SECOND ORDER CIRCUIT. Revision of 1 st order circuit Second order circuit Natural response...

SECOND ORDER CIRCUITSECOND ORDER CIRCUIT

• Revision of 1st order circuit

• Second order circuit

• Natural response (source-free)

• Forced response

SECOND ORDER CIRCUITSECOND ORDER CIRCUIT

Revision of 1Revision of 1stst order circuit order circuit

NATURAL RESPONSE (SOURCE-FREE)NATURAL RESPONSE (SOURCE-FREE)

- initial energy in capacitor

Solution:

- i.e. vC(0) = Vo

KCL 0ii RC

0RC

v

dt

dv CC

Solving this first order differential equation gives:

RCtoC eV)t(v

+vC

iR iC+vR CR

Solution: KCL 0ii RC

0R

Vv

dt

dvC sCC

Solving this first order differential equation gives:

RCtsC e1V)t(v

+vC

Vsu(t)

R

C+

FORCED RESPONSEFORCED RESPONSE

- no initial energy in capacitor

- i.e. vC(0) = 0

RC

V

RC

v

dt

dv sCC


RCts

RCtoC e1VeV)t(v

COMPLETE RESPONSECOMPLETE RESPONSE

Complete response = natural response + forced response

v(t) = vn(t) + vf(t)

RCtsosC eVVV)t(v

Complete response = Steady state response + transient response

v(t) = vss(t) + vt(t)


COMPLETE RESPONSECOMPLETE RESPONSE

In general, this can be written as:

te)(x)0(x)(x)t(x

- can be applied to voltage or current

- x() : steady state value

- x(0) : initial value

For the 2nd order circuit, we are going to adopt the same approach


RCtsosC eVVV)t(v

To successfully solve 2To successfully solve 2ndnd order equation, need to know how to order equation, need to know how to get the initial condition and final values CORRECTLY get the initial condition and final values CORRECTLY

In 1st order circuit •need to find initial value of inductor current (RL circuit) OR capacitor voltage (RC circuit): iL(0) or vC(0)•Need to find final value of inductor current OR capacitor voltage: iL(∞) or vC(∞)

Before we begin …..Before we begin …..

INCORRECT initial conditions /final values will result in a wrong INCORRECT initial conditions /final values will result in a wrong solutionsolution

In 2nd order circuit •need to find initial values of iL and/or vC : iL(0) or vC(0)•Need to find final values of inductor current and/or capacitor voltage: iL(∞) , vC(∞)•Need to find the initial values of first derivative of iL or vC : diL(0)/dt dvC(0)/dt

Section 8.2 of Alexander/SadikuSection 8.2 of Alexander/Sadiku

Finding initial and final valuesFinding initial and final values

Example 8.1Example 8.1

Switch closed for a long time and open at t=0. Find:Switch closed for a long time and open at t=0. Find:

i(0i(0++), v(0), v(0++), ),

di(0di(0++)/dt, dv(0)/dt, dv(0++)/dt, )/dt,

i(∞), v(∞)i(∞), v(∞)

Finding initial and final valuesFinding initial and final values

PP 8.2PP 8.2

Find:Find:

iiLL(0(0++), v), vCC(0(0++), v), vRR(0(0++) )

didiLL(0(0++)/dt, dv)/dt, dvCC(0(0++)/dt, dv)/dt, dvRR(0(0++)/dt, )/dt,

iiLL(∞), v(∞), vCC(∞), v(∞), vRR(∞) (∞)

Natural Response of Series RLC CircuitNatural Response of Series RLC Circuit(Source-Free Series RLC Circuit)(Source-Free Series RLC Circuit)

Second order circuitSecond order circuit

R L

C

i

Applying KVL,

0dtiC

1

dt

diLRi

t

Differentiate once,

0C

i

dt

idL

dt

diR

2

2

0LC

i

dt

di

L

R

dt

id2

2

This is a second order differential equation with constant coefficients

We want to solve for i(t).


0LC

i

dt

di

L

R

dt

id2

2

AssumingstAe)t(i

0eLC

Ase

L

AReAs ststst2

0LC

1s

L

RsAe 2st

Since stAe cannot become zero,

0LC

1s

L

Rs2

This is known as the CHARACTERISTIC EQUATION of the diff. equation


0LC

1s

L

Rs2

Solving for s,

LC

1

L2

R

L2

Rs

2

1

LC

1

L2

R

L2

Rs

2

2

Which can also be written as

2o

21s 2

o2

2s

where LC

1,

L2

Ro

s1, s2 – known as natural frequencies (nepers/s)

– known as neper frequency, o – known as resonant frequency


2o

21s 2

o2

2s

Case 1

o

ts2

ts1

21 eAeA)t(i

A1 and A2 are determined from initial conditions

Overdamped solution

Case 2

o Critically damped solution

Case 3

o Underdamped solution


ts2

ts1

21 eAeA)t(i

Overdamped responseCase 1

o

Roots to the characteristic equation are real and negative

A1 and A2 are determined from initial conditions:

(i) At t = 0, 21 AA)0(i

(ii) At t = 0, 2211 AsAsdt

)0(di


Overdamped responseCase 1

o100 0.05H

0.5mF 200)0005.0(05.0

1

LC

1

,1000)05.0(2

100

L2

R

o

+vc

Initial condition vc(0) =100V


t3

t2

t1 eAeAeA)t(i

Critically damped responseCase 2

o

A3 is determined from 22 initial conditions: NOT POSSIBLE

(i) At t = 0, 2A)0(i

(ii) At t = 0, 21 AAdt

)0(di

solution should be in different form: t2

t1 eAteA)t(i

A1 and A2 are determined from initial conditions:


Critically damped responseCase 2

o20 0.05H

0.5mF 200)0005.0(05.0

1

LC

1

,200)05.0(2

20

L2

R

o

+vc



Underdamped responseCase 3

o

ts2

ts1

21 eAeA)t(i

Roots to the characteristic equation are complex

22o1 js 22

o2s

dj dj

22od - known as damped natural frequency



o

Using Euler’s identity: ej = cos + jsin

tsinjtcosAtsinjtcosAe)t(i dd2dd1t

tsinAAjtcosAAe d21d21t

tsinBtcosBe)t(i d2d1t

212211 AAjBandAAB where



o

tsinBtcosBe)t(i d2d1t

(i) At t = 0, 1B)0(i

(ii) At t = 0, 2d1 BBdt

)0(di



o10 0.05H

0.5mF 200)0005.0(05.0

1

LC

1

,100)05.0(2

10

L2

R

o

+vc



Underdamped, overdamped and critically damped responses

SECOND ORDER CIRCUIT. Revision of 1 st order circuit Second order circuit Natural response...

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