10: Sine waves
⊲ and phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
E1.1 Analysis of Circuits (2016-7787)
10: Sine waves and phasors
Phasors: 10 – 1 / 11
Sine Waves
Usually differentiation changes the
shape of a waveform.
1
0
-1
0
For bounded waveforms there is
only one exception:
v(t) = sin t ⇒ dv
dt = cos t
same shape but with a time shift.
1
2
t
3
4
1
2
t
3
4
5
0
-5
0
1
0
-1
0
5
10
15
10
15
t
sin t completes one full period every
time t increases by 2π.
1
dv/dt
⊲
di
For inductors and capacitors i = C dv
and
v
=
L
dt
dt so we need to
differentiate i(t) and v(t) when analysing circuits containing them.
v(t)
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
0
-1
0
5
t
sin 2πf t makes f complete repetitions every time t increases by 1; this
gives a frequency of f cycles per second, or f Hz.
We often use the angular frequency , ω = 2πf instead.
ω is measured in radians per second. E.g. 50 Hz ≃ 314 rad.s−1 .
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – 2 / 11
Rotating Rod
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
⊲
A useful way to think of a cosine wave is as the
projection of a rotating rod onto the horizontal axis.
For a unit-length rod, the projection has length cos θ.
If the rod is rotating at a speed of f revolutions per
second, then θ increases uniformly with time:
θ = 2πf t.
The only difference between cos and sin is the starting position of the rod:
1
1
0
0
-1
0
5
10
t
v = cos 2πf t
15
-1
0
5
10
v = sin 2πf t = cos 2πf t −
sin 2πf t lags cos 2πf t by 90◦ (or π2 radians) because its peaks occurs
a cycle later (equivalently cos leads sin) .
E1.1 Analysis of Circuits (2016-7787)
15
t
π
2
1
4
of
Phasors: 10 – 3 / 11
Phasors
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
⊲
If the rod has length A and starts at an angle φ then the projection onto
the horizontal axis is
A cos (2πf t + φ)
= A cos φ cos 2πf t − A sin φ sin 2πf t
= X cos 2πf t − Y sin 2πf t
At time t = 0, the tip of the rod has coordinates
(X, Y ) = (A cos φ, A sin φ).
If we think of the plane as an Argand Diagram (or complex plane), then the
complex number X + jY corresponding to the tip of the rod at t = 0 is
called a phasor .
√
The magnitude of the phasor, A = X 2 + Y 2 , gives the amplitude (peak
value) of the sine wave.
Y
, gives the phase shift relative
The argument of the phasor, φ = arctan X
to cos 2πf t.
If φ > 0, it is leading and if φ < 0, it is lagging relative to cos 2πf t.
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – 4 / 11
Phasor Examples
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
⊲
V = 1, f = 50 Hz
v(t) = cos 2πf t
1
0
-1
0
0.02
0.04
0.06
0.04
0.06
t
V = −j
v(t) = sin 2πf t
1
0
-1
0
0.02
t
V = −1 − 0.5j = 1.12∠ − 153◦
v(t) = − cos 2πf t + 0.5 sin 2πf t
= 1.12 cos (2πf t − 2.68)
V = X + jY
v(t) = X cos 2πf t − Y sin 2πf t
Beware minus sign.
V = A∠φ = Aejφ
v(t) = A cos (2πf t + φ)
A phasor represents an entire waveform (encompassing all time) as a single
complex number. We assume the frequency, f , is known.
A phasor is not time-varying, so we use a capital letter: V .
A waveform is time-varying, so we use a small letter: v(t).
Casio: Pol(X, Y ) → A, φ, Rec(A, φ) → X, Y . Saved → X & Y mems.
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – 5 / 11
[Algebraic Phasor↔Waveform Mapping]
A phasor is a complex number, V , that uniquely defines a waveform, v(t), via the mapping V =
Aejφ ←→ v(t) = A cos (2πf t + φ). It is sometimes conveninet to give an algebraic formula for this.
For the direction V −→ v(t) the mapping is easy:
v(t) = ℜ V ej2πf t =
1
2
(V + V ∗ ) cos 2πf t + 21 j (V − V ∗ ) sin 2πf t.
The reverse mapping, V ←− v(t) is a bit more complicated and we use a technique that you will also
use in the Maths of Fourier transforms. The mapping is given by
Z 1
f
v(t)e−j2πf t dt.
V = 2f
0
To confrm that this is true, we can substitute v(t) = A cos (2πf t + φ) and do the integration:
2f
Z
1
f
−j2πf t
v(t)e
dt
=
Af
Z
Af
Z
0
1
f
0
=
1
f
0
=
jφ
Ae
E1.1 Analysis of Circuits (2016-7787)
j(2πf t+jφ
−j2πf t−jφ
e
+e
e−j2πf t dt
Z
ejφ + e−j4πf t−jφ dt = Aejφ + Af e−jφ
Af e−jφ −j4πf t
e
+
−j4πf
h
i1
f
0
jφ
= Ae
1
f
e−j4πf t dt
0
Af e−jφ −j4π
e
− 1 = Aejφ
+
−j4πf
Phasors: 10 – note 1 of slide 5
Phasor arithmetic
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
⊲
Phasors
Waveforms
V = X + jY
v(t) = X cos ωt − Y sin ωt
where ω = 2πf .
aV
a × v(t) = aX cos ωt − aY sin ωt
V1 + V2
v1 (t) + v2 (t)
Adding or scaling is the same for waveforms and phasors.
V̇ = (−ωY ) + j (ωX)
= jω (X + jY )
= jωV
dv
dt
= −ωX sin ωt − ωY cos ωt
= (−ωY ) cos ωt − (ωX) sin ωt
Differentiating waveforms corresponds to multiplying
phasors by jω.
Rotate anti-clockwise 90◦ and scale by ω = 2πf .
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – 6 / 11
Complex Impedances
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex
Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
⊲
Resistor:
v(t) = Ri(t) ⇒ V = RI
⇒
V
I
=R
⇒
V
I
= jωL
⇒
V
I
=
Inductor:
di
⇒ V = jωLI
v(t) = L dt
Capacitor:
i(t) = C dv
dt ⇒ I = jωCV
1
jωC
For all three components, phasors obey Ohm’s law if we use the complex
1
as the “resistance” of an inductor or capacitor.
impedances jωL and jωC
If all sources in a circuit are sine waves having the same frequency, we can
do circuit analysis exactly as before by using complex impedances.
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – 7 / 11
Phasor Analysis
(1) Find capacitor complex impedance
1
1
= 6.28j×10
Z = jωC
−4 = −1592j
(2) Solve circuit with phasors
Z
VC = V × R+Z
−1592j
1000−1592j
= −10j ×
= −4.5 − 7.2j = 8.47∠ − 122◦
vC = 8.47 cos (ωt − 122◦ )
10
v
R
⊲
Given v = 10 sin ωt where ω = 2π × 1000, find
vC (t).
vR
vC
0
C
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
-10
0
0.5
1
t (ms)
1.5
2
(3) Draw a phasor diagram showing KVL:
V = −10j
VC = −4.5 − 7.2j
VR = V − VC = 4.5 − 2.8j = 5.3∠ − 32◦
Phasors add like vectors
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – 8 / 11
[Differential Equation Analysis]
To solve the problem form the previous slide without using phasors, we define i to be the current flowing
C
clockwise and use the capacitor equation i = C dv
.
dt
From KVL, we have v = vR + vC = iR + vC .
Differentiating and applying the capacitor equation gives
dv
dt
di
= 10ω cos ωt = R dt
+
1
i.
C
We need to find the particular integral for the above equation. To do so, we guess that the answer will
be of the form i = A cos ωt + B sin ωt and substitute it into the equation (multiplied by C).
10Cω cos ωt
=
RC (−Aω sin ωt + Bω cos ωt) + (A cos ωt + B sin ωt)
=
(A + RCBω) cos ωt + (B − RCAω) sin ωt
which gives two siultaneous equations: A + RCωB = 10Cω and −RCωA + B = 0. Substituting
values for R, C and ω gives A + 0.628B = 0.00628 and −0.628A + B = 0. Solving these simultaneous
equations gives A = 4.5 mA and B = 2.8 mA.
The resistor voltage is therefore vR = iR = 4.5 cos ωt + 2.8 sin ωt and therefore, from KVL, the
capacitor votage is vC = v − vR = −4.5 cos ωt + 7.2 sin ωt.
Thus we get the same answer as using phasors but with more work even for a simple circuit like this.
For more complicated circuits the difference is much much bigger.
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – note 1 of slide 8
CIVIL
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
⊲
Capacitors: i = C dv
dt
⇒ I leads V
di
⇒ V leads I
Inductors: v = L dt
Mnemonic: CIVIL = “In a capacitor I lead V but V leads I in an inductor”.
COMPLEX ARITHMETIC TRICKS:
(1) j × j = −j × −j = −1
(2) 1j = −j
(3) a + jb = r∠θ = √
rejθ
where r = a2 + b2 and θ = arctan ab (±180◦ if a < 0)
(4) r∠θ = rejθ = (r cos θ) + j (r sin θ)
a∠θ
= ab ∠ (θ − φ).
(5) a∠θ × b∠φ = ab∠ (θ + φ) and b∠φ
Multiplication and division are much easier in polar form.
(6) All scientific calculators will convert rectangular to/from polar form.
Casio fx-991 (available in all exams except Maths) will do complex
arithmetic (+, −, ×, ÷, x2 , x1 , |x|, x∗ ) in CMPLX mode.
Learn how to use this: it will save lots of time and errors.
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – 9 / 11
Impedance and Admittance
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
⊲
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× Reactance
Z = R + jX (Ω)
|Z|2 = R2 + X 2
∠Z = arctan X
R
1
= Conductance + j× Susceptance
(2) Admittance = Impedance
Y = Z1 = G + jB Seimens (S)
1
2
2
∠Y = −∠Z = arctan B
|Y |2 = |Z|
2 = G + B
G
Note:
Y = G + jB =
So
G
1
1
R
=
=
2
Z
R+jX
R +X 2
R
R
= R2 +X
2 =
|Z|2
B=
Beware: G 6=
E1.1 Analysis of Circuits (2016-7787)
1
R
−X
R2 +X 2
=
+ j R2−X
+X 2
−X
|Z|2
unless X = 0.
Phasors: 10 – 10 / 11
Summary
10: Sine waves and
phasors
Sine Waves
Rotating Rod
Phasors
Phasor Examples
Phasor arithmetic
Complex Impedances
Phasor Analysis
CIVIL
Impedance and
Admittance
Summary
⊲
• Sine waves are the only bounded signals whose shape is unchanged by
differentiation.
• Think of a sine wave as the projection of a rotating rod onto the
horizontal (or real) axis.
◦ A phasor is a complex number representing the length and position
of the rod at time t = 0.
◦ If V = a + jb = r∠θ = rejθ , then
jωt
v(t) = a cos ωt − b sin ωt = r cos (ωt + θ) = ℜ V e
◦ The angular frequency ω = 2πf is assumed known.
• If all sources in a linear circuit are sine waves having the same
frequency, we can use phasors for circuit analysis:
1
◦ Use complex impedances: jωL and jωC
◦ Mnemonic: CIVIL tells you whether I leads V or vice versa
(“leads” means “reaches its peak before”).
◦ Phasors eliminate time from equations ,, converts simultaneous
differential equations into simultaneous linear equations ,,,.
◦ Needs complex numbers / but worth it.
see Hayt Chapter 10
E1.1 Analysis of Circuits (2016-7787)
Phasors: 10 – 11 / 11