Circuit Networks (Lecture Notes)

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

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%)

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

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

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

what’s the complete response (CR)?

Complete response = transient response + steady state response

OR….

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 [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

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.

RC and RL circuits

RC circuit complete response:

RL circuit complete response:

)/())0(()( CRtOCOC

teVvVtv −−+=

tLRSCSC

teIiIti )/())0(()( −−+=

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

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 kohmsR3 = 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

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=+

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+∫−=∫

−

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 −=

SOLUTION CONTINUED

( ) oct

oc VCR

texpV)0(v)t(v +⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−+=

CRtexpV)0(vV)t(vt

ococ

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

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.

WITH AN INDUCTOR

+–

t = 0

R1 R2

R3 Lvs

( )21

sRR

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(vIt

sc =−−+

0)t(idt

)t(diLR1It

sc =−−+ sctt I

LR

)t(iLR

dt)t(di

+=+

Why ?

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

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

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.

Figure E8.3-1 (p. 308)

Figure E8.3-2 (p. 309)

Figure E8.3-3 (p. 309)

Figure E8.3-4 (p. 309)

Figure E8.3-5 (p. 309)

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

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+ )

Unit step or pulse signal

vo (t) = A + Be-at

for t > 0

Example

8.6-2, p. 321-323

Figure 8.6-12 (p. 322) The circuit considered in Example 8.6-2

Figure 8.6-13 (p. 322) Circuits used to calculate the steady-state response (a) before t = 0 and (b) after t = 0.

HANDY CHARTELEMENT CURRENT VOLTAGE

R

C

L

RVI = RIV ∗=

dtdvCi c

c = dtiC1v

tcc ∫=

∞−

dtdiLv L

L =dtvL1i

tLL ∫=

∞−

IMPORTANT CONCEPTS FROM CHAPTER 8

determining Initial Conditionsdetermining T or N equivalent to simplifysetting up differential equationssolving for v(t) or i(t)

Don’t forget HW 8 (test review)

Thursday 5.15 pm 11 Dec after lab