Transcript Analogue-Theory-and-Circuit-Analysis

2. Analogue Theory and Circuit
Analysis
2.1 Steady-State (DC) Circuits
2.2 Time-Dependent Circuits
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Electrical systems have two
main objectives:
• To gather, store, process, transport, and present
information
• To distribute and convert energy between various
forms
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Electrical Engineering Subdivisions
•
•
•
•
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Communication
systems
Computer systems
Control systems
Electromagnetics
• Electronics
• Power systems
• Signal processing
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Electrical Current
Electrical current is the time rate of flow of
electrical charge through a conductor or
circuit element. The units are amperes (A),
which are equivalent to coulombs per
second (C/s).
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Electrical Current
dq(t )
i (t ) 
dt
t
q(t )   i (t )dt  q(t0 )
t0
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Direct Current
Alternating Current
When a current is constant with time, we
say that we have direct current,
abbreviated as dc. On the other hand, a
current that varies with time, reversing
direction periodically, is called alternating
current, abbreviated as ac.
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.
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Voltages
The voltage associated with a circuit
element is the energy transferred per unit
of charge that flows through the element.
The units of voltage are volts (V), which are
equivalent to joules per coulomb (J/C).
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Transients
The time-varying currents and voltages
resulting from the sudden application of
sources, usually due to switching, are
called transients. By writing circuit
equations, we obtain integrodifferential
equations.
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DC STEADY STATE
The steps in determining the forced
response for RLC circuits with dc sources
are:
1. Replace capacitances with open
circuits.
2. Replace inductances with short
circuits.
3. Solve the remaining circuit.
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CAPACITANCE
ic
dvc
ic C
dt

vc
C[Farads]

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d
I
n
D
C
S
te
a
d
y
S
ta
te
; 
0
d
t
iC
0
O
p
e
n
C
ir
c
u
it
S
S
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CAPACITANCE
q  Cv
t
qt    i t dt  qt0 
t0
dv
iC
dt
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t
1
v t    i t dt  v t0 
C t0
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INDUCTANCE
iL
diL
vL L
dt

vL

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L[Henries]
d
I
n
D
C
S
te
a
d
y
S
ta
te
; 
0
d
t
v
0
S
h
o
r
tC
ir
c
u
it
L
S
S
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INDUCTANCE
di
v t   L
dt
t
1
i t    v t dt  i t0 
L t0
1 2
wt   Li t 
2
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SWITCHED CIRCUITS
•
•
•
•
•
Circuits that Contain Switches
Switches Open or Close at t = t0
to = Switching Time
Often choose to = 0
Want to Find i’s and v’s in Circuit
Before and After Switching Occurs
• i(to-), v(t0-); i(to+), v(t0+)
• Initial Conditions of Circuit
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INITIAL CONDITIONS
•
•
•
•
•
•
•
C’s and L’s Store Electrical Energy
vC Cannot Change Instantaneously
iL Cannot Change Instantaneously
In DC Steady State; C => Open Circuit
In DC Steady State; L => Short Circuit
Use to Find i(to-), v(t0-); i(to+), v(t0+)
Let’s do an Example
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EXAMPLE
i1
S
w
itc
h
O
p
e
n
sa
tt
0
 v1 
12 V
2
i3
 v3 
i2
2
 4
1 F
v2

 iC
vC

A
s
s
u
m
e
S
w
i
t
c
h
h
a
s
b
e
e
n
C
l
o
s
e
dF
i
n
d
I
n
i
t
i
a
lC
o
n
d
i
t
i
o
n
s

+
f
o
r
a
l
o
n
g
t
i
m
e
b
e
f
o
r
e
t

0
i
'
s
a
n
d
v
'
s
a
tt
0
a
n
d
t
0
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EXAMPLE
At t  0 :
i3
C
S
te
a
d
yS
ta
te
i1 D
 v1  Sw
itchC
losed  v3 
2
iC
i2
2
12 V
 4 
v C C
v2
 Open Ckt



1
2
i
(0
) 0
i
(0
)

0
3
C
i
(
0
)

i
(
0
)


3
A
1
2

2

2
v
(0
) 0
3


v
(
0
)

v
(
0
)

3
x
2

6
V 


1
2
v
(
0
)v

(
0
)v

(
0

V
C
2
3 )6

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
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EXAMPLE

At t  0 :
i1


i(
0)

0

v
(
0)
1
1
 v1  Sw
itchO
pen
2
i2
2
12 V
i3  i C
 v3 
 4
v 2 1 F


vC

6


v
(
0
)

v
(
0

6
V
i
(
0
)


i
(
0
)


i
(
0
)

1
A
C
C )
4

2


v
(
0
)

2
x
1

2
V
v
(
0

4
x
(1

)
4
V
2
3 )

2
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
3

C
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EXAMPLE
In
itia
lC
o
n
d
itio
n
s
t  0
i1  3 A
t  0
i1  0 A
i2  3 A
i2  1 A
i3  0 A
i3   1 A
iC  0 A
iC   1 A
v1  6 V
v1  0 V
 6 V
v2  2 V
v
2
v3  0 V
v
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C
 6 V
v3  4 V
v
C
 6 V
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1ST ORDER SWITCHED DC
CIRCUITS
W
illL
ookat1 O
rderC
ircuits(C
ircuitsw
ith
1Cor1L
)w
ithS
w
itchedD
CInputsT
om
orrow
W
illU
seInitialC
onditionstoH
elpU
sS
olvethe
st
st
1O
rderD
ifferentialE
quationR
elatingthe
O
utputtotheInput
T
odayW
eW
illL
ookata1 O
rderC
ircuitusing
P
S
pice
st
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ACTIVITY 13-1
R
100 V
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20 nF
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
vC

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ACTIVITY 13-1
• Charge a 20 nF Capacitor to 100 V
thru a Variable Resistor, Rvar:
• Let’s Use a Switch that Closes at t = 0
• Rvar = 250k, 500k, 1 M
• Circuit File Has Been Run:
• C:/Files/Desktop/CE-Studio/Circuits/act_52.dat
• But Let’s Practice Using Schematics
and Take a Quick Look
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ACTIVITY 13-1
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Circuit File
v 1 0 dc 100
R 1 2 {R}
C 2 0 20n ic=0
.param R=250k
.step param R list 250k 500k 1meg
.tran .1 .1 uic
.probe
.end
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ACTIVITY 13-1
P
r
in
tG
r
a
p
h
so
fv
s
.tim
e
Cv
F
illin
T
a
b
le
f
o
rA
c
tiv
ity
1
3
1
H
a
n
d
I
n
f
o
rG
r
a
d
in
g
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Transient Behaviour
 Introduction
 Charging Capacitors and Energising Inductors
 Discharging Capacitors and De-energising Inductors
 Response of First-Order Systems
 Second-Order Systems
 Higher-Order Systems
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Introduction
 So far we have looked at the behaviour of systems
in response to:
– fixed DC signals
– constant AC signals
 We now turn our attention to the operation of
circuits before they reach steady-state conditions
– this is referred to as the transient response
 We will begin by looking at simple RC and RL
circuits
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Charging Capacitors and
Energising Inductors
• Capacitor Charging
 Consider the circuit shown here
– Applying Kirchhoff’s voltage law
iR

v

V
– Now, in a capacitor
dv
i C
dt
– which substituting gives
d
v
CR

v
V
d
t
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 The above is a first-order differential equation
with constant coefficients
 Assuming VC = 0 at t = 0, this can be solved to give
t
t
CR

v

V
(
1

e
)

V
(
1

e
)
 Since i = Cdv/dt this gives (assuming VC = 0 at t = 0)
– where I = V/R
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t
t
CR

i
I
e

I
e
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 Thus both the voltage and current have an
exponential form
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• Inductor energising
 A similar analysis of this circuit gives
Rt
t
L

v

V
e
V
e
Rt
t
L

i
I
(
1

e
)
I
(
1

e
)
where I = V/R
–
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 Thus, again, both the voltage and current have
an exponential form
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Discharging Capacitors and
De-energising Inductors
• Capacitor discharging
 Consider this circuit for
discharging a capacitor
– At t = 0, VC = V
– From Kirchhoff’s voltage law
iR

v
0
– giving
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d
v
CR

v
0
d
t
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 Solving this as before gives
t
t
CR

v

V
e

V
e
t
t
CR

i

I
e


I
e
– where I = V/R
–
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 In this case, both the voltage and the current
take the form of decaying exponentials
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• Inductor de-energising
 A similar analysis of this
circuit gives
Rt
t
L

v


V
e


V
e
Rt
t

i
IeL
Ie
– where I = V/R
– see Section 18.3.1
for this analysis
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 And once again, both the voltage and the
current take the form of decaying exponentials
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 A comparison of the four circuits
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Response of First-Order Systems
 Initial and final value formulae
– increasing or decreasing exponential waveforms
(for either voltage or current) are given by:
–
–
–
–
–
where Vi and Ii are the initial values of the voltage and current
where Vf and If are the final values of the voltage and current
the first term in each case is the steady-state response
the second term represents the transient response
the combination gives the total response of the arrangement
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• The input voltage to the following CR network
undergoes a step change from 5 V to 10 V at time t =
0. Derive an expression for the resulting output
voltage.
•
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•
Here the initial value is 5 V and the final value is 10 V. The time
constant of the circuit equals CR = 10  103 20  10-6 = 0.2s.
Therefore, from above, for t  0
•

t/
vV

(
V

V
)
e
f
i
f

t/0
.2
10
(510
)e

t/0
.2
10
5
e
volts
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• The nature of exponential curves
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•
Response of first-order
systems to a square
waveform
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•
Response of first-order
systems to a square
waveform of different
frequencies
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Key Points
 The charging or discharging of a capacitor, and the energising
and de-energising of an inductor, are each associated with
exponential voltage and current waveforms
 Circuits that contain resistance, and either capacitance or
inductance, are termed first-order systems
 The increasing or decreasing exponential waveforms of firstorder systems can be described by the initial and final value
formulae
 Circuits that contain both capacitance and inductance are usually
second-order systems. These are characterised by their
undamped natural frequency and their damping factor
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