To find the step response of an RC circuit
0
)]()([)()( 0
tt
evtvvtv
The final capacitor voltage v()
The initial capacitor voltage v(t0)
The time constant = RC
To find the step response of an RL circuit
0
)]()([)()( 0
tt
eitiiti
The final capacitor voltage v()
The initial capacitor voltage v(t0)
The time constant = L/R
2
Lecture 6 DC Circuits Transient Circuits – 2nd Order circuits
Contents
• Examples of 2nd order RLC circuit
• The source-free series RLC circuit
• The source-free parallel RLC circuit
• Step response of a series RLC circuit
• Step response of a parallel RLC
3
Examples of Second Order RLC circuits
What is a 2nd order circuit?
4
A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements of different type or the same type.
Source-Free Series RLC Circuits
5
• The solution of the source-free series RLC circuit is called as the natural response of the circuit.
• The circuit is excited by the energy initially stored in the capacitor and inductor.
02
2
LC
i
dt
di
L
R
dt
idThe 2nd order of expression
How to derive and how to solve?
Source-Free Series RLC Circuits: to derive the equation
vR=Ri
KVL:
To eliminate the integral
02
2
C
i
dt
idL
dt
diR
dt
diLvL
t
C idtC
v1
01
t
idtCdt
diLRi
Source-Free Series RLC Circuits: to solve the equation
02
2
LC
i
dt
di
L
R
dt
id
Assume: the solution has exponential form, stAei
To be determined
To be determined
012
LCs
L
RsAe st
Define: L
R
2
LC
10
Neper frequency/ damping factor
Resonant frequency/ undamped natural factor 02 2
0
2 ss
Characteristic equation
Solution: 2
0
2
2,1 s Natural Frequencies
Linear equation: tstseAeAti 21
21)(
Still solving… tsts
eAeAti 21
21)( How to get A?
Initial conditions:
The inductor: 0)0( Ii The capacitor: 0
01
)0( VidtC
v
021)0( IAAi 0)0( 0
0
Vdt
diLRi
t
tstsesAesA
dt
tdi21
2211
)(
0022110 VsAsALRIA1 and A2
What if s1 = s2 (when ) 0 Ls
VRIAA 00
21
I0 and V0
not independent
02 2
2
2
idt
di
dt
id 0
i
dt
dii
dt
di
dt
d teAi
dt
di 1
tetAAti 12)(To get Ai
02 2
02
2
idt
di
dt
id
9
There are three possible solutions for the following 2nd order differential equation:
1. If > o, over-damped case
tstseAeAti 21
21)( 2
0
2
2,1 swhere
2. If = o, critical damped case
tetAAti )()( 122,1swhere
3. If < o, under-damped case
)sincos()( 21 tBtBeti dd
t where
22
0 d
Source-Free Series: RLC Circuits
Damped natural frequency
Finding initial and final values
• Initial values: – Passive convention
• The polarity of voltage
• The direction of current
– Variables that cannot change abruptly • The capacitor:
• The inductor:
10
00
)(,
)(),0(),0(
tt dt
tdi
dt
tdviv
)(),( vi
)0()0( vv
)0()0( ii Focus on these variables
• Final values: – DC steady state
• The capacitor: Open circuit
• The inductor: short circuit
The time just before a switching event which
takes place at t = 0
Example
If R = 10 Ω, L = 5 H, and C = 2 mF in 8.8, find α, ω0, s1 and s2.
What type of natural response will the circuit have?
11
Answer: underdamped
12
L
R 10
10
LC 95.912
0
2
2,1 js
Example
The circuit shown below has reached steady state at t = 0-.
If the make-before-break switch moves to position b at t = 0, calculate i(t) for
t > 0.
12
Answer: i(t) = e–2.5t[5cos1.6583t – 7.538sin1.6583t] A
Step-Response Series: RLC Circuits
13
• The step response is obtained by the sudden application of a dc source.
The 2nd order of expression
LC
v
LC
v
dt
dv
L
R
dt
vd s2
2
The above equation has the same form as the equation for source-free series RLC circuit.
• The same coefficients (important in determining the frequency parameters).
• Different circuit variable in the equation.
Step-Response Series: RLC Circuits
14
The solution of the equation should have two components: the transient response vt(t) & the steady-state response vss(t):
)()()( tvtvtv sst
The transient response vt is the same as that for source-free case
The steady-state response is the final value of v(t).
• vss(t) = v(∞)
The values of A1 and A2 are obtained from the initial conditions:
• v(0) and dv(0)/dt.
tsts
t eAeAtv 21
21)( (over-damped)
t
t etAAtv )()( 21 (critically damped)
)sincos()( 21 tAtAetv dd
t
t
(under-damped)
Example
Having been in position for a long time, the switch in the circuit below is moved to position b at t = 0. Find v(t) and vR(t) for t > 0.
15
Answer: v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e–2t]} V
vR(t)= [2.31sin3.464t]e–2t V
Lecture 7 AC Circuits Sinusoids and Phasors
Appendix
Source-Free Parallel: RLC Circuits
18
The 2nd order of expression
011
2
2
vLCdt
dv
RCdt
vd
0
0 )(1
)0( dttvL
IiLet
v(0) = V0
Apply KCL to the top node:
t
dt
dvCvdt
LR
v0
1
Taking the derivative with
respect to t and dividing by C
Source-Free Parallel: RLC Circuits
LCRCv
dt
dv
dt
vd 1and
2
1 where0 2 0
2
02
2
19
There are three possible solutions for the following 2nd order differential equation:
1. If > o, over-damped case
tstseAeAtv 21 )( 21 2
0
2
2,1 swhere
2. If = o, critical damped case
tetAAtv )( )( 12 2,1swhere
3. If < o, under-damped case
)sincos()( 21 tBtBetv dd
t where
22
0 d
Example
Refer to the circuit shown below. Find v(t) for t > 0.
20
• Please refer to lecture or textbook for more detail elaboration.
Answer: v(t) = 66.67(e–10t – e–2.5t) V
Step-Response Parallel: RLC Circuits
21
• The step response is obtained by the sudden application of a dc source.
The 2nd order of expression
It has the same form as the equation for source-free parallel RLC circuit. • The same coefficients (important in determining the frequency
parameters). • Different circuit variable in the equation.
LC
I
LC
i
dt
di
RCdt
id s1
2
2
Step-Response Parallel : RLC Circuits
22
The solution of the equation should have two components: the transient response vt(t) & the steady-state response vss(t):
)()()( tititi sst
The transient response it is the same as that for source-free case
The steady-state response is the final value of i(t).
iss(t) = i(∞) = Is
The values of A1 and A2 are obtained from the initial conditions:
i(0) and di(0)/dt.
tsts
t eAeAti 21
21)( (over-damped)
t
t etAAti )()( 21 (critical damped)
)sincos()( 21 tAtAeti dd
t
t
(under-damped)
Example
Find i(t) and v(t) for t > 0 in the circuit shown in circuit shown below:
23
• Please refer to lecture or textbook for more detail elaboration.
Answer: v(t) = Ldi/dt = 5x20sint = 100sint V