Transcript CSCI 2980: Introduction to Circuits, CAD, and Instrumentation

EENG 2610: Circuit Analysis
Class 14: Sinusoidal Forcing Functions, Phasors,
Impedance and Admittance
Oluwayomi Adamo
Department of Electrical Engineering
College of Engineering, University of North Texas
AC Steady-State Analysis

Sinusoidal forcing function (f(t) is forcing function)
dx (t )
 ax (t )  f (t ),
dt

x(t )  x p (t )  xc (t )
The natural response xc(t) is a characteristics of the circuit
network and it is independent of the forcing function.
 The forced response xp(t) depends on the type of forcing function.
Why study sinusoidal forcing function?
 This is the dominant waveform in electric power industry.
 Any periodic signal can be represented by a sum of sinusoids
(you will learn it in Fourier analysis)
We will only concentrate on the steady-state forced response of
networks with sinusoidal forcing functions.
 We will ignore the initial conditions and the transient or natural
response


d 2 x(t )
dx(t )

a
 a2 x(t )  f (t ),
1
2
dt
dt
Sinusoids

Definition





x(t )  X M sin(t   )
XM is the amplitude
ω is radian or angular frequency (unit: radian/second)
θ is phase angle (unit: radian),  radian  180 
T=2π/ω is period (unit: second)
2
f=1/T is frequency (unit: Hertz),  
 2 f
T

In-phase and out-of-phase

Any point on the waveform XMsin(ωt+θ) occurs θ radians
earlier in time than the corresponding point on the
waveform XMsin(ωt).


We say XMsin(ωt+θ) leads XMsin(ωt) by θ radians, or
XMsin(ωt) lags XMsin(ωt+θ) by θ radians.

In the more general situation, if

Then,



 x1 (t )  X M sin(t  1 )

 x2 (t )  X M sin(t   2 )
x1(t) leads x2(t) by (θ1 - θ2) radians, or,
x2(t) lags x1(t) by (θ1 - θ2) radians.
If θ1 = θ2 ,the waveforms are identical and the functions are said
in phase; otherwise, it is said out of phase.
Important Trigonometric Identities


cos(t )  sin(t  2 )

sin(t )  cos(t   )

2
 radian  180 
Rectangular form
cos(t   )   cos(t )

sin(t   )   sin(t )
cos( )  cos( )

sin( )   sin( )
sin(   )  sin  cos   cos sin 
sin(   )  sin  cos   cos sin 


cos(   )  cos cos   sin  sin 
cos(   )  cos cos   sin  sin 
Polar form
x  jy  re j
 r  x2  y2

y

  tan1

x
 x  r cos

 y  r sin 
1
 e  j
j
e
 sin(2 )  2 sin  cos

2
2
cos(
2

)

cos


sin



2

2
cos
 1

2


1

2
sin


Sinusoidal and Complex Forcing
Functions

Forcing function and circuit response



If we apply a constant forcing function (i.e., source function)
to a network, the steady state circuit response is also a
constant.
If we apply a sinusoidal forcing function to a linear network,
the steady state circuit response will also be sinusoidal.
Sinusoidal source function


If the input source is v(t)=Asin(ωt+θ), then the output will be
in the same sinusoidal form. For example, i(t)=Bsin(ωt+φ).
That means: if the input source is sinusoidal function, we
know the form of the output response, and therefore the
solution involves simply determining the values of the two
parameters B and φ.
Learning Example
di
(t )  Ri (t )  v (t )
dt
In steady state i (t )  A cos( t   ), or
i (t )  A1 cos  t  A2 sin  t
KVL : L
di
(t )   A1 sin  t  A2 cos  t
dt
( LA1  RA2 ) sin  t  ( LA2  RA1 ) cos t 
 VM cos t
 LA1  RA2  0 algebraic problem
LA2  RA1  VM
A1 
RVM
LVM
,
A

2
R 2  (L) 2
R 2  (L) 2
Determining the steady state solution can
be accomplished with only algebraic tools!
FURTHER ANALYSIS OF THE SOLUTION
The solution is i (t )  A1 cos t  A2 sin  t
The applied voltage is v (t )  VM cos t
For comparison purposes one can write i (t )  A cos( t   )
A1  A cos , A2   A sin 
A1 
RVM
LVM
,
A

2
R 2  (L) 2
R 2  (L) 2
A
i (t ) 
A  A12  A22 , tan   
A2
A1
VM
1 L
,


tan
R
R 2  (L) 2
VM
1 L
cos(

t

tan
)
2
2
R
R  (L)
For L  0 the current ALWAYS lags the voltage
If R  0 (pure inductor) the current lags the voltage by 90
Phasors

If the forcing function for a linear network is of the form v(t)=VMejωt,


Then every steady-state voltage or current in the network will have the
same form and the same frequency ω; for example, a current will be of
the form i(t)=IMej(ωt+φ).
In our circuit analysis, we can drop the factor ejωt, since it is common to
every term in the describing equations.
Phasors are defined as:
Sinusoidal signal:
V  VM   VM e j  VM (cos  j sin  )
v(t )  VM cos(t   )
 Re{VM e j ( t  ) }
 Re{VM e j e j t }
 Re{VM  e
j t
}
I  I M   I M e j  I M (cos  j sin  )
The magnitude of phasors are positive !
V  VM 
 VM (   )
Phasor Analysis
(or Frequency Domain Analysis)




The circuit analysis after dropping ejωt term is called phasor analysis
or frequency domain analysis.
By phasor analysis, we have transformed a set of differential
equations with sinusoidal forcing functions in the time domain into a
set of algebraic equations containing complex numbers in the
frequency domain.
The phasors are then simply transformed back to the time domain to
yield the solution of the original set of differential equations.
Phasor representation:
Time Domain
Frequency Domain
A cos(t   )  A  
A sin(t   )  A(  90)
Phasor Relationships for Circuit Elements

We will establish the phasor relationships between voltage and
current for the three passive elements R, L, C.
v(t )  Ri(t )
VM e j ( t  v )  RIM e j ( t i )
VM e
j v
 RIM e
j i
V  RI
V  VM e j v  VM  v

I  I M e ji  I M  i
Phasor diagram
di(t )
dt
d
 L {I M e j ( t i ) }
dt
 jLI M e j i
v(t )  L
VM e j ( t  v )
VM e j v
V  jLI
V  VM e j v  VM  v

I  I M e ji  I M  i
j  1
 1e j 90  190
Voltage leads current by 90°
dv(t )
dt
d
 C {VM e j ( t  v ) }
dt
 jCVM e j v
i (t )  C
I M e j ( t i )
I M e j i
I  jCV
j

V  VM e v  VM  v

j i

I  I M e  I M  i
j  1
 1e j 90  190
Current leads voltage by 90°

Definition of Impedance (unit: ohms):

Impedance is defined as the ratio of the phasor voltage V to the
phasor current I at the two terminals of the element related to one
another by the passive sign convention:
Z  Z z

V VM  v VM


( v   i )
I I M  i I M
Z( )  R( )  jX ( )
R : resistance,
2
2

Z  R  X

X : reactance
1

 z  tan ( X / R)

 R  Z cos  z

 X  Z sin  z
It’s important to note that:


Resistance R and reactance X are real function of the frequency
of the forcing function ω, thus Z(ω) is frequency dependent.
Impedance Z is a complex number; however, it is not a phasor,
since phasors denote sinusoidal functions.
Passive element impedance:
Element PhasorEquation Impedance
V  RI
V  jLI
1
V
I
jC
R
L
C
ZR
Z  jL
1
Z
jC
Equivalent impedance if impedances
are connected in series:
Zs  Z1  Z2  ... Zn
Equivalent impedance if impedances
are connected in parallel:
1
1
1
1


 ... 
Z p Z1 Z 2
Zn
Two terminal input admittance:
Y  G  jB 
1
1

Z R  jX
KVL and KCL are both
valid in frequency domain
1 I

(unit : siemens)
Z V
Y( )  G ( )  jB( )
G : conductance, B : susceptance
Y
G
R
X
,
B

R2  X 2
R2  X 2
In general, R and G are not reciprocals of one another.
The purely resistive case is an exception.
Example 8.9: Determine the equivalent impedance of the network.
Then compute i(t) for f = 60 Hz and f = 400 Hz.
v(t )  50cos( t  30) V
SPECIAL APPLICATION:
IMPEDANCES CAN BE COMBINED USING THE SAME RULES
DEVELOPED
FOR RESISTORS
I
 V1 
I
 V2 
I
Zs  Z1  Z2
Z2
Z1

Z2 V
Z1
Z s   k Zk
LEARNING EXAMPLE
f  60 Hz , v (t )  50 cos( t  30)
I

V


1
1
k
Zp
Zk
Zp 
Z1Z 2
Z1  Z 2
Compute equivalent impedance and current
  120 , V  5030, Z R  25
ZR  R
1
j120  50  106
Z L  j 7.54, ZC   j 53.05
Z L  j120  20  103  , Z C 
Z s  Z R  Z L  ZC  25  j 45.51
Z L  jL
I
ZC 
1
j C
V
5030
5030

( A) 
( A)
Z s 25  j 45.51
51.93  61.22
I  0.9691.22( A)  i (t )  0.96 cos(120 t  91.22)( A)
LEARNING EXAMPLE SERIES-PARALLEL REDUCTIONS
Z3  4  j 2
1
2  j4

2  j 4 ( 2) 2  ( 4 ) 2
1
4  j2
Y34 

4  j2
20
Y2 
Y4   j 0.25  j 0.5  j 0.25
Z 4  1 / Y4   j 4
1  (  j 2)
1  j2
1
Z1 
1  j 0 .5
Z1 
1  j 0.5
Z1 
1  (0.5) 2
Z1  0.8  j 0.4()
Z4 
j 4  (  j 2) 8

j4  j2
j2
Y2  0.1  j 0.2( S )
Y34  0.2  j 0.1
Z2  2  j6  j 2  2  j 4
Z34  4  j 2
Z 234 
Y234  0.3  j 0.1( S )
Z 234 
1
Y234

1
0.3  j 0.1

0.3  j 0.1
0.1
Z 2 Z 34
 3  j1
Z 2  Z 34
Zeq  Z1  Z234  3.8  j 0.6  3.8478.973
AC Steady-State Analysis


For relatively simple circuits (e.g., those with single source), use:
 Ohm’s law for AC analysis, i.e., V=IZ
 The rules for combining impedance Z (or admittance Y)
 KCL and KVL
 Current division and voltage division
For more complicated circuits with multiple sources, use:
 Nodal analysis
 Loop or mesh analysis
 Superposition
 Source exchange
 Thevenin’s and Norton’s theorem
 Software tools: MATLAB, PSPICE, …
LEARNING EXAMPLE
COMPUTE ALL THE VOLTAGES AND CURRENTS
Compute I1
Use current divider for I2 , I3
Ohm's law for V1 , V2
V1  690 I 2
Zeq  4  ( j 6 || 8  j 4)
Z eq
V1  16.2678.42(V )
24  j 48 32  j8  24  j 48
 4

8  j2
8  j2
V2  7.2815(V )
56  j 56
79.19645

 9.60430.964()
8  j2
8.24614.036
V
2460
I1  S 
 2.49829.036( A)
Z eq 9.60430.964
j6
690
I3 
I1 
2.49829.036( A)
8  j2
8.24614.036
Z eq 
I2 
8  j4
8.944  26.565
I1 
2.49829.036( A)
8  j2
8.24614.036
I1  2.529.06
I 2  2.71  11.58
V2  4  90 I3
I 3  1.82105
Steady-State Power Analysis

Here we study powers in AC circuits:






Instantaneous power
Average power
Maximum power transfer,
Power factor,
Complex power.
Device power ratings:

Typically, electrical and electronic devices have peak
power or maximum instantaneous power ratings that
cannot be exceeded without damaging the devices.
Instantaneous Power
Steady-state voltage and current:
v(t )  VM cos(t  v )
i(t )  I M cos(t  i )
Instantaneous power:
With passive sign convention
p(t )  v(t )i (t )
 VM cos(t   v ) I M cos(t   i )

VM I M
[cos( v   i )  cos(2t   v   i )]
2
independent of time
a function of time
LEARNING EXAMPLE
Instantane ous
Power Supplied
to Impedance
p( t )  v ( t ) i ( t )
p(t )  4 cos 30  4 cos( 2 t  90)
i (t )  I M cos( t   i )
p(t )  VM I M cos( t   v ) cos( t   i )
p( t ) 
1
cos(1  2 )  cos(1  2 )
2
VM I M
cos( v   i )  cos(2 t   v   i )
2
constant
V 460

 230( A)
Z 230
i (t )  2 cos( t  30)( A)
VM  4, v  60
I
I M  2, i  30
In steady State
v (t )  VM cos( t   v )
cos1 cos2 
Assume: v (t )  4 cos( t  60),
Z  230
Find : i (t ), p(t )
Twice the
frequency
Average Power
Since p(t) is a periodic function of time, the average power:
1 t 0 T
P 
p(t )dt
t
0
T
1 t 0 T
  VM cos(t   v ) I M cos(t   i )dt
T t0
1 t0 T VM I M
 
[cos( v   i )  cos(2t   v   i )]dt
t
0
T
2
For passive sign convention.
1
 VM I M cos( v   i )
2
In the equation, t0 is arbitrary, T=2π/ω is the period of the voltage or current.
2
1
1
1
1
V
2
M
For purely resistive circuit (i.e., Z = R+j0 ): P  V I cos(   )  V I  I R 
M M
v
i
M M
M
2
2
2
2 R
For purely reactive circuit (i.e., Z = 0 + jX ):
1
P  VM I M cos( 90)  0
2
That’s why reactive elements are called lossless elements
LEARNING EXAMPLE
Determine the average power absorbed by each resistor,
the total average power absorbed and the average power
supplied by the source
Inductors and capacitors do not absorb
power in the average
Ptotal  18  28.7W
Psupplied  Pabsorbed  Psupplied  46.7W
If voltage and current are in phase
2
1 2
1
1
V
M
 v   i  P  VM I M  RI1M 
2
2
2 R
1245
I1 
 345( A)
4
1
P4  12  3  18W
2
1245
1245
I2 

 5.3671.57( A)
2  j1
5  26.37
1
P2   2  5.362 (W )  28.7W
2
Verification
I  I1  I 2  345  5.3671.57
I  8.1562.10( A)
P
VM I M
cos( v   i )
2
1
Psupplied  12  8.15  cos(45  62.10)
2
Maximum Average Power Transfer
Average power at the load:
1
1
P  VL I L cos(  vL   iL )  I L2 RL
2
2
Voc

I

L

ZTh  Z L


V  Voc Z L
 L ZTh  Z L
Voc2 RL
1
PL 
2 ( RTh  RL ) 2  ( X Th  X L ) 2
If the load is purely resistive (i.e., XL = 0):
dPL
0
dRL
Z Th  RTh  jX Th

Z L  RL  jX L
For maximum average power transfer:
I L  Voc /(2 RTh )
*

ZL  ZTh  RTh  jXTh
Voc2
1 2

 PL ,max  2 I L RTh  8 R
Th

Z L  RL  jX L  RL
For maximum average
power transfer:
RL  RTh2  X Th2
PL ,max 
1 2
I L RL
2
Effective or RMS Values



The effective value of a periodic current (or voltage) is defined as a constant
or DC value, which would deliver the same average power to a resistor R.
The 120 V AC electrical outlets in our
120 2  170  Vrms  120 V
home is the rms value of the voltage: v(t )  170cos(377t ) 2  60  377  f  60 Hz
It is common practice to specify the voltage rating of AC electrical devices
(such as light bulb) in terms of the rms voltage.
The average power delivered to a
resistor by DC effective current: P 
The average power delivered to a
resistor by a periodic current:
P
On using the rms values for
the sinusoidal voltage and
current, the average power:
Effective (or rms) value of a
periodic current:
2
I eff
R
I eff  I rms
1
T
Vrms 

t 0 T
t0
VM
,
2
1 t0 T 2

i (t )dt

t
0
T
i 2 (t ) Rdt
I rms 
IM
2
P  Vrms I rms cos(v  i )
The power absorbed by a
resistor:
2
2
P  I rms
R
Vrms
R
LEARNING EXAMPLE
Compute the rms value of the voltage waveform
T 3
X rms
1

T
 4t 0  t  1

v (t )  
0 1 t  2
 4(t  2) 2  t  3

T
v
1
2
3
(t )dt   (4t ) dt   (4(t  2)) 2 dt
0
2
0
2
1
3
16 3  32
v
(
t
)
dt

2


 3 t   3
0
0
2
Vrms 
1 32

 1.89(V )
3 3
t 0 T
x
t0
2
(t )dt