Signals And Networks

Thevenin's Theorem and Norton's Theorem To be replaced

LOAD i

P

Network A

v

Network B

Q

Two networks to describe Thevenin's theorem

Thevenin's Theorem : Characteristics of Network A To be replaced P

Network A Q

1. Linear elements 2. Sources : dependent (controlled) or independent 3. Initial conditions on passive elements 4. No magnetic or controlled-source coupling to network B

Thevenin's Theorem : Characteristics of Network A Nonlinear Elements

Linear Elements

i L

R

C

v

T

D

M

Yes Independent Sources

No Controlled Sources

v1

i

i1 Yes

v Yes

Thevenin's Theorem : Characteristics of Network A M L1 Vg

C1

Vo

L2

R C2

(1-D)(Vg-Vo) -D(Vg-Vo) Network A

M

M L1

P

C1

Network B L2

R C2

Vg

(1-D)(Vg-Vo) -D(Vg-Vo) Q

NO

Vo

Thevenin's Theorem : Characteristics of Network A L1 Vg

C1

Vo

L2

R C2

(1-D)(Vg-Vo) -D(Vg-Vo) Network A

M

M L1

P

C1

Network B L2

R C2

Vg (1-D)(Vg-Vo) Q

NO

Vo

-D(Vg-Vo)

Thevenin's Theorem: Network A is replaced by C and Vth LOAD M

P

Network C Vth

Network B

v Q

Network A

i

Thevenin's Theorem: How to find Vth? LOAD P Vth

Network A Q Open Ckt. Voltage

Disconnect

Network B

Thevenin's Theorem: How to determine C?

P Network A

1. Set Initial Conditions to Zero 2. Independent Sources Turned Off 3. Dependent Sources are Operating

Q P Network C M

Determination of Vth : An Example Vcc Rc Ic

R1 B IB R2

IB+IC Re

Rc Ic

R1 B

Vcc

N

Network B

Network A

Vth

R2

B

Vcc IB+IC Re

Vth=Vcc[R2/(R1+R2)] Simplification of the base circuit

Thevenin's Theorem : Characteristics of Network B Nonlinear Elements

Linear Elements

i L

R

C

v

T

D

M

Yes Independent Sources

Yes Controlled Sources

v1

i

i1 Yes

v Yes

Determination of Network C : Example I Note : No initial condition and No dependent source Network C

Network A

Vcc

R1

R1

B

B

R2

N

R2

Shorted

M

Simplification of the base circuit

Network C B R1 R2 M

Rb=(R1||R2)

Application of Thevenin's Theorem: An Example Network B

Vcc Rc Ic

R1 B IB R2

Rb=(R1||R2) Rb B M

IB+IC Re N

Rc Ic Vcc IB+IC

Vth

Vth=Vcc[R2/(R1+R2)]

Simplification of the base circuit

Re N

Application of Thevenin's Theorem: An Example Network B

Vcc

Linear Model Rc Ic

Rc Ic

R1 B IB R2

IB IB+IC Re

Vth

 IB Rb B

Vth=Vcc[R2/(R1+R2)] Rb=(R1||R2) Simplification of the base circuit N

Vcc IB+IC Re N

Application of Thevenin's Theorem: Example II Common Source Amplifier : Small Signal Gain Vdd

Cb

vi

vo Av= vi

RL D Id

R1 G

S

R2

Is Rs

N

Vo=VO +vo

Application of Thevenin's Theorem: Example II Linear Small Signal Equivalent Circuit of the CS Amplifier D Cb

G

gm vgs

rd vo

S

R2 R1 vi

Rs N

RL

Application of Thevenin's Theorem: Example II Simplified Small Signal Equivalent Circuit of CS Amplifier Network A G

D

gm vgs S

vi

Rs N

Network B M Network C D

rd vo

Vth RL

RL N

Application of Thevenin's Theorem: Example II Determination of Vth Network A G

D

gm.vgs S

vi

Rs N

rd Vth=-(gm.vgs)rd =-(gm.rd)vi

Application of Thevenin's Theorem: Example II Determination of Netwok C Network C G

D

gm.vgs

rd

S Shorted

Rs M

M Rth

D

Application of Thevenin's Theorem: Example II Determination of Rth D vx=(ix-gm.vgs)rd-vgs gm.vgs

ix rd

S Rs

vgs

G,M

vx

vgs=-ix.Rs Therefore, vx=ix.rd+gm.ix.Rs.rd +ix.Rs vx Rth= ix =rd+Rs(1+gm.rd)

Application of Thevenin's Theorem: Example II The equivalent circuit of the CS amplifier Rth M  vi

rd

Rs(1+  )

 =gm. rd D

RL vo

Vth

N

Application of Thevenin's Theorem: Example III RLC Network : Zero Initial Conditions on L and C Network B Network A C L C L R vo vi

RL

R vi L

Zth = 1/sC + sL||R Vth=vi R R+sL

Vth

C R

Zth

Norton's Theorem: Network A is replaced by C and Ith LOAD Network A M

Ith

P

Network C

i Network B

v Q

P

Network C as in Thevenin's Theorem

Q

Ith =>Norton's Theorem is dual of Thevenin's Theorem

Network A

v1

B

R1 Vg

C

v2

D

R2

Loop v3 Direction R3

A

Fig. 1 One-loop resistive network to which Kirchoff's Voltage Law is applied

− vg + v1 + v2 + v3 = 0 v1 + v2 + v3 = vg

Fig.2 Ground

b

ia

c

i1

1

i2

v2

v1

R2

R1

vg d i4

i3

1 ib

a Fig. 3 i1+i2=i3+i4

Fig. 4(a)

1 vg 1 Equivalent Network

Req

Fig. 4(b)

ia 1

R1

R2

Rn

vg 1 Fig. 5 n R eq = R 1 + R 2 + ........... + R n =  j=1 Rj

Series Connected Resistive Network

ib

ia 1

i1

va L1

i2 L2

vg

Leq

1 Fig. 5(a)

Fig. 5(b) Leq= 1/L1+1/L2

Equivalent Inductive One-Port Network

Number of Network Equations Formulation: 3i1 + 2i2 - i3=4 (1) -i1 + 5i2 + 3i3 =-2 (2) i1 + 12 i2 + 5i3 = 0 (3) Solve: Ax=b

e'

i3 No solution

f'

e Unique Solution

g'

g

Note: 2*(2)+(1)=(3) i2

i1 family of solutions f

* Equations must be linearly independent * Formulate the network using minimum no of variables

How to formulate the network using linearly independent equations? => How many unknowns in a network with 'n' nodes and 'b' branches? C1

b6

4 b7 C3

2

R2

R1 1 b4

b1

b5

3 RL

R3

C2 b2

b3

0 : Ref Node n=5 and b=8

b8

=> 2*b unknowns (branch voltage and branch current) => effectively b unknowns example : if voltage across a capacitor is known current can be found out from the expression CdVc/dt = Ic

How many KCLs and how many KVLs?

C1

b6

4 b7 C3

2

R2

R1 11 b4

b1

b5

V

b2

3

example : n-1=4 => 4 KCLs b-(n-1)=8-4=4 => 4 KVLs

RL R3

C2

(1) (n-1) nodes => (n-1) KCL equations (2) b-(n-1) KVL equations must be used

b3

0 : Ref Node

note: easy to apply KCL but difficult to apply KVL because there are so many b8 loops (b4,b1,b2), (b2,b5,b8), (b3,b7,b8), (b4,b1,b5,b8), (b4,b6,b3), (b4,b6,b7,b8), (b1,b6,b7,b5)

Problem: With a wrong choice the equations will be linearly dependent

How to define voltages and currents in KCL and KVL? VC3 C1

b6

4 b7 C3

2

R2

R1 11 b4

b1

b5

Node Voltage : Voltage of a node with respect to the reference node (0) 3 RL

R3

C2 b2

Example : V4 (i.e. voltage across R1 in branch b3)

b3

0 : Ref Node

Branch Voltage : Voltage across a branch of the network b8 Example : VC3 (i.e. voltage across branch b7)

Note: Vc3 = V4-V3 => KVL It is enough if we know the node voltages and vice versa

How to define voltages and currents in KCL and KVL?

C1 1

4 b7 C3

b6

R1

R2

2 b1

Loop Current : Current in a loop of the network 3

b5

b4 i1 C2 V Ib2 b2

R3 b3

0 : Ref Node

i2

Example : i2 (i.e. current in loop consisting of branches b2,b5,b8)

RL Branch Current : Current in a branch of the network b8 Example : Ib2 (i.e. current in branch b2, through capacitor C1)

Note: Ib2 = i1-i2 => KCL It is enough if we know the loop currents and vice versa

Formulate network equations with explicit application of KVL C1 4 b7 C3 4 b7 b6 b6 R1 1 b4

R2

2 b1

b5

3

b1 b4

V b2

1

RL R3

C2

b3

b8

0 : Ref Node Choice of variables: Loop Currents and Branch Voltages

3

2 b5 b2 b3 0 : Ref Node Graph of the network

Independent Equations: => Draw a graph : Figure with connection information

b8

Formulate network equations with explicit application of KVL 4 b7

b6 1

b1 b4

b6 3

2

0 : Ref Node Graph of the network

b8

3

b1 b4

b2

b7

2

1

b5

b3

4

Tree: (b1,b2,b5,b7)

b2

b5 b3

b8 Co-tree: (b4,b6,b3,b8)

0 : Ref Node A tree of the network

Construct a tree => A connected graph spanning all nodes that does not have a circuit. A tree has n nodes and (n-1) branches. => Graph has more than one tree

Formulate network equations with explicit application of KVL 4

b6

b7 3

2

1

b1 b4

Tree: (b1,b2,b5,b7)

b2

=>The co-tree has b-(n-1) branches

b5 b3

b8 Co-tree: (b4,b6,b3,b8)

0 : Ref Node Tree of the network

Every branch of the co-tree (known as chord) is responsible for \ formation of a loop (circuit). This is known as fundamental circuit. Example : (b6 , b1,b5,b7) forms a fundamental circuit No of fundamental circuits = no of chords = b-(n-1)

If we apply KVL to these fundamental circuits then the equations are linearly independent

Formulate network equations with explicit application of KVL b6 i3 2 i1 b1

1 b4

Tree: (b1,b2,b5,b7)

b2

4

b7 i4

C1 3

b5 i2 b3 b8 Co-tree: (b4,b6,b3,b8)

KVLs: 0 : Ref Node =>(b4: chord) V=Vb1+Vb2 =>(b8: chord) Vb2=Vb5+Vb8 Vb6=Vb1+Vb5-Vb7 =>(b6: chord) Vb7=-Vb3-Vb2+Vb5 =>(b3: chord)

R1 1 b4

4 b7 C3

b6

R2

2 b1

3

b5 R3

C2 V b2

b3

b8

0 : Ref Node Original Network 7 voltages and current in V are unknown (7+1=8)

Formulate network equations with explicit application of KVL C1 R1 1 b4

4 b7 C 3

b6

R2

2 b1

V-I relationship of the elements 3

b5

R4 R3

C2 V b2

b3

b8

b4

4 b7 b6 I4 I3 2 b1 b5 I1 b3 I2 b2 b8

0

Vb2=1/C2 Ib2 dt Vb3=Ib3*R3 Vb4=V

0 : Ref Node Original Network

1

Vb1=Ib1*R1

Vb5=Ib5*R2 3

Vb6=1/C1 Ib6 dt Vb7=1/C3 Ib7 dt Vb8=R4*Ib8 Expressed in terms of 8 (branch) currents

Formulate network equations using KVL Apply KCL at b1:

4

1 b4

b7 b6 I4 I3 2 b1 b5 I1 b3 I2 b2 b8

Ib1= (I1-I3)

3

Apply KCL at b2: Ib2=I1-I2+I4 Apply KCL at b5: Ib5=I2-I3-I4 Apply KCL at b7:

Ib4

0 Note: Ib4=-I1 ; Ib6=I3; Ib8=I2; Ib3=I4

Ib7=I4+I3 => enough if we know the loop currents => chords

Formulate network equations using KVL: All the equations KCLs: Ib1= (I1-I3) Ib2=I1-I2+I4 Ib5=I2-I3-I4 Ib7=I4+I3 Ib4=-I1 ; Ib6=I3; Ib8=I2; Ib3=I4 KVLs: =>(b4: chord) V=Vb1+Vb2 =>(b8: chord) Vb2=Vb5+Vb8 Vb6=Vb1+Vb5-Vb7 =>(b6: chord) Vb7=-Vb3-Vb2+Vb5 =>(b3: chord)

Elements: Vb1=Ib1*R1 Vb2=1/C2 Ib2 dt Vb3=Ib3*R3 Vb4=V Vb5=Ib5*R2 Vb6=1/C1 Ib6 dt Vb7=1/C3 Ib7 dt Vb8=Ib8*R4

Network equations using KVL Unknowns: Loop Currents: I1,I2,I3 and I4

Therefore the analysis is called Loop Variable Analysis

V=Vb1+Vb2 => V = (I1-I3)R1+ 1/C2 (I1-I2+I4)dt Vb2=Vb5+Vb8 => 1/C2 (I1-I2+I4)dt = (I2-I3-I4)R2+I2*R4 Vb7=-Vb3-Vb2+Vb5 => 1/C3 (I4+I3)dt=-I4*R3-1/C2 (I1-I2+I4)dt+(I2-I3-I4)R2 Vb6=Vb1+Vb5-Vb7 1/C1 I3dt = (I1-I3)R1 + (I2-I3-I4)R2 -1/C3 (I4+I3)dt

Loop Variable Analysis: Matrix Representation V

R1+1/C2 dt

-1/C2 dt

-R1

-1/C2 dt

I1

0

-1/C2 dt

1/C2 dt

-R2

-R2 -1/C2 dt

I2

+R2+R4

= 0 0

-R1

-R2

1/C2 dt -1/C2 dt -R2

1/C3 dt +1/C1 dt

R2

+R1+R2

+1/C3 dt

1/C3 dt +1/C2 dt dt +R2 1/C2 +R3+R2

I3 I4

Loop Variable Analysis: Generalization I1

I4

I2 I3

IL

Graph of a network with L independent loop currents

Rkj = total resistance common to loops k and j Lkj = Ckj =

Voltage drop in loop k produced by current Ij Rkj*Ij+Lkj*dIj/dt +1/Ckj Ij dt = (Rkj+Lkj d/dt +1/Ckj dt)Ij =akjIj akj: symbol to summarize the operation on Ij

Loop Variable Analysis: Generalization The general form of KVL L  akjIj=Vk , Vk is the active voltage source in loop k j=1 General form of KVL for L loop L  akjIj=Vk , k =1,....,L j=1 Loop 1 => Loop 2 => Vector Equation: Loop L =>

V1

Operator on I1 I2 I3 I4 a11 a12 a13 ---- a1L I1

V2

I2

VL

= ---

--- ---

aL1 ---

---- ---

--- --- aLL IL

Loop Variable Analysis: Matrix Representation V

R1+1/C2 dt =a11

-1/C2 dt =a12

0

-1/C2 dt

1/C2 dt

=a21

+R2+R4 =a22

= 0 0

-R1 =a31

-R2 =a32

1/C2 dt -1/C2 dt -R2 =a41

-R1 =a13 -R2 =a23

-1/C2 dt =a14

I1

-R2 -1/C2 dt =a24

I2

1/C3 dt +1/C1 dt

R2

+R1+R2

+1/C3 dt

1/C3 dt +1/C2 dt 1/C2 dt +R2 +R3+R2

Solution : Find the inverse of the matrix (square: b-(n-1)*b-(n-1)) May use Gauss Elimination Method : A process called trangularization

I3 I4

Formulate network equations with explicit application of KCL C1 4 b7 C3 4 b7 b6 b6 R1 1 b4

R2

2 b1

b5

3

b1 b4

V b2

1

RL R3

C2

b3

b8

0 : Ref Node Choice of variables: Node Voltages and Branch Currents

3

2 b5 b2 b3 0 : Ref Node Graph of the network

Independent Equations: => Draw a graph : Figure with connection information

b8

Formulate network equations with explicit application of KCL 4 b7

b6 1

b1 b4

b6 3

2

0 : Ref Node Graph of the network

b8

3

b1 b4

b2

b7

2

1

b5

b3

4

Tree: (b1,b2,b5,b7)

b2

b5 b3

b8 Co-tree: (b4,b6,b3,b8)

0 : Ref Node A tree of the network

Construct a tree => A connected graph spanning all nodes that does not have a circuit. A tree has n nodes and (n-1) branches. => Graph has more than one tree

Formulate network equations with explicit application of KCL 4

b6 1

2

b7

=>The tree has (n-1) branches 3

Every branch of the tree b5 is responsible for the b3 formation of a cut-set. This b2 b4 b8 is known as fundamental cut-set. Tree: Co-tree: (b1,b2,b5,b7) (b4,b6,b3,b8) A cut-set is the set of branches whose removal disconnects the graph 0 : Ref Node Example : (b1 ,b4, b6) forms a cut-set Tree of the network b1

No of fundamental cut-sets = no of branches in a tree = (n-1) If we apply KCL to these fundamental cut-sets then the equations are linearly independent

Identify the fundamental cut-sets to apply KCL 4

b7

b1 b4

b6

This node is disconnected

3

1

b5

0 This is a fundamental cut-set

b8 0

b6

b6

b8 (b1,b6,b4)

4 b7

b1

b7

b3

0

2

1

4

3 b5

0

b3

b7

2

b2

b4

2

b2

b1 2

1

3 (b5,b3,b8,b7) 4

2

1

4

4

b6

3 4

b5

3

1 b3

b4

b2 b3

(b7,b3,b6) 0

0

b8 (b2,b3,b4,b8)

Formulate network equations with application of KCL 4 b6 C1 C3 4 b7 b6 1 R2 R1 2 2 3 b5 1 b1 b5 b3 b4 RL b8 R3 C2 V b2 b3 b8 0 Cut-set of b5: Ib6+Ib3+Ib5-Ib8=0 0 : Ref Node Conservation of Charges I1

A In I1+ I2 + ... + In =0

B

Cut-set of b7:

-Ib4-Ib1-Ib6=0 Ib6-Ib7+Ib3=0

Cut-set of b2:

Ib2+Ib4-Ib3+Ib8=0

Cut-set of b1:

3

V-I relationship of the elements C1 R1 1 b4

4 b7 C 3

b6

R2

2 b1

In terms of branch voltage and branch currents 3

b5

R4

Ib1=Vb1/R1

R3

C2

Ib2=C2 dVb2/dt

V b2

b3

0 : Ref Node Original Network

b8

Ib3=Vb3/R3 Vb4=V ; Ib4=? Ideal voltage source Ib5=Vb5/R2 Ib6=C1dVb6/dt Ib7=C3dVb7/dt Ib8=Vb8/R4 Expressed in terms of 8 (branch) voltages

Formulate network equations using KCL

4

b6 Ib6 1

b7 Ib7

b1

Ib3

b3

b2

b4

Vb1= (V1-V2)

3

2 Ib1

Apply KVL at all branches:

Ib4

0

Ib5

b5 Ib8

b8

Vb2=V2-V3 Vb3= V4 Vb4=V1 Vb5=V2-V3 Vb6=V1-V4 Vb7=V4-V3 Vb8=V3 => enough if we know the node voltages

Formulate network equations using KCL: All the equations KCLs: Ib6+Ib3+Ib5-Ib8=0

Elements:

-Ib4-Ib1-Ib6=0

Ib2=C2 dVb2/dt

Ib6-Ib7+Ib3=0 Ib2+Ib4-Ib3+Ib8=0

Ib1=Vb1/R1

Ib3=Vb3/R3 Vb4=V ; Ib5=Vb5/R2

V2,V3,V4 and Ib4

Ib6=C1dVb6/dt

Note: V1 is given (V1=V) Ib7=C3dVb7/dt However, Ib4 is unknown Ib8=Vb8/R4

KVLs: Vb1= (V1-V2) Vb2=V2-V3 Vb3= V4 Vb4=V1 Vb5=V2-V3 Vb6=V1-V4 Vb7=V4-V3 Vb8=V3

Note: You can write KCLs at all nodes except the reference node and obtain (n-1) linearly independent equations

Network equations using KCL Unknowns: Node Voltages: V2,V3, V4 and Ib4

Therefore the analysis is called Node Variable Analysis

No. of Unknowns: n-1 =4 Solve a system of linear integro-differential equations: C1d/dt (V-V4) + V4/R3 + V2-V3/R2 - V3/R4 = 0 -Ib4 - (V1-V2))/R1 -C1d/dt ( V-V4) = 0 C1d/dt (V-V4) -C3d/dt (V4-V3) + V4/R3 = 0 C2d/dt (V2-V3) + Ib4 - V4/R3 + V3/R4 = 0 Note: If (n-1) < b- (n-1) then it is advantageous to use nodal analysis as we need to solve less number of linearly independent equations

Node Variable Analysis:Matrix Representation C1dV/dt

0

1/R2

V/R1 + C1dV/dt

-1

1/R1

-1/R2-1/R4

0

-C1d/dt +1/R3

Ib4

C1d/dt

V2

= C1dV/dt

0

0

1

0

C2d/dt

C3d/dt

1/R4 + C2d/dt

1/R3 + C3d/dt

-1/R3

V3

V4

Note : There is no active current source. The voltage source V is equivalent to a current source Ib4 (unknown) with terminal voltage V (known)

Node Variable Analysis: Generalization

1

node j

C1 C2 R1 R2 L1 L2

Ikx

node x

Ikj

1/Rkj = (1/R1+1/R2+...) 1/Lkj = (1/L1+1/L2+...)

node k Ik (Active Source)

Ckj = C1+C2+...

Elements Connecting Nodes j and k

Current flowing out of node k due to voltage in node j= Vj*(1/R1 + 1/R2 + ...) + (1/L1+1/L2 + ...) Vjdt + (C1+C2 +..)dVj dt = (1/Rkj + 1/Lkjbkj dt + Ckjd/dt)Vj = bkjVj bkj: symbol to summarize the operation on Vj

Loop Variable Analysis: Generalization The general form of KCL L  bkjVj=Ik , Ik is the active current source connected to node k j=1 General form of KCL for N nodes L  bkjVj=Ik , k =1,....,N j=1 Node 1 => Node 2 => Vector Equation: Node N =>

I1

Operator of I1 I2 I3 VN b11 b12 b13 ---- b1L V1

I2

V2

IN

= ---

--- ---

bL1 ---

---- ---

--- --- bLL VN

Node Variable Analysis: Identify the operators V/R1 + C1dV/dt

-1

-V/R1

0 =

0

0

1/R1 =b12 -1/R2-1/R1-C2d/dt =b22 1/R2 =b32

1/R2 =b23

-1/R2-1/R4-C3d/dt =b33

C1d/dt =b14

Ib4

0 =b24

V2

C3d/dt =b34

V3

0 C3d/dt -1/R3-(C3+C1)d/dt V4 =b42 =b43 =b44 Note : There is no active current source. The voltage source V is equivalent to a current source Ib4 (unknown) with terminal voltage V (known) -C1d/dt

0

0 =b13

Apply KCL at each node: Formulation by observation C1 R1 1 b4

4 b7 C3

b6

R2

2 b1

KCL at Node 1: -Ib4 - (V-V2)/R1-C1d/dt (V1-V4) = 0

3

b5 R4 R3

C2

(V-V2)/R1 - C2d/dt (V2) - (V2-V3)/R2 = 0

V b2

b3

KCL at Node 2 :

b8

0 : Ref Node

Note: For a DC source dV/dt = 0 Even without identification of the trees it will be possible to formulate the network equations

KCL at Node 3: (V2-V3)/R2 -V3/R4 + C3d/dt (V4-V3) = 0 KCL at Node 4: C1d/dt(V-V4) - V4/R3 - C3d/dt (V4-V3) = 0

v1 v1+v2=v v2

i1

i2

i1+i2=i

Sources can be combined

v

R

v

R i i No Difference

R v(t)

v1(t)

i1(t)

i1(t) v/R

Source Transformation

R

L

v1(t)

i1(t) i=1/L vdt

v(t)

C v(t)

v1(t) i1(t)

v1(t)

i1(t)

v1(t) i1(t) i=Cdv/dt

C

Source Transformation

R1

L1

R1

R2

R2 v1

L1

v1

v1

Equivalent Networks

Network Functions

I V

I

One-Port Network

Port : A pair of terminals in which the current into one terminal equals the current out of other A one-port network is completely specified if voltage-current relationship at the terminals of the port is known. I

I I

10

5 I

10

Network Functions Two-Port Network : =>Defined by two pairs of voltage-current relationships =>The variables are V1,I1,V2,I2.

Dependent

=>Two dependent variables Two independent variables I1 V1

Two-Port Network

Independent

1.

V1,V2 z parameters

I1,I2

2.

I1,I2 y parameters

V1,V2

3.

V1,I2 I1,V2 h (hybrid) parameters

4.

V1,I1 V2,-I2 A,B,C,D parameters

I2 V2

Network Functions Mathematical description using dependent and independent variables I1=y11*V1+y12*V2 I2=y21*V1 + y22*V2 Short-Ckt. Admittance Definition of y parameters:

V1=z11*I1+z12*I2 V2=z21*I1 + z22*I2 Open-Ckt. Impedance Definition of z parameters: V1 z12 = I2

V1 z11 = I1

I2=0 Open-Circuit Parameters

V2 z21 = I1

I2=0

V2 z22 = I2

I1=0

I1 y11 = V1

V2=0

I1 y12 = V2

V1=0

Short- Circuit Parameters

I1=0

I2 y21 = V1

V2=0

I2 y22 = V2

V1=0

Network Functions Mathematical description using dependent and independent variables V1=A*V2 + B*-I2 I1= C*V2 + D*-I2

V1=h11*I1+h12*V2 I2= h21*I1 + h22*V2

Definition of h (hybrid) parameters: Definition of A,B,C,D parameters: V1 h11 = I1

V2=0 Short-Ckt

I2 h21 = I1

V2=0

V1 h12 = V2

A= I1=0

V1 V2

B=

Open_Ckt

I2=0 Open-Ckt

I2 h22 = V2

I1 V2

C= I1=0

V2=0

Short-Ckt D=

I2=0

-V1 I2

-I1 I2

V2=0

Network Functions Example: Find the open-circuit parameters for the T- circuit 2

V1=z11*I1+z12*I2 V2=z21*I1 + z22*I2

Transfer V2 z21 = I1

I2=0

V1 z12 = I2

I1=0

Driving Point I2=0

V2 z22 = I2

1

2

1

2

Rc I2 I1 Ra V1 V2 Rb 10

Definition of z parameters: V1 z11 = I1

7

I1=0

z11= Ra+Rb=12

Note: Z12=Z21 I2*Rb = 10 Reciprocal z12= I2 Network => z22 = Rb+Rc = 17 All passive networks I1*Rb z21 = = 10 I1

How do we obtain the z parameters of any two port (active or passive) network? A set of Node Equations: I1 = b11V1 + b12V2 + b13V3 +.........+ b1nVn I2 = b21V1 + b22V2 + b23V3 + ........+ b2nVn 0 = b31V1 + b32V2 + b33V3 + ........+ b3nVn ........................................................................ ........................................................................ 0 = bn1V1 + bn2V2 + bn3V3 + ....... + bnnVn Vector Equation: I= GV Solution: -1 V=G I=ZI

z21 Z= G -1 =

For passive network:  12 =  21 z12 = z21

z11

z12

z22

V1,V2 : terminal voltages V3,....,Vn : Node Voltages : dependent variables I1,I2, .... : Current Sources : independent Variables bjk : operator due to Vj on node k

 11  21  n1     12  22 ---------------------------  ---------------------------------- nn  2n  1n   

 ij = co-factor of bij  = determinant of G

Example: Open-circuit (Z parameters) of the Pi Circuit 1

V2

V1

2

I1=(GA+GC)V1-GCV2 I2=-GCV1+(GB+GC)V2

GC I1

GA

GB

1

I2 2

Delta-star(wye) Transformation: z12 = Rb = GC/ z22 = Rb+Rc = (GA+GC)/ => Rc = GA/ z11 = Ra+Rb = (GB+GC)/ => Ra = GB/

= GA*GB+GA*GC+GB*GC  11 = GB+GC  12= GC  21= GC

 22 = GA+GC

GB+GC z11= GA*GB+GB*GC+GC*GA GC z21= z12= GA*GB+GB*GC+GC*GA GA+GC z22= GA*GB+GB*GC+GC*GA

How do we obtain the y parameters of any two port (active or passive) network? A set of Loop Equations: V1 = a11I1 + a12I2 + a13I3 +.........+ a1nIn V2 = a21I1 + a22I2 + a23I3 + ........+ a2nIn 0 = a31I1 + a32I2 + a33V3 + ........+ a3nIn ........................................................................ ........................................................................ 0 = an1I1 + an2I2 + In3I3 + ....... + annIn Vector Equation: V= RI Solution: -1 I=R V=YI

y21 Y= R -1 =

For passive network:  12 =  21 y12 = y21

y11

y12

y22

I1,I2 : Port input currents I1,I2,I3,....,In : Loop Currents : dependent variables V1,V2, .... : Voltage Sources : independent Variables ajk : operator due to Vj on node k

 11  21  n1     12  22 ---------------------------  ---------------------------------- nn  2n  1n   

 ij = co-factor of aij  = determinant of R

Network Functions Example: Find the short-circuit parameters for the bridged T- circuit 5

 = 12(17*14 - 7*7) -(10)(10*14 - 7*-2) +(-2)(10*7 - (17)*-2) = 520

I3 1 V1 1

2 I1

7 10 I2

2 V2

11 =17*14 - 7*7=189

2 y11=

 11

189 = 520 = 0.363

 V1 = (10+2)I1 +10I2 - 2I3  21 -154 V2 = 10I1 +(10+7)I2 +7I3 y12= = 520 = -0.296 =y12  0 = -2I1+7I2 +(5+7+2)I3  22 164 y22= = 520 = 0.315 12 10 -2  R = 10 17 7 When y11=y22 or z11=z12, the network -2 7 14 is symmetrical

The h parameters : comparison Extremely useful for describing bipolar junction transistor circuits V1= Vbe = Base to emitter voltage I2 = Ic = Collector current I1= Ib = Base current Definition of h (hybrid) parameters: V2 = Vce = Collector to emitter voltage I1 y11 = => h11= 1/y11 V1 V1 V1 h12 = h11 = V2=0 V2 I1 I1=0 V2=0 V2 => h22 = 1/z22 z22 = Short-Ckt I2 Open_Ckt I1=0 I2 I2 V1 h21 = h22 = => h12 = z12/z22 z12 = I1 V2 I2 V2=0 I1=0 I1=0 I2 => h21 = y21/y11 y21 = V1 V2=0 V1=h11*I1+h12*V2 I2= h21*I1 + h22*V2

Example: h parameters of the Pi Circuit 1

V1 I2

GC I1

GA

2 h11 = 1/(GA+GC) V2

GB

1

2

V1 h11 = I1

V1 h12 = V2

I2 h21 = I1

I2 h22 = V2

V2=0 Short-Ckt

V2=0

h21 = -GC/(GA+GC) h12 = GC/(GA+GC) h22 = GB + GC*GA/(GC+GA)

I1=0 Open_Ckt

I1=0

Note: h12=-h21 => Reciprocity Condition

The A,B,C,D parameters : comparison Extremely useful for describing transmission network Transmission engineers convention V1=A*V2 + B*-I2 I1= C*V2 + D*-I2 V1, I1 are sending end voltage and current variables A B <= Transmission V2, I2 are receiving end voltage C D Matrix and current variables Definition of A,B,C,D parameters: A=

V1 V2

B=

I2=0 Open-Ckt C=

I1 V2

V2=0

Short-Ckt D=

I2=0

-V1 I2

-I1 I2

V2=0

V1 z11 = I1 V2 z21 = I1

I2=0 => A =

=> C = I2=0

z11 z21 1 z21

The A,B,C,D parameters : comparison Extremely useful for describing transmission network Transmission engineers convention V1=A*V2 + B*-I2 I1= C*V2 + D*-I2 V1, I1 are sending end voltage and current variables A B <= Transmission (T) V2, I2 are receiving end voltage C D Matrix and current variables Definition of A,B,C,D parameters: I1 -y11 -V1 V1 y11 = => D = B= A= V1 y21 I2 V2 V2=0 V2=0 I2=0 Open-Ckt Short-Ckt I2 -1 -I1 I1 y21 = => B = D= C= V1 y21 I2 V2 V2=0 V2=0 I2=0

Example: T parameters of the Pi Circuit 1 I1

2

I2

GC

I1 V1

GA

V1

GB

V2

1 A=

2 V1 V2

B=

I2=0 Open-Ckt C=

I1 V2

V2=0

Short-Ckt D=

I2=0

-V1 I2

-I1 I2

V2=0

V2

1

2

GC GA

GB

1

Ix 2

GB+GC 1/GB+1/GC A= = GC 1/GB 1/GA Ix = I * 1/GA + 1/GB + 1/GC V2 = Ix*1/GB = GA*GB +GB*GC+GC*GA = GC GA*GB+GB*GC+GC*GA C= GC

Example: T parameters of the Pi Circuit 1 I1

I2

GC GA

V1

2

GB

I2 1

2 I1

V2

V1

GC GA

1

1 A=

2 V1 V2

B=

I2=0 Open-Ckt C=

I1 V2

V2=0

Short-Ckt D=

I2=0

-V1 I2

-I1 I2

2

I2 = - V1*GC 1 B= GC I2 = -

1/GA I1 1/GA+1/GC

GC+GA GC Note : AD-BC = 1 Equivalent condition for reciprocity D=

V2=0

shorted

GB

Relationship Between Two Port Parameters The z and y relationships can be obtained by using matrix notation z11 z12 y11 y12 Z= ; I = YV ; V = ZI Y= z21 z22 y21 y22 V = ZI = ZYV

Therefore, Y = Z

1

y22 = z11/ z

z11 = y22/ y z22 = y11/ y

y12 = -z12/ z y21 = -z21/ z

z12 = -y12/ y z21 = -y21/ y

y11 = z22/ z

z = z11*z22-z12*z21

y = y11*y22-y12*y21

h and ABCD parameters can be expressed in terms of y and z parameters

Interconnection of Two Port Parameters I1 Na

V1

I2a

I2b

V2a

V2b

I2 Nb

V2

Cascade connection of Two Ports A

B

Aa

Ba

Ab

Bb

Ca

Da

Cb

Db

= C

D

The transmission matrix of overall two port network is simply the product of the transmission matrices of the individual two-ports

Interconnection of Two Port Parameters

V2a

Na

V1a

V1a=V1b = V1 V2a = V2b = V2 I1 = I1a + I1b I2 = I2a + I2b

I2a I2 1:1

I1a

I1

I1a

I1 I1b V1b

I2b Nb

I1b Ib =

Ia =

I= I2

I2a

I2b

V2b Ib =

Va =

Parallel Connection of two ports V = V2 I = Ia + Ib = YaVa + YbVb = (Ya+Yb)V Y = (Ya+Yb)

V1

V1

V1

V2

V2

Interconnection of Two Port Parameters I1

V1a

Na

V2a

V1 I1b V1b

V1a +V1b = V1 V2a + V2b = V2 I1 = I1a = I1b I2 = I2a = I2b

I2a I2 1:1

I1a

I2b Nb

V2 I1 I1 I1 Ib = Ia = I= I2 I2 I2

V2b

Series Connection of two ports

V2 V = Va + Vb = ZaIa + ZbIb = (Za+Zb)I Z = (Za+Zb)

Ib =

Va =

V=

V1b

V1a

V1

V2a

V2b

Interconnection of Two Port Parameters 1. When the two ports are connected in parallel, find the y parameters first, and, from the y parameters derive the other two-port parameters 2. When two-ports are connected in series, it is usually easiest to find the z parameters 3. When two ports are connected in tandem, it is generally easiest to find the transmission matrix.