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Thevenin Theorem in Sinusoidal Steady Analysis
Aim: To obtain a simple equivalent circuit for a 1-port circuit that
consists of linear, time-invariant resistors, capacitors, inductors
and independent sources.
Thevenin Theorem in
Thevenin Equivalent: V  Z Th I  VTH
sinusoidal steady
analysis: A 1-port circuit
i
I
that consists of linear
ZTH
+
resistor, capacitor,
+
1-port
+
inductors and
v
_ VTH
V
circuit
independent sources has
_
_
a Thevenin equivalent
circuit in sinusoidal
steady state if the port
ZTh
Thevenin impedance
voltage phasor can be
Equivalent impedance between terminals uniquely determined for a
given port current
when sources are set to zero.
phasor, in other words, if
VTh
Open circuit voltage
the 1-port is currentcontrolled.
The voltage of the port when the port is
left as open circuit.

Norton Theorem in Sinusoidal Steady Analysis
Aim: To obtain a simple equivalent circuit for a 1-port circuit that
consists of linear, time-invariant resistors, capacitors, inductors
and independent sources.
Norton Theorem in
I  GN V  I N
Norton Equivalent:
sinusoidal steady
analysis: A 1-port circuit
i
I
that consists of linear
+
resistor, capacitor,
1-port
+
inductors and
v
circuit
V
YN
independent sources has
_
IN
_
a Norton equivalent
circuit in sinusoidal
GN
Norton conductance
steady state if the port
current phasor can be
Equivalent conductance between
uniquely determined for a
terminals when sources are set to zero.
given port voltage phasor,
in other words, if the 1IN
Short circuit current
port is voltageThe current through the port when the
controlled.
port is short-circuited.

How to obtain Thevenin equivalent circuit?
There exist two methods for this!
I
1-port
circuit
+
_
V
V*
_
+
I
+
1-port
circuit
V
_
I*
•
Connect a sinusoidal current source to the port.
•
Solve the circuit using sinusoidal steady analysis
and obtain a relation between phasors I* and V*.
•
Use I=I* and V=-V* to obtain a relation
between I and V.
•
Set the values of independent sources to
zero.
•
Calculate the equivalent impedance ZTh = V / I.
•
Assume that I=0 (open-circuit the port) and
calculate Vth=V taking into account all
independent sources .
How to obtain Norton equivalent circuit?
There exist two methods for this!
I
+
1-port
circuit
+
V* +-
V
_
_
I
+
1-port
circuit
V
_
I*
•
Connect a sinusoidal voltage source to the port.
•
Solve the circuit using sinusoidal steady analysis
and obtain a relation between phasors I* and V*.
•
Use I=-I* and V=V* to obtain a relation
between I and V.
•
Set the values of independent sources to
zero.
•
Calculate the equivalent admitance GN = I / V.
•
Assume that V=0 (short-circuit the port) and
calculate IN=I taking into account all
independent sources .
Interchange between Thevenin and Norton
V  ZTH I  VTH
• Thevenin Equivalent:
If 1-port is not current-controlled there is no Thevenin eq..
• Norton Equivalent:
I  YNV  I N
If 1-port is not voltage-controlled there is no Norton eq..
From Thevenin to Norton:
• Z TH  0
1
VTH
I
V
Z TH
Z TH
YN
ZTH  0, No Norton equivalent!
IN
From Norton to Thevenin:
• YN  0
IN
1
V
I
YN
YN
ZTH
VTH
YN  0, No Thevenin equivalent!
Example 1: Find the Thevenin equivalent circuit for the following circuit!
Example 2: Find the Norton equivalent circuit for the following circuit!
Circuit(Network) Functions in Sinusoidal Steady
Analysis
+
IS
E1
_
linear,
timeindependent
elements
Assume that
there is only one
source.
0
 AT

 0
0
0
A  Vd    
 



I
0  V   I s 
 
M ( jw) N ( jw)  I    
 0 
T ( jw)
How does Vd affect Is ? Depending on w!
k
Vd k ( w) 
cofactork , s T ( w)
det T ( w)
≠0
IS
Circuit(Network) Functions in Sinusoidal Steady
Analysis
+
E1
_
IS
linear,
timeindependent
elements
How does Vd affect Is ? Depending on w!
k
Vd k ( w)
IS
Vd k ( w)
IS

P( jw)
Q( jw)
P( jw) are Q( jw) polynomials
in (jw) with real coefficients.
a1 ( jw)  a2 ( jw) 2  ...  an ( jw) n

b1 ( jw)  b2 ( jw) 2  ...  bm ( jw) m
This depends on the
circuit but not on the
value of Is .
One can define many other circuit functions:
Vd k ( w)
I s ( w)
V1 ( w)
I s ( w)
Vd k ( jw)
V1 ( jw)
I k ( jw)
I s ( jw)
Impedance Function
Input Impedance Function
Voltage Transfer Function
Current Transfer Function
Symmetries of Circuit Functions
Lemma: Let n( s ), s    jw be a polynomial in complex variable s with
real coefficients.
1) n( s )  n( s )
2) n( z )  0  n( z )  0 (z is called as a root of n(z).)
Proof:
n(s)  nk s k  nk 1s k 1  ...  n1s  n0
nk , nk 1,...n1, n0  R
1) n( s )  n s k  n s k 1  ...  n s  n
k
k 1
1
0
 nk s k  nk 1s k 1  ...  n1s  n0
 nk s k  nk 1 s k 1  ...  n1 s  n0
 n(s )
2) n( z )  0  n( z )  0  0
From (1) n( z )  n( z )  n( z )  0  n( z )  0
Theorem: For a circuit in sinusoidal steady state, any circuit function is
well-defined and is the ratio of two polynomials in (jw) with real
coefficients if det(T(jw)) is nonzero.
Circuit function: H ( jw) 
n( jw)
d ( jw)
H ( jw)  H ( jw) e jH ( jw)
Symmetry Property: The magnitude of any circuit function is an even
function of w and its phase is an odd function of w.
Proof:
 n( jw)  n( jw)

H ( jw)  

 d ( jw)  d ( jw)
n( jw)
 H ( jw)
jw   jw and from Lemma H ( jw) 
d ( jw)
H ( jw) 
w
n( jw)
d ( jw)
H ( jw)  H ( jw)  H ( jw)  H ( jw)
Since the phase of z is  z.
 H ( jw)  H ( jw)
Vs  Vs  e jVs
+
_
Vs (t)
Vo  Vo  e jVo
Vo  H ( jw)Vs
1-port
circuit
Vo  H ( jw)  e H ( jw)  Vs  e Vs
Vo  H ( jw)  Vs  e H ( jw)  Vs
Vo  H ( jw)  Vs
Vo  H ( jw)  Vs
vo t   H ( jw)  Vs  coswt  H ( jw)  Vs 
Result:
In order to find the behaviour of the circuit for the frequency w, one
should find H ( jw) , Vs and H ( jw), Vs .