rl and rc circuits first- order response electric circuits ent 161/4
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RL and RC circuits first-order response
Electric circuitsENT 161/4
RL and RC circuit original response
A first-order circuit is characterized by a first-order differential equation. This circuit contain resistor and capacitor or inductor in one close circuit.
The natural response of a circuit refers to the behaviour ( in terms of voltages and currents) of the circuit itself, with no external sources of excitation.
RL circuit: circuit that have resistor and inductor.
RC circuit: circuit that have resistor and capacitor.
Natural response RC circuit
Consider these three condition :
1. At initially, t=0 -, switch doesn’t change for some time
2. At initial, t=0 +, switch doesn’t change for some time
3. At final condition, t→∞, switch doesn’t change for some time
Known 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
Therefore t ≥ 0:
For t > 0,
RCt
eVtv 0)(RCt
eR
V
R
tvtiR
0)()(
RCt
eVCtvCtW2
20
2
2
1)(
2
1)(
Natural response RC circuit graph
0
0)(
0
0
teV
tVtv
RCt
This show that the voltage response of the RC circuit is an exponential decay of the initial voltage. constant, τ = RC
t
eVtv 0)(
Constant τ define how fast voltage reach stable condition :
Natural response RL circuit
Consider these three condition :
1. At initially, t=0 -, switch doesn’t change for some time
2. At initial, t=0 +, switch doesn’t change for some time
3. At final condition, t→∞, switch doesn’t change for some time
Known at t ≤ 0, i(t) = I0
Therefore 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
For t > 0,
LRteIti 0)(
LRteRI
Rtitv
0
)()(
LRteLI
tiLtw
220
2
2
1
)(2
1)(
EXAMPLESwitch in circuit for some time before
open at t=0. Calculate
a) IL (t) at t ≥ 0
b) I0 (t) at t ≥ 0+
c) V0 (t) at t ≥ 0+
d) Total energy percentage that stored in inductor 2H that absorb by 10Ω resistor.
Answer
a) Switch close for some time until t=0, known voltage at inductor should be zero at t = 0-. Therefore, initial current at inductor was 20A at t = 0-. Thus iL (0+) also become 20A, because immidiate changes for current didn’t exist in inductor.
Equivalent resistance from inductor and constant time
1010402eqR
saatR
L
eq
2.010
2
Therefore, current iL (t)
020
)0()(5
tAe
eitit
L
t
b) Current at resistor 40Ω could be calculate by using current divider law,
4010
100 Lii
This current was at t ≥ 0+ because i0 = 0 at t = 0-. Inductor will
become close circuit when switch open immediately and produce changes immediately at current i0.
Therefore, 04)( 5
0 tAeti t
c) V0 could be calculate by using Ohm’s Law,
0160
40)(5
00
tVe
itVt
d) Total power absorb by 10Ω resistor
02560
10)(
10
20
10
tWe
Vtp
t
Total energy absorb by 10Ω resistor
J
dtetW t
256
2560)(0
1010
Initial energy stored at 2H inductor
J
iLW
40040022
1
)0(2
1)0( 2
Therefore, energy percentage that absorb by 10Ω resistor
%64100400
256
Step response RC circuit The step response of a circuit is its
behaviour when the excitation is the step function, which may be a voltage or a current source.
Consider these three condition :
1. At initially, t=0 -, switch doesn’t change for some time
2. At initial, t=0 +, switch doesn’t change for some time
3. At final condition, t→∞, switch doesn’t change for some time
Known at 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)(
voltan
Current for step response RC circuit
t
t
t
eR
V
R
V
eVVR
eVVC
dt
dvCti
s
s
s
0
0
0
1
)(1
)(
t
eiti )0()(
Then, for t >0
t
t
eVVV
VV
VV
eVVVV
sn
sf
nf
ss
0
0
Where
Vf = Force voltage or known as steady-state response
Vn = known as transient response is the circuit’s temporary response that will die out with time.
Step response RC circuit graph
force
Natural
total
Step Response RL circuitStep Response RL circuit
Consider these three condition :
1. At initially, t=0 -, switch doesn’t change for some time
2. At initial, t=0 +, switch doesn’t change for some time
3. At final condition, t→∞, switch doesn’t change for some time
known i(t)=I0 at 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
Finally,
0
0)(
0
0
teI
tIti
LR
ss tR
VR
V
0
0)(
)(
0
teIRV
tdt
tdiLtv
LRt
s
Question Switch in those circuit was at x
position for some time. At t=0, switch move to position y immediately. Calculate,
(a) Vc(t) at t ≥ 0(b) V0 (t) at t ≥ 0+(c) i0 (t) at t ≥ 0+(d) Total energy absorb by 60kΩ resistor.
Answer (a)
Constant for circuit
ms40)1080)(105.0( 36
VC (0)=100V
equivalent resistor = 80kΩ.
Then, VC(t) for t ≥ 0:
0100)( 25 tVetV tC
Answer (b)V0 (t) could be calculate by using
voltage divider law.
060
)(80
48)(
25
0
tVe
tVtV
t
C
Answer (c)
current i0 (t) can be calculated by using ohm’s law
01060
)()( 25
30
0 tmAetV
ti t
Answer (d)
Power absorb by 60kΩ resistor
060
1060)()(50
32060
tmWe
titpt
k
Total energy
mJ
dttiW k
2.1
1060)( 3
0
2060
Second-order RLC circuit
RLC circuit : circuit that contain resistor, inductor and capacitor
Second-order response : response from RLC circuit
Type of RLC circuit:1. RLC series circuit2. RLC parallel circuit
Original response for parallel RLC circuit
Take total current flows out from node
01
0 0 t
dt
dvCIvd
LR
V
differential of t,
01
2
2
dt
vdC
L
v
dt
dv
R
01
2
2
LC
v
dt
dv
RCdt
vd
Take steAv
02 ststst eLC
Ae
RC
AseAs
012
equationsticcharacteri
st
LCRC
sseA
Characteristic equation known as zero :
012
LCRC
ss
The root of the characteristic equation are
LCRCRCs
1
2
1
2
12
1
LCRCRCs
1
2
1
2
12
2
Response for RLC parallel circuit
tsts eAeAv 2121
The root of the characteristic equation are
RC2
1
20
21 s
20
22 s
where:
LC
10
summarize
20
22
20
21
s
s
RC2
1
0 LC
10
Parameter Terminology Value in natural response
s1, s2characteristic
equation
α frequency Neper
resonant radian frequency
Roots solution s1 and s2 depend on α and
Consider these cases saperately: 1. If < α , voltage response was
overdamped 2. If > α , voltage response was
underdamped3. If = α , voltage response was
critically damped
0
0
0
0
Overdamped voltage responseOverdamped voltage response
Solution for overdamped voltage
tsts eAeAv 2121
constant A1 and A2 can be determined from the initial conditions v(0+) and
Known, dt
dv )0(
21)0( AAv
2211
)0(AsAs
dt
dv
Here v(0+) = V0 and initial value for
dv/dt was
C
i
dt
dv C )0()0(
Solution for overdamped natural response, v(t) :
1. Calculate characteristic equation, s1 and s2, using R, L and C value.
2. Calculate v(0+) and
using circuit analysis. dt
dv )0(
3. Calulate A1 and A2 by solve those equation
4. Insert s1, s2, A1 and A2 value to calculate overdamped natural response for t ≥ 0.
21)0( AAv
2211
)0()0(AsAs
C
i
dt
dv C
Example for overdamped natural response for v(0) = 1V and
i(0) = 0
Underdamped voltage response
At > α2, root of the characteristic equation was complex number and those response called underdamped.
0
Therefore
ωd : damped radian frequency
dj
j
s
220
2201 )(
djs 2
underdamped voltage response for RLC parallel circuit was
teB
teBtv
dt
dt
sin
cos)(
2
1
constant B1 and B2 was real number.
10)0( BVv
211
)0()0(BB
C
i
dt
dvd
C
Solve those two linear equation
to calculate B1 and B2,
Example for underdamped voltage response for v(0) = 1V and i(0) = 0
Critically Damped Voltage Response
Second-order circuit was critically damped when = α . When circuit was critically damped, two characterictic root equation was real and same,
RCss
2
121
0
Solution for voltage tt eDetDtv 21)(
21
20
)0()0(
)0(
DDC
i
dt
dv
DVv
C
•Linear equation to calculate D1 and D2 value
Example for critically damped voltage response at v(0) = 1V and i(0) = 0
Step response RLC parallel Step response RLC parallel circuitscircuits
Step response RLC parallel Step response RLC parallel circuitscircuits
From Kirchhoff current law
Idt
dvC
R
vi
Iiii
L
CRL
Known
Therefore
dt
diLv
2
2
dt
idL
dt
dv L
Have,
Idt
idLC
dt
di
R
Li LLL
2
2
LC
I
LC
i
dt
di
RCdt
id LLL 1
2
2
There are two solution to solve the equation, direct approach and indirect approach.
Indirect approachIndirect approach
From Kirchhoff’s current law:
Idt
dvC
R
vvd
L
t0
1
Differential
01
2
2
dt
vdC
dt
dv
RL
v
01
2
2
LC
v
dt
dv
RCdt
vd
Depend on characteristic equation root :
tsts eAeAv 2121
teB
teBv
dt
dt
sin
cos
2
1
tt eDetDv 21
Insert in Kirchhoff’s current law eq:
tstsL eAeAIi 21
21
teB
teBIi
dt
dt
L
sin
cos
2
1
ttL eDetDIi 21
Direct approach
It’s simple to calculate constant for the equation
directly by using initial value response function.
212121 ,,,B,, DDBAA
Constant of the equation could be calculate from
and dt
diL )0()0(Li
The solution for a second-order differential equation with a constant forcing function equals the forced response plus a response funtion identical in form to the natural response.
If and Vf represent the final value of the response function. The final value may be zero,
responsenaturaltheas
formsametheoffunctionIi f
responsenaturaltheas
formsametheoffunctionVv f
Natural response for RLC Series circuit
The procedures for finding the natural or step responses of a series RLC circuit are the same as those used to find the natural or step responses of a parallel RLS circuit, because both circuits are described by differential equations that have the same form.
RLC series circuit
Summing the voltages around the closed path in the circuit,
01
00 Vdi
Cdt
diLRi
t
differential
02
2
C
i
dt
idL
dt
diR
02
2
LC
i
dt
di
L
R
dt
id
Characteristic equation for RLC series circuit
012 LC
sL
Rs
Characteristic equation root
LCL
R
L
Rs
1
22
2
2,1
@
20
22,1 s
Neper frequency (α) for RLC series circuit
sradL
R/
2
and resonant radian frequency was,
sradLC
/1
0
Current response
Overdamped
Underdamped
critically damped
220
220
220
Three kind of solution
tsts eAeAti 2121)(
teB
teBti
dt
dt
sin
cos)(
2
1
tt eDetDti 21)(
Step response for RLC series circuit
The procedures for finding the step responses of series RLC circuit are the same as those used to find the step response of a parallel RLC circuit.
RLC series circuit
Cvdt
diLiRv
Using Kirchhoff’s voltage law,
Current known as,
dt
dvCi C
Differential for current
2
2
dt
vdC
dt
di C
Insert in Voltage current law equation
LC
V
LC
v
dt
dv
L
R
dt
vd CCC 2
2
Three solution that possibly for vC
tstsfC eAeAVv 21
21
teB
teBVv
dt
dt
fC
sin
cos
2
1
ttfC eDetDVv 21
Contoh 1 Tenaga awal yang disimpan oleh litar
berikut adalah sifar. Pada t = 0, satu punca arus DC 24mA diberikan kepada litar. Nilai untuk perintang adalah 400Ω.
1. Apakah nilai awal untuk iL?2. Apakah nilai awal untuk ?
3. Apakah punca-punca persamaan ciri?4. Apakah ungkapan numerik untuk iL(t)
pada t ≥ 0?
dt
diL
Jawapan1. Tiada tenaga yang disimpan
dalam litar sebaik sahaja punca arus digunakan, maka arus awal bagi induktor adalah sifar. Induktor mencegah perubahan yang serta-merta pada arus induktor, oleh itu iL (0)=0 sebaik sahaja suis dibuka.
2. Nilai awal voltan kapasitor adalah sifar sebelum suis dibuka, oleh itu ia akan sifar sebaik sahaja suis dibuka. Didapati:
dt
diLv L maka 0
)0(
dt
diL
3. Dari elemen-elemen dalam litar, diperolehi
812
20 1016
)25)(25(
101
LC
srad
RC
/105
)25)(400)(2(
10
2
1
4
9
82 1025
Oleh kerana , maka punca-punca persamaan ciri adalah nyata
srad
s
srad
s
/00080
103105
/00020
103105
442
441
4. sambutan arus induktor adalah overdamped dan persamaan penyelesaian adalah
tstsfL eAeAIi 21
21
Dua persamaan serentak:
0)0(
0)0(
2211
21
AsAsdt
di
AAIi
L
fL
mAAmAA 832 21
Penyelesaian numerik:
0
8
3224)(
80000
20000
tuntuk
mAe
eti
t
t
L
Contoh 2
Tiada tenaga disimpan dalam inductor 100mH atau kapasitor 0.4µF apabila suis di dalam litar berikut ditutup. Dapatkan vC (t) untuk t ≥ 0.
JawapanPunca-punca persamaan ciri:
sradjs
sradj
s
/48001400
/48001400
4.01.0
10
2.0
280
2.0
280
2
62
1
Punca-punca adalah kompleks, maka sambutan voltan adalah underdamped. Oleh itu, diperolehi voltan vC :
04800sin
4800cos48
14002
14001
tteB
teBv
t
tC
Pada awalnya, tiada tenaga tersimpan dalam litar, maka:
12
1
140048000)0(
480)0(
BBdt
dv
Bv
C
C
Selesaikan untuk dan 1B
2B
VB
VB
14
48
2
1
penyelesaian untuk vC (t)
0
4800sin14
4800cos4848)(
1400
1400
tuntuk
Vte
tetv
t
t
C
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