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

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)

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

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

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

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

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

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

Thevenin Equivalent at t=0+

Rt

C+–

Voc

+v(t)-

i(t)

+ -

Norton equivalent at t=0+

RtIsc

+v(t)-

L i(t)

1st ORDER CIRCUITS WITH CONSTANT INPUT

+–

t = 0

R1 R2

R3 Cvs

+v(t)-

s321

3 vRRR

R0v

Example (before switch closes)

If vs = 4V, R1 = 20kohms,

R2 = 20 kohms R3 = 40 kohms

What is v(0-) ?

s321

3 vRRR

R0v

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)

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

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

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

SOLUTION CONTINUED

oct

oc VCR

texpV)0(v)t(v

CR

texpV)0(vV)t(v

tococ

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

Example

8.3-1, p. 315

WITH AN INDUCTOR

+–

t = 0

R1 R2

R3 Lvs

21

s

RR

v0i

i(t)

Why ?

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 ?

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

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

Example

8.3-2, p. 316

for more practice: Exercises

8.3-1, p. 321 8.3-2, p. 321

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+)

Example

8.7-2, p. 336

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

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)

Don’t forget HW 8 (test review)

Course Evals……