SJTU 1 Chapter 9 Sinusoids and Phasors
SJTU1 Chapter 9 Sinusoids and Phasors. SJTU2 Sinusoids A sinusoid is a signal that has the form of the sine or cosine function.
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SJTU 1
Chapter 9
Sinusoids and Phasors
SJTU 2
Sinusoids
A sinusoid is a signal that has the form of the sine or cosine function.
anglephase
umentt
frequencyangular
amplitudeVmwhere
tVv m
arg
)cos(
SJTU 3
)cos( tVv m
t
fT
22
radians/second
(rad/s)
f is in hertz(Hz)
SJTU 4
)cos()(
)cos()(
222
111
tVtv
tVtv
m
m
Phase difference:
byvlagsv
byvleadsv
phaseinarevandv
phaseofoutarevandvif
tt
210
210
210
210
)()(
21
21
SJTU 5
Complex Number
form lexponentia
form sinusoidal
formpolar
formr rectangula
jrez
jrsinrcosz
rz
jyxz
SJTU 6
Phasor
a phasor is a complex number representing the amplitude and phase angle of a sinusoidal voltage or current.
Eq.(8-1)
and
Eq. (8-2)
Eq.(8-3)
SJTU 7
When Eq.(8-2) is applied to the general sinusoid we obtain
Eq.(8-4)
The phasor V is written as
Eq.(8-5)
SJTU 8
Fig. 8-1 shows a graphical representation commonly called a phasor diagram.
Fig. 8-1: Phasor diagram
Two features of the phasor concept need emphasis:
1. Phasors are written in boldface type like V or I1 to distinguish them from signal waveforms such as v(t) and i1(t).
2. A phasor is determined by amplitude and phase angle and does not contain any information about the frequency of the sinusoid.
SJTU 9
In summary, given a sinusoidal signal , the corresponding phasor representation is . Conversely, given the phasor , the corresponding sinusoid is found by multiplying the phasor by and reversing the steps in Eq.(8-4) as follows:
Eq.(8-6)
)cos()( tVtv m VmVTime domain representation
Phase-domain representation
SJTU 10
Properties of Phasors
• additive property
Eq.(8-7)
Eq.(8-8)
Eq.(8-9)
SJTU 11
• derivative property
Eq.(8-10)
Vjdt
dv Time domain representation
Phase-domain representation
SJTU 12
• Integral property
Time domain representation
Phase-domain representation
j
Vvdt
The differences between v(t) and V:
1. V(t) is the instantaneous or time-domain representation, while V is the frequency or phasor-domain representation.
2. V(t) is a real signal which is time dependent, while V is just a supposed value to simplify the analysis
SJTU 13
The complex exponential is sometimes called a rotating phasor, and the phasor V is viewed as a snapshot of the situation at t=0.
Fig. 8-2: Complex exponential
SJTU 14
+ j
+ r e a l
- r e a l
- j
t
V m
= 0
= 9 0 o r / 2
= - 9 0 o r - / 2
= 1 8 0 o r
15 10 5 0 5 10 1515
13.5
12
10.5
9
7.5
6
4.5
3
1.5
010V rms ac s igna l a t 0.5 Hz
voltage in volts
angular freq
uency times
time in radian
s
0
12 .566
t n
14.14214.142 v real t( ) n
0 5 10 1515
12
9
6
3
0
3
6
9
12
1510V rms ac s igna l a t 0.5 Hz
angular frequency times time in radians
Voltage in v
olts
14 .142
14.142
v imag t n
12.5660 t( )n
)sin(
is axis )(imaginary j on thephasor
rotating theof projection The
tVv mima g
)cos(
is axis real on thephasor
rotating theof projection The
tVv mrea l
case particular In this
5.02cos102cos ttfVv m
SJTU 15
EXAMPLE 8-1(a) Construct the phasors for the following signals:
(b) Use the additive property of phasors and the phasors
found in (a) to find v(t)=v1(t)+v2(t).
SOLUTION
(a) The phasor representations of v(t)=v1(t)+ v2(t) are
SJTU 16
(b) The two sinusoids have the same frequent so the additive property of phasors can be used to obtain their sum:
The waveform corresponding to this phasor sum is
V1
V2
1
j
V
SJTU 17
EXAMPLE 8-2 (a) Construct the phasors representing the following signals:
(b) Use the additive property of phasors and the phasors found i
n (a) to find the sum of these waveforms.SOLUTION:
(a) The phasor representation of the three sinusoidal currents are
SJTU 18
(b) The currents have the same frequency, so the additive property of phasors applies. The phasor representing the sum of these current is
Fig. 8-4
SJTU 19
EXAMPLE 8-3Use the derivative property of phasors to find the time derivative of v(t)=15 cos(200t-30° ).
The phasor for the sinusoid is V=15∠-30 ° . According to the derivative property, the phasor representing the dv/dt is found by multiplying V by j .
SOLUTION:
The sinusoid corresponding to the phasor jV is
SJTU 20
Device Constraints in Phasor Form
Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.
Resistor:
Re
jIm
I
V
0
IV
mm RIV
SJTU 21
Device Constraints in Phasor Form
Inductor:
Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.
90IV
mm LIV
SJTU 22
Device Constraints in Phasor Form
Capacitor:
Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain.
90VI
mm CVI
SJTU 23
Connection Constraints in Phasor Form
KVL in time domain
Kirchhoff's laws in phasor form (in frequency domain)
KVL: The algebraic sum of phasor voltages around a loop is zero.
KCL: The algebraic sum of phasor currents at a node is zero.
SJTU 24
The Impedance Concept
The IV constraints are all of the form
V=ZI or Z= V/I Eq.(8-16)
where Z is called the impedance of the element
The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms()
reactance. theis ZImX and resistance theis ZReR where jXRZ
The impedance is inductive when X is positive
is capacitive when X is negative
SJTU 25
The Impedance Concept
sin,cos
tan, where 122
ZXZRandR
XXRZ
ZZ
SJTU 26
EXAMPLE 8-5
Fig. 8-5
The circuit in Fig. 8-5 is operating in the sinusoidal steady state with and . Find the impedance of the elements in the rectangular box.
SOLUTION:
SJTU 27
37.90.278/RVI L23
SJTU 28
The Admittance Concept
The admittance Y is the reciprocal of impedance, measured in siemens (S)
V
I
ZY
1
Y=G+jB
Where G=Re Y is called conductance and B=Im Y is called the susceptance
2222,
1
XR
XB
XR
RG
jXRjBG
How get Y=G+jB from Z=R+jX ?
SJTU 29
CjYcapacitor
LjYinductor
GR
Yresistor
C
L
R
:
1:
1:
SJTU 30
Basic Circuit Analysis with Phasors
Step 1: The circuit is transformed into the phasor domain by representing the input and response sinusoids as phasor and the passive circuit elements by their impedances.
Step 2: Standard algebraic circuit techniques are applied to solve the phasor domain circuit for the desired unknown phasor responses.
Step 3: The phasor responses are inverse transformed back into time-domain sinusoids to obtain the response waveforms.
SJTU 31
Series Equivalence And Voltage Division
where R is the real part and X is the imaginary part
SJTU 32
EXAMPLE 8-6
Fig. 8-8
The circuit in Fig. 8 - 8 is operating in the sinusoidal steady state with
(a) Transform the circuit into the phasor domain. (b) Solve for the phasor current I. (c) Solve for the phasor voltage across each element. (d) Construct the waveforms corresponding to the phasors found in (b) and (c)
SJTU 33
SOLUTION:
SJTU 34
PARALLEL EQUIVALENCE AND
CURRENT DIVISION
Rest of the circuit
Y1Y1 Y2 YN
I
VI1 I2 I3
phasor version of the current division principle
SJTU 35
EXAMPLE 8-9
Fig. 8-13
The circuit in Fig. 8-13 is operating in the sinusoidal steady state with iS(t)=50cos2000t mA.(a) Transform the circuit into the phasor domain. (b) Solve for the phasor voltage V.(c) Solve for the phasor current through each element. (d) Construct the waveforms corresponding to the phasors found in (b) and (c).
SJTU 36
SOLUTION:
(a) The phasor representing the input source current is Is=0.050°∠ A. The impedances of the three passive elements are
Fig. 8-14
SJTU 37
And the voltage across the parallel circuit is
The sinusoidal steady-state waveforms corresponding to the phasors in (b) and (c) are
The current through each parallel branch is
SJTU 38
EXAMPLE 8-10
Fig. 8-15
Find the steady-state currents i(t), and iC(t) in the circuit of Fig. 8-15 (for Vs=100cos2000t V, L=250mH, C=0.5 F, and R=3k ).
SOLUTION:
Vs=100 0°∠
SJTU 39
SJTU 40
SJTU 41
Y←→ TRANSFORMATIONS△
The equations for the to Y △transformation are
SJTU 42
The equations for a Y-to- transformation are △
when Z1=Z2=Z3=ZY or ZA=ZB=ZC=ZN.
ZY=ZN /3 and ZN =3ZY balanced conditions
SJTU 43
EXAMPLE 8-12Use a to Y transformation to solve for the phasor current I△ X in Fig. 8-18.
Fig. 8-18
SOLUTION:
ABC to Y△
SJTU 44
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