BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

1 . Identify the frequency, angular frequency, peak value, rms value, and phase of a sinusoidal signal.
2 . Solve steady-state ac circuits using phasors and complex impedances
.
3 . Compute power for steady-state ac circuits.
4 . Find Thévenin and Norton equivalent circuits.
5 . Determine load impedances for maximum power transfer.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

The sinusoidal function v(t) = V
M sin

t is plotted (a) versus

t and (b) versus t.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

The sine wave V
M radian sin (

t +

) leads V
M sin
 t by

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Frequency f

1
Angular frequency
T
 
 f
 
2

T sin z
 cos( z

90 o
) sin
 t
 cos(
 t

90 o
) sin(
 t

90 o
)
 cos
 t
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

A graphical representation of the two sinusoids v
1 and v
2
.
The magnitude of each sine function is represented by the length of the corresponding arrow, and the phase angle by the orientation with respect to the positive x axis.
In this diagram, v1 leads v2 by 100 o + 30 o = 130 o , although it could also be argued that v
2 leads v
1 by 230 o .
It is customary, however, to express the phase difference by an angle less than or equal to 180 o in magnitude.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Euler’s identity
   t
 
2
 f
A cos
 t

A cos 2
 f t
In Euler expression,
A cos
 t = Real (A e j
 t
A sin
 t = Im ( A e j
 t )
)
Any sinusoidal function can be expressed as in Euler form.
4-6
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

The complex forcing function V m complex response I m e j (
 t +

) . e j (
 t +

) produces the
It is a matter of concept to make use of the mathematics of complex number for circuit analysis. (Euler Identity)
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

The sinusoidal forcing function V m steady-state response I m cos (

t + θ) produces the cos (

t + φ).
The imaginary sinusoidal forcing function j V m produces the imaginary sinusoidal response j I m sin (

t +
θ
) sin (

t +
φ
).
Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING

Re(V m e j (
 t +

) )

Re(I m e j (
 t +

) )
Im(V m e j (
 t +

) )

Im(I m e j (
 t +

) )
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

v
1

V
1

ω t
 θ
1

V
1
V
1


V
1
 
1
e j (
 t
 
1
)
V
1

e j (

1
)
by dropping
 t
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

A phasor diagram showing the sum of
V
1
= 6 + j 8 V and V
2
= 3 – j 4 V,
V
1
+ V
2
= 9 + j 4 V = Vs
Vs = Ae j
θ
A = [9 2 + 4 2 ] 1/2
θ
= tan -1 (4/9)
Vs = 9.85

24.0
o V.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Step 1: Determine the phasor for each term.
Step 2: Add the phasors using complex arithmetic.
Step 3: Convert the sum to polar form.
Step 4: Write the result as a time function.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

v
1

 t


v
2
t

 t


V

20
 
45

V

10
 
30

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

V s

V
1

V
2



20
14 .
14
23 .

06



45
 j j

14
19 .
10
.
14
14



30

8 .
660


29 .
97
 
39 .
7
 j 5
V s

Ae j

A
 v s
23 .
06
2


29 .
97
(

19 .
14 )
2 cos

 t


29 .
96 ,

39 .
7


 tan 
1

19 .
14
 
39 .
7

23 .
06
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

To determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise.
Then when standing at a fixed point, if V
1 arrives first followed by V
2 say that V
1 leads V
2 by
θ
. after a rotation of
θ
, we
Alternatively, we could say that V
2 lags V
1 by
θ
. (Usually, we take
θ as the smaller angle between the two phasors.)
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

To determine phase relationships between sinusoids from their plots versus time, find the shortest time interval t p of the two waveforms. between positive peaks
Then, the phase angle is
θ
= ( t p
/T ) × 360 ° .
If the peak of v
1
( t ) occurs first, we say that v
1
( t ) leads v
2
( t ) or that v
2
( t ) lags v
1
( t ).
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

V
L
 j

L

I
L
Z
L
 j

L
 
L


V

L
Z I
L L
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

(a)
(c)
(b)
In the phasor domain,
(a) a resistor R is represented by an impedance of the same value;
(b) a capacitor C is represented by an impedance 1/j

C;
(c) an inductor L is represented by an impedance j

L.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

V

Ve j
 t
I

C d V dt

C d Ve j
 t dt
I
 j

C V

V
I

 j

1
C j

CVe j
 t

Zc
Zc is defined as the impedance of a capacitor
The impedance of a capacitor is 1/j

C. It is simply a mathematical expression. The physical meaning of the j term is that it will introduce a phase shift between the voltage across the capacitor and the current flowing through the capacitor.
As I
 j

C V , if v

V cos
 t , then i
 
CV cos (
 t

90

)
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

As I
I
M
 j

C V ,
 
CV
M if v

V
M cos
 t , then i
 
CV
M cos (
 t

90

)
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

I

Ie j
 t
V

L d dt
I

L d Ie j
 t dt
V
 j

L I

V
I

 j

LIe j

L

Z
L j
 t
Z
L is defined as the impedance of an inductor
The impedance of a inductor is j

L. It is simply a mathematical expression. The physical meaning of the j term is that it will introduce a phase shift between the voltage across the inductor and the current flowing through the inductor.
As V
 or i
 j

C I ,
I cos (
 t if i

I cos
 t , then v
 
LI cos (
 t

90

)

90

), and v
 
LI cos
 t .
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

As V or i


I
M j

C I , cos (
 t if i

I
M cos
 t , then v

90

), and v
 
LI
M
 
LI
M cos
 t , V
M cos (
 t


90

)

LI
M
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Complex Impedance in Phasor Notation
V
C

Z
C
I
C
Z
C
V
L
Z
L

1 j

C
  j
1

C


Z
L
I
L j

L
 
L

90


1

C
 
90

V

R
RI
R
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Vm

Im

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.
The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Circuit Analysis Using Phasors and Impedances
1. Replace the time descriptions of the voltage and current sources with the corresponding phasors.
(All of the sources must have the same frequency.)
2. Replace inductances by their complex impedances
Z
L
= jωL.
Replace capacitances by their complex impedances
Z
C
= 1/(jωC).
Resistances remain the same as their resistances.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

3 . Analyze the circuit using any of the techniques studied earlier performing the calculations with complex arithmetic.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Z total

100


100
 j ( 150

50 ) j 100

141 .
4

45

I

V
Z total

100

30

141 .
4

0 .
707
 
15


45


0 .
707

30
 
45

V
R
V
L

100

I

 j 150

I

70 .
7

150

15


90

, v
R
( t )

70 .
7

0 .
707
 
15
 cos( 500 t


106 .
05

15

90

)

15


106 .
05

75

, v
L
( t )

106 .
05 cos( 500 t

75

)
V
C
  j 50

I

50
 
90
 
0 .
707
 
15
 
35 .
35
 
90
 
15


35 .
35
 
105

, v
L
( t )

35 .
35 cos( 500 t

105

)
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Z
RC

1 / R
1

1 / Zc

1 / 100
1

1 /(
 j 100 )
As

1 j 100


1 j 100
 j j
 j 0 .
01
 j
2

Z
RC

0 .
01

1 j 0 .
01


50
 j 50
1

0

0 .
01414

45
 j 0 .
01

70 .
71
 
45

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Vc

Vs
Z
L
Z

RC
Z
RC

10
 
90

( voltage division j
70 .
71

100

50


45
 j 50

)
10
 
90

70 .
71
 
50
 j
45
50
 v c
( t

10
 
45

)

10 cos (
70 .
71
 
45

70 .
71

45

1000 t

135

)

10
 
135

 
10 cos 1000 t V
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

I


Z
L
Vs

Z
RC j
10

100


50
90

 j 50

10

50


90
 j 50 i ( t )
10
 
90



70 .
71

45

0 .
414 cos

0 .
414
( 1000 t

 
135

135

)
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

I
I
R i
R
(

V
C

R
10
 
135

100 t )

0 .
1 cos

0
( 1000 t
.
1



135

135

)
C

V
C
Zc

10
 
135

 j 100

10
 
135

100
 
90
 i
R
( t )

0 .
1 cos ( 1000 t

45

) A

0 .
1
 
45

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Solve by nodal analysis
V
1
10
V
2 j 10

V
1

 j
V
2
5

V
2


V j 5
1

2
 
90

1 .
5

0

  

1 .
5 j 2 eq ( 1 ) eq ( 2 )
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

V
1
10

V
1


V j 5
2

2
 
90
   j 2
From eq ( 1 ) eq ( 1 )
V
2 j 10

V
2


V j 5
1

1 .
5

0
 
1 .
5
0 .
1 V
1
 j 0 .
2 V
1
 j 0 .
2 V
2
  j 2 As
1 j

1
 j j j
 j

1
  j ,

1 j
 j
( 0 .
1
 j 0 .
2 ) V
1
 j 0 .
2 V
2
  j 2
From eq ( 2 )
 j 0 .
Solving V
1
( 0 .
1

2 V j
1
0 .

2 j 0 .
1 V
2 by
) V
1

3


1 eq ( 1 )
 j
.
5
2
2
 eq ( 2 )
V
1

3

0 .
1
 j j 2
0 .
2

3 .
6
 
0 .
2236

33 .
69


63 .
43


16 .
1

29 .
74
 v
1

16 .
1 cos( 100 t

29 .
74

) V

16 .
1
 
33 .
69
 
63 .
43
 eq ( 2 )
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

Vs= - j10, Z
L
=j

L=j ( 0.5×500 )= j250
Use mesh analysis,
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering


Vs

(


V
R j 10 )

V
Z

I


0
250

I

( j 250 )

0
I


250
 j 10 j 250

10
 
90

353 .
33

45

I

0 .
028
 
135


0 .
028
 
90
 
45
 i

0 .
028 cos( 500 t

135

) A
V
L

I

Z
L


7
 
45

( 0 .
028
 
135

)

250

90
 v
L
( t )

7 cos( 500 t

45

) V
V
R

I

R

( 0 .
028
 
135

)

250 v
R
( t )

7 cos( 500 t

135

)

7
 
135

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

5
-j50 j200
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

5
-j50 j200
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

j100 -j200
100
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering j100

100 j100 -j200 j100
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

P

V rms
PF

I cos
  rms cos
 
   v
  i
Q

V rms
I rms sin
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering


V rms
I rms
P
2 
Q
2 

V rms
I rms

2
P
 2
I rms
R
Q
 2
I rms
X
P

2
V
R rms
R
Q

2
V
X rms
X
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.
V
 t
V oc
We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.
Z t

V oc
I sc

V
I sc t
I
 n
I sc
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

If the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.
If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.
BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering

BASIC ELECTRONIC ENGINEERING Department of Electronic Engineering