RESPONSE OF FIRST-ORDER RC AND RL CIRCUITS · 2011. 3. 3. · C.T. Pan 5 7.1 The Natural Response...
Transcript of RESPONSE OF FIRST-ORDER RC AND RL CIRCUITS · 2011. 3. 3. · C.T. Pan 5 7.1 The Natural Response...
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RESPONSE OF RESPONSE OF FIRSTFIRST--ORDER RC ORDER RC AND RL CIRCUITS AND RL CIRCUITS
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit
7.2 The Natural Response of an RL Circuit 7.2 The Natural Response of an RL Circuit
7.3 Singularity Functions7.3 Singularity Functions
7.4 The Step Response of RC and RL Circuit7.4 The Step Response of RC and RL Circuit
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit
Resistive Circuit => RC CircuitResistive Circuit => RC Circuit
algebraic equations => differential equationsalgebraic equations => differential equations
Same Solution Methods (a) Nodal AnalysisSame Solution Methods (a) Nodal Analysis
(b) Mesh Analysis(b) Mesh Analysis
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit
The solution of a linear circuit, called dynamic response, can be decomposed into
Natural Response + Forced Response
or in the form of
Steady Response + Transient Response
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit
nThe natural response is due to the initial condition of the storage component ( C or L).nThe forced response is resulted from external input ( or force).nIn this chapter, a constant input (DC input) will be considered and the forced response is called step response.
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit
Example 1 : Two forms of the first order circuit for Example 1 : Two forms of the first order circuit for natural responsenatural response
Find i(t)Find v(t)
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit
Example 1 : (cont.) Four forms of the first orderExample 1 : (cont.) Four forms of the first ordercircuit for step response circuit for step response
TH
TH
RV
A capacitor connected to a Thevenin equivalent
A capacitor connected to a Norton equivalent
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit
Example 1 : (cont.)Example 1 : (cont.)
TH
TH
RV
An inductor connected to a Thevenin equivalent
An inductor connected to a Norton equivalent
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC CircuitExample 2 Example 2
0t ≥
nodal analysis 0
1 0
c Ri idvC vdt R
+ =
+ =
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC CircuitExample 2 (cont.)Example 2 (cont.)
R
ic iR
v
v0t ≥
1
+ -
0
1 0
1
( ) , 0
(0 ) = (0 ) =
( )
tRC
0t
RC
characteristic root S,
CSR
SRC
v t Ke t
From the initial conditionv v V
v t V e
−
−
+ =
= −
∴ = ≥
∴ =
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC CircuitExample 2 (cont.)Example 2 (cont.)
0
(t) = , 0( ) , 0.36788
3 , 4.979% 5%
5 , 0.674% 1%
t
0
RC time constant
v V e tv ttV
t
t
τ
τ
τ
τ
τ
−
=
≥
= =
= <
= <
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC CircuitExample 2 (cont.)Example 2 (cont.)It is customary to assume that the capacitor is fully discharged after five time constants.
0( )( ) , 0t
RVv ti t e t
R Rτ
−
= = ≥
The power dissipated in R is 22
0( )t
RVp t vi eR
τ−
= =
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC CircuitExample 2 (cont.)Example 2 (cont.)
The energy dissipated in R is2
20
0
20
1( ) (1 )2
1( )2
tt
R
R
w t pdt CV e
w CV
τ−
= = −
∞ =
∫
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC CircuitExample 3 :Example 3 : Find Find vvcc , , vvxx , and , and iix x for t > 0for t > 0
This circuit contains only one energy storage element.Step 1. Use Thevenin theorem to find the equivalent RTH
looking into a-b terminals.
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC CircuitExample 3 :Example 3 : Find Find vvcc , , vvxx , and , and iix x for t > 0for t > 0
Step 2. Find vc(t)Step 3. Replace vc(t) as a voltage source in the original
circuit and solve the resistive circuit.
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit
1. (8 12) 5 4THStep R = + = ΩP
2.Step
2.5
c
(0) 15V
0.1 040.4 s
(t) = 15 V, 0t
c
c c
v
dv vdtRC
v e t
τ−
=
+ =
= =
≥
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7.1 The Natural Response of an RC Circuit7.1 The Natural Response of an RC Circuit 3.Step
2.5
2.5
12( ) ( )8 12
9 , 0
( )( ) 0.7512
t
t
x c
xx
By using voltage divider principle
v t v t
e V t
v ti t e A
−
−
=+
= × ≥
= =
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL CircuitExample 4 Example 4
Li
Remember that vRemember that vcc(t) and i(t) and iLL(t) are continuous functions (t) are continuous functions for bounded inputs.for bounded inputs.
00 , (0 ) IS
S
Vt i iR
−≤ = = @
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL Circuit
0t ≥
0
0
L R
mesh analysis
v v
diL Ridt
+ =
+ =
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL Circuit
0
0
( ) , 0
(0 ) (0 ) I
( ) I , 0
,
t
R tLi t Ke t
i i
i t e t
L time constantR
τ
τ
−
−
+ −
= ≥
= =
∴ = ≥
@
0
characteristic equation
LS R
RSL
+ =
∴ = −
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2
2
0
20
200
( ) I
I
1( ) ( ) LI (1 )2
1( )2
t
t
t
R
R R
t
R R
2R 0
v t Ri Re
p v i Re
w t p t dt e
w LI
τ
τ
τ
−
−
−
= =
= =
= = −
∞ =
∫
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL Circuit
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL CircuitExample 5 : Example 5 : ii(0)=10A , find (0)=10A , find ii(t) and (t) and iixx(t) (t)
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL CircuitStep 1. Find the Thevenin equivalent circuit looking
into a-b terminals.
Apply is , find e. Then RTH = e / is
si i= −3
4 23
4 4 2
s
s s
e i e i
e ei i
−+ =
+ + =
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL Circuit3 14 4
13
s
s
e i
ei
=
∴ = Ω
Step 2.
13
2-3
10.5 030.5 1.5 S1/ 3
(t)=10e A , t 0t
mesh equationdi idtLR
i
τ
+ =
= = =
∴ ≥
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL CircuitStep 3 Replace the inductor with an equivalent Step 3 Replace the inductor with an equivalent
current source with i(t) and solve the current source with i(t) and solve the resistive circuit.resistive circuit.
-2 t3(t) = 10 e Ai
-2 tL 3x
v -5 = = e V , t 02 3
i∴ ≥Ω
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL CircuitFor this problem, since 2For this problem, since 2ΩΩ is in parallel with the is in parallel with the inductor, it is trivial to getinductor, it is trivial to get
L
-2 t3
-2 t3
-2 t3
d(t) = L dt
d = 0.5 (10 e )dt
-2 = 0.5 10 ( )e3
-10 = e V3
iv
× ×
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL CircuitExample 6 : The switch has been closed for a long Example 6 : The switch has been closed for a long
time. At t=0, it is opened. Find time. At t=0, it is opened. Find iixx ..
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL Circuit
-1
-x
- -1
+ -
40V(0 ) = = 8 A2+(12//4)
(0 ) = 012 3(0 ) = (0 ) = 8 = 6 A
12+4 4 Initial codition (0 ) = (0 ) = 6 A
i
i
i i
i i
× ×
∴
-t=0
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL Circuit
+t 0≥
Step 1Step 1 RRTHTH = (4+12) // 16 = 8 = (4+12) // 16 = 8 ΩΩ
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL Circuit
Step 2Step 2 mesh analysismesh analysis
-4t
+
-4t
d2 + 8 = 0dt
(t) = K e , t 0
(0 ) = 6 A
(t) = 6 e A , t 0
i i
i
ii
≥
∴ ≥
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7.2 The Natural Response of an RL Circuit7.2 The Natural Response of an RL CircuitStep 3Step 3 Replace L with an equivalent current source, Replace L with an equivalent current source,
and find and find iixx’’ solve the resistive circuit.solve the resistive circuit.
x
-4t
4+12(t) = (t)12+4+16
= 3e A , t 0
i i∴
≥
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7.3 Singularity Functions 7.3 Singularity Functions Switching functions are convenient for describing Switching functions are convenient for describing the switching actions in circuit analysis.the switching actions in circuit analysis.
They serve as good approximations to the switching They serve as good approximations to the switching signals.signals.
unit step function unit step function unit impulse function unit impulse function unit ramp function unit ramp function
uu(t)(t)δδ(t)(t)rr(t)(t)
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7.3 Singularity Functions 7.3 Singularity Functions
00
0
Definition 0 , t<0
u(t) = 1 , t>0
undefined at t=0
or more general0 , t<t
u(t-t ) = 1 , t>t
Example 7 Example 7
VV0 0 uu((tt--tt00)) is applied is applied to ato a--b terminalsb terminals
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7.3 Singularity Functions 7.3 Singularity Functions
0 , t<0d(t) = u(t) = undefined , t=0dt
0 , t>0
Definition
δ
Also known as the delta functionAlso known as the delta function+
-
+0-
0
00
t0t
(t) dt = 1
(t-t ) dt = 1
or
or
δ
δ
∫
∫
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7.3 Singularity Functions 7.3 Singularity Functions The unit impulse function is not physically The unit impulse function is not physically realizable but is a very useful mathematical tool.realizable but is a very useful mathematical tool.
An approximation as d 0→
Another approximation as d 0→
12d
1d
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7.3 Singularity Functions 7.3 Singularity Functions
b0 0a
b b0 0 0a a
b0 0a
0
f(t) (t-t )dt = f(t )
f(t) (t-t )dt = f(t ) (t-t )dt
= f(t ) (t-t )dt = f(t )
δ
δ δ
δ
∫
∫ ∫
∫
Q
0Shifting property, if t [a , b]∈
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7.3 Singularity Functions 7.3 Singularity Functions
t- r(t) = u(t)dt = t u(t)
or0 , t 0
r(t) = t , t 0
Definition
∞
≤ ≥
∫
t
t
d u(t)(t) = , u(t) = (t)dtdt
d r(t)u(t) = , r(t) = u(t)dtdt
Note that
δ δ∫
∫
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7.3 Singularity Functions 7.3 Singularity Functions
( )aExample 8 Example 8
( )b
(t) = 10 [u(t-2) - u(t-5)]v
d (t) = 10 (t-2) - 10 (t-5)dtv
δ δ
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7.3 Singularity Functions 7.3 Singularity Functions ( )c
(t) = 10 u(t) -20 u(t-2)+10u(t-4)i
( )d
(t) = 5 (t)-5 (t-2)-10u(t-2)v r r
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
When a dc voltage (current) source is suddenly
applied to a circuit , it can be modeled as a step
function , and the resulting response is called
step response .
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL CircuitsExample 9 Example 9
( ) 00 , , dvV V i C choose v as unknowndt
− = =
Step 1 Mesh Analysis
, 0SdvRC v V tdt
+ = ≥
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL CircuitsStep 2 Solving the differential equation
( )
( )
0
, 0( )
hh
tRC
h
pp S
p S
advRC vdt
v t Ke tb
dvRC v V
dtv V
−
+ =
= ≥
+ =
=
homogeneous solution
particular solution
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
( )
( ) ( )
( )( )
0
0
0
0
(c) com plete so lu tion + In itial condition v t
0 0
, 0
, 0
h p
tR C
s
s
t
s s
v v
K e V
v v V
K V VV t
v tV V V e tτ
−
+ −
−
= +
= +
= =
∴ = −
<= + − ≥
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
i(t)sV
R1( ) ( )( )
, 0
0 , ( )
.
tRC
o s
ts o RC
So
dvi t C C V V edt RC
V V e tR
VFor V then i oR
C is initially short circuited
−
−
+
−= = −
−= ≥
= =
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
Example 10 Example 10 Before t=0 , the circuit is under steady state . At t=0 , the switch is moved to B . Find v(t) , t > 0
3kΩ 4kΩ
5kΩ
5kΩ
3kΩ
(0 )v −
For t<0 , the circuit shown:
( ) 50 24V 15V3 5
v − = × =+
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
4kΩ
( )( )
( ) ( )
( ) ( )
3 3
3 3
4 10 0.5 10 30
30 , 04 10 0.5 10 2
0 0 15
30 15 30 , 0
p h
t
t
mesh analysisdv v Vdt
v t v v
Ke tRC second
v v V
v t e t
τ
τ
τ
−
−
−
+ −
−
× × × + =
= +
= + >
= = × × × =
= =
∴ = + − ≥
For t>0
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
10Ω
20Ω14
F
Example 11 Example 11 Before the switch is open at t=0, the circuit is in steady state. Find i(t) & v(t) for all t.
For t<0 , then
20Ω
10Ω
( )
( )
10 , 010 1 , 010
v t V t
i t A A t
= <
= − = − <
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
( ) ( )0 0 10v v V+ −= =
( )
( ) ( )3 35 5
14
20 1 20 , 03 4
20 20 10 20 , 0
c
t t
dv dvi t Cdt dt
dv v V tdt
v t Ke e t− −
= =
× + = >
= + = + − >
14
F
14
F
203
Ω
For t>0
Thevenin equivalent circuit
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
( ) ( )
( ) ( )
0.6
0.6
0.6
: ( )20
1 , 010 , 0
20 10 , 0
1 , 0
1 , 0
(0 ) 1 (0 ) 2
t
t
t
From the original circuitv dvKCL i t C
dte t
V tv t
e V t
A ti t
e A t
Note thati A i A
−
−
−
− +
= +Ω
= + >
<= − ≥− <= + ≥
= − ≠ =
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
Example 12 Example 12
( )
, 0
0 0
TH THdiL R i V tdt
i
+ = ≥
=
Mesh analysisMesh analysis
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
( ) ( ) ( ) ( )
(0 ) (0 ) 00 :
1
tTH
h pTH
TH
TH
Vi t i t i t i t KeR
characteristic equation i iLS R Note L is equivalent to open circuit
RSL
τ
τ
−
+ −
= + ∴ = +
= =+ =
= − = − ( ) (1 ) , 0
( ) ( ) ( )
( )
tTH
THt t
h TH
THp
TH
Vi t e tR
dii t Ke V t L V e u tdt
Vi tR
τ
τ τ
−
− −
= − ≥
= = =
=
Solution:Solution:
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
( ) (1 ) , 0 t
TH
TH
Vi t e tR
τ−
= − ≥ ( ) ( )t
THdiV t L V e u tdt
τ−
= =
TH
TH
VR
THV
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
Example 13 Example 13
t<0 , steady state
t=0 , switch is opened
10 (0 ) 52
i A− = =
13
H
For t<0For t<0
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
13
H 13
H
For t For t ≧ 00
1 5 10 , 0 ( )3
15 2 , 0
1 ( ) , 15
( ) 2
h p
t
t
h
p
Mesh Analysisdi i V t i t i idt
s Ke A t
i t Ke
i t A
τ
τ τ
−
−
+ = ≥ = +
= − = + ≥
= =
=
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
(0 ) (0 ) 5initial condition
i i A+ −= =
( ) 2 3 , 0t
i t e tτ−
∴ = + ≥
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
( ) :
0 0
: (0 ) 10 (2 3)5 15
1 (2 3 )3
1 , 0
(0 ) 15 , .
t
t
Note that if V is chosen for solution
V V
KVL V V V
di dor V L edt dt
e t
V V same answer
τ
τ
τ
−
+
−
− +
+
=
= − + = −
= = +
= − ≥
= −
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
Example 14 :Example 14 : SS11 is closed at t=0 Find is closed at t=0 Find ii(2)=(2)=??
SS22 is closed at t=4 is closed at t=4 ii(5)=?(5)=?
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL CircuitsFor t For t < 00
t t < 0 , 0 , ii=0 , open circuit=0 , open circuit
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL CircuitsFor For 0+≦ t < 4S
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL CircuitsFor For 0+≦ t < 4S
2
8
5 10 40 , 0 4
(0 ) (0 ) 05 1
10 2( ) 4(1 ) , 0 4
(4 ) 4(1 ) 4
t
t
di i V t Sdt
i i ALR
i t e A t Si e A
τ
+
+ −
− +
− −
+ = ≤ ≤
= =
= = =
∴ = − ≤ ≤
= − ≈
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL CircuitsFor For t ≧ 4+
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL CircuitsFor For t ≧ 4+
(4 ) 440 10 2 10 204 2
4 224 2 6 63 3
5 1522 / 3 22
TH
TH
TH
i A
V V
R
L SR
τ
− =−
= × + =+
= Ω Ω + Ω = + = Ω
= = =
P
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL CircuitsFor For t ≧ 4+
223
Ω
( )22 ( 4)15
22 ( 4)15
30(4 ) 4 4 ( ) , 411
22 305 20 (4 ) 43 11
30 14 4 411 11
( ) , 4
t
t
h
i i A i t Ke tdi i V i A KdT
K
i t
whe T t
Ke t
re
− −+ − +
+
− −
= = = + ≥
+ = = = +
= − =
=
=
≥
−22 ( 4)1530 14 ( ) , 4
11 1120 60 30( ) 2.727
22 / 3 22 11
t
p
i t e t
i t
− −+ +∴ = + ≥
= = = =
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7.4 The Step Response of RC and RL Circuits7.4 The Step Response of RC and RL Circuits
2
22( 4)
0 , 0( ) 4(1 ) , 0 4
30 14 , 411 11
t
t
In Summary
ti t e t
e t
−
− −
≤
= − ≤ ≤
+ ≥
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n Objective 1 : Be able to determine the natural response of both RC and RL circuits.
n Objective 2 : Be able to find the step response of both RC and RL circuits.
n Objective 3 : Know and be able to use the singularity functions.
n Objective 4 : Be able to analyze circuits with sequential switching.
C.T. PanC.T. Pan 6565
SummarySummary
C.T. PanC.T. Pan 6666
SummarySummary
Chapter Problems : 7.77.147.237.327.477.63
Due within one week.