv 21
21
teB
teBVv
dt
dt
fC
sin
cos
2
1
ttfC eDetDVv 21
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 isapplied to the circuit. The value of the
resistoris 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?
Solution1. 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
dt
diLv L thus 0
)0(
dt
diL
3. From the circuit elements,
812
20 1016
)25)(25(
101
LC
srad
RC
/105
)25)(400)(2(
10
2
1
4
9
82 1025
•Thus the roots of the characteristic equation are real,
srad
s
srad
s
/00080
103105
/00020
103105
442
441
4. The inductor current response will be overdamped.
tstsfL eAeAIi 21
21
•Two simultaneous equation:
0)0(
0)0(
2211
21
AsAsdt
di
AAIi
L
fL
mAAmAA 832 21
• Numerical solution:
0
8
3224)(
80000
20000
tfor
mAe
eti
t
t
L
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:
sradjs
sradj
s
/48001400
/48001400
4.01.0
10
2.0
280
2.0
280
2
62
1
•The roots are complex, so the voltage response is underdamped. Thus:
04800sin
4800cos48
14002
14001
tteB
teBv
t
tC
•No energy is stored in the circuit initially, so both vC(0) and dvC(0+)/dt are zero. Then:
12
1
140048000)0(
480)0(
BBdt
dv
Bv
C
C
•Solving for B1’and B2’yields,
VB
VB
14
48
2
1
•Thus, the solution for vC(t),
0
4800sin14
4800cos4848)(
1400
1400
tfor
Vte
tetv
t
t
C