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), 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 d 2 x/dt 2 = A d 2 f/dt 2 + B d 2 g/dt 2 Example: Example: x(t)=cos(  t-  )  dx/dt = –  sin(  t-  ) x(t)=Acos(  t-  )  dx/dt = –  Asin(  t-  )

3. Review: Spring Mass System Newton’s Second Law: Newton’s Second Law: f(t) = m d 2 x/dt 2 Force of a Spring: Force of a Spring: f(t) = – k x(t) The Spring-Mass System Equation with no external forces: The Spring-Mass System Equation with no external forces: d 2 x/dt 2 = - (k/m) x(t)

4. Solution: Cosine First and Second Derivatives of a Cosine: First and Second Derivatives of a Cosine: x(t) = cos(  t-  ) dx/dt = –  sin(  t-  ) d 2 x/dt 2 = –  2 cos(  t–  ) Spring-Mass System: Spring-Mass System: d 2 x/dt 2 = –(k/m) x(t) –  2 cos(  t–  ) = – (k/m) cos(  t-  ) It only works at the “natural frequency:” It only works at the “natural frequency:”  0  √(k/m) Linearity means that you can multiply by “A” here, if you want to, and the same “A” will appear here.

5. The Other Function We Know First and Second Derivatives of an Exponential: First and Second Derivatives of an Exponential: x(t) = e  t dx/dt =  e  t d 2 x/dt 2 =  2 e  t Could x(t)= e  t also solve the spring-mass system? Could x(t)= e  t also solve the spring-mass system? d 2 x/dt 2 = –(k/m) x(t)  2 e  t = – (k/m) e  t It only works if: It only works if:  √(-k/m)

6. Imaginary Numbers Definition (the part that you memorize): Definition (the part that you memorize): j  √-1 Linearity: Linearity: 2j  √-4  √4  √-1 3j  √-9  √9  √-1 Solution to the spring-mass system: Solution to the spring-mass system:  √(-k/m)  √-1  √(k/m)  j   x(t) = e  t = e j  t 0

7. Complex Exponentials Definition (another bit to memorize): Definition (another bit to memorize): x(t) = e j(  t  ) = cos(  t  ) + j sin(  t  ) Then the “real part” of x(t) is defined to be: Then the “real part” of x(t) is defined to be: Re{x(t)} = cos(  t  ) So x(t)=e j  t and x(t)=cos(  t) are actually exactly the same solution! So x(t)=e j  t and x(t)=cos(  t) are actually exactly the same solution! The “imaginary part,” Im{x(t)}=sin(  t), doesn’t change the solution. The “imaginary part,” Im{x(t)}=sin(  t), doesn’t change the solution. If you like, visualize x(t) as a movement in two dimensions: If you like, visualize x(t) as a movement in two dimensions: Re(x(t)) is movement in the horizontal direction Re(x(t)) is movement in the horizontal direction Im(x(t)) is movement in Buckaroo Banzai’s mysterious 8 th dimension. Im(x(t)) is movement in Buckaroo Banzai’s mysterious 8 th 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. 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! Answer: you can’t! What you CAN do: plot either the real part or the imaginary part What you CAN do: plot either the real part or the imaginary part x(t) = e j2t Re{x(t)} = cos(  t) Im{x(t)} = sin(  t)

9. Special Numbers e j  = cos(  ) + j sin(  ) 1 = cos(  ) + j sin(  ) = e j0 j = cos(  ) + j sin(  ) = e j  /2  1 = cos(  ) + j sin(  ) = e j  (e j  /2 ) 2 = e j  (j) 2 =  1

10. Linearity Again “Real Part” and “Imaginary Part” are linear operators: “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 Example x(t) = 2 e j  t + 3 e j(  t  ) Re{x(t)} = 2cos(  t) + 3cos(  t  ) Im{x(t)} = 2sin(  t) + 3sin(  t  )

11. Amplitude, Phase, and Frequency of Cosines vs. Exponentials Exponential: Exponential: x(t) = Ae j(  t  ) = A e  j  e j  t Cosine: Cosine: Re{x(t)} = A cos(  t  )

12. The Life Story of a Cosine Vocal fold motion is a cosine at (say)  400 , with amplitude of A=0.001m: Vocal fold motion is a cosine at (say)  400 , with amplitude of A=0.001m: x(t) = 0.001 + 0.001 e j400  t m Notice this looks like an exponential, but it’s “really” a cosine. Notice this looks like an exponential, but it’s “really” a cosine. Re{x(t)} = 0.001 + 0.001cos(400  t) Air puffs come through when the vocal folds are open, with a maximum rate of 0.001 m 3 /second: Air puffs come through when the vocal folds are open, with a maximum rate of 0.001 m 3 /second: u Glottis (t)=0.0005+0.0005e j400  t m 3 /s The same air puffs reach the lips 0.5ms later: The same air puffs reach the lips 0.5ms later: u Lips (t)=0.0005+0.0005e j400  (t  0.0005) =0.0005 + 0.0005 e  0.2j  e j400  t liter/s Air pressure at the lips is the derivative of u(t), times 0.003 kg/s: Air pressure at the lips is the derivative of u(t), times 0.003 kg/s: p Lips (t)= 0 + 0.0003j  e  j  e j400  t = 0.0003  e  j  e j400  t Pascals Acoustic wave reaches the listener’s ear 2ms later: Acoustic wave reaches the listener’s ear 2ms later: p Ear (t)= 0.0003  e 0.3j  e j400  t  0.02) = 0.0003  e  j7.7  e j400  t

13. Look Closer: x(t) = 0.001 e j400  t u Glottis (t) = 0.0005 e j400  t u Lips (t) = 0.0005 e  0.2j  e j400  t p Lips (t) = 0.0003  e  j  e j400  t p Ear (t) = 0.0003  e  j7.7  e j400  t Amplitude, frequency, phase. Which one stays the same? Amplitude, frequency, phase. Which one stays the same?

14. Phasor Notation x = 0.001 u Glottis = 0.0005 u Lips = 0.0005 e  0.2j  p Lips = 0.0003  e  j  p Ear = 0.0003  e  j7.7  Phasor notation: write down only the amplitude and phase, not the frequency. 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. A phasor specifies the amplitude and phase of a cosine, but not its frequency. x = A e j  To get x(t) back, you look back through the problem definition in order to find , then write To get x(t) back, you look back through the problem definition in order to find , then write x(t) = Re{ x e j  t } (Written in boldface if possible)

16. Example Phasor x, u, p Complex x(t), u(t), p(t) Real Vocal Fold x(t) 10 10 e j400  t 10cos(400  t) Vocal Fold u(t) 5 5 e j400  t 5 cos(400  t) Lips u(t) 5 e  0.2j  5 e  0.2j  e j400  t 5 cos(400  t  0.2  Lips p(t) 3  e  j  3  e  j  e j400  t 3  cos(400  t+  Ear p(t) 3  e  j7.7  3  e  j7.7  e j400  t 3  cos(400  t  7.7 

17. The Main Purpose of Phasors: They Turn Derivatives into Multiplication x(t) = Re{ x e j  t } x(t) = Re{ x e j  t } dx/dt = Re{ j  x e j  t } dx/dt = Re{ j  x e j  t } d 2 x/dt 2 = Re{ (j  ) 2 x e j  t } = Re{  2 x e j  t } d 2 x/dt 2 = Re{ (j  ) 2 x e j  t } = Re{  2 x e j  t }

18. The Main Purpose of Phasors: They Turn Derivatives into Multiplication In regular notation: In regular notation: v(t) = dx/dt v(t) = dx/dt x(t) = A cos(  t  )  v(t) =  A sin(  t  ) x(t) = A cos(  t  )  v(t) =  A sin(  t  ) In phasor notation: In phasor notation: v =  j  x v =  j  x x = Ae  j   v =  j  Ae -j  x = Ae  j   v =  j  Ae -j  In regular notation: In regular notation: d 2 x/dt 2 =  (k/m) x(t) d 2 x/dt 2 =  (k/m) x(t) In phasor notation: In phasor notation:  2 x =  (k/m) x  2 x =  (k/m) x Solution:  =√(k/m) !!! Solution:  =√(k/m) !!!

19. Summary Linearity means you can: Linearity means you can: add two solutions, or add two solutions, or scale by a constant. scale by a constant. x(t)=e  t could solve the spring-mass system, but only if  =√(-k/m) x(t)=e  t could solve the spring-mass system, but only if  =√(-k/m) j 2 =  1 j 2 =  1 e  j  = cos(  ) + j sin(  ) e  j  = cos(  ) + j sin(  ) We don’t really need the imaginary part; Re{e j  t } = cos(  t) We don’t really need the imaginary part; Re{e j  t } = cos(  t) We don’t really need the frequency part either; it is never changed by any linear operation (e.g., scaling, time shift, derivative) 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: Phasor notation encodes just the amplitude and phase of a cosine: x = Ae  j  To get back to the time domain: To get back to the time domain: x(t) = Re{ xe j  t }