EE40 Lecture 12 Venkat Anantharamee40/sp08/lectures/EE40_Spring08_Lecture12.pdfEE40 Spring 2008...

EE40 Spring 2008 Venkat Anantharam

EE40Lecture 12

Venkat Anantharam

2/20/08 Reading: Chap. 4: first order

circuits


First Order Circuits

R+

-Cvs(t)

+-vc(t)

+ -vr(t)ic(t)

vL(t)is(t) R L

+

-

iL(t)

KVL around the loop: KCL at the node:

( ) ( ) ( )cc s

dv tRC v t v tdt

+ = ( ) ( ) ( )LL s

di tL i t i tR dt

+ =


Response of a Circuit• Transient response of a circuit is the portion of the total response

that dies to zero as time goes to infinity.• Steady-state response is what remains of the total response after the

transient response is removed.• Natural response of a circuit is the portion of the response due to the

initial conditions. This is generally part of the transient response, except in cases where there is some energy in the initial conditions that cannot dissipate through resistors (e.g. a pure LC oscillator).

• Forced response is the portion of the response due to the forcing function (the right hand side of the differential equation that comes from the sources). In general this will also have a transient part.

• We will later be mostly interested in the steady state response of circuits to a DC forcing function or more generally a sinusoidal forcing function

• Note: some of this terminology is at variance with your book. Your book’s definition of ``forced response”. for instance, is inconsistent.


Natural Response of an RC Circuit• Consider the following circuit, for which the switch is

closed for t < 0, and then opened at t = 0:

Notation:0– is used to denote the time just prior to switching0+ is used to denote the time immediately after switching

• The voltage on the capacitor at t = 0– is Vo

C

Ro

Vo R

t = 0+−

+v–


Solving for the Voltage (t ≥ 0)

• For t > 0, the circuit reduces to

• Applying KCL to the RC circuit:

• Solution:

+

v

–

RCtetv /o V)( −=

CRo

Vo R+−

i


Solving for the Current (t > 0)

• Note that the current changes abruptly:

RCtoeVtv /)( −=

RVi

eRV

Rvtit

i

o

RCto

=⇒

==>

=

+

−

−

)0(

)( 0,for

0)0(

/

+

v

–

CRo

Vo R+−

i


Solving for Power and Energy Delivered (t > 0)

( )RCto

tRCxo

t

RCto

eCV

dxeRVdxxpw

eRV

Rvp

/22

0

/22

0

/222

121

)(

−

−

−

−=

==

==

∫∫

RCtoeVtv /)( −=

+

v

–

CRo

Vo R+−

i


Natural Response of an RL Circuit• Consider the following circuit, for which the switch is

closed for t < 0, and then opened at t = 0:

Notation:0– is used to denote the time just prior to switching0+ is used to denote the time immediately after switching

• t<0 the entire system is at steady-state; and the inductor is like short circuit

• The current flowing in the inductor at t = 0– is Io and V across is 0.

LRo

t = 0 i +

v

–Io R


Solving for the Current (t ≥ 0)• For t > 0, the circuit reduces to

• Applying KVL to the LR circuit:• v(t)=i(t)R• At t=0+, i=I0, • At arbitrary t>0, i=i(t) and

• Solution:

( )( ) di tv t Ldt

=

LRo

i +

v

–Io R

-

= I0e-(R/L)ttLReiti )/()0()( −=


Solving for the Voltage (t > 0)

tLRoeIti )/()( −=

• Note that the voltage changes abruptly:

LRo

+

v

–Io R

I0Rv

ReIiRtvtv

tLRo

=⇒

==>

=

+

−

−

)0(

)(0,for 0)0(

)/(


Solving for Power and Energy Delivered (t > 0)tLR

oeIti )/()( −=

LRo

+

v

–Io R

( )tLRo

txLR

o

t

tLRo

eLI

dxReIdxxpw

ReIRip

)/(22

0

)/(22

0

)/(222

121

)(

−

−

−

−=

==

==

∫∫


Natural Response SummaryRC Circuit

• Capacitor voltage cannot change instantaneously

• time constant

RL Circuit

• Inductor current cannot change instantaneously

• time constantRL

=τ

τ/)0()()0()0(teiti

ii−

+−

=

=

R

i

L

+

v

–

RC

τ/)0()()0()0(tevtv

vv−

+−

=

=

RC=τ


Digital SignalsWe compute with pulses.

We send beautiful pulses in:

time

volta

ge

time

volta

ge

But we receive lousy-looking pulses at the output:

Capacitor charging effects are responsible!

• Every node in a real circuit has capacitance; it’s the charging of these capacitances that limits circuit performance (speed)


Pulse DistortionR The input voltage pulse

width must be large enough; otherwise the output pulse is distorted.(We need to wait for the output to reach a recognizable logic level, before changing the input again.)

0123456

0 1 2 3 4 5Time

Vout

Pulse width = 0.1RC

0123456

0 1 2 3 4 5Time

Vout

0123456

0 5 10 15 20 25Time

Vout

+

Vout–

Vin(t) C+

–

Pulse width = RC Pulse width = 10RC


Example

Vin

RVout

C

Suppose a voltage pulse of width5 µs and height 4 V is applied to theinput of this circuit beginning at t = 0:

R = 2.5 kΩC = 1 nFτ = RC = 2.5 µs

• First, Vout will increase exponentially toward 4 V.• When Vin goes back down, Vout will decrease exponentially

back down to 0 V.

What is the peak value of Vout?

The output increases for 5 µs, or 2 time constants.It reaches 1-e-2 or 86% of the final value.

0.86 x 4 V = 3.44 V is the peak value


First Order Circuits: Forced Response

R+

-Cvs(t)

+-vc(t)

+ -vr(t)ic(t)

vL(t)is(t) R L

+

-

iL(t)

KVL around the loop:vr(t) + vc(t) = vs(t) )()(1)( tidxxv

LRtv

s

t

=+ ∫∞−

KCL at the node:

( ) ( ) ( )LL s

di tL i t i tR dt

+ =

( ) ( ) ( )cc s

dv tRC v t v tdt

+ =


Complete Solution• Voltages and currents in a 1st order circuit satisfy a

differential equation of the form

– f(t) is called the forcing function.• The complete solution is the sum of any particular solution and the

associated complementary solution

– Particular solution is any solution that satisfies the original equation.– Complementary solution is a solution of the homogeneous equation (the

one with zero forcing function) used to satisfy the initial conditions. In effect it also compensates for the value of the particular solution at 0.

( )( ) ( )dx tx t f tdt

τ+ =

/

( )( ) 0

( )

cc

tc

dx tx tdt

x t Ke τ

τ

−

+ =

=

( )( ) ( )pp

dx tx t f t

dtτ+ = Homogeneous

equation

( ) ( ) ( )p cx t x t x t= +

Original equation Solution to the homogeneous equation.


The Time Constant

• The complementary solution for any 1st order circuit looks like, for some K,

• For an RC circuit, τ = RC• For an RL circuit, τ = L/R

/( ) tcx t Ke τ−=


What Does Xc(t) Look Like?

τ = 10-4/( ) tcx t e τ−=

• τ is the amount of time necessary for an exponential to decay to 36.7% of its initial value.

• -1/τ is the initial slope of an exponential with an initial value of 1.


Particular Solution• A particular solution xp(t) is usually found as a

weighted sum of f(t) and its first derivative.• For example, if f(t) is constant (DC forcing

function), then xp(t) can be taken to be a constant.

• If f(t) is sinusoidal, then xp(t) can be taken to be sinusoidal.

• Note: The notion of ``particular solution” is not unique. Any solution of the homogeneous equation can be added to a particular solution to give another particular solution! The corresponding complementary solution would then adjust accordingly to give the same total response for the given initial conditions.


Particular Solution: F(t) Constant

Fdttdxtx P

P =+)()( τ

Guess a solutionBtAtxP +=)(

FdtBtAdBtA =

+++

)()( τ

Equation holds for all time and thus each coefficient is

individually zero

FBBtA =++ τ)(

0)()( =+−+ tBFBA τ

0)( =−+ FBA τ0)( =B

FA =0=B


The Particular Solution: F(t) Sinusoid)cos()sin()()( wtFwtF

dttdxtx BA

PP +=+τ

Guess a solution )cos()sin()( wtBwtAtxP +=

)cos()sin())cos()sin(())cos()sin(( wtFwtFdt

wtBwtAdwtBwtA BA +=+

++ τ

Equation holds for all time and \thus each coefficient is

individually zero

( )sin( ) ( )cos( ) 0A BA B F t B A F tτω ω τω ω− − + + − =( ) 0BB A Fτω+ − =( ) 0AA B Fτω− − =

2( ) 1A BF FA τωτω+

=+ 2( ) 1

A BF FB τωτω

−= −

+

2 2 2

1

2

1 1( ) sin( ) cos( )( ) 1 ( ) 1 ( ) 1

1 cos( ); tan ( )( ) 1

Px t t t

t where

τω ω ωτω τω τω

ω θ θ τωτω

−

⎡ ⎤= +⎢ ⎥

⎢ ⎥+ + +⎣ ⎦

= − =+


The Particular Solution: F(t) Exp.

21)()( FeF

dttdxtx tP

P +=+ −ατGuess a solution

21)()( FeF

dtBeAdBeA t

tt +=

+++ −

−− α

αα τt

P BeAtx α−+=)(

Equation holds for all time and thus each coefficient is

individually zero21)( FeFBeBeA ttt +=−+ −−− ααα ατ

0)()( 12 =−−+− − teFBFA αατ

0)( 2 =− FA0)( 1 =−− FB ατ

2FA =1FB +=ατ


The Total Solution: F(t) Sinusoid

)cos()sin()( wtBwtAtxP +=

)cos()sin()()( wtFwtFdttdxtx BA

PP +=+τ

2( ) 1A BF FA τωτω+

=+ 2( ) 1

A BF FB τωτω

−= −

+

τt

C Ketx −=)(

τt

T KewtBwtAtx −++= )cos()sin()(

Only K is unknown and is determined by the

initial condition at t =0 Example: xT(t=0) = VC(t=0)

)0()0cos()0sin()0(0

==++=− tVKeBAx CT

τ

)0()0( ==+= tVKBx CT BtVK C −== )0(


Example

C

RVs

t = 0+−

+vc–

• Given vc(0-)=1, Vs=2 cos(ωt), ω=200.• Find i(t), vc(t)=?

EE40 Lecture 12 Venkat Anantharamee40/sp08/lectures/EE40_Spring08_Lecture12.pdfEE40 Spring 2008...

Documents

Transcript of EE40 Lecture 12 Venkat Anantharamee40/sp08/lectures/EE40_Spring08_Lecture12.pdfEE40 Spring 2008...