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(wt-f)  dx/dt = –w sin(wt-f)
x(t)=Acos(wt-f)  dx/dt = –w Asin(wt-f)
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(wt-f)
dx/dt = –w sin(wt-f)
d2x/dt2 = –w2 cos(wt–f)

Spring-Mass System:
d2x/dt2 = –(k/m) x(t)
–w2 cos(wt–f) = – (k/m) cos(wt-f)

It only works at the “natural frequency:”
w = w0 = √(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) = eat
dx/dt = a eat
d2x/dt2 = a2 eat

Could x(t)= eat also solve the spring-mass system?
d2x/dt2 = –(k/m) x(t)
a2 eat = – (k/m) eat

It only works if:
a = √(-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:
a = √(-k/m) = √-1  √(k/m) = jw0
x(t) = eat = ejw t
0
Complex Exponentials

Definition (another bit to memorize):
x(t) = ej(wt-f) = cos(wt-f) + j sin(wt-f)

Then the “real part” of x(t) is defined to be:
Re{x(t)} = cos(wt-f)

So x(t)=ejwt and x(t)=cos(wt) are actually exactly the same solution!

The “imaginary part,” Im{x(t)}=sin(wt), 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
ejf = cos(f) + j sin(f)
1 = cos(0) + j sin(0) = ej0
j = cos(p/2) + j sin(p/2) = ejp/2
-1 = cos(p) + j sin(p) = ejp
(ejp/2)2 = ejp
(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 ejwt + 3 ej(wt-p/2)
Re{x(t)} = 2cos(wt) + 3cos(wt-p/2)
Im{x(t)} = 2sin(wt) + 3sin(wt-p/2)
Amplitude, Phase, and Frequency
of Cosines vs. Exponentials

Exponential:
x(t) = Aej(wt-f) = A e-jf ejwt

Cosine:
Re{x(t)} = A cos(wt-f)
The Life Story of a Cosine

Vocal fold motion is a cosine at (say) w=400p, with amplitude of A=0.001m:
x(t) = 0.001 + 0.001 ej400pt m

Notice this looks like an exponential, but it’s “really” a cosine.
Re{x(t)} = 0.001 + 0.001cos(400pt)

Air puffs come through when the vocal folds are open, with a maximum
rate of 0.001 m3/second:
uGlottis(t)=0.0005+0.0005ej400pt m3/s

The same air puffs reach the lips 0.5ms later:
uLips(t)=0.0005+0.0005ej400p(t-0.0005) =0.0005 + 0.0005 e-0.2jp ej400pt liter/s

Air pressure at the lips is the derivative of u(t), times 0.003 kg/s:
pLips(t)= 0 + 0.0003jp e-0.2jp ej400pt = 0.0003p e0.3jp ej400pt Pascals

Acoustic wave reaches the listener’s ear 2ms later:
pEar(t)= 0.0003p e0.3jp ej400p(t-0.02) = 0.0003p e-j7.7p ej400pt
Look Closer:
x(t) = 0.001 ej400pt
uGlottis(t) = 0.0005 ej400pt
uLips(t) = 0.0005 e-0.2jp ej400pt
pLips(t) = 0.0003p e0.3jp ej400pt
pEar(t) = 0.0003p e-j7.7p ej400pt

Amplitude, frequency, phase. Which one
stays the same?
Phasor Notation
x = 0.001
uGlottis = 0.0005
uLips = 0.0005 e-0.2jp
pLips = 0.0003p e0.3jp
pEar = 0.0003p e-j7.7p

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 ejf
To get x(t) back, you look back through the
problem definition in order to find w, then
write
x(t) = Re{ x ejwt }
Example
Phasor
x, u, p
Vocal Fold x(t)
10
Complex
x(t), u(t), p(t)
Real
x(t), u(t), p(t)
10 ej400pt
10cos(400pt)
Vocal Fold u(t)
5
5 ej400pt
5 cos(400pt)
Lips u(t)
5 e-0.2jp
5 e-0.2jp ej400pt
5 cos(400pt-0.2p)
Lips p(t)
3p e0.3jp
3p e0.3jp ej400pt
3p cos(400pt+0.3p)
Ear p(t)
3p e-j7.7p
3p e-j7.7p ej400pt
3p cos(400pt-7.7p)
The Main Purpose of Phasors: They
Turn Derivatives into Multiplication

x(t) = Re{ x ejwt }

dx/dt = Re{ jwx ejwt}

d2x/dt2 = Re{ (jw)2x ejwt } = Re{ -w2x ejwt }
The Main Purpose of Phasors: They
Turn Derivatives into Multiplication



In regular notation:

v(t) = dx/dt

x(t) = A cos(wt-f)
v(t) = -w A sin(wt-f)

v = -jw Ae-jf
In phasor notation:

v = -jwx

x = Ae-jf
In regular notation:



d2x/dt2 = -(k/m) x(t)
In phasor notation:

-w2 x = -(k/m) x

Solution: w=√(k/m)
!!!
Summary

Linearity means you can:

add two solutions, or
 scale by a constant.
x(t)=eat could solve the spring-mass system, but only if a=√(-k/m)






j2 = -1
e-jf = cos(-f) + j sin(-f)
We don’t really need the imaginary part; Re{ejwt} = cos(wt)
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-jf

To get back to the time domain:
x(t) = Re{ xejwt }