1
U1. Transient circuits
response
Circuit Analysis, Grado en Ingeniería de Comunicaciones
Curso 2016-2017
Philip Siegmann ([email protected])
Departamento de Teoría de la Señal y Comunicaciones
Index
• Recall
• Goals and motivation
• Linear differential equation (Recall)
• The homogeneous solution
• 1st and 2nd order linear differential equations
• Examples of 2nd order circuits
• Initial conditions
• Transient responses (example on 2nd order circuit)
• Cause of the transient responses
• Analysis using Laplace transform
Recall
• Relation between i(t) and v(t) for the
passive elements R,L,C
– For R:
– For C:
– For L:
• Energy in these elements:
– Dissipated in R:
– Stored in C and L:
Circuit Analysis / Transient circuits response / Recall
)()( tRitv
dt
tdvCtitCvtq C
CC
)()()()(
dt
tdiLtv L
L
)()(
)(tiL
L vL(t)
)(tiC
C vC(t)
)(tiR
R vR(t)
2)(2
1)( tvCtW CC 2)(
2
1)( tiLtW LL
dttiRtW RR )()( 2
dt
tdqti
dq
tdWtv
)()(,
)()(
Goals
• We want to solve circuits for whatever applied
source (not only DC and Sinusoidal Steady
State)
– Direct resolution in the time domain
– Resolution using Laplace transforms
• In particular we want to understand what
happens when an abrupt change takes place in
the circuit, which will produce the transient
response.
4
Circuit Analysis / Transient circuits response / Goals
Motivation
• Signal transmission
Transient response
Circuit Analysis / Transient circuits response / Motivation
S.S.S.
6
Examples of 2nd order circuits
• RLC-serial (http://en.wikipedia.org/wiki/RLC_circuit )
• RLC-parallel
dt
tde
Lti
LCdt
tdi
L
R
dt
tid
tetqCdt
tdiLtRi
tetvtvtv
g
g
gCLR
)(1)(
1)()(
)()(1)(
)(
)()()()(
2
2
dt
tdi
Ctv
LCdt
tdv
RCdt
tvd
dt
tdi
dt
tvdCtv
Ldt
tdv
R
titititidt
d
g
g
gCLR
)(1)(
1)(1)(
)()()(
1)(1
)()()()(
2
2
2
2
(Energy conservation)
(Charge conservation)
Circuit Analysis / Transient circuits response / Example 2nd order circuits
7
Transient response
• Response of a circuit (voltage or current) when
an abrupt change happens (e.g. switching)
• Time evolution until achieving a new equilibrium
• The transition function follows exponential
variations (decreasing or increasing, fluctuating
or no fluctuating)
• They are solutions of linear differential equations
Circuit Analysis / Transient circuits response / Introduction
8
General solution
),()()(
),()(...)()(
01
1
1
tytyty
tgtyadt
tyda
dt
tyd
ph
n
n
nn
n
yh(t) is the solution for g(t)=0: the Complementary, natural or homogeneous
solution. Gives the transient behavior of the circuit due to the passive elements.
It is dependent of the initial conditions.
yp(t) is a particular solution for the given source or forcing function g(t). The
particular solution looks like the forcing function, e.g.:
- If g(t) is constant, then yp(t) is constant
- If g(t) is sinusoidal, then yp(t) is sinusoidal (i.e. the S.S.S.)
The homogeneous solution decreases exponentially so that
y(t) yp(t)
Linear differential equation of order n:
Circuit Analysis / Transient circuits response / Linear differential equation
9
The homogeneous solution
yh(t) is the solution for g(t)=0 (external energy
supply =0)
The solution has the form: yh(t)=A·exp(st) (“Ansatz”)
Circuit Analysis / Transient circuits response / The homogeneous solution
stkst
k
k
AsAdt
dee :since
n
k
ts
k
ts
n
tsts
h
n
n
n
n
kn AAAAty
ssss
asas
1
21
21
0
1
1
ee...ee)(
},...,,{
equation polynomial sticcharacteri ,0...
21
A1, A2, … are obtained with the initial (and/or boundary) conditions
characteristic polynomial equation
(“Eigenwerte”)
10
1st and 2nd order linear
differential equations
2
0
1
012
2
2
),()()()(
n
n
a
a
tgtyadt
tdya
dt
tyd
is called the damping ratio (accounts for the energy loss)
n is called the natural frequency (maximum energy storage)
(http://en.wikipedia.org/wiki/Damping)
1
),()()(
0
0
a
tgtyadt
tdy
is a time constant (how fast yh(t) decreases)
Circuit Analysis / Transient circuits response / 1st nd 2nd order differential equations
1st order
2nd order
For circuits containing
one energy storage
element (C or L)
For circuits containing
two independent energy
storage element
11
Example of 2º orden circuits
• RLC-serial (http://en.wikipedia.org/wiki/RLC_circuit )
• RLC-parallel
L
R
LC
dt
tde
Lti
LCdt
tdi
L
R
dt
tid
n
n
g
2,
1
)(1)(
1)()(2
2
CRLC
dt
tdi
Ctv
LCdt
tdv
RCdt
tvd
n
n
g
2
1,
1
)(1)(
1)(1)(2
2
Circuit Analysis / Transient circuits response / Ejemplos circuitos 2º orden
12
Initial conditions
• For each energy storage element we need an
initial condition (at t=t0):
– For C: vC(t=t0) = V0
– For L: iL(t=t0) = I0
Circuit Analysis / Transient circuits response / Initial conditions
2)(2
1)( tiLtW LL 2)(
2
1)( tvCtW CL
vC(t) iL(t)
(t)
E(t)
13
Condiciones iniciales
• When an abrupt change happens at t=t0 in a
circuit, there is always continuity in the
variation of the energies in C and L
– There is continuity in the voltage at C:
– There is continuity in the current trough L:
Circuit Analysis / Transient circuits response / Initial conditions
)()( 00
tvtv CC
)()( 00
titi LL
Justo después de t0 Justo antes de t0
(not so for the current iC(t))
(not so for the voltage vL(t))
14
Example of transient response
01
:equationsticCharacteri
,e)(
form theof solutionsth equaion wi free) (source Homogeneus
,0)(1)()(
,0)(
)(1
)(
0For
)0(
0)0(
:0atconditionsInitial
2
2
LCs
L
Rs
Ati
tiLCdt
tdi
L
R
dt
tid
dt
tdiLtq
CtRi
t
Ev
i
t
2
st
gC
Discharge of the capacitor
Characteristic equation:
Circuit Analysis / Transient circuits response / Transient responses
dt
tdiLtRitvtvtv LRC
)()()()()(
Once i(t) is known we can obtain vC(t):
For t ≥0
15
Type of solutions of the
homogeneous equation
.
)(
ee)()(
)0()0(
ee)(0)0(
Overdamped:ee)(
)1(,4If
)s :(units ,12
4
1,2being,0
21
21
1
0
121
21
0
2
1
1-20
2
11
2,1
2
0101
2
21
21
21
ssL
Eti
ssL
EAE
dt
tdiLRiv
AtiAAi
AAti
aa
aaas
LCa
L
Raasas
tsts
g
g
g
t
C
tsts
tsts
nn
nn
Overdamped
s2+a1s +a0= 0
Circuit Analysis / Transient circuits response / Transient responses
0 2 4 6
x 10-3
-0.025
-0.02
-0.015
-0.01
-0.005
0
t (s)
i(t
), (
A)
0 0,002 0,004 0.006
t (s)
vc(t
), (
V)
Eg
The voltage at C decreases. The
current first increases (in opposite
direction) then decreases to zero
because of the energy dissipated in R.
16
Type of solutions of the
homogeneous equation
.2
4sine
4
2)(
4
2)(0)0(
,...π2,π,00sin)0(
frequency natural undamped theis 2
4: here
dUnderdampe:2
4sine)(
)1(,4If
2
102
2
102
10
1
0
1
2
10
d
2
1021
210
2
1
1
1
taa
aaL
Eti
aaL
EAE
dt
tdiLv
Ai
aa
taa
Ati
ssaa
ta
gg
g
t
C
ta
damped natual frequency
Underdamped
Circuit Analysis / Transient circuits response / Transient responses
The oscillation is a consequence of the
energy exchange between C and L. First
it moves from C to L, on the way some
energy is dissipated by R. Once the
remaining energy is stored in L it moves
back to C dissipating again some energy
in R and so on until all the energy is
dissipated by R.
17
Type of solutions of the
homogeneous equation
.1
sin)()(0)0(
,...π2,π,00sin)0(
Undamped:sin)(
)0(,0If
.e)()(0)0(
e)(0)0(
damped Critically:e)(
22)1(,4If
0
1
0
1
01
1
2
2
0
21
21
1210
2
1
1
1
1
tLCL
CEti
aL
EAE
dt
tdiLv
Ai
taAti
a
tL
Eti
L
EAE
dt
tdiLv
tAtiAi
tAAti
L
Rassaa
gg
g
t
C
ta
g
g
g
t
C
ts
ts
n
Undamped
Critically damped
Circuit Analysis / Transient circuits response / Transient responses
undamped natual frequency
18
Type of solutions of the
homogeneous equation
Discharge of the capacitor
Circuit Analysis / Transient circuits response / Transient responses
19
Cause of the transient respinse
• RLC-serie. The resulting damping ratio is:
• RLC-parallel. The resulting damping ratio is:
L
R
n
2
CR n
2
1
Circuit Analysis / Transient circuits response / Causa of the transient response
L C
)(ti
R
v(t)
Is proportional to R
because the energy
dissipated in R increases
with R:
)(ti L C R
Is inversely proportional
to R since the energy
dissipated in R decreases
with R:
R
tvtRitp RR
)()()(
22
)()( 2 tRitpR
20
Transient circuit’s analysis
using Laplace transforms
• By using Laplace transforms the circuits
can be solved much easily:
– No differential equation has to be obtained
– We will solve algebraic instead of differential
equations
– No need to perform the tedious operations to
calculate the constants (A1,A2,…) of the
solution
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
21
Laplace transform (L )
• The solutions are superposition's of
exponential decreasing functions starting from the initial instant (t=0)
• The Laplace transform (L) allows to transform
the differential equation into an algebraic
equation with coefficients A(s)
0
e)()()( dttytysA stL
n
ts
nnAty e)(
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
22
Some properties of L
L is lineal, F(s)= L[f(t)]
L of a derivation
Time translation
Translation in s domain
Theorem of the final and initial value
)()()()()()( 213213 sbFsaFsFtbftaftf
)0(...)0(')0()()( )1(21)( nnnnn ffsfstfstf LL
0)(if,)(e)( 000
ttftfttf
st LL
)(e)( tfasF taL
.)(lim)(lim
,)(lim)(lim
0
0
ssFtf
ssFtf
st
st
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
Ohm’s and Kirchhoff law’s are still valid in L-domain
Differential equation are transformed into algebraic equations
23
L of some functions • Step function displaced in time
• Slope
• Rec function
• Dirac delta function
• Periodic functions with period T
stst
s
AtuAttAu 00 e)(e)( 0
LL
stst
s
AttuAttuttA 00 e)(e)()(
200
LL
stst
s
AtuA
s
AttutuA 00 e1)(e)()( 0
LL
ststttt 00 e1)(e)( 0
LL
Ts
nTs
n
tgtgtf
e1
)(e)()(
LLL
)()()()(,)()( TtututftgnTtgtfn
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
t0
A
t0 0
A
t0
t0
24
L of some functions
• Exponential
• Sine
• Cosine
• In practice we will use a table with the most common inverse Laplace transforms (L-1) used for the resolution of the proposed problems
as
at
1
eL
22
cosas
sat
L
22
sinas
aat
L
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
25
Resolution using Laplace
• The circuits will be solved in the Laplace
domain:
– Draw the circuit in the transformed domain, for this
the initial conditions are deeded:
– The transformed circuit is then solved using the
known methods, thus, by applying the Kirchhoff laws
to the transformed currents and voltages:
– Ones you know the Laplace transformed voltage or
current, the inverse Laplace transform is applied to
get the currents and voltages in the time domain
)()(,)()( 11 sVtvsIti LL
).0(),0( CL vi
).(),( sVsI
(*): Convention: Laplace transformed variables in capital letters
(*)
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
26
L transform of the inductor
.)0(
)()(
)()(
)],([)(
)],([)(
00
s
isILsdte
dt
tdiLdtetvsV
tvsV
tisI
LL
stLst
LL
LL
LL
L
L
L
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
27
L transform of the capacitor
.)0(
)(1
)(1)0(1
)(1
)()(
)],([)(
)],([)(
00
s
vsI
sC
sIss
q
Cdtetq
CdtetvsV
tvsV
tisI
CC
C
stst
CC
CC
CC
L
L
L
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
28
L transform of the generator
• After the switching:
For example:
• If the switching happens at t00, perform a time translation by defining: t’=t-t0. This has to be taken into account by performing the inverse transform.
L
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
G(s)
t=t0
22
cosas
sat
L
L sinat
a
s2 a2,
L E
E
s,
29
Initial conditions
• The initial condition we need are:
– The currents trough each of the coils just after
the switching takes place: iL(t0+)
– The voltages at the terminals of the capacitors
after the switching takes place: vC(t0+)
• If they are not known, then they have to be
calculated by a previous (before the
switching) resolution of the circuit,
– Remember: iL(t0+)=iL(t0
), vC(t0+)=vC(t0
)
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
30
Resolution in the L-domain
• The transformed circuit is solved by applying the mesh or nod methods of the transformed voltages or the currents which depend on the variable s.
• The following expression for the voltage or current has to be obtained:
where the denominator is the characteristic equation:
From which the roots (s1 and s2) are obtained, these allows: – Predict the kind of solution
– Find the inverse in the Laplace transform table
01
2
)()(
asas
sfsY
0
2
112,101
2 42
10 aaasasas
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
31
Predicting of the kind of solution
• If the roods are real and different
• For equal and real roots
• For complex conjugated roots: s1,2=pjq
• Imaginary roots: s1,2=jq
tsts
AAtyssss
sf
asas
sfsY 21
1
ee)()()(
)( 11
2101
2
L
qtAetyqps
sf
asas
sfsY pt sin)(
)()()(
1
22
01
2
L
ts
tAAtyss
sf
asas
sfsY 1
1
e)()()(
)( 212
101
2
L
qtAtyqs
sf
asas
sfsY sin)(
)()()(
1
22
01
2
L
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
32
Inverse Laplace transform (L-1)
• The roots of the characteristic equation allows:
– To know beforehand the kind of solution
– It makes easier to find the inverse Laplace transform
in the inverse Laplace transform table.
• Do not forget: if there was a time shifting,
substitute t’ by t-t0 in the inverse transform.
• The obtained solution is defined for a time
interval after the switching.
• Check the initial condition.
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
33
Example 1
• In the circuit of the figure, the switcher is in position (1) since t = -. At t =/2 the switcher changes to position (2). Obtain the temporal evolution of iL(t).
Data: eg(t)=2cos2t V, R=2, L=1H, C=0.5F.
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
34
Example 2
• In the circuit of the figure, the switcher is in position (1) since t = -. At t =0 the switcher switches to position (2). Obtain the temporal evolution of vC(t).
Data: e(t)=10V, Rg=R1=1, R2=2, L=1H, C=2F.
Circuit Analysis / Transient circuits response / Analysis using Laplace transform
35
Simulation with 5Spice
Circuit Analysis / Transient circuits response / Simulation
36
Simulación con 5Spice
Circuit Analysis / Transient circuits response
