Chapter 3 · Area of parallelogram described by points (0,0), (a,b) and (c,d) is given by: ad bc c...

05 second order 1TUE/EE 5DD17 netwerk analyse 04/05 - © NvdM

Chapter 3Interconnects and RC Ladder Circuits


Interconnect

Wires are not ideal interconnectionsThey may have non-negligible capacitance, resistance, inductanceThese are called wire parasiticsCan dominate performance of chipMust be accounted for during designUsing approximate models

L

§3.2


Wires on an Integrated Circuit

← IBM CMOS 7S copper process, 0.16 µm


Interconnect Hierarchy

Example Interconnect Hierarchy for typical 0.25µ process (Layer Stack, cross-sectional view)

≈1µm


On-Chip Wire Capacitance

l

w

h hwlC rεε= 0

ε0 = 8.85 pF/m

εr = 3.9 (SiO2)wire

w

h C

substrate (Si)

Typical value for on-chip interconnect: Cint = 100pF/m

CfringeCfringe+ fringe factor)( h

wlC = ε0εr


Wire Resistance

Proportional to lInversely proportional to w and t (cross-sectional area)Proportional to ρ: specific resistivity, material property [Ωm]R = ρl/wtAluminum: ρ = 2.7x10-8 Ωm

lw

t

Typical value for on-chip interconnect: Rint = 1000Ω/cm

But MUCH greater variation then Cint


Delay

Wire is an RC network (two-port)

Model driver as linearized (Thévenin)source V, Rs, assume step input

V

Rs

Model load as CL

CL

delay from here to here

§3.3


Capacitive Wire Model

( ) 0=−

++s

inoutoutLw R

VVdt

dVCC

( ) int

out VeV τ−−= 1

( )Lws CCR +=τ

Cw CL

VoutRs

Vint = 0

Assume wire behaves purely capacitive


Capacitive-Resistive Wire Model

Is this a good model?

( )( )Lwws CCRR ++=τ

NoR and C are distributed along the wire

Rs

CLCw

Rw

CLV

Rs

Now, assume wire capacitance and resistance

§3.4


Cw/2N Cw/N Cw/N Cw/2N

Rw/N

∞→Nlim

=

Cw/N

Uniform RC Line

CLV

Rs

SymbolR

C Cw/2

Rw1-lump π-model (N=1) Cw/2

Below, we have a 1-lump π-model

More lumps as on the left means greater

accuracy (in principle)


Basic (1-lump) π-model of wire

RC Delay (Uniform RC Line)

τ?

CLV

Rs

Cw/2V

Rs

CL

Rw

Cw/2

There is not one τ!

Now, we will develop new theory for such circuits!


Prototypical 2nd Order RC Circuit

Cw/2

Rs

CL

Rw

Cw/2

Vs

+

-C1

R1C2

R2

t=0 vC1 vC2=vo

iC1 iC2

Vs

t=0

What is the waveform at output Vo after switch closes?

(initial conditions needed)§3.5-3.6


+

-Vs

R

t=0

C vC

+

-

+ -vR

one capacitorone ‘state’ node

first order circuitone d/dt equation

Vs

+

-C1

R1C2

R2

t=0 vC1 vC2=vo

iC1 iC2

two independent capacitorstwo ‘state’ nodes

second order circuittwo d/dt equations

Solution of second order circuit will showMany similarities to first order

Beautiful connection to linear algebra

Introduction to 2nd Order Circuits


+

-Vs C1

R1

C2

R2

t=0 vC1 vC2=vo

iC1iC2

0. Circuit:

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

011

1111

112

1

2222

122112

1

CRV

vv

CRCR

CRRRC

dtdv

dtdv

S

C

C

c

c

3. Resulting set of equations:

KCL and Element Equations for 2 D.E.’s

02

211

11 =

−++

−R

vviR

Vv CCC

SC

dtdvCi C

C1

11 =

02

2111

11 =

−++

−R

vvdt

dvCR

Vv CCCSC

1. KCL for VC1:

2. KCL for VC2:

022

12 =+−

CCC i

Rvv


Detailled Derivation of First D.E.

02

211

11 =

−++

−R

vviR

Vv CCC

SCdt

dvCi CC

111 =

02

2111

11 =

−++

−R

vvdt

dvCR

Vv CCCSC

1221

1111

CRvv

CRvV

dtdv CCCSC −

−−

=

S11

2C12

1C211

VCR1v

CR1v

R1

R1

C1 +⎟

⎠

⎞⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛+−=

212

112

11111

1111CCCS v

CRv

CRv

CRV

CR+−−=

§3.6.1


⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

011

1111

112

1

2222

122112

1

CRV

vv

CRCR

CRRRC

dtdvdt

dvS

C

C

c

c

Second order

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

2

1

2221

1211

2

1

KK

vv

aaaa

dtdvdt

dv

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

2

1

2

1

2

1KK

vv

vv

dtd A

KAvv +=dtd

Comparison to First Order

csc vV

dtdvRC −=

+

-Vs

R

t=0

C vC

+

-

+ -vR

Kavdtdv

+=

First order

Thick, non-italic symbols for matrix/vector


Powerful

Kavdtdv

+= KAvv +=dtd

Linear algebra

Useful

Linear Algebra

We have seen a transition from a scalar equation to a matrix equation

We are (re-)entering the field of Linear Algebra


w/o constant term, RHS=0

Linear AlgebraSolving sets of linear equationsTheory about existence and nature of solutionsWe need to know when a 2 x 2 homogeneous equation has a non-trivial solution

0Ax = ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⇔

00

2

1

2221

1211xx

aaaa

⎩⎨⎧

=+=+

⇔00

222121

212111xaxaxaxa

Trivial solution⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⇔=

00

02

1xx

x

Trivial solution does always exist.But, does a non-trivial solution with x1 a/o x2 ≠ 0 exist ?If yes, how can we compute it ?

w/o constant term

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⇒

00

00

2221

1211aaaa


Geometric Interpretation

x1

x2

3. This point satisfies both equationsTrival Solution: x1=x2=0

2221

aa

−

2. All points on this line satisfy equation of 2nd row

12221

2

222121 0xaxa

xaax −=

=+

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡00

2

1

2221

1211xx

aaaa

4. Non-trivial solution if and only ifTwo lines are on top of each otherequivalent to

2221

1211

aa

aa

= or a11a22 = a12a21

1211

aa

−

1. All points on this line satisfy equation of 1st row

11211

2

212111 0xa xa

xaax −=

=+

1

3a. RHS=0 ⇔all lines pass through origin


Determinant

Sneak Preview of Kolman/Hill Chapter 5

With each square (n x n) matrix, a characteristic number called its determinant is associated

Notation: det(A) = determinant of matrix A

Theorem: For a square matrix A, the homogeneous system Ax=0 has a non-trivial solution if and only if det(A) = 0

P. 141 Ch. 5

211222112221

1211 aaaaaaaa

det −=⎥⎦

⎤⎢⎣

⎡(brief explanation follows)

This is the formula for determinant of 2x2 matrixGeneralized formula exists for any square matrixWill be introduced with Linear Algebra


Geometric Interpretation of Determinant (2D)

(a,b)

(c,d)

Area of parallelogram described by points (0,0), (a,b) and (c,d) is given by:

bcaddcba

det −=⎥⎦

⎤⎢⎣

⎡ Show that ad-bc is the area of parallelogram

Conclusion:area = 0 ⇔ determinant = 0 ⇔ thick lines on top of each other ⇔ non-trivial solution exists

The determinant of a 2x2 matrix equals the area of a (2D) parallelogram constructed from its rows! (apart from possibly a minus sign)


Summary

1. This 2x2 matrix equation: Ax = 0

2. Has non-trivial solution iff

a11a22 = a12a21

3. This is equivalent to

det(A) = 0


Eigenvalues and Eigenvectors

But suppose we now have a matrix A, a vector x and a scalar λCan we have non-trivial solutions for Ax = λx ?

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

2

1

2

1

2221

1211xx

xx

aaaa

λ

Answer: yes, rewrite to see connection to determinant:

Ax = λIx ⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

2

1

2

1

2221

12110

0xx

xx

aaaa

λλ

(A - λI)x = 0 ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−00

2

1

2221

1211xx

aaaa

λλ

Non-trivial solution iff det(A-λI) = 0λ is an eigenvalue of A, x is an associated eigenvector of A

P. 142

Ch. 6

Sneak Preview of Kolman/Hill Chapter 6Now we know about non-trivial solutions for Ax = 0


Characteristic Equation

Characteristic Equation

( ) 02221

1211 =⎥⎦

⎤⎢⎣

⎡−

−=−

λλ

λaa

aadetdet IA

( )( ) 021122211 =−−−⇔ aaaa λλ

( ) 02112221122112 =−++−⇔ aaaaaa λλ

02 =++⇔ qpλλ with p = -(a11 + a22) and q = a11a22 - a12a21

⎪⎪⎩

⎪⎪⎨

⎧

−−−=

−+−=

⇒

24

24

22

21

qpp

qpp

λ

λFor Chapter 3, we only need the case p2 > 4q

⇒ λ1 ≠ λ2

Ax = λx ⇔ (A - λI)x = 0 ⇔ det(A-λI) = 0


Summary1. This 2x2 matrix

equation: Ax = 0

2. Has non-trivial solution iff

a11a22 = a12a21

3. This is equivalent to det(A) = 0

4. This 2x2 matrix equation

Ax = λx

5. Has a non-trivial solution iff

det(A - λI) = 0

6. In general, 2 eigenvalues which are the roots of the characteristic equation

242

21qpp

,−±−

=λ

p = -(a11 + a22) and q = a11a22 - a12a21


+

-Vs C1

R1

C2

R2

t=0 vC1 vC2=vo

iC1iC2

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

011

1111

112

1

2222

122112

1

CRV

vv

CRCR

CRRRC

dtdv

dtdv

S

C

C

c

c

Circuit Analysis: 2nd Order RC Circuit

Back to our core business

How does interconnectinfluence the switching speed?

How can we solve the associated 2nd order D.E.?


Plan (Global)

Compare to first order case (again)

Total Solution = Transient

SolutionSteady State

Solution+

+

-Vs C1

R1

C2

R2

t=0 vC1 vC2

iC1iC2

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

011

1111

112

1

2222

122112

1

CRV

vv

CRCR

CRRRC

dtdvdt

dvS

C

C

c

c

Again separate solution in transient part and steady state

⇒ vci(t) = vti(t) + vci (∞) i=1,2

Multiple time constants follow from eigenvaluesare equal to 1/


Plan (More Detailed)

B. Compute Vci(∞) from DC equivalent circuit

A. Separate Transient and Steady State: vci(t) = vti(t) + vci (∞)

C. Compute transient solution vti(t) by assumingexponential solutions

Procedure for vti(t) involves linear algebra

+

-Vs C1

R1

C2

R2

t=0 vC1 vC2

iC1iC2

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

011

1111

112

1

2222

122112

1

CRV

vv

CRCR

CRRRC

dtdvdt

dvS

C

C

c

c


A. Separate Transient and Steady-State

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0CR

V

vv

CR1

CR1

CR1

R1

R1

C1

dtdv

dtdv

11S

2C

1C

2222

122112c

1c

vdtd A v K= +

KAvv +=dtd

)(∞+= vvv t

[ ] [ ] KvAv +∞+=∞+ )()( vvdtd

tt

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∞

∞

∞

∞

2

1

2

1

2

12

1

2

1

KK

VV

vv

dtdV

dtdV

dtdv

dtdv

t

tt

t

AA

KAvAvvv +∞+=∞+ )()( tt dtd

dtd Because d/dt is a

linear operation


+

-Vs C1

R1

t=0 vC1

iC1iC2

C2

R2

vC2

vC1(∞) =

vC2(∞) =

Result:

B. Compute Vci(∞) from DC equivalent circuit

When t → ∞, we have reached steady state

Voltages are constant ⇒ time derivatives are zero ⇒capacitor currents are zero ⇒ DC equiv. ckt

Use DC equivalent circuit to compute vCi(∞)

Vs

Vs

vc1(t) = vt1(t) + Vs


Now we know this

Next steps will compute this


C. Compute Transient SolutionBut how? vc1(t) = vt1(t) + Vs


Now we start to compute this

Watch this:Let t → ∞ and simplify

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∞

∞

∞

∞

2

1

2

1

2

12

1

2

1

AAKK

VV

vv

dtdV

dtdV

dtdv

dtdv

t

tt

t

Cancels in steady state

Always cancels because constant

Cancels in steady state, because transient decays to zero

§3.6.4

Sum must be zero at t = ∞

But these are independent of time

This sum is constant 0


⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∞

∞

∞

∞

2

1

2

1

2

12

1

2

1

AAKK

VV

vv

dtdV

dtdV

dtdv

dtdv

t

tt

t

Now, t isarbitrary again

C. Compute Transient Solution

This term cancels because time derivative of constant voltage is 0

These terms TOGETHER are ALLWAYS ZERO: see previous slide

⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

12

1

t

tt

t

vv

dtdv

dtdv

A

This is what remains for vt



Now we start to compute this

Transient solution obeys HomogeneousD.E.


Kavdtdv

+=

Comparison

KAvv +=dtd

ttdtd Avv = Homogoneous eq for vt

Non-homogoneouseq for v

⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

12

1

t

tt

t

vv

dtdv

dtdv

A

This is what remains for vtTransient solution obeys HomogeneousD.E.


Assume Exponential Solution1. We have the homogenous

equation for the transient solution2. We know from first order case

that transient solutions can be of the form est

3. Try it: vti(t)=Viest and vti(t)’=sViest

for i = 1,2!

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

dtvdtv

dtdv

dtdv

t

t

t

t

2

1

2

1

A

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=⎥⎦

⎤⎢⎣

⎡st

st

st

st

eVeV

CRCR

CRRRCesVesV

2

1

2222

122112

111

1111Result:

§3.6.4Eq. 3.31


Solve Homogeneous Equation

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=⎥⎦

⎤⎢⎣

⎡st

C

stC

stC

stC

eVeV

CRCR

CRRRCesVesV

2

1

2222

122112

111

1111

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛+−

=⎥⎦

⎤⎢⎣

⎡

2

1

2222

122112

111

1111

C

C

C

CVV

CRCR

CRRRCsVsV

sV=AVScalar times vector = matrix times same vectorThis is an eigenvalue equation with s playing the role of λ!Solution for s follows from det(sI-A) = 0

Cancel common factor est in LHS and RHS

P. 142


Solve Homogeneous Equation

sV=AVScalar times vector = matrix times same vector

This is an eigenvalue equation with s playing the role of λ!

Solution for s follows from det(sI-A) = 0

Solution of det(sI-A) = 0 for 2x2 matrix A gives two solutions s1 and s2, which are the eigenvalues of A

For RC circuits, it can be shown that s1,2 < 0 and s1 != s2

Remark: det(sI-A) = - det(A -sI)


Linear Combinations1 and s2 are generalizations of (one-over (één gedeelddoor)) time constant of first-order circuitThey are not the inverse of the time constant of the first or second capacitor, respectively!

THEY ARE NOT the inverse of the time constant of the first or second capacitor, respectively!They are called the natural frequenciesIn fact, we need a linear combination of 2 exponentials, for each natural frequency, for both capacitor voltages.

t2s12

t1s111t eUeU)t(v +=

t2s22

t1s212t eUeU)t(v +=


Total Response

t2s12

t1s111t eUeU)t(v +=

t2s22

t1s212t eUeU)t(v +=

Solution for transient response:

Solution for steady-state response:

Vci (∞): has been computed from DC equivalent

vci(t) = vti(t) + Vci (∞)Total response is sum of both:

)(12121111 ∞++= ctsts

C VeUeU)t(v

)(22221212 ∞++= ctsts

C VeUeU)t(v

Total Response:


Determination of U

)(12121111 ∞++= ctsts

C VeUeU)t(v

)(22221212 ∞++= ctsts

C VeUeU)t(v

Total Response:

We know Vci(∞), s1 and s2

What remains is to compute Uij

Symbolic computation possible but unattractive, we will later demonstrate procedure using numerical example (as in book)


Natural Frequency

)(12121111 ∞++= ctsts

C VeUeU)t(v

)(22221212 ∞++= ctsts

C VeUeU)t(v

Total Response:

s1 and s2 have dimension s-1 (1/second). (Must have, why?)That is the same as Hz (frequency), but they are not genuine frequencies, keep using s-1.Value is characteristic for speed of energy exchange with environment: natural frequency

Example: swing, car that vibrates at certain speed, guitar string, pendulum, …(RC circuits of this chapter will not oscilate, but wait until chapter 5, RLC circuits)


General Procedure

1. Use KCL and element equations to construct system of D.E.

3. Determine steady-state response V(∞)

Use DC equivalent circuit

2. Write assumed solutions in general form

)(∞+= V(t)VV(t) t

)(∞+= VUest


These steps differ from book

5. Determine U from initial valuesneed 4 equations for 4 coefficients2 equations from vC (0)2 equations from v’C (0)Much easier compared to procedure in book. Recommended!

4. Determine natural frequenciesUse homogeneous equation

Substitute into assumed solution0)( =− AIsdet

General Procedure


Numeric, Complete Example

Vs

+

-1F

1/3Ω

t=0 vC1(0) = 0V

iC1 iC2

vC2(0) = 0V

1F1/2Ω

Ex. 3.7

Now, let us demonstrate the procedure

Determine VC1(t) and VC2(t)

Given (IC): VC1(0) = VC2(0) = 0

Use outline from previous slides

04 second or der 41TUE/EE 5DD17 netwerk analyse 04/05 - © NvdM

Method: General Procedure for Higher Order Circuits

1. Use KCL and element equations to construct system of D.E.

3. Determine steady-state response V(∞)

Use DC equivalent circuit

2. Write assumed solutions in general form

)(∞+= V(t)VV(t) t

)(∞+= VUest


These steps differ from book

5. Determine U from initial valuesneed 4 equations for 4 coefficients2 equations from vC (0)2 equations from v’C (0)Much easier compared to procedure in book. Recommended!

4. Determine natural frequenciesUse homogeneous equation

Substitute into assumed solution0)( =− AIsdet

General Procedure


1. Use KCL and Element Equations

KCL at vC1: 0)1(2

1211

311 =

−++

−

/vv

dtdv

/Vv CCCSC

SCCC Vvv

dtdv 325 21

1 ++−= See book for KCL at vC2

Vs

+

-1F

1/3Ω

t=0 vC1(0) = 0V

iC1 iC2

vC2(0) = 0V

1F1/2Ω

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

03

2225

2

12

1S

C

CC

CV

vv

dtdv

dtdvResult:


2. Assume General Solution

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

03

2225

2

12

1S

C

CC

CV

vv

dtdv

dtdv

)(12121111 ∞++= ctsts

C VeUeU)t(v

)(22221212 ∞++= ctsts

C VeUeU)t(v

Result:

Remember: second-order, so two independent natural frequencies, take linear combination of es1t

and es2t , add steady state value


3. Determine Steady State Response

use DC equivalent circuit

+

-Vs 1F

1/3Ω

t=0 vC1

iC1iC2

1F1/2Ω

vC2

VC1(∞) = VC2(∞) = VsResult:


4. Determine Natural Frequencies

[ ] 0=− AIsdet

022

251001

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−−⎥⎦

⎤⎢⎣

⎡sdet

022

25=⎥⎦

⎤⎢⎣

⎡+−−+

ss

det

( )( ) ( )( ) 02225 =−−−++ ss

0672 =++ ss

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

03

2225

2

12

1S

C

CC

CV

vv

dtdv

dtdv

A

11 −=s62 −=s

Result:

224497 −±−

=s

Unit of s1, s2: 1/s (1/seconde)


4. Determine Natural Frequencies

11 s1 −−=s

12 s6 −−=s

Stt

C VeUeU)t(v ++= −− 612111

Stt

C VeUeU)t(v ++= −− 622212

Result:

Ststs

C VeUeU)t(v ++= 21 12111

Ststs

C VeUeU)t(v ++= 21 22212

General Solution:

Note: s-1 (roman type) is unit, 1/seconde

si (italic type) is parameter (natural frequency)


together with (derivative at t=0) )0(')0(CiCi v

dtdv

=

Now, we need to determine the constants Uij

This procedure departs from method in book

It is simpler: vC1 and vC2 can be determined independently

(2 independent pairs of eq. with 2 variables)

For vCi : use vCi (0) (initial condition)

5. Determine Uij

Stt

C VeUeU)t(v ++= −− 612111

Stt

C VeUeU)t(v ++= −− 622212

Result:


5a. Get Derivatives at t=0

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

03

2225

2

12

1S

C

CC

CV

vv

dtdv

dtdv

0)0(0)0( 21 == cc vv

⎥⎦

⎤⎢⎣

⎡=⎥⎦

⎤⎢⎣

⎡+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

03

03

00

2225

)0(

)0(

2

1SS

C

CVV

dtdv

dtdv sC V

dtdv 3)0(1 =

0)0(2 =dt

dvC

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

03

)0()0(

2225

)0(

)0(

21

2

1S

CC

C

CV

vv

dtdv

dtdv

use t=0


Stt

C VeUeU)t(v ++= −− 612111

5b. Determine U1i From Initial Condition2 equations for vC1, from vc1(0)=0 and vc1’(0)=3Vs

sxx

C VeUeU)(v ++= −− 0612

01111 0 sVUU −=+⇒ 1211

ttC eUeUdt

tdv 612111 61)( −− −−=

12111 61)0( UUdt

dvC −−= sVUU 36 1211 =−−⇒

Next: solve these two equations

sV3=

0=


Stt

C VeUeU)t(v ++= −− 612111

5c. Solve for U1i

ss

vv

UU

UU

36 1212

1111 −

==

−+

−

sVU 25 12 =−+

sVU52

12 −=⇒ sVU53

11 −=⇒

st

st

sC VeVeVtV +−−= −− 61 5

253)(


5d. Determine U2i using same procedure

0)0(2 =dt

dvC

Stt

C VeUeUtv ++= −− 622212 )(

0)0(2 =Cv

Show this yourself!

st

st

sC VeVeVtV ++−= −− 62 5

156)(


Final Solution

+

-Vs 1F

1/3Ω

t=0 vC1

iC1iC2

1F1/2Ω

vC2

0

0.2

0.4

0.6

0.8

1

1.2

0

0.5 1

1.5 2

2.5 3

3.5 4

4.5 5

v/Vs

t

vc2

vc1

0

0.2

0.4

0.6

0.8

1

1.2

0

0.5 1

1.5 2

2.5 3

3.5 4

4.5 5

v/Vsv/Vs

tt

vc2

vc1This graph shows TOTAL solutions

st

st

sC VeVeVtV +−−= −− 61 5

253)(

st

st

sC VeVeVtV ++−= −− 62 5

156)(

Final Result:


-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

00

0.55 1.1

1.65 2.2

2.75 3.3

3.85 4.4

4.95

Dominant Natural Frequency

C )t(V sV+= tseV/− −6

52

1t

seV/ −5

3−

0tseV/− −6

52

The higher the natural frequency, the faster the transient will decay!

Fast terms may be ignored for computing delay, one (or more) dominant ones remain

Ex. 3.9

This graph shows TRANSIENT solutions

Behavior for hypothetical, much faster, transient term


Summary of Current Procedure for Ui

use vCi(0)

sVUU 36 1211 =−−

sCCC V)(v)(vUU'v 302056)0( 2112111 ++−=−−=

write v’C1 in terms of vCi

Solve

sVUU −=+ 1211

use vC1(0)

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0V3

vv

2225

dtdv

dtdv

S

2C

1C

2C

1C

Stt

C VeUeU)t(v ++= −− 612111

)0(2Cv

)0(1Cv

Use t=0 vci(0) are known, both 0 in this ex.

sc VUUv ++= 12111(0)use t=0


Summary of procedure in book

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0V3

vv

2225

dtdv

dtdv

S

2C

1C

2C

1C

Stt

C VeUeU)t(v ++= −− 612111

Stt

C VeUeU)t(v ++= −− 622212

Book: equate coefficients per natural frequency

0UU2 2111 =−

0U2U 2212 =−−

Combine with other equations and solve

+ ( ) teUU 25 62212 +− −( ) teUU 25 2111 +− −=− − − t6

12eU611−teU

Comprehend differences of both approaches


Higher Order Circuits

§3.7,3.9

Generalization:

First order circuit: One capacitor, one exponential term, one time constant, one IC, one DE, scalar equation

Second order circuit: Two capacitors, two exponential terms, twonatural frequencies, two IC’s, two DE’s, 2x2 matrix

N-th order circuit: n capacitors, n exponential terms, n natural frequencies, n IC’s, n DE’s, nxn matrix

Solving is mostly a direct generalization of 2x2 procedure (needhigher order derivatives for Ui, or use procedure of book).

Higher order means higher accuracy (in principle)

Diminishing returns from increasing order A model should be as simple as possible, but not simpler….


Conclusion

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∞

∞∞

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

)(

)()(

2

12

1

21

22221

141211

2

1

ntns

ts

ts

nnnn

n

n v

vv

e

ee

VVV

VVVVVV

v

vv

KK

K

KKKK

K

K

K

Circuit with n independent capacitors has following transient solution:

1. Use KCL and branch relations for system of DE2. Determine DC response3. Write assumed solution in general form4. Determine value of natural frequencies5. Determine coefficients of exponential terms6. (Check the solution)

System with RC interconnect can be modeled as RC ladder circuit

Solution methodology:

Ch. 3

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