Circuit Networks (Lecture Notes)
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![Page 1: Circuit Networks (Lecture Notes)](https://reader033.fdocuments.in/reader033/viewer/2022051208/5467ec4eaf79596e458b5640/html5/thumbnails/1.jpg)
CHAPTER 8CHAPTER 8
NETWORKS 1: NETWORKS 1: 09092010909201--03/0403/0410 December 2003 –
Lecture 8b
ROWAN UNIVERSITYROWAN UNIVERSITYCollege of EngineeringCollege of Engineering
Dr Peter Mark Jansson, PP PEDr Peter Mark Jansson, PP PEDEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERINGDEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING
Autumn Semester 2003Autumn Semester 2003
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adminhw 7 due today, hw 8 due at finaltest review 5.15pm thurs. at end of lablast lab 6 due by end of next week’s normal lab day (no later than 5 PM)final exam: Next Mon 15 Dec 2:45pm
Rowan Hall Auditorium
take – home portionAssignment 8 (15%)Tool Kit (10%)
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networks I
Today’s learning objectives –master first order circuitsbuild knowledge of the complete responseuse Thevenin and Norton equivalents to
simplify analysis of first order circuitscalculate the natural (transient) response
and forced (steady-state) response
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new concepts from ch. 8
response of first-order circuitsto a constant input
the complete responsestability of first order circuitsresponse of first-order circuits
to a nonconstant (sinusoidal) source
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What does First Order mean?
circuits that contain capacitors and inductors can be defined by differential equations
circuits with ONLY ONE capacitor OR ONLY ONE inductor can be defined by a first order differential equation
such circuits are called First Order Circuits
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what’s the complete response (CR)?
Complete response = transient response + steady state response
OR….
Complete response = natural response + forced response
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finding the CR of 1st Ord. Cir
1) Find the forced response before the disturbance. Evaluate at t = t(0-) to determine initial conditions [v(0) or i(0)]
2) Find forced response (steady state) after the disturbance t= t(∞) [Voc or Isc ]
3) Add the natural response (Ke-t/τ) to the new forced response. Use initial conditions to calculate K
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Figure 8.0-1 (p. 290) A plan for analyzing first-order circuits. (a) First, separate the energy storage element from the rest of the circuit. (b) Next, replace the circuit connected to a capacitor by its Thévenin equivalent circuit, or replace the circuit connected to an inductor by its Norton equivalent circuit.
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RC and RL circuits
RC circuit complete response:
RL circuit complete response:
)/())0(()( CRtOCOC
teVvVtv −−+=
tLRSCSC
teIiIti )/())0(()( −−+=
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simplifying for analysis
Using Thevenin and Norton Equivalent circuits can greatly simplify the analysis of first order circuits
We use a Thevenin with a Capacitorand a Norton with an Inductor
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Thevenin Equivalent at t=0+
Rt
C+–Voc
+v(t)-
i(t)
+ -
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Norton equivalent at t=0+
RtIsc
+v(t)-
L i(t)
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1st ORDER CIRCUITS WITH CONSTANT INPUT
+–
t = 0
R1 R2
R3 Cvs
+v(t)-
( ) s321
3 vRRR
R0v
++=−
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Example (before switch closes)
If vs = 4V, R1 = 20kohms, R2 = 20 kohmsR3 = 40 kohms
What is v(0-) ?
( ) s321
3 vRRR
R0v
++=−
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as the switch closes…
THREE PERIODS emerge….. 1. system change (switch closure)2. (immediately after) capacitor or inductor
in system will store / release energy (adjust and/or oscillate) as system moves its new level of steady state (a.k.a. transient or natural response) …. WHY??? 3. new steady state is then achieved (a.k.a.
the forced response)
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Thevenin Equivalent at t=0+
Rt
C+–Voc
+v(t)-
32
32t RR
RRR
+= s
32
3oc v
RRR
V+
=
KVL 0)t(vR)t(iV toc =−−+
i(t)
+ -
0)t(vdt
)t(dvCRV toc =−−+ CRV
CR)t(v
dt)t(dv
t
oc
t=+
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SOLUTION OF 1st ORDER EQUATION
CRV
CR)t(v
dt)t(dv
t
oc
t=+
CR)t(v
CRV
dt)t(dv
tt
oc −= dtCR
)t(vV)t(dv
t
oc −=
dtCR
1)t(vV
)t(dv
toc=
−dt
CR1
V)t(v)t(dv
toc−=
−
DdtCR
1V)t(v)t(dv
toc+∫−=∫
−
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SOLUTION CONTINUED
( ) DCR
tV)t(vlnt
oc +−=−
DdtCR
1V)t(v)t(dv
toc+∫−=∫
−
⎟⎟⎠
⎞⎜⎜⎝
⎛+−=− D
CRtexpV)t(vt
oc
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−=−
CRtexpDexpV)t(vt
oc ( ) oct
VCR
texpDexp)t(v +⎟⎟⎠
⎞⎜⎜⎝
⎛−=
( ) oct
VCR
0expDexp)0(v +⎟⎟⎠
⎞⎜⎜⎝
⎛−= ( ) ocV)0(vDexp −=
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SOLUTION CONTINUED
( ) oct
oc VCR
texpV)0(v)t(v +⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−−+=
CRtexpV)0(vV)t(vt
ococ
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so complete response is…
complete response = v(t) =forced response (steady state) = Voc
+natural response (transient) =
(v(0-) –Voc ) * e -t/RtC) NOTE: τ
=Rt C
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−−+=
CRtexpV)0(vV)t(vt
ococ
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Figure 8.3-4 (p. 301) (a) A first-order circuit and (b) an equivalent circuit that is valid after the switch opens. (c) A plot of the complete response, v(t), given in Eq. 8.3-8.
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WITH AN INDUCTOR
+–
t = 0
R1 R2
R3 Lvs
( )21
sRR
v0i
+=−
i(t)
Why ?
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Norton equivalent at t=0+
RtIsc
+v(t)-
L i(t)32
32t RR
RRR
+=
2
ssc R
vI =
KCL 0)t(iR
)t(vIt
sc =−−+
0)t(idt
)t(diLR1It
sc =−−+ sctt I
LR
)t(iLR
dt)t(di
+=+
Why ?
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SOLUTIONsc
tt IL
R)t(i
LR
dt)t(di
+=+CR
VCR)t(v
dt)t(dv
t
oc
t=+
CR1
LR
t
t ⇔
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−−+=
CRtexpV)0(vV)t(vt
ococ
( ) ⎟⎠
⎞⎜⎝
⎛−−+= tLR
expI)0(iI)t(i tscsc
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so complete response is…
complete response = i(t) =forced response (steady state) = Isc
+natural response (transient) =
(i(0-) –isc ) * e –t(Rt/L)) NOTE: τ
=L/Rt
( ) ⎟⎠
⎞⎜⎝
⎛−−+= tLR
expI)0(iI)t(i tscsc
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Figure 8.3-5 (p. 302) (a) A first-order circuit and (b) an equivalent circuit that is valid after the switch closes. (c) A plot of the complete response, i(t), given by Eq. 8.3-9.
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Figure E8.3-1 (p. 308)
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Figure E8.3-2 (p. 309)
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Figure E8.3-3 (p. 309)
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Figure E8.3-4 (p. 309)
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Figure E8.3-5 (p. 309)
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Stability of 1st order circuits
when τ>0 the natural response vanishes as t ∞
THIS IS A STABLE CIRCUIT
when τ<0 the natural response grows without bound as t ∞
THIS IS AN UNSTABLE CIRCUIT
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forced response summary
Forcing function y(t) (steady-state before)
Forced response xf (t)(steady-state after)
Constant y(t) = M Constant: xf (t) = N
Exponential y(t) = Me-bt
Exponential xf (t) = Ne-bt
Sinusoid y(t) = M sin (ωt + )
Sinusoid xf (t) = Asin (ωt+ ) + Bcos(ωt+ )
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Unit step or pulse signal
vo (t) = A + Be-at
for t > 0
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Example
8.6-2, p. 321-323
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Figure 8.6-12 (p. 322) The circuit considered in Example 8.6-2
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Figure 8.6-13 (p. 322) Circuits used to calculate the steady-state response (a) before t = 0 and (b) after t = 0.
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HANDY CHARTELEMENT CURRENT VOLTAGE
R
C
L
RVI = RIV ∗=
dtdvCi c
c = dtiC1v
tcc ∫=
∞−
dtdiLv L
L =dtvL1i
tLL ∫=
∞−
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IMPORTANT CONCEPTS FROM CHAPTER 8
determining Initial Conditionsdetermining T or N equivalent to simplifysetting up differential equationssolving for v(t) or i(t)
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Don’t forget HW 8 (test review)
Thursday 5.15 pm 11 Dec after lab