chapter 8
DESCRIPTION
CHAPTER 8. NETWORKS 1: 0909201-01 11 December 2002 – Lecture 8b ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002. admin. hw 7 due today, hw 8 due at final - PowerPoint PPT PresentationTRANSCRIPT
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CHAPTER 8CHAPTER 8
NETWORKS 1: NETWORKS 1: 0909201-010909201-01 11 December 2002 – Lecture 8b
ROWAN UNIVERSITYROWAN UNIVERSITY
College of EngineeringCollege of Engineering
Professor Peter Mark Jansson, PP PEProfessor Peter Mark Jansson, PP PEDEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERINGDEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING
Autumn Semester 2002Autumn Semester 2002
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admin
hw 7 due today, hw 8 due at final hw 8 posted end of today last lab 6 due by end of day monday final exam: Next Wed 18 Dec 10:15am take – home portion
Pick up at end of day from office window Yellow and Blue folders (by Lab Section)
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networks I
Today’s learning objectives – understand first order circuits build your knowledge of the concept of
complete response learn how Thevenin and Norton equivalents
help simplify analysis of first order circuits learn how to calculate the natural
(transient) response and forced (steady-state) response
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new concepts from ch. 8
response of first-order circuits to a constant input
the complete response stability of first order circuits response 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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complete response?
Complete response = transient response + steady state response
For circuits we will review today: both – constant source p.315 and sinusoidal source p. 304…
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
2) Find forced response (steady state) after the disturbance t= t(0+)
3) Add the natural response (Ke-t/) to the new forced response. Use initial conditions to calculate K
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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 Capacitor and 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 kohms R3 = 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
RR
RV
KVL 0)t(vR)t(iV toc
i(t)
+ -
0)t(vdt
)t(dvCRV toc CR
V
CR
)t(v
dt
)t(dv
t
oc
t
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SOLUTION OF 1st ORDER EQUATION
CR
V
CR
)t(v
dt
)t(dv
t
oc
t
CR
)t(v
CR
V
dt
)t(dv
tt
oc dtCR
)t(vV)t(dv
t
oc
dtCR
1
)t(vV
)t(dv
toc
dt
CR
1
V)t(v
)t(dv
toc
DdtCR
1
V)t(v
)t(dv
toc
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SOLUTION CONTINUED
DCR
tV)t(vln
toc
DdtCR
1
V)t(v
)t(dv
toc
D
CR
texpV)t(v
toc
CR
texpDexpV)t(v
toc oc
tV
CR
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
CR
texpV)0(vV)t(v
tococ
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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/RC)
CR
texpV)0(vV)t(v
tococ
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Example
8.3-1, p. 315
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WITH AN INDUCTOR
+–
t = 0
R1 R2
R3 Lvs
21
s
RR
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(vI
tsc
0)t(idt
)t(diL
R
1I
tsc sc
tt IL
R)t(i
L
R
dt
)t(di
Why ?
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SOLUTIONsc
tt IL
R)t(i
L
R
dt
)t(di
CR
V
CR
)t(v
dt
)t(dv
t
oc
t
CR
1
L
R
t
t
CR
texpV)0(vV)t(v
tococ
tL
RexpI)0(iI)t(i t
scsc
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so complete response is…
complete response = i(t) = forced response (steady state) = Isc
+ natural response (transient) = (i(0-) –isc) * e -tR/L)
tL
RexpI)0(iI)t(i t
scsc
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Example
8.3-2, p. 316
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for more practice: Exercises
8.3-1, p. 321 8.3-2, p. 321
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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 (t + )
Sinusoid xf(t) = Asin (t+) + Bcos(t+)
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Example
8.7-2, p. 336
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HANDY CHARTELEMENT CURRENT VOLTAGE
R
C
L
R
VI RIV
dt
dvCi c
c dtiC
1v
t
cc
dt
diLv L
L dtvL
1i
t
LL
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IMPORTANT CONCEPTS FROM CHAPTER 8
determining Initial Conditions determining T or N equivalent to simplify setting up differential equations solving for v(t) or i(t)
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Don’t forget HW 8 (test review)
Course Evals……