RLC Circuits - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

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nd

Order

RLC Circuits

Cap & Ind Physics Summary

• Under Steady-State (DC) Conditions
  • Caps act as OPEN Circuits
  • Inds act as SHORT Circuits
• Under Transient (time-varying) Conditions
  • Cap VOLTAGE can NOT Change Instantly
    • Resists Changes in Voltage Across it
  • Ind CURRENT can NOT change Instantly
    • Resists Changes in Curring Thru it

Transient Response

• The VI Relations can be Combined with KCL

and/or KVL to solve for the Transient (time- varying) Response of RL, RC, and RLC circuits

 Kirchoff’s Current Law  The sum of all Currents entering any Circuit-Node is equal to Zero

 Kirchoff’s Voltage Law  The sum of all the Voltage-Drops around any Closed Circuit-Loop is equal to Zero

∑^ ( )

=

N k

ik t 1

(^0) ∑ ( )

=

N k

vk t 1

0

Second Order Circuits

 Single Node-Pair

i (^) R iL^ iC

− iS + iR + iL + iC = 0 ( ); (^1) ( ) ( 0 ); ( ) 0

i vRt i L v x dx iL t iC Cdvdt t

t R L t = = ∫ + =

L S

t t Rv^ +^ L^1 ∫^ v ( x ) dx + i ( t^0 )+ Cdvdt ( t )= i 0

• By K C L

• vR −

• vC −

−

v L

• By K V L − vS + vR + vC + vL = 0 ; 1 ( ) ( 0 ); ( ) 0

v Ri v C i x dx vC t vL Ldtdi t

t R C t = = ∫ + =

C S

t t

Ri + (^) C^1 ∫ i ( x ) dx + v ( t 0 )+ Ldtdi ( t )= v 0

 Single Loop

dt^ Differentiating

di L

v dt

dv dt R

C d v + 1 + = S 2

2 dt

dv C

i dt Rdi dt

L d i + + = S 2

2

ODE for iL ( t ) in SNP

• Single-Node Ckt
• By

KCL

• Note That
• Use Ohm & Cap Laws
  • Recall v-i Relation for Inductors
  • Sub Out vL in above

i (^) R i^ L iC

i (^) R +iL +iC = i S

v ( ) t = vL( )t

S

L L

L (^) i dt

dv i C R

v

• • =

dt

di v (^) L = L L

S

L L

L (^) i dt

di L dt

d i C dt

di L R

ODE Derivation Alternative

• Take Derivative and ReArrange
• Make CoEff of 2nd Order Term = 1 - Use Similar Method to vC ( t ) for Single LOOP Circuit - Then

S

L L

L (^) i dt

di L dt

d i C dt

di L R

( ) L S

L L i i dt

di R

L

dt

d i LC 2 + + =

2

( ) LC

i i dt LC

di dt RC

d i S L

L + 1 L +^1 =

2

2

C

R C L S

i i

v v v v

and

dt

dv

i C

v

dt

di

Ri v L

C C

S

C C C

but

Illustration

• Write The Differential Eqn for v(t) & i(t) Respectively



 > = < 0 ( )^00 I t i t t S S

dt

di L

v dt

dv dt R

C d v+ 1 + = S 2

2

didt (^) S(t )= 0 ;t> 0



 > = < 0 0 ( )^0 t v (^) S t VS t

dt

dv C

i dt

Rdi dt

L d i+ + = S 2

2

dvdt (^) S( t)= 0 ;t> 0

i S

−

v S

 The Forcing Function

 Parallel RLC Model

 In This Case  So (^10) 2

2

• • = L

v dt

dv dt R

C d v

 The Forcing Function

 Series RLC Model

 In This Case  So (^0) 2

2

• • = C

i dt

R di dt

Ld i

2 nd^ Order Response Equation

• Need Solutions to the 2nd Order ODE

 As Before The Solution Should Take This form

 If the Forcing Fcn is a Constant, A , Then Discern a Particular Soln

 Verify xp

1 2 ( ) 1 ( ) 2 ( ) ( )

2 t a x t f t dt

t a dx dt

⋅d^ x + + =

x ( t) = xp (t) + xc(t )

 Where

• xp ≡ Particular Solution
• xc ≡ Complementary Solution

a 2

A

f t = A⇒ xp =

A a

a x a A

dt

d x dt

dx a

x A

p

p p p

⇒ = =

= ⇒ = =

2

2 2

2

2

2

0

 For Any const Forcing Fcn, f(t) = A ( ) ( ) 2

x t a

x t A = + c



Complementary Solution cont

• Sub Assumed Solution (x = Ke st^ ) into the Homogenous Eqn

 A value for “s” That SATISFIES the CHARACTERISTIC Eqn ensures that Kest^ is a SOLUTION to the Homogeneous Eqn  Units Analysis

 Canceling Kest

 The Above is Called the Characteristic Equation

( )

• 1

2 2 2

S

Also Unitless

A/S; A/S

Let ( ) ( ) Amps[A]

⇒

⇒ ≡

⇒ ⇒

= ⇒

s

e st

di dt d i dt

x t i t

st

s^2 Kest^ + 2 α sKest + ω 02 Kest = 0

2 (^ )^2 ( )^02 ( )^0

2

• t + x t = dt

t dx dt

d x α ω

2 0

2 0

2 s + αs +ω =

Complementary Solution cont.

• Recall Homog. Eqn. (^)  Short Example: Given Homogenous Eqn Determine - Characteristic Eqn - Damping Ratio, ζ - Natural frequency, ω 0  Given Homog. Eqn

 Discern Units after Canceling Amps

2 (^ )^2 ( )^02 ( )^0

2

• t + i t = dt

t di dt

d i α ω

(^1) UnitLess

(^1) ( ) 1 1

1 / radians sec

(^1) ( ) 1

1 1 /

1

0 2 0

0 1

02 02 2 2

1

∴ ≡ ≡

← → ⇒ ≡

∴ ≡ =

← → ⇒ =

← → ⇒ ≡ =

−

−

−

S S

dt dt

t dt

S S

t S dt

st s S S

SameUnits

SameUnits

SameUnits

ζ

ω

ζω ζ

ω

ω ω 4 2 ( ) 8 ( ) 16 ( ) 0

2

• t + x t = dt

t dx dt

d x

 Coefficient of 2nd^ Order Term MUST be 1

2 ( )^2 ( )^4 ( )^0

2

• t + x t = dt

t dx dt

d x

Complementary Solution cont.

• If Ke st^ is a Solution Then Need

 Solve By Completing the Square

• The CHARACTERISTIC Equation

 Solve For s by One of

• Quadratic Eqn
• Completing The Square
• Factoring (if we’re REALLY Lucky)

 The Solution for s Generates 3 Cases

1. ζ>
2. ζ<
3. ζ=

s^2 + 2 α s+ω 02 = 0

2 1 , 2 0 0

2 0

2 1 , 2 0

2 0

2

2 2 0

2

ζω ω ζ

α ω α ω

α α ω

α ω α

s

s

s

s

Aside: Completing the Square

• Start with:
• ReArrange:
• Add Zero → 0 = y−y:
• ReArrange:
• Grouping
  • The First Group is a PERFECT Square
• ReWriting:

s^2 + 2 α s+ω 02 = 0

s^2 + 2 α s+ ω 02 = 0

s^2 + 2 α s+ (α 2 −α^2 ) +ω 02 = 0

s^2 + 2 α s+α^2 +ω 02 −α^2 = 0

(s 2 + 2 α s+α^2 ) +(ω 02 −α^2 ) = 0

( s+α)^2 +(ω 02 −α^2 ) = 0

Case 1: ζ>1 → OVERdamped

• The Damped Natural Frequencies, s 1 and s 2 , are REAL and UNequal
• The Natural Response Described by the Relation
• The TOTAL Natural Response is thus a Decaying Exponential plus a Constant

s t s t xc t K e K e

1 2 ( ) = 1 + 2

( )

2 0

1 2

1 1

0 0 2 0 0 2

( ) ( )

ω

ζω ω ζ ζω ω ζ K e K e A

x t x t x t

t t

TOT c p

= + +

= +

−^ − −  −^ + − 

0

2 0

2 0

2

as s 1 , 2 = −ζω 0 ± ζ ω −ω and α = ζω

Case 2: ζ<1 → UNDERdamped

• Since ζ<1 The Characteristic Eqn Yields COMPLEX Roots as Complex Conjugates - So with j=√(-1)

 Then The UnderDamped UnForced (Natural) Response Equation

 Where

• ωn ≡ Damped natural Oscillation Frequency
• α ≡ Damping Coefficient

n

n

s j j

s j j

ζω ω ζ α ω

ζω ω ζ α ω

= − − − = − −

= − + − = − + 2 2 0 0

2 1 0 0

1

1

x ( )t^ e (^ A nt A nt)

t c ω^ ω

α = 1 cos + 2 sin

−