chapter 3 · area of parallelogram described by points (0,0), (a,b) and (c,d) is given by: ad bc c...
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05 second order 1TUE/EE 5DD17 netwerk analyse 04/05 - © NvdM
Chapter 3Interconnects and RC Ladder Circuits
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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
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Wires on an Integrated Circuit
← IBM CMOS 7S copper process, 0.16 µm
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Interconnect Hierarchy
Example Interconnect Hierarchy for typical 0.25µ process (Layer Stack, cross-sectional view)
≈1µm
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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
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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
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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
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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
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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
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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)
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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!
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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
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+
-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
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+
-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
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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
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⎥⎥
⎦
⎤
⎢⎢
⎣
⎡+⎥⎦
⎤⎢⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
⎟⎠
⎞⎜⎝
⎛+−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
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
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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
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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
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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
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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
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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)
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Summary
1. This 2x2 matrix equation: Ax = 0
2. Has non-trivial solution iff
a11a22 = a12a21
3. This is equivalent to
det(A) = 0
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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
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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
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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
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+
-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.?
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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/
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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
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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
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+
-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
vc2(t) = vt2(t) + Vs
Now we know this
Next steps will compute this
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C. Compute Transient SolutionBut how? vc1(t) = vt1(t) + Vs
vc2(t) = vt2(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
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⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
∞
∞
∞
∞
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
vc1(t) = vt1(t) + Vs
vc2(t) = vt2(t) + Vs
Now we start to compute this
Transient solution obeys HomogeneousD.E.
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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.
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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
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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
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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)
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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 +=
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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:
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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)
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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)
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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
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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
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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
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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
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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
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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:
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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
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3. Determine Steady State Response
use DC equivalent circuit
+
-Vs 1F
1/3Ω
t=0 vC1
iC1iC2
1F1/2Ω
vC2
VC1(∞) = VC2(∞) = VsResult:
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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)
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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)
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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:
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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
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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=
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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)(
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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)(
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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:
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-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
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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
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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
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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….
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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