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					    Steady-State Sinusoidal
           Analysis




BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
Steady-State Sinusoidal Analysis
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) = VM sin  t is
    plotted (a) versus  t and (b) versus t.



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


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

 sin(t  90 )  costo


   BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
                                           A graphical representation of
                                           the two sinusoids v1 and v2.
                                           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 100o + 30o = 130o, although
it could also be argued that v2 leads v1 by 230o.
It is customary, however, to express the phase difference by
an angle less than or equal to 180o 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.


       BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering

4-6
The complex forcing function Vm e j ( t + ) produces the
complex response Im e j (t + ).
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 Vm cos ( t + θ) produces the
steady-state response Im cos ( t + φ).




 The imaginary sinusoidal forcing function j Vm sin ( t + θ)
 produces the imaginary sinusoidal response j Im sin ( t + φ).

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



BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
                Phasor Definition
Time function : v1 t   V1 cosωt  θ1 
Phasor : V1  V11
                                     j (t 1 )
              V1  Re(e                                           )
                                     j (1 )
              V1  Re(e                              ) by dropping t

   BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
                               A phasor diagram showing the sum of
                               V1 = 6 + j8 V and V2 = 3 – j4 V,
                               V1 + V2 = 9 + j4 V = Vs
                               Vs = Ae j θ
                               A = [9 2 + 4 2]1/2
                               θ = tan -1 (4/9)
                               Vs = 9.8524.0o V.




BASIC ELECTRONIC ENGINEERING      Department of Electronic Engineering
      Adding Sinusoids Using
             Phasors
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
Using Phasors to Add Sinusoids

               v1t   20 cos(t  45 )                                  

                                                                          
               v2 (t )  10 cos(t  30 )
                                                                      
                      V1  20  45
                                                                      
                      V2  10  30

BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
Vs  V1  V2
    20  45 10  30
    14.14  j14.14  8.660 j5
    23.06  j19.14
    29.97  39.7
Vs  Ae j
                                             19 .14
A  23 .06  ( 19 .14 )  29 .96 ,   tan
               2                    2
                                                      39 .7                 1

                                             23 .06
vs t   29 .97 cost  39 .7  
     BASIC ELECTRONIC ENGINEERING       Department of Electronic Engineering
            Phase Relationships
To determine phase relationships from a phasor diagram,
consider the phasors to rotate counterclockwise.
Then when standing at a fixed point,
if V1 arrives first followed by V2 after a rotation of θ , we
say that V1 leads V2 by θ .

Alternatively, we could say that V2 lags V1 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 tp between positive peaks
of the two waveforms.

Then, the phase angle isθ = (tp/T ) × 360°.

If the peak of v1(t) occurs first, we say that v1(t)
leads v2(t) or that v2(t) lags v1(t).


    BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
  COMPLEX IMPEDANCES
                     VL  jL  I L

               Z L  jL  L90                                      




                        VL  Z L I L


BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
                (a)                                                   (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)            (c) an inductor L is represented by
                                an impedance jL.

BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
                                        V  Ve jt
                                              dV    d Ve jt
                                        IC      C            jCVe jt
                                              dt      dt
                                                    V        1
                                        I  j CV               Zc
                                                    I     j C

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  j C V, if v  VM cos t, then i   CVM cos( t  90 )
I M   CVM



    BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
                                        I  Ie jt
                                              dI     d Ie jt
                                         VL      L           jLIe jt
                                               dt      dt
                                                     V
                                         V  j LI   j L  Z L
                                                     I
ZL 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  j C I, if i  I cos t, then v   LI cos( t  90 )
or i  I cos( t  90 ), and v   LI cos t.

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

       BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
Complex Impedance in Phasor Notation

   VC  ZC IC
          1         1   1
   ZC         j         90

        jC        C C
   VL  Z LI L
   Z L  jL  L90
   VR  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
    Kirchhoff’s Laws in Phasor Form


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
   ZL= jωL.
   Replace capacitances by their complex
   impedances
   ZC = 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
Ztotal  100  j (150  50)
      100  j100  141.445
    V      10030
I                   0.70730  45
   Ztotal 141.445 


   0.707  15
VR  100  I  70.7  15 , vR (t )  70.7 cos(500t  15 )
VL  j150  I  15090  0.707  15  106.0590  15
    106.0575 , vL (t )  106.05 cos(500t  75 )
VC   j50  I  50  90  0.707  15  35.35  90  15
    35.35  105 , vL (t )  35.35 cos(500t  105 )


      BASIC ELECTRONIC ENGINEERING             Department of Electronic Engineering
BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
              1                 1
Z RC                
       1 / R  1 / Zc 1 / 100  1 /( j100)
       1           1      j j 0.01
As                                 j 0.01
     j100  j100 j  j            2


               1           10
Z RC                              70.71  45
         0.01  j 0.01 0.0141445
        50  j50

       BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
           Z RC
Vc  Vs            ( voltage division)
        Z L  Z RC
                70.71  45               70.71  45
    10  90                  10  90
               j100  50  j50               50  j50
                  70.71  45
     10  45                   10   135
                   70.7145
vc (t )  10 cos (1000t  135 )   10 cos 1000t V


      BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
                 Vs
        I
             Z L  Z RC
               10  90       10  90
                            
             j100  50  j50    50  j50
                10  90
                            0.414  135
                70.7145
        i (t )  0.414 cos (1000t  135 )


BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
        VC
IR 
         R
        10  135
                     0.1  135
            100
 iR (t )  0.1 cos (1000t  135 )
     VC 10  135 10  135
IC                           0.1  45
     Zc    j100    100  90
iR (t )  0.1 cos (1000t  45 ) A


  BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
Solve by nodal analysis

        V1 V1  V2
                     2  90   j 2 eq(1)
                               

        10      j5
         V2 V2  V1
                      1.50  1.5
                              
                                       eq(2)
         j10     j5

      BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
V1 V1  V2                                       V2 V2  V1
            2  90   j 2 eq(1)                        1.50  1.5   eq( 2)
10    j5                                        j10    j5
From eq (1)
                                         1 1 j  j        1
0.1V1  j 0.2V1  j 0.2V2   j 2   As           j,     j
                                         j j j 1        j
(0.1  j 0.2)V1  j 0.2V2   j 2
From eq ( 2)
 j 0.2V1  j 0.1V2  1.5
Solving V1 by eq(1)  2  eq( 2)
(0.1  j 0.2)V1  3  j 2
       3  j2      3.6  33.69
V1                               16.1  33.69  63.43
     0.1  j 0.2 0.2236  63.43
   16.129.74
v1  16.1cos( t  29.74 ) V
            100

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

Use mesh analysis,




   BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
 Vs  VR  VZ  0
 (  j10)  I  250  I  ( j 250)  0
      j10      10  90
I                        0.028  90  45
   250  j 250 353.3345
I  0.028  135
i  0.028 cos(500t  135 ) A
VL  I  Z L  (0.028  135 )  25090
    7  45
vL (t )  7 cos(500t  45 ) V
VR  I  R  (0.028  135 )  250  7  135
vR (t )  7 cos(500t  135 )



        BASIC ELECTRONIC ENGINEERING      Department of Electronic Engineering
BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
    5                                                        j200
                               -j50




BASIC ELECTRONIC ENGINEERING          Department of Electronic Engineering
5                                 j200
                    -j50




BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
                       j100                            -j200




        100                                                           j100




BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
             j100               -j200




 100                                                      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
     AC Power Calculations
                  P  Vrm sI rm s cos 
                       PF  cos 
                            v  i
                   Q  Vrm sI rm s sin  


BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
         apparent power  VrmsI rms

            P  Q  VrmsI rms 
                  2        2                                          2



                                                                               2
PI       2
          rm s    R                                   P
                                                                          V   Rrm s
                                                                              R
                                                                          2
QI        2
           rm s   X                                   Q
                                                                      V   Xrm s
                                                                          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
  THÉVENIN EQUIVALENT
       CIRCUITS




BASIC ELECTRONIC ENGINEERING   Department of Electronic Engineering
The Thévenin voltage is equal to the open-circuit
phasor voltage of the original circuit.


                         Vt  Voc

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.


                           Voc Vt
                      Z t      
                           I sc   I sc

                                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
    Maximum Power Transfer
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

				
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