ECE 598: The Speech Chain
Lecture 3: Phasors
A Useful One-Slide Idea: Linearity
Derivatives are “linear,” meaning that, for any
functions f(t) and g(t),
x(t) = A f(t) + B g(t)
Implies
dx/dt = A df/dt + B dg/dt
d2x/dt2 = A d2f/dt2 + B d2g/dt2
Example:
x(t)=cos(ωt-φ) ↔ dx/dt = –ω sin(ωt-φ)
x(t)=Acos(ωt-φ) ↔ dx/dt = –ω Asin(ωt-φ)
Review: Spring Mass System
Newton’s Second Law:
f(t) = m d2x/dt2
Force of a Spring:
f(t) = – k x(t)
The Spring-Mass System Equation with no
external forces:
d2x/dt2 = - (k/m) x(t)
Solution: Cosine
First and Second Derivatives of a Cosine:
x(t) = cos(ωt-φ)
dx/dt = –ω sin(ωt-φ)
d2x/dt2 = –ω2 cos(ωt–φ)
Spring-Mass System:
d2x/dt2 = –(k/m) x(t)
–ω2 cos(ωt–φ) = – (k/m) cos(ωt-φ)
It only works at the “natural frequency:”
ω = ω0 = √(k/m)
Linearity
means that
you can
multiply by
“A” here, if
you want to,
and the same
“A” will
appear here.
The Other Function We Know
First and Second Derivatives of an Exponential:
x(t) = eαt
dx/dt = α eαt
d2x/dt2 = α2 eαt
Could x(t)= eαt also solve the spring-mass system?
d2x/dt2 = –(k/m) x(t)
α2 eαt = – (k/m) eαt
It only works if:
α = √(-k/m)
Imaginary Numbers
Definition (the part that you memorize):
j = √- 1
Linearity:
2j = √-4 = √4 × √-1
3j = √-9 = √9 × √-1
Solution to the spring-mass system:
α = √(-k/m) = √-1 × √(k/m) = jω0
x(t) = eαt = ejω t
0
Complex Exponentials
Definition (another bit to memorize):
j(ωt−φ) = cos(ωt−φ) + j sin(ωt−φ)
x(t) = ej(ω
Then the “real part” of x(t) is defined to be:
Re{x(t)} = cos(ωt−φ)
So x(t)=ejωt and x(t)=cos(ωt) are actually exactly the same solution!
The “imaginary part,” Im{x(t)}=sin(ωt), doesn’t change the solution.
If you like, visualize x(t) as a movement in two dimensions:
Re(x(t)) is movement in the horizontal direction
Im(x(t)) is movement in Buckaroo Banzai’s mysterious 8th
dimension.
All we really care about is the movement in the horizontal
direction; the movement in the Buckaroo Banzai direction is a
convenient fiction that just happens to make the math work out.
How do you Plot a Complex
Exponential?
Answer: you can’t!
What you CAN do: plot either the real part or the imaginary part
x(t) = ej2t
Re{x(t)} = cos(2t)
Im{x(t)} = sin(2t)
4
3
2
1
0
-10.00 0.79 1.57 2.36 3.14 3.93 4.71 5.50 6.28
-2
-3
-4
3cos(2t)
3cos(2t-pi/2)
Special Numbers
ejφ = cos(φ) + j sin(φ)
1 = cos(0) + j sin(0) = ej0
j = cos(π/2) + j sin(π/2) = ejπ/2
−1 = cos(π) + j sin(π) = ejπ
(ejπ/2)2 = ejπ
(j)2 = −1
Linearity Again
“Real Part” and “Imaginary Part” are linear operators:
x(t) = A f(t) + B g(t)
Re{x(t)} = A Re{f(t)} + B Re{g(t)}
Im{x(t)} = A Im{f(t)} + B Im{g(t)}
Example
x(t) = 2 ejωt + 3 ej(ωt−π/2)
Re{x(t)} = 2cos(ωt) + 3cos(ωt−π/2)
Im{x(t)} = 2sin(ωt) + 3sin(ωt−π/2)
Amplitude, Phase, and Frequency
of Cosines vs. Exponentials
Exponential:
x(t) = Aej(ωt−φ) = A e−jφ ejωt
Cosine:
Re{x(t)} = A cos(ωt−φ)
The Life Story of a Cosine
Vocal fold motion is a cosine at (say) ω=400π, with amplitude of A=0.001m:
j400πt m
x(t) = 0.001 + 0.001 ej400π
Notice this looks like an exponential, but it’s “really” a cosine.
Re{x(t)} = 0.001 + 0.001cos(400πt)
Air puffs come through when the vocal folds are open, with a maximum
rate of 0.001 m3/second:
j400πt m3/s
uGlottis(t)=0.0005+0.0005ej400π
The same air puffs reach the lips 0.5ms later:
j400π(t−
(t−0.0005) =0.0005 + 0.0005 e−0.2jπ
0.2jπ ej400π
j400πt liter/s
uLips(t)=0.0005+0.0005ej400π
Air pressure at the lips is the derivative of u(t), times 0.003 kg/s:
0.2jπ ej400π
j400πt = 0.0003π e0.3j
0.3jπ ej400π
j400πt Pascals
pLips(t)= 0 + 0.0003jπ e−0.2j
Acoustic wave reaches the listener’s ear 2ms later:
0.3jπ ej400π(
j400π(tt−0.02) = 0.0003π e−j7.7π
j7.7π ej400π
j400πt
pEar(t)= 0.0003π e0.3jπ
Look Closer:
x(t) = 0.001 ej400πt
uGlottis(t) = 0.0005 ej400πt
uLips(t) = 0.0005 e−0.2jπ ej400πt
pLips(t) = 0.0003π e0.3jπ ej400πt
pEar(t) = 0.0003π e−j7.7π ej400πt
Amplitude, frequency, phase. Which one
stays the same?
Phasor Notation
x = 0.001
uGlottis = 0.0005
uLips = 0.0005 e−0.2jπ
pLips = 0.0003π e0.3jπ
pEar = 0.0003π e−j7.7π
Phasor notation: write down only the
amplitude and phase, not the frequency.
Definition of Phasor Notation
A phasor specifies the amplitude and phase
of a cosine, but not its frequency.
(Written in boldface if possible)
x = A ejφ
To get x(t) back, you look back through the
problem definition in order to find ω, then
write
x(t) = Re{ x ejωt }
Example
Phasor
x, u, p
Vocal Fold x(t)
10
Complex
x(t), u(t), p(t)
Real
x(t), u(t), p(t)
j400πt
10 ej400π
10cos(400πt)
Vocal Fold u(t)
5
j400πt
5 ej400π
5 cos(400πt)
Lips u(t)
0.2jπ
5 e−0.2jπ
0.2jπ ej400π
j400πt
5 e−0.2jπ
5 cos(400πt−0.2π)
Lips p(t)
0.3jπ
3π e0.3j
0.3jπ ej400π
j400πt
3π e0.3j
3π cos(400πt+0.3π)
Ear p(t)
j7.7π
3π e−j7.7π
j7.7π ej400π
j400πt
3π e−j7.7π
3π cos(400πt−7.7π)
The Main Purpose of Phasors: They
Turn Derivatives into Multiplication
x(t) = Re{ x ejωt }
dx/dt = Re{ jωx ejωt}
d2x/dt2 = Re{ (jω)2x ejωt } = Re{ −ω2x ejωt }
The Main Purpose of Phasors: They
Turn Derivatives into Multiplication
In regular notation:
v(t) = dx/dt
x(t) = A cos(ωt−φ)
v(t) = −ω A sin(ωt−φ)
↔
v = −jω Ae-jφ
In phasor notation:
v = −jωx
x = Ae−jφ
In regular notation:
↔
d2x/dt2 = −(k/m) x(t)
In phasor notation:
−ω2 x = −(k/m) x
Solution: ω=√(k/m)
!!!
Summary
Linearity means you can:
add two solutions, or
scale by a constant.
x(t)=eαt could solve the spring-mass system, but only if α=√(-k/m)
j2 = −1
e−jφ = cos(−φ) + j sin(−φ)
We don’t really need the imaginary part; Re{ejωt} = cos(ωt)
We don’t really need the frequency part either; it is never changed by
any linear operation (e.g., scaling, time shift, derivative)
Phasor notation encodes just the amplitude and phase of a cosine:
x = Ae−jφ
To get back to the time domain:
x(t) = Re{ xejωt }