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EE 221 Circuits II
Chapter 9
Sinusoids and Phasors
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Sinusoids and Phasors
9.1
9.2
9.3
9.4
9.5
9.6
9.7
Motivation
Sinusoids’ features
Phasors
Phasor relationships for circuit elements
Impedance and admittance
Kirchhoff’s laws in the frequency domain
Impedance combinations
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9.1 Motivation
How to determine v(t) and i(t) at steady-state?
First order circuit (Chap. 7), apply KVL to
determine the equation, write the general
expression of the solution, interested only in the
steady-state or forced response.
Is there a quicker way?
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9.2 Sinusoids y A sinusoid is a signal that has the form of the
sine or cosine function.
y A general expression for the sinusoid,
v(t ) = Vm sin(ωt + φ )
where
Vm = the amplitude of the sinusoid
ω = the angular frequency in radians/s
Ф = the phase angle.
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9.2 Sinusoids A periodic function is one that satisfies v(t) = v(t + nT),
for all t and for all integers n.
T=
2π
ω
f =
1
Hz
T
ω = 2πf
• Only two sinusoidal values with the same frequency
can be compared by their amplitude and phase
difference.
• If phase difference is zero, they are in phase; if phase
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difference is not zero, they are out of phase.
9.2 Sinusoids Example 1
Given a sinusoid, 5 sin( 4 π t − 60 o ), calculate its
amplitude, phase, angular frequency, period,
and frequency.
Solution:
amplitude = 5,
phase = –60o,
angular frequency = 4π rad/s,
period = 0.5 s,
Linear frequency = 2 Hz.
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9.2 Sinusoids Example 2
Find the phase angle between i1 = −4 sin( 377 t + 25 o )
and i2 = 5 cos(377t − 40 o ) , does i1 lead or lag i2?
Solution:
Since sin(ωt+90o) = cos ωt
i2 = 5sin(377t − 40o + 90o ) = 5sin(377t + 50o )
i1 = −4sin(377 t + 25o ) = 4sin(377 t +180o + 25o ) = 4sin(377 t + 205o )
therefore, i1 leads i2 by 155o.
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9.3 Phasors • A phasor is a complex
number that represents the
amplitude and phase of a
sinusoid.
• It can be represented in
one of the following three
forms:
a. Rectangular z = x + jy = r (cos φ + j sin φ )
b. Polar z = r ∠ φ
c. Exponential z = re
jφ
where
r=
x2 + y2
φ = tan −1
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y
x
9.3 Phasors Mathematic operation of complex number:
1. Addition
z1 + z 2 = ( x1 + x2 ) + j ( y1 + y 2 )
z1 − z2 = ( x1 − x2 ) + j ( y1 − y2 )
2. Subtraction
3. Multiplication
z1 z 2 = r1r2 ∠ φ1 + φ2
4. Division
5. Reciprocal
6. Square root
1 1
= ∠ −φ
z
r
z1 r1
= ∠φ1 − φ2
z 2 r2
z = r ∠φ 2
∗
− jφ
z
=
x
−
jy
=
r
∠
−
φ
=
re
7. Complex conjugate
8. Euler’s identity
e ± jφ = cos φ ± j sin φ
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9.3 Phasors Example 3
• Evaluate the following complex numbers:
a.
[(5 + j2)(−1 + j4) − 5∠ 60o ]
o
10
+
j5
+
3
∠
40
b.
+ 10 ∠30o
− 3 + j4
Solution:
a. –15.5 + j13.67
b. 8.293 + j2.2
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9.3 Phasors y Transform a sinusoid to and from the time domain to the
phasor domain:
v(t ) = Vm cos(ωt + φ )
(time domain)
V = Vm ∠φ
(phasor domain)
• Amplitude and phase difference are two principal
concerns in the study of voltage and current sinusoids.
• Phasor will be defined from the cosine function in all our
proceeding study. If a voltage or current expression is in
the form of a sine, it will be changed to a cosine by
subtracting 90 deg from the phase.
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9.3 Phasors
Example 4
Transform the following sinusoids to phasors:
i = 6cos(50t – 40o) A
v = –4sin(30t + 50o) V
Solution:
a. I = 6∠ − 40°
A
b. Since –sin(A) = cos(A+90o);
v(t) = 4cos (30t+50o+90o) = 4cos(30t+140o) V
Transform to phasor → V
= 4∠140° V
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9.3 Phasors
Example 5:
Transform the phasors to corresponding sinusoids:
a. V = − 10∠30° V
b. I = j(5 − j12) A
Solution:
a) v(t) = 10cos(ωt + 210o) V
b) Since I = 12 + j5 = 12 2 + 5 2 ∠ tan −1 ( 5 ) = 13∠ 22.62 °
12
o
i(t) = 13cos(ωt + 22.62 ) A
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9.3 Phasors
The differences between v(t) and V:
•
v(t) is instantaneous or time-domain representation
V is the frequency or phasor-domain representation.
•
v(t) is time dependent, V is not.
•
v(t) is always real with no complex term, V is
generally complex.
Note: Phasor analysis applies only when frequency is
constant; when it is applied to two or more sinusoid
signals only if they have the same frequency.
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9.3 Phasors (derivatives and integrals) v(t ) = Vm cos(ωt + φ ) = ℜ{Vm e j (ωt +φ ) }
= ℜ{Vm e jωt e jφ } = ℜ{e jωtVm ∠φ } ⇒ Vm ∠φ
dv (t ) / dt = d [Vm cos(ωt + φ )] / dt = d [ℜ{Vm e j (ωt +φ ) }] / dt
= d [ℜ{Vm e jωt e jφ }] / dt = jωℜ{e jωtVm ∠φ } ⇒ jωVm ∠φ
j ( ω t +φ )
v
(
t
)
dt
=
V
cos(
ω
t
+
φ
)
dt
=
ℜ
{
V
e
}dt
∫
∫ m
∫ m
1
= ∫ ℜ{Vm e e }dt = ℜ{e Vm ∠φ } / jω ⇒
Vm ∠φ
jω
jω t
jφ
jω t
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9.3 Phasors Example 6
Use phasor approach, determine the current i(t) in a
circuit described by the integro-differential equation.
di
4i + 8∫ idt − 3 = 50 cos(2t + 75°)
dt
Answer:
i(t) = 4.642cos(2t + 143.2o) A
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9.3 Phasors
y We can derive the differential equations for the following
circuit, then transform it in the phasor domain, solve for Vo,
then transform back in the time domain to find vo(t).
However, the derivation may sometimes be very tedious
d 2 vo 5 dv0
400
o
+
+
20
=
−
−
v
sin(
4
t
15
)
0
2
3 dt
dt
3
Is there any quicker and more systematic methods to solve
the voltage across the inductor?
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9.3 Phasors
The answer is YES!
Instead of first deriving the differential
equation and then transforming it into
phasor to solve for Vo, we can transform
all the RLC components into phasor the
phasor domain first, then apply the KCL
laws and other theorems to set up a
phasor equation involving Vo directly.
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9.4 Phasor Relationships for Circuit Elements Resistor:
Inductor:
Capacitor:
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9.4 Phasor Relationships for Circuit Elements Summary of voltage-current relationship
Element
Time domain
R
v = Ri
L
v=L
C
dv
i=C
dt
di
dt
Frequency domain
V = RI
V = jωLI
V =
20
I
jωC
9.4 Phasor Relationships for Circuit Elements Example 7
If voltage v(t) = 6cos(100t – 30o) is applied to a 50 μF
capacitor, calculate the current, i(t), through the
capacitor.
Answer: i(t) = 30 cos(100t + 60o) mA
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9.5 Impedance and Admittance • The impedance Z of a circuit is the ratio of the phasor voltage V to
the phasor current I, measured in Ω.
V
Z = = R ± jX
I
where R = Re (Z) is the resistance and X = Im (Z) is the
reactance. Positive (+) is for inductance and negative (-) is for
capacitance.
• The admittance Y is the reciprocal of impedance, measured in (S).
1 I
Y = = = G ± jB
Z V
where G = Re (Y) is the conductance and B = Im (Y) is the
suceptance. Positive (+) is for capacitance and negative (-) is for
inductance.
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9.5 Impedance and Admittance Impedances and admittances of passive elements
Element
R
L
C
Impedance
Z=R
Z = j ωL
Z =
1
jω C
Admittance
1
Y=
R
Y=
1
jωL
Y = jω C
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9.5 Impedance and Admittance ω = 0; Z = 0
Z = j ωL
ω → ∞; Z → ∞
ω = 0; Z → ∞
Z=
1
jω C
ω → ∞; Z = 0
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9.5 Impedance and Admittance After we know how to convert RLC
components from the time domain to thw
phasor domain, we can transform a time
domain circuit into a phasor/frequency
domain circuit.
Hence, we can apply the KCL laws and
other theorems to directly set up phasor
equations involving our target variable(s)
for solving.
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9.5 Impedance and Admittance Example 8
Refer to Figure below, determine v(t) and i(t).
vs = 5 cos(10t )
Answers: i(t) = 1.118cos(10t – 26.56o) A;
v(t) = 2.236cos(10t + 63.43o) V
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9.6 Kirchhoff’s Laws in the Frequency Domain • Both KVL and KCL are hold in the phasor
domain or more commonly called frequency
domain.
• Moreover, the variables to be handled are
phasors, which are complex numbers.
• All the mathematical operations involved
are now in complex form.
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9.7 Impedance Combinations • All principles used for DC circuit analysis also
apply to AC circuit.
• For example:
a. voltage division
b. current division
c. Series – parallel combinations
d. Y-Δ transformation
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9.7 Impedance Combinations Example 9
Determine the input impedance of the circuit in figure
below at ω =10 rad/s.
Answer: Zin = 32.38 – j73.76 Ω
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