10: sine waves and phasors - imperial college london · a useful way to think of a cosine wave is...
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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
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 – 2 / 11
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
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 – 2 / 11
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.
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
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 – 2 / 11
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.
Usually differentiation changes theshape of a waveform.
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
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 – 2 / 11
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.
Usually differentiation changes theshape of a waveform.
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
0 5 10 15-1
0
1
t
v(t)
0 5 10 15-1
0
1
t
dv/d
t
Sine Waves
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 – 2 / 11
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.
Usually differentiation changes theshape of a waveform.
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 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
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 – 2 / 11
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.
Usually differentiation changes theshape of a waveform.
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 tsame shape but with a time shift.
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
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 – 2 / 11
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.
Usually differentiation changes theshape of a waveform.
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 tsame shape but with a time shift.
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
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
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 – 2 / 11
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.
Usually differentiation changes theshape of a waveform.
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 tsame shape but with a time shift.
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
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
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 – 3 / 11
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
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 – 3 / 11
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.
If the rod is rotating at a speed of f revolutions persecond, then θ increases uniformly with time:θ = 2πft.
Rotating Rod
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 – 3 / 11
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.
If the rod is rotating at a speed of f revolutions persecond, then θ increases uniformly with time:θ = 2πft.
The only difference between cos and sin is the starting position of the rod:
Rotating Rod
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 – 3 / 11
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.
If the rod is rotating at a speed of f revolutions persecond, then θ increases uniformly with time:θ = 2πft.
The only difference between cos and sin is the starting position of the rod:
0 5 10 15-1
0
1
t
v = cos 2πft
Rotating Rod
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 – 3 / 11
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.
If the rod is rotating at a speed of f revolutions persecond, then θ increases uniformly with time:θ = 2πft.
The only difference between cos and sin is the starting position of the rod:
0 5 10 15-1
0
1
t
v = cos 2πft
0 5 10 15-1
0
1
t
v = sin 2πft
Rotating Rod
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 – 3 / 11
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.
If the rod is rotating at a speed of f revolutions persecond, then θ increases uniformly with time:θ = 2πft.
The only difference between cos and sin is the starting position of the rod:
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
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 – 3 / 11
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.
If the rod is rotating at a speed of f revolutions persecond, then θ increases uniformly with time:θ = 2πft.
The only difference between cos and sin is the starting position of the rod:
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
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 – 4 / 11
If the rod has length A and starts at an angle φ then the projection onto thehorizontal axis is
A cos (2πft+ φ)
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 – 4 / 11
If the rod has length A and starts at an angle φ then the projection onto thehorizontal axis is
A cos (2πft+ φ)= A cosφ cos 2πft−A sinφ sin 2πft
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 – 4 / 11
If the rod has length A and starts at an angle φ then the projection onto thehorizontal axis is
A cos (2πft+ φ)= A cosφ cos 2πft−A sinφ sin 2πft= X cos 2πft− Y sin 2πft
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 – 4 / 11
If the rod has length A and starts at an angle φ then the projection onto thehorizontal axis is
A cos (2πft+ φ)= A cosφ cos 2πft−A sinφ sin 2πft= X cos 2πft− Y sin 2πft
At time t = 0, the tip of the rod has coordinates(X, Y ) = (A cosφ, A sinφ).
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 – 4 / 11
If the rod has length A and starts at an angle φ then the projection onto thehorizontal axis is
A cos (2πft+ φ)= A cosφ cos 2πft−A sinφ sin 2πft= X cos 2πft− Y sin 2πft
At time t = 0, the tip of the rod has coordinates(X, Y ) = (A cosφ, A sinφ).
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
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 – 4 / 11
If the rod has length A and starts at an angle φ then the projection onto thehorizontal axis is
A cos (2πft+ φ)= A cosφ cos 2πft−A sinφ sin 2πft= X cos 2πft− Y sin 2πft
At time t = 0, the tip of the rod has coordinates(X, Y ) = (A cosφ, A sinφ).
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 .
The magnitude of the phasor, A =√X2 + Y 2, gives the amplitude (peak
value) of the sine wave.
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 – 4 / 11
If the rod has length A and starts at an angle φ then the projection onto thehorizontal axis is
A cos (2πft+ φ)= A cosφ cos 2πft−A sinφ sin 2πft= X cos 2πft− Y sin 2πft
At time t = 0, the tip of the rod has coordinates(X, Y ) = (A cosφ, A sinφ).
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 .
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 +
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 – 5 / 11
V = 1, f = 50Hz
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
0 0.02 0.04 0.06-1
0
1
t
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
0 0.02 0.04 0.06-1
0
1
t
V = −j
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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 +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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 +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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 +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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 +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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 +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
v(t) = X cos 2πft− Y sin 2πft
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
v(t) = X cos 2πft− Y sin 2πftBeware minus sign.
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
v(t) = X cos 2πft− Y sin 2πftBeware minus sign.
V = A∠φ
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
v(t) = X cos 2πft− Y sin 2πftBeware minus sign.
V = A∠φ
v(t) = A cos (2πft+ φ)
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
v(t) = X cos 2πft− Y sin 2πftBeware minus sign.
V = A∠φ = Aejφ
v(t) = A cos (2πft+ φ)
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
v(t) = X cos 2πft− Y sin 2πftBeware minus sign.
V = A∠φ = Aejφ
v(t) = A cos (2πft+ φ)
A phasor represents an entire waveform (encompassing all time) as asingle complex number. We assume the frequency, f , is known.
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
v(t) = X cos 2πft− Y sin 2πftBeware minus sign.
V = A∠φ = Aejφ
v(t) = A cos (2πft+ φ)
A phasor represents an entire waveform (encompassing all time) as asingle 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).
Phasor Examples +
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 – 5 / 11
V = 1, f = 50Hzv(t) = cos 2πft
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)
V = X + jY
v(t) = X cos 2πft− Y sin 2πftBeware minus sign.
V = A∠φ = Aejφ
v(t) = A cos (2πft+ φ)
A phasor represents an entire waveform (encompassing all time) as asingle 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.
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
a× v(t)
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
a× v(t) = aP cosωt− aQ sinωt
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
v1(t) + v2(t)
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
dvdt
= −ωP sinωt− ωQ cosωt
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
dvdt
= −ωP sinωt− ωQ cosωt= (−ωQ) cosωt− (ωP ) sinωt
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
V = (−ωQ) + j (ωP )
dvdt
= −ωP sinωt− ωQ cosωt= (−ωQ) cosωt− (ωP ) sinωt
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
V = (−ωQ) + j (ωP )= jω (P + jQ)
dvdt
= −ωP sinωt− ωQ cosωt= (−ωQ) cosωt− (ωP ) sinωt
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
V = (−ωQ) + j (ωP )= jω (P + jQ)= jωV
dvdt
= −ωP sinωt− ωQ cosωt= (−ωQ) cosωt− (ωP ) sinωt
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
V = (−ωQ) + j (ωP )= jω (P + jQ)= jωV
dvdt
= −ωP sinωt− ωQ cosωt= (−ωQ) cosωt− (ωP ) sinωt
Differentiating waveforms corresponds to multiplyingphasors by jω.
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
V = (−ωQ) + j (ωP )= jω (P + jQ)= jωV
dvdt
= −ωP sinωt− ωQ cosωt= (−ωQ) cosωt− (ωP ) sinωt
Differentiating waveforms corresponds to multiplyingphasors by jω.
Phasor arithmetic
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 – 6 / 11
Phasors
V = P + jQ
Waveforms
v(t) = P cosωt−Q sinωtwhere ω = 2πf .
aV a× v(t) = aP cosωt− aQ sinωt
V1 + V2 v1(t) + v2(t)
Adding or scaling is the same for waveforms and phasors.
V = (−ωQ) + j (ωP )= jω (P + jQ)= jωV
dvdt
= −ωP sinωt− ωQ cosωt= (−ωQ) cosωt− (ωP ) sinωt
Differentiating waveforms corresponds to multiplyingphasors by jω.
Rotate anti-clockwise 90 and scale by ω = 2πf .
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t)
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Inductor:
v(t) = L didt
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Inductor:
v(t) = L didt
⇒ V = jωLI
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Inductor:
v(t) = L didt
⇒ V = jωLI ⇒ VI= jωL
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Inductor:
v(t) = L didt
⇒ V = jωLI ⇒ VI= jωL
Capacitor:
i(t) = C dvdt
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Inductor:
v(t) = L didt
⇒ V = jωLI ⇒ VI= jωL
Capacitor:
i(t) = C dvdt
⇒ I = jωCV
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Inductor:
v(t) = L didt
⇒ V = jωLI ⇒ VI= jωL
Capacitor:
i(t) = C dvdt
⇒ I = jωCV ⇒ VI= 1
jωC
Complex Impedances
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Inductor:
v(t) = L didt
⇒ V = jωLI ⇒ VI= jωL
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
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 – 7 / 11
Resistor:
v(t) = Ri(t) ⇒ V = RI ⇒ VI= R
Inductor:
v(t) = L didt
⇒ V = jωLI ⇒ VI= jωL
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 +
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 – 8 / 11
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 +
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 – 8 / 11
Given v = 10 sinωt where ω = 2π × 1000,find vC(t).
(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 +
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 – 8 / 11
Given v = 10 sinωt where ω = 2π × 1000,find vC(t).
(1) Find capacitor complex impedanceZ = 1
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 +
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 – 8 / 11
Given v = 10 sinωt where ω = 2π × 1000,find vC(t).
(1) Find capacitor complex impedanceZ = 1
jωC= 1
6.28j×10−4 = −1592j
(2) Solve circuit with phasorsVC = V × Z
R+Z
= −10j × −1592j
1000−1592j
0 0.5 1 1.5 2-10
0
10
t (ms)
v
Phasor Analysis +
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 – 8 / 11
Given v = 10 sinωt where ω = 2π × 1000,find vC(t).
(1) Find capacitor complex impedanceZ = 1
jωC= 1
6.28j×10−4 = −1592j
(2) Solve circuit with phasorsVC = V × Z
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 +
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 – 8 / 11
Given v = 10 sinωt where ω = 2π × 1000,find vC(t).
(1) Find capacitor complex impedanceZ = 1
jωC= 1
6.28j×10−4 = −1592j
(2) Solve circuit with phasorsVC = V × Z
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 +
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 – 8 / 11
Given v = 10 sinωt where ω = 2π × 1000,find vC(t).
(1) Find capacitor complex impedanceZ = 1
jωC= 1
6.28j×10−4 = −1592j
(2) Solve circuit with phasorsVC = V × Z
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 +
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 – 8 / 11
Given v = 10 sinωt where ω = 2π × 1000,find vC(t).
(1) Find capacitor complex impedanceZ = 1
jωC= 1
6.28j×10−4 = −1592j
(2) Solve circuit with phasorsVC = V × Z
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
(3) Draw a phasor diagram showing KVL:V = −10jVC = −4.5− 7.2jVR = V −VC = 4.5− 2.8j = 5.3∠− 32
Phasor Analysis +
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 – 8 / 11
Given v = 10 sinωt where ω = 2π × 1000,find vC(t).
(1) Find capacitor complex impedanceZ = 1
jωC= 1
6.28j×10−4 = −1592j
(2) Solve circuit with phasorsVC = V × Z
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
(3) Draw a phasor diagram showing KVL:V = −10jVC = −4.5− 7.2jVR = V −VC = 4.5− 2.8j = 5.3∠− 32
Phasors add like vectors
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
Mnemonic: CIVIL = “In a capacitor I lead V but V leads I in an inductor”.
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
Mnemonic: CIVIL = “In a capacitor I lead V but V leads I in an inductor”.
COMPLEX ARITHMETIC TRICKS:
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
Mnemonic: CIVIL = “In a capacitor I lead V but V leads I in an inductor”.
COMPLEX ARITHMETIC TRICKS:
(1) j × j = −j ×−j = −1
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
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) 1
j= −j
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
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) 1
j= −j
(3) a+ jb = r∠θ = rejθ
where r =√a2 + b2 and θ = arctan b
a(±180 if a < 0)
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
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) 1
j= −j
(3) a+ jb = r∠θ = rejθ
where r =√a2 + b2 and θ = arctan b
a(±180 if a < 0)
(4) r∠θ = rejθ = (r cos θ) + j (r sin θ)
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
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) 1
j= −j
(3) a+ jb = r∠θ = rejθ
where r =√a2 + b2 and θ = arctan b
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
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
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) 1
j= −j
(3) a+ jb = r∠θ = rejθ
where r =√a2 + b2 and θ = arctan b
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.(6) All scientific calculators will convert rectangular to/from polar form.
CIVIL
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 – 9 / 11
Capacitors: i = C dvdt
⇒ I leads V
Inductors: v = L didt
⇒ V leads I
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) 1
j= −j
(3) a+ jb = r∠θ = rejθ
where r =√a2 + b2 and θ = arctan b
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.(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
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 – 10 / 11
For any network (resistors+capacitors+inductors):
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2
∠Z = arctan XR
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2
∠Z = arctan XR
(2) Admittance = 1
Impedance= Conductance + j× Susceptance
Y = 1
Z= G+ jB Siemens (S)
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2
∠Z = arctan XR
(2) Admittance = 1
Impedance= Conductance + j× Susceptance
Y = 1
Z= G+ jB Siemens (S)
|Y |2 = 1
|Z|2= G2 +B2
∠Y = −∠Z = arctan BG
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2
∠Z = arctan XR
(2) Admittance = 1
Impedance= Conductance + j× Susceptance
Y = 1
Z= G+ jB Siemens (S)
|Y |2 = 1
|Z|2= G2 +B2
∠Y = −∠Z = arctan BG
Note:Y = G+ jB = 1
Z
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2
∠Z = arctan XR
(2) Admittance = 1
Impedance= Conductance + j× Susceptance
Y = 1
Z= G+ jB Siemens (S)
|Y |2 = 1
|Z|2= G2 +B2
∠Y = −∠Z = arctan BG
Note:Y = G+ jB = 1
Z= 1
R+jX
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2
∠Z = arctan XR
(2) Admittance = 1
Impedance= Conductance + j× Susceptance
Y = 1
Z= G+ jB Siemens (S)
|Y |2 = 1
|Z|2= G2 +B2
∠Y = −∠Z = arctan BG
Note:Y = G+ jB = 1
Z= 1
R+jX= R
R2+X2 + j −XR2+X2
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2
∠Z = arctan XR
(2) Admittance = 1
Impedance= Conductance + j× Susceptance
Y = 1
Z= G+ jB Siemens (S)
|Y |2 = 1
|Z|2= G2 +B2
∠Y = −∠Z = arctan BG
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
Impedance and Admittance
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 – 10 / 11
For any network (resistors+capacitors+inductors):
(1) Impedance = Resistance + j× ReactanceZ = R + jX (Ω)|Z|2 = R2 +X2
∠Z = arctan XR
(2) Admittance = 1
Impedance= Conductance + j× Susceptance
Y = 1
Z= G+ jB Siemens (S)
|Y |2 = 1
|Z|2= G2 +B2
∠Y = −∠Z = arctan BG
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
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
Summary
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
Summary
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
The angular frequency ω = 2πf is assumed known.
Summary
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
The angular frequency ω = 2πf is assumed known.
• If all sources in a linear circuit are sine waves having the samefrequency, we can use phasors for circuit analysis:
Summary
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
The angular frequency ω = 2πf is assumed known.
• 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
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
The angular frequency ω = 2πf is assumed known.
• 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
Mnemonic: CIVIL tells you whether I leads V or vice versa(“leads” means “reaches its peak before”).
Summary
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
The angular frequency ω = 2πf is assumed known.
• 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
Mnemonic: CIVIL tells you whether I leads V or vice versa(“leads” means “reaches its peak before”).
Phasors eliminate time from equations ,
Summary
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
The angular frequency ω = 2πf is assumed known.
• 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
Mnemonic: CIVIL tells you whether I leads V or vice versa(“leads” means “reaches its peak before”).
Phasors eliminate time from equations ,, converts simultaneousdifferential equations into simultaneous linear equations ,,,.
Summary
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
The angular frequency ω = 2πf is assumed known.
• 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
Mnemonic: CIVIL tells you whether I leads V or vice versa(“leads” means “reaches its peak before”).
Phasors eliminate time from equations ,, converts simultaneousdifferential equations into simultaneous linear equations ,,,.
Needs complex numbers / but worth it.
Summary
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 – 11 / 11
• Sine waves are the only bounded signals whose shape is unchanged bydifferentiation.
• 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. If V = a+ jb = r∠θ = rejθ, then
v(t) = a cosωt− b sinωt = r cos (ωt+ θ) = ℜ(
V ejωt)
The angular frequency ω = 2πf is assumed known.
• 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
Mnemonic: CIVIL tells you whether I leads V or vice versa(“leads” means “reaches its peak before”).
Phasors eliminate time from equations ,, converts simultaneousdifferential equations into simultaneous linear equations ,,,.
Needs complex numbers / but worth it.See Hayt Ch 10 or Irwin Ch 8