ECE 598: The Speech Chain

1. ECE 598: The Speech Chain Lecture 3: Phasors

2. 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)

3. 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)

4. Linearity means that you can multiply by “A” here, if you want to, and the same “A” will appear here. 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)

5. 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)

6. 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

7. 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.

8. 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)

9. 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

10. 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)

11. Amplitude, Phase, and Frequency of Cosines vs. Exponentials • Exponential: x(t) = Aej(wt-f) = A e-jfejwt • Cosine: Re{x(t)} = A cos(wt-f)

12. 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-0.2jpej400pt liter/s • Air pressure at the lips is the derivative of u(t), times 0.003 kg/s: pLips(t)= 0 + 0.0003jp e-0.2jpej400pt = 0.0003p e0.3jpej400pt Pascals • Acoustic wave reaches the listener’s ear 2ms later: pEar(t)= 0.0003p e0.3jpej400p(t-0.02) = 0.0003p e-j7.7pej400pt

13. Look Closer: x(t) = 0.001 ej400pt uGlottis(t) = 0.0005 ej400pt uLips(t) = 0.0005 e-0.2jpej400pt pLips(t) = 0.0003p e0.3jpej400pt pEar(t) = 0.0003p e-j7.7pej400pt • Amplitude, frequency, phase. Which one stays the same?

14. 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.

15. Definition of Phasor Notation • A phasor specifies the amplitude and phase of a cosine, but not its frequency. 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 } (Written in boldface if possible)

16. Example

17. 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 }

18. 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) • In phasor notation: • v = -jwx • x = Ae-jf v = -jw Ae-jf • In regular notation: • d2x/dt2 = -(k/m) x(t) • In phasor notation: • -w2x = -(k/m) x • Solution: w=√(k/m) !!!

19. 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 }