Ckt Transients n Tellegen Theorem

Electric Circuits: Transient AnalysisGeneral & Particular Solutions using Differential

Equations & Laplace

Vineet Sahula

[email protected]

3rd Sem. B. Tech. (ECE) V. Sahula Transient Analysis – p. 1/32

Linear Diff. Eq.

a0di(t)

dt+ a1i(t) = v(t)

a0dni

dtn+ a1

dn−1i

dtn−1+ ... + an−1

di

dt+ ani = v(t)

v(t) is forcing function or excitation


Integrating Factor

di

dt+ Pi = Q

Using Integrating Factor (I.F.) ePt we get

ePtdi

dt+ PiePt = QePt

d

dt(iePt) = QePt

Solving leads to

iePt =

Z

QePtdt + K

⇒ i = e−Pt

Z

QePtdt + Ke−Pt

For P being a function of time, I.F. will be eR

Pdt


Network solution

I part is Particular integral & II part is Complementary function

i = e−Pt

Z

QePtdt + Ke−Pt

Q is forcing function & K is arbitrary constant

Thus, with t → ∞ i.e. Steady State

limt→∞

Ke−Pt = 0

i(∞) = limt→∞

i(t) = limt→∞e−Pt

Z

QePtdt

Whereas, with t → 0 i.e. Initial condition

i(0) = limt→0

i(t) = limt→0e−Pt

Z

QePtdt + K

In case, P & Q are constants,

i(0) =Q

P+ K = K2 + K

i(∞) =Q

P+ K = K2

In general,

i(t) = iP + iC = iss + it


Example

For an RL circuit under switched-on condition

Ldidt

+ Ri = V i.e. didt

+ RL

i = VL

with P = RL

&Q = VL

i = e−Pt∫

QePtdt + Ke−Pt

→ i = VR

+ Ke−Rt

L

In general, when P & Q are constants, i = K2 + K1e−

t

T

In case, P &Q are constants,

K2 = i(∞)

K2 + K1 = i(0)

⇒ K1 = i(0) − i(∞)

⇒ i = i(∞) − [i(∞) − i(0)]e−t

T


Example-2

L R

R

K1

2

Vi

Determine current when K is CLOSED at t = 0 and later after

steady state is reached when K is OPENED

at t = 0 i(∞) = VR1

i(0) = VR1+R2

& T = LR1

⇒ i = VR1

(

1 −R1

R1+R2e−

R1t

L

)


More Complicated Networks

Networks described by one time-constant ?

Simple circuits having simple RC or RL combinations

Containg single L or C, but in combination of any number

of resistors, R

Networks, which can be simplified by using equivalence

conditions so as to represented by a single equivalent

L/C/R

Solve many examples !!


Initial Conditions in Networks

Resistor: VR = iR the current changes instanteneously, if the voltagechanges instanteneously

Inductor: vL = L ·diL

dt, diL

dtfor L is finite, hence current CANNOT

change instanteneously; BUT an arbitrary voltage may appearacross it

Inductor: iC = C ·dvC

dt, dvC

dtfor C is finite, hence voltage

CANNOT change instanteneously; BUT an arbitrary current mayappear across it

Element Equivalant ckt at t = 0

R R

L Open Ckt (OC)

C Short Ckt (SC)

L, I0 Current source I0 in parallel with OC

C, V0 Voltage source V0 in series with SC


Final Conditions in Networks

Element, IC Equivalant ckt at t = ∞

R R

L Open Ckt (OC)

C Short Ckt (SC)

L, I0 Current source I0in parallel with SC

C, V0 Voltage source V0 in series with OC


Two special cases- Initial Conditions

A loop or mesh containing a VOLTAGE source Vs with only

capacitors,

implying a virtual short-circuit across Vs;

Imagine infinite current to flow through capacitors so as to

charge them to appropriate voltages instanteneously

In a dual situation: A node connected with a CURRENT

source Is with only Inductors in other branches

implying a virtual open-circuit across Is;

Imagine infinite voltage across Is to exist so as to drive

finite FLUX in all the inductors to bring appropriate current

in them instanteneously


Second Order Diff. Equations

a0d2i

dt2+ a1

di

dt+ a2i = v(t)

To satisfy the equation, the solution function MUST be of such

form that all three terms are of SAME form.

i(t) = kemt

a0m2kemt + a1mkemt + a2kemt = 0

Charateristic Equation, a0m2 + a1mk + a2k = 0

m1, m2 = −a1

2a0±

12a0

√

a21 − 4a0a2

i(t) = k1em1t + k2e

m2t


Solving Second order Diff. Eqns.

roots may simple (real), equal OR complex (conjugate)

Simple roots i(t) = k1em1t + k2e

m2t

Equal roots m1 = m2 = m

⇒ i(t) = k1emt + k2te

mt

Complex (conjugate) roots m1, m2 = −σ ± j · ω

i(t) = k1e−σte+ωt + k2e

−σte−ωt

i(t) = e−σt(k1e+ωt + k2e

−ωt)

i(t) = e−σt(k3 cos ωt + k4 sinωt)

i(t) = e−σtk5 cos(ωt + φ)


Solving 2nd order Diff. Eqns.

Initial Conditions

Two constants k1 & k2 need be evaluated

This requires two IC to be formulated

First IC is computed as either i(0+) OR v(0+), whichever

is independent/unknown [i is independent in a series

circuit; v is independent in a parallel circuit ]

Second IC is based on first order differential of the same

parameter,di

dt(0+) or

dv

dt(0+)



Example-1

A series RLC ckt- V=1 volts, R=3 Ω; L=1 H; C=12 F

Ldi

dt+ Ri +

1

C

∫

idt = V (1)

d2idt2

+ 3didt

+ 2i = 0 Ch. Eqn., m2 + 3mk + 2 = 0

m1 = −1,m2 = −2

i(t) = k1e−t + k2e

−2t

R1

C1

v1

L



Example-1 . . .

Ldi

dt+ Ri +

1

C

∫

idt = V (2)

The two Initial ConditionsFirst IC: i(0+) = 0 as CURRENT in inductorCANNOT change instanteneously

Second IC:di

dt(0+) = V

L= 1 as second & third terms

in expression (2) are zero at t = 0

k1 + k2 = 0 and − k1 − 2k2 = 1 ⇒ i(t) = e−t− e−2t



Example-2

A fully parallel RLC ckt- R = 18 Ω; L = 1

8H; C = 2 F

Cdv

dt+ Gv +

1

L

∫

vdt = Is (3)

2d2vdt2

+ 8dvdt

+ 8v = 0 Ch. Eqn., 2m2 + 8mk + 8 = 0

m1 = −2,m2 = −2

v(t) = k1e−2t + k2te

−2t

v1

R CL



Example-2 . . .

Cdv

dt+ Gv +

1

L

∫

vdt = Is (4)

The two Initial ConditionsFirst IC: v(0+) = 0 as VOLTAGE across capacitorCANNOT change instanteneouslySecond IC: second & third terms in expression (4)are zero as v(0+) = 0 as well as CURRENT in L

CANNOT change instanteneously at

t = 0 ⇒dv

dt(0+) = Is

C= 1

2

k1 = 0 and − 2k1 + k2 = 12 ⇒ v(t) = 1

2te−2t



Example-3A series RLC ckt- V=1 volts, R=2 Ω; L=1 H; C= 1

2 F

Ldi

dt+ Ri +

1

C

∫

idt = V (5)

d2idt2

+ 2 didt

+ 2i = 0 Ch. Eqn., m2 + 2mk + 2 = 0

m1, m2 = −1 ± 1

i(t) = k1e(−1+1)t + k2e

(−1−1)t

i(t) = e−t (k1et + k2e

−t)

As e±t = cos ωt ± sinωt, Euler’s identity

⇒ i(t) = e−t (k3 cos t + k4 sin t) withk3 = k1 + k2 & k4 = (k1 − k2)



Example-3 . . .

Ldi

dt+ Ri +

1

C

∫

idt = V (6)

The two Initial ConditionsFirst IC: i(0+) = 0 as CURRENT in inductorCANNOT change instanteneously

Second IC:di

dt(0+) = V

L= 1 as second & third terms

in expression (6) are zero at t = 0

i(0+) = e−0 (k3 cos 0 + k4 sin 0) = k3

di

dt(0+) = 1 = k4(e

−0 cos 0 + sin 0 e−0)

k3 = 0 and k4 = 1 ⇒ i(t) = e−t sin t


Solving Higher order Diff. Eqns.

Internal Excitation

a0sn + a1s

n−1 + · · · + an−1s + an=0

Can be expanded into first order and second order FACTORS

Find all n roots, one-by-one finding a root into respective

simple FACTOR and complex root into second order FACTOR

Formulate charateristic equation

Using solutions already derived for I − order and II − order,

the solution can be arrived

i = (k1 + tk2)em1 + k3e

m3t + k4e−σt cos ωt + k5e

−σt sinωt



External Excitation

a0d2idt2

+ a1didt

+ a2i = v(t)

i = iP + iC

iC is found as in the case of Internal ExcitationIn order to find iP , the form of forcing function v(t)need be known, likeV (constant) sin ωt, kt, e−αt, e−αt cos ωt

thereafter TRail functions based on forncingfunction form are predicted & the list of choice forParticular Integral may be compiled vis-avisforcing function factor in v(t), in a table



Procedure with External Excitation1. Determine complementary function iC ; compare each part of iC with

form of v(t)

2. Write trial form of Particular Integral using Table; each trial solutionassigned different letter coefficient

3. Substitute trial solution into differential equation; Form algebraicequations in unknown coefficients by equating coefficients of liketerms in the substituted-differential equation

4. Solve for the unknown (undetermined) coefficients; there will be NOarbitrary coefficients in Particular Integral

5. CAUTION The initial conditions MUST always be applied to the totalsolution

and NEVER to the Complementary function alone, unlessiP = 0 [when v(t) = 0]


Solving Diff. Eqns.- Laplace Method

Advantages of Laplace method

1. Solution progresses systematically

2. provides TOTAL solution- the particular integral andcomplementary function - in one operation

3. Integro-differential equation is transfomed intoalgebraic realtions

4. Initial conditions are automatically specified i thetransformed equations


Solving Diff. Eqns.- Laplace Method

Preliminaries

Laplace transform, L[f(t)] = F (s) =∫

∞

0− f(t)e−stdt

Transform pair, F (s) ⇔ f(t)

Some basic theorems

Initial conditions are automatically specified i thetransformed equations


Laplace Method

Xm of derivative

L[ ddt

f(t)] =∫

∞

0−ddt

f(t)e−stdt

⇒ L[ ddt

f(t)] = sF (s) − f(0−)


Laplace Method

Xm of Integral

With zero IC, L

[

∫ t

0− f(t)dt]

L

[

∫ t

0− f(t)dt]

=∫

∞

0−

[

∫ t

0− f(t)dt]

e−stdt

⇒ L

[

∫ f

0−(t)dt]

= F (s)s

In general conditions, L

[

∫ t

−∞f(t)dt

]

∫ t

−∞f(t)dt =

∫ 0−−∞

f(t)dt +∫ t

0− f(t)dt

L

[

∫ f

0−(t)dt]

= F (s)s

and

L[∫ 0−−∞

f(t)dt] = q(0−)s


Example Solution- Laplace Method

RC Network

Switching of key at t = 0 is accounted with Unit-stepfunction, u(t)

1C

∫

idt + Ri = V u(t)

I(s)(

1Cs

+ R)

= Vs

I(s) =V

R

s+ 1

RC

taking inverse transform, i(t) = L−1[I(s)] = L

−1[

V

R

s+ 1

RC

]

i(t) = VR

e−t

RC for t > 0


Partial Fraction Expansion

a0dni

dtn+ a1

dn−1i

dtn−1+ ... + an−1

di

dt+ ani = v(t)

Taking Laplace on both sides and manipulating,

I(s) = L[v(t)]+initial conditions termsa0sn+a1sn−1+···+an−1s+an

I(s) = P (s)Q(s) = B0 + B1s + B2s

2 + · · · + Bm−nsm−n + P1(s)Q(s)

The i(t) can be found by taking inverse transform ofI(s)


Partial Fraction Expansion

Heaviside Expansion Theorem

Expand P1(s)Q(s) into partial fractions using Heaviside

expansion theorem,P1(s)

(s−s1)(s−s2)= K1

(s−s1)+ K2

(s−s2)

P1(s)(s−s1)r = K11

(s−s1)+ K12

(s−s1)2+ · · · + K1r

(s−s1)r

P1(s)(s+α+ω)(s+α−ω) = K1

(s+α+ω) + K2

(s+α−ω)


Tellegens Theorem

Instanteneous Power in a NetworkStatement : In a network having b branches, for arbitrarily chosenvoltages vk and currents ik;

b∑

k=1

vkik = 0

with the condition that these vk MUST satisfy KirchoffsVoltage Law and currents ik MUST satisfy Kirchoffs Current Law


Tellegens Theorem

Instanteneous Power . . .

∑b

k=1 vkik =

vBiAB +(vB−vC)iBC +vCiCA+(vC−vD)iCD+vDiDA+(vD−vB)iDB

re-arranging,∑b

k=1 vkik =

vB(iAB +iBC −iDB)+vC(−iBC +iCA +iCD)+vD(−iCD +iDA +iDB)

⇒= vB(KCL at node B)+ vC(KCL at node C)+vD(KCL atnode D)=0

R1 R2

L

v

2C

C1

1

A

B CD


Tellegens Theorem

Instanteneous Power VerificationElement

Item 1 2 3 4 5 6

vk

ik

vkik

i,k

vki,k

Develop table entries,

The Tellegens Theorem is satisfied ONLY for vki,k, where i

,k satisfy

KCL


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Ckt Transients n Tellegen Theorem