10: Sine waves and phasors

10: Sine waves and phasors


• Sine Waves

• Rotating Rod

• Phasors

• Phasor Examples +

• Phasor arithmetic

• Complex Impedances

• Phasor Analysis +

• CIVIL• Impedance andAdmittance

• Summary

E1.1 Analysis of Circuits (2017-10213) Phasors: 10 – 1 / 11

Sine Waves


• Sine Waves

• Rotating Rod

• Phasors






• Summary


For inductors and capacitors i = C dvdt

and v = L didt

so we need todifferentiate i(t) and v(t) when analysing circuits containing them.

Sine Waves


• Sine Waves

• Rotating Rod

• Phasors






• Summary



and v = L didt


Usually differentiation changes theshape of a waveform.

0 1 2 3 4-1

0

1

t

0 1 2 3 4-5

0

5

t

Sine Waves


• Sine Waves

• Rotating Rod

• Phasors






• Summary



and v = L didt



For bounded waveforms there isonly one exception:

0 1 2 3 4-1

0

1

t

0 1 2 3 4-5

0

5

t

v(t) = sin t ⇒ dvdt

= cos t

Sine Waves


• Sine Waves

• Rotating Rod

• Phasors






• Summary



and v = L didt




0 1 2 3 4-1

0

1

t

0 1 2 3 4-5

0

5

t


= cos t

0 5 10 15-1

0

1

t

v(t)

0 5 10 15-1

0

1

t

dv/d

t

Sine Waves


• Sine Waves

• Rotating Rod

• Phasors






• Summary



and v = L didt




0 1 2 3 4-1

0

1

t

0 1 2 3 4-5

0

5

t


= cos tsame shape but with a time shift.

0 5 10 15-1

0

1

t

v(t)

0 5 10 15-1

0

1

t

dv/d

t

Sine Waves


• Sine Waves

• Rotating Rod

• Phasors






• Summary



and v = L didt




0 1 2 3 4-1

0

1

t

0 1 2 3 4-5

0

5

t



sin t completes one full periodevery time t increases by 2π.

0 5 10 15-1

0

1

t

v(t)

0 5 10 15-1

0

1

t

dv/d

t

Sine Waves


• Sine Waves

• Rotating Rod

• Phasors






• Summary



and v = L didt




0 1 2 3 4-1

0

1

t

0 1 2 3 4-5

0

5

t




0 5 10 15-1

0

1

t

v(t)

0 5 10 15-1

0

1

t

dv/d

t

sin 2πft makes f complete repetitions every time t increases by 1; thisgives a frequency of f cycles per second, or f Hz.

Sine Waves


• Sine Waves

• Rotating Rod

• Phasors






• Summary



and v = L didt




0 1 2 3 4-1

0

1

t

0 1 2 3 4-5

0

5

t




0 5 10 15-1

0

1

t

v(t)

0 5 10 15-1

0

1

t

dv/d

t

sin 2πft makes f complete repetitions every time t increases by 1; thisgives 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. 50Hz ≃ 314 rad.s−1.

Rotating Rod


• Sine Waves

• Rotating Rod

• Phasors






• Summary


A useful way to think of a cosine wave is as theprojection of a rotating rod onto the horizontal axis.

For a unit-length rod, the projection has length cos θ.

Rotating Rod


• Sine Waves

• Rotating Rod

• Phasors






• Summary




If the rod is rotating at a speed of f revolutions persecond, then θ increases uniformly with time:θ = 2πft.

Rotating Rod


• Sine Waves

• Rotating Rod

• Phasors






• Summary





The only difference between cos and sin is the starting position of the rod:

Rotating Rod


• Sine Waves

• Rotating Rod

• Phasors






• Summary






0 5 10 15-1

0

1

t

v = cos 2πft

Rotating Rod


• Sine Waves

• Rotating Rod

• Phasors






• Summary






0 5 10 15-1

0

1

t

v = cos 2πft

0 5 10 15-1

0

1

t

v = sin 2πft

Rotating Rod


• Sine Waves

• Rotating Rod

• Phasors






• Summary






0 5 10 15-1

0

1

t

v = cos 2πft

0 5 10 15-1

0

1

t

v = sin 2πft = cos(

2πft− π2

)

Rotating Rod


• Sine Waves

• Rotating Rod

• Phasors






• Summary






0 5 10 15-1

0

1

t

v = cos 2πft

0 5 10 15-1

0

1

t

v = sin 2πft = cos(

2πft− π2

)

sin 2πft lags cos 2πft by 90 (or π2

radians) because its peaks occurs 1

4

of a cycle later (equivalently cos leads sin) .

Phasors


• Sine Waves

• Rotating Rod

• Phasors






• Summary


If the rod has length A and starts at an angle φ then the projection onto thehorizontal axis is

A cos (2πft+ φ)

Phasors


• Sine Waves

• Rotating Rod

• Phasors






• Summary



A cos (2πft+ φ)= A cosφ cos 2πft−A sinφ sin 2πft

Phasors


• Sine Waves

• Rotating Rod

• Phasors






• Summary



A cos (2πft+ φ)= A cosφ cos 2πft−A sinφ sin 2πft= X cos 2πft− Y sin 2πft

Phasors


• Sine Waves

• Rotating Rod

• Phasors






• Summary




At time t = 0, the tip of the rod has coordinates(X, Y ) = (A cosφ, A sinφ).

Phasors


• Sine Waves

• Rotating Rod

• Phasors






• Summary





If we think of the plane as an Argand Diagram (or complex plane), then thecomplex number X + jY corresponding to the tip of the rod at t = 0 iscalled a phasor .

Phasors


• Sine Waves

• Rotating Rod

• Phasors






• Summary






The magnitude of the phasor, A =√X2 + Y 2, gives the amplitude (peak

value) of the sine wave.

Phasors


• Sine Waves

• Rotating Rod

• Phasors






• Summary






The magnitude of the phasor, A =√X2 + Y 2, gives the amplitude (peak

value) of the sine wave.

The argument of the phasor, φ = arctan YX

, gives the phase shift relativeto cos 2πft.If φ > 0, it is leading and if φ < 0, it is lagging relative to cos 2πft.

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary


V = 1, f = 50Hz

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary


V = 1, f = 50Hzv(t) = cos 2πft

0 0.02 0.04 0.06-1

0

1

t

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5jv(t) = − cos 2πft+0.5 sin 2πft

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153

v(t) = − cos 2πft+0.5 sin 2πft

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153

v(t) = − cos 2πft+0.5 sin 2πft= 1.12 cos (2πft− 2.68)

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY

v(t) = X cos 2πft− Y sin 2πft

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY

v(t) = X cos 2πft− Y sin 2πftBeware minus sign.

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY


V = A∠φ

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY


V = A∠φ

v(t) = A cos (2πft+ φ)

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY


V = A∠φ = Aejφ


Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY


V = A∠φ = Aejφ


A phasor represents an entire waveform (encompassing all time) as asingle complex number. We assume the frequency, f , is known.

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY


V = A∠φ = Aejφ



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).

Phasor Examples +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



0 0.02 0.04 0.06-1

0

1

t

V = −j

v(t) = sin 2πft0 0.02 0.04 0.06

-1

0

1

t

V = −1−0.5j = 1.12∠−153


V = X + jY


V = A∠φ = Aejφ



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.

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms

v(t) = P cosωt−Q sinωtwhere ω = 2πf .

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms


a× v(t)

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms


a× v(t) = aP cosωt− aQ sinωt

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms


aV a× v(t) = aP cosωt− aQ sinωt

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



v1(t) + v2(t)

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)

Adding or scaling is the same for waveforms and phasors.

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)


dvdt

= −ωP sinωt− ωQ cosωt

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)


dvdt

= −ωP sinωt− ωQ cosωt= (−ωQ) cosωt− (ωP ) sinωt

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)


V = (−ωQ) + j (ωP )

dvdt


Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)


V = (−ωQ) + j (ωP )= jω (P + jQ)

dvdt


Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)


V = (−ωQ) + j (ωP )= jω (P + jQ)= jωV

dvdt


Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)


V = (−ωQ) + j (ωP )= jω (P + jQ)= jωV

dvdt


Differentiating waveforms corresponds to multiplyingphasors by jω.

Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)


V = (−ωQ) + j (ωP )= jω (P + jQ)= jωV

dvdt



Phasor arithmetic


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Phasors

V = P + jQ

Waveforms



V1 + V2 v1(t) + v2(t)


V = (−ωQ) + j (ωP )= jω (P + jQ)= jωV

dvdt



Rotate anti-clockwise 90 and scale by ω = 2πf .

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t)

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Inductor:

v(t) = L didt

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Inductor:

v(t) = L didt

⇒ V = jωLI

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Inductor:

v(t) = L didt

⇒ V = jωLI ⇒ VI= jωL

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Inductor:

v(t) = L didt


Capacitor:

i(t) = C dvdt

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Inductor:

v(t) = L didt


Capacitor:

i(t) = C dvdt

⇒ I = jωCV

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Inductor:

v(t) = L didt


Capacitor:

i(t) = C dvdt

⇒ I = jωCV ⇒ VI= 1

jωC

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Inductor:

v(t) = L didt


Capacitor:

i(t) = C dvdt

⇒ I = jωCV ⇒ VI= 1

jωC

For all three components, phasors obey Ohm’s law if we use the compleximpedances jωL and 1

jωCas the “resistance” of an inductor or capacitor.

Complex Impedances


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Resistor:

v(t) = Ri(t) ⇒ V = RI ⇒ VI= R

Inductor:

v(t) = L didt


Capacitor:

i(t) = C dvdt

⇒ I = jωCV ⇒ VI= 1

jωC

For all three components, phasors obey Ohm’s law if we use the compleximpedances jωL and 1

jωCas the “resistance” of an inductor or capacitor.

If all sources in a circuit are sine waves having the same frequency, we cando circuit analysis exactly as before by using complex impedances.

Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Given v = 10 sinωt where ω = 2π × 1000,find vC(t).

0 0.5 1 1.5 2-10

0

10

t (ms)

v

Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary



(1) Find capacitor complex impedanceZ = 1

jωC= 1

6.28j×10−4 = −1592j

0 0.5 1 1.5 2-10

0

10

t (ms)

v

Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary




jωC= 1

6.28j×10−4 = −1592j

(2) Solve circuit with phasorsVC = V × Z

R+Z

0 0.5 1 1.5 2-10

0

10

t (ms)

v

Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary




jωC= 1

6.28j×10−4 = −1592j


R+Z

= −10j × −1592j

1000−1592j

0 0.5 1 1.5 2-10

0

10

t (ms)

v

Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary




jωC= 1

6.28j×10−4 = −1592j


R+Z

= −10j × −1592j

1000−1592j

= −4.5− 7.2j = 8.47∠− 1220 0.5 1 1.5 2

-10

0

10

t (ms)

v

Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary




jωC= 1

6.28j×10−4 = −1592j


R+Z

= −10j × −1592j

1000−1592j

= −4.5− 7.2j = 8.47∠− 122

vC = 8.47 cos (ωt− 122)0 0.5 1 1.5 2

-10

0

10

t (ms)

C

vv

C

Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary




jωC= 1

6.28j×10−4 = −1592j


R+Z

= −10j × −1592j

1000−1592j

= −4.5− 7.2j = 8.47∠− 122

vC = 8.47 cos (ωt− 122)0 0.5 1 1.5 2

-10

0

10

t (ms)

C

vv

C

(3) Draw a phasor diagram showing KVL:V = −10jVC = −4.5− 7.2jVR = V −VC = 4.5− 2.8j = 5.3∠− 32

Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary




jωC= 1

6.28j×10−4 = −1592j


R+Z

= −10j × −1592j

1000−1592j

= −4.5− 7.2j = 8.47∠− 122

vC = 8.47 cos (ωt− 122)0 0.5 1 1.5 2

-10

0

10

t (ms)

CR v

vCv

R


Phasor Analysis +


• Sine Waves

• Rotating Rod

• Phasors






• Summary




jωC= 1

6.28j×10−4 = −1592j


R+Z

= −10j × −1592j

1000−1592j

= −4.5− 7.2j = 8.47∠− 122

vC = 8.47 cos (ωt− 122)0 0.5 1 1.5 2

-10

0

10

t (ms)

CR v

vCv

R


Phasors add like vectors

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary


Capacitors: i = C dvdt

⇒ I leads V

Inductors: v = L didt

⇒ V leads I

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I

Mnemonic: CIVIL = “In a capacitor I lead V but V leads I in an inductor”.

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I


COMPLEX ARITHMETIC TRICKS:

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I



(1) j × j = −j ×−j = −1

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I



(1) j × j = −j ×−j = −1(2) 1

j= −j

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I



(1) j × j = −j ×−j = −1(2) 1

j= −j

(3) a+ jb = r∠θ = rejθ

where r =√a2 + b2 and θ = arctan b

a(±180 if a < 0)

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I



(1) j × j = −j ×−j = −1(2) 1

j= −j



a(±180 if a < 0)

(4) r∠θ = rejθ = (r cos θ) + j (r sin θ)

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I



(1) j × j = −j ×−j = −1(2) 1

j= −j



a(±180 if a < 0)

(4) r∠θ = rejθ = (r cos θ) + j (r sin θ)(5) a∠θ × b∠φ = ab∠ (θ + φ) and a∠θ

b∠φ= a

b∠ (θ − φ).

Multiplication and division are much easier in polar form.

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I



(1) j × j = −j ×−j = −1(2) 1

j= −j



a(±180 if a < 0)


b∠φ= a

b∠ (θ − φ).

Multiplication and division are much easier in polar form.(6) All scientific calculators will convert rectangular to/from polar form.

CIVIL


• Sine Waves

• Rotating Rod

• Phasors






• Summary



⇒ I leads V


⇒ V leads I



(1) j × j = −j ×−j = −1(2) 1

j= −j



a(±180 if a < 0)


b∠φ= a

b∠ (θ − 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 complexarithmetic (+,−,×,÷, x2, 1

x, |x|, x∗) in CMPLX mode.

Learn how to use this: it will save lots of time and errors.

Impedance and Admittance


• Sine Waves

• Rotating Rod

• Phasors






• Summary


For any network (resistors+capacitors+inductors):



• Sine Waves

• Rotating Rod

• Phasors






• Summary



(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)



• Sine Waves

• Rotating Rod

• Phasors






• Summary



(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2

∠Z = arctan XR



• Sine Waves

• Rotating Rod

• Phasors






• Summary




∠Z = arctan XR

(2) Admittance = 1

Impedance= Conductance + j× Susceptance

Y = 1

Z= G+ jB Siemens (S)



• Sine Waves

• Rotating Rod

• Phasors






• Summary




∠Z = arctan XR

(2) Admittance = 1


Y = 1


|Y |2 = 1

|Z|2= G2 +B2

∠Y = −∠Z = arctan BG



• Sine Waves

• Rotating Rod

• Phasors






• Summary




∠Z = arctan XR

(2) Admittance = 1


Y = 1


|Y |2 = 1

|Z|2= G2 +B2


Note:Y = G+ jB = 1

Z



• Sine Waves

• Rotating Rod

• Phasors






• Summary




∠Z = arctan XR

(2) Admittance = 1


Y = 1


|Y |2 = 1

|Z|2= G2 +B2


Note:Y = G+ jB = 1

Z= 1

R+jX



• Sine Waves

• Rotating Rod

• Phasors






• Summary




∠Z = arctan XR

(2) Admittance = 1


Y = 1


|Y |2 = 1

|Z|2= G2 +B2


Note:Y = G+ jB = 1

Z= 1

R+jX= R

R2+X2 + j −XR2+X2



• Sine Waves

• Rotating Rod

• Phasors






• Summary




∠Z = arctan XR

(2) Admittance = 1


Y = 1


|Y |2 = 1

|Z|2= G2 +B2


Note:Y = G+ jB = 1

Z= 1

R+jX= R

R2+X2 + j −XR2+X2

So G = RR2+X2 = R

|Z|2

B = −XR2+X2 = −X

|Z|2



• Sine Waves

• Rotating Rod

• Phasors






• Summary




∠Z = arctan XR

(2) Admittance = 1


Y = 1


|Y |2 = 1

|Z|2= G2 +B2


Note:Y = G+ jB = 1

Z= 1

R+jX= R

R2+X2 + j −XR2+X2

So G = RR2+X2 = R

|Z|2

B = −XR2+X2 = −X

|Z|2

Beware: G 6= 1

Runless X = 0.

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary


• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary



• Think of a sine wave as the projection of a rotating rod onto thehorizontal (or real) axis. A phasor is a complex number representing the length and position

of the rod at time t = 0.

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary




of the rod at time t = 0. If V = a+ jb = r∠θ = rejθ, then

v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(

V ejωt)

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary






V ejωt)

The angular frequency ω = 2πf is assumed known.

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary






V ejωt)


• If all sources in a linear circuit are sine waves having the samefrequency, we can use phasors for circuit analysis:

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary






V ejωt)


• If all sources in a linear circuit are sine waves having the samefrequency, we can use phasors for circuit analysis: Use complex impedances: jωL and 1

jωC

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary






V ejωt)



jωC

Mnemonic: CIVIL tells you whether I leads V or vice versa(“leads” means “reaches its peak before”).

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary






V ejωt)



jωC


Phasors eliminate time from equations ,

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary






V ejωt)



jωC


Phasors eliminate time from equations ,, converts simultaneousdifferential equations into simultaneous linear equations ,,,.

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary






V ejωt)



jωC



Needs complex numbers / but worth it.

Summary


• Sine Waves

• Rotating Rod

• Phasors






• Summary






V ejωt)



jωC



Needs complex numbers / but worth it.See Hayt Ch 10 or Irwin Ch 8

Documents

10: Sine waves and phasors