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ECE 598: The Speech Chain

ECE 598: The Speech Chain. Lecture 2: Sound Pressure Level; Stiffness and Mass. Review: The Five Fundamental Units (MKS). Length: 1 meter Mass: 1 kilogram Time: 1 second Temperature: 1 degree Kelvin Angle: 1 radian. Review: Useful Composite Units in Acoustics.

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ECE 598: The Speech Chain

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  1. ECE 598: The Speech Chain Lecture 2: Sound Pressure Level; Stiffness and Mass

  2. Review: The Five Fundamental Units (MKS) • Length: 1 meter • Mass: 1 kilogram • Time: 1 second • Temperature: 1 degree Kelvin • Angle: 1 radian

  3. Review: Useful Composite Units in Acoustics • Frequency: 1 Hz = 1 cycle/second (Hertz) • Audible sound is roughly 10 kHz > f > 10 Hz • Density: 1 kg/m3 = 1 mg/(cm)3 • Air is about 1.29 mg/(cm)3 • Force: 1 N = 1 kg m/s2 (Newtons) • The force of gravity on a pencil • Pressure: 1 Pa = 1 N/m2 (Pascals) • Air pressure at sea level is about 100,000 Pa • Audible pressures are usually 20mPa < p(t) < 20Pa • Energy: 1 J = 1 N m (Joules) • Burning 1kcal of sugar yields 4184 J • Power: 1 W = 1 J/s = 1 N m/s (Watts) • The heat given off by one human ~ 100W

  4. Frequency = 1/PeriodAngular Velocity = 1/(Time Constant) x(t) x(t) x(t) = cos(t) • Period: • T = 2p ~ 6.28 seconds/cycle • Frequency: • f = 1/T = 1/2p cycles/second • Angular Velocity: • w = 1 radian/second • Time Constant: • 1/w = 1 second/radian t t x(t) = 3cos(2t) • Period: • T = p ~ 3.14 seconds/cycle • Frequency: • f=1/T = 1/p cycles/second • Angular Velocity: • w = 2 radians/second • Time Constant: • 1/w = 0.5 seconds/radian

  5. Review: Root Mean Square (RMS) • x(t) = A cos(wt) • A is called the “amplitude” or “maximum amplitude” of x(t). • A has the same units as x(t). • The average value of a cosine is always E[x(t)]=0. x(t) t t • A more useful “average amplitude” is the “root mean square,” defined as the square root of the mean of the square of x(t), xRMS = √E[x2(t)] • For a cosine, xRMS turns out to be xRMS = A/√2 ≈ 0.707 A

  6. Review: Phase Shift x(t) t • “Phase Shift” and “Time Shift” are different words for exactly the same thing: x(t) = A cos( w(t-t) ) = A cos( wt – f ),f=wt • What are the units of f? (w radians/second) (t seconds) = f radians

  7. Review: Quadrature Phase Shifts cos(wt-p) = -cos(wt) !!! cos(wt-p/2) = sin(wt) !!! cos(wt-3p/2) = -sin(wt) !!! cos(wt-2p) = cos(wt) !!!

  8. Review: All the Calculus You’ll Ever Need to Know, on One Slide A, w, f, and a can be replaced by any expression, no matter how horrible and complicated, as long as the expression doesn’t contain any t.

  9. Review: x(t)=10t? x(t) (in meters) • Define the “natural logarithm” as follows: y=ln(x) if and only if x=ey • For example, 10=ey if and only if y = ln(10) ≈ 2.3 • So x(t) = 10t = (eln(10))t ≈ e2.3t is a special case of x(t)=eat, for the constant a=ln(10) ≈ 2.3, and its derivative is dx/dt = ln(10) eln(10)t = ln(10) 10t t x(t) = 10t = e(ln10)t dx/dt (in m/s) t dx/dt = ln(10) 10t

  10. Review: What is Sound? • When you hear a sound, it is because the air pressure at your ear drum is changing in small rapid fluctuations • Example: pure tone p(t) = P0 + A cos(wt) P0 = Atmospheric pressure ≈ 105 Pa (we will often leave this out of the equation) A = Sound pressure, in Pascals

  11. Review: The Semitone Scale • Let’s choose a global “reference note.” For example, we could choose A0, the lowest note on a piano. A0 rings at a frequency of f=22.5Hz, or w=2pf=55p radians/second. • Any other note, p(t)=cos(2pft), is N semitones above A1: N = . 12 (ln(f/22.5)) ln(2)

  12. Example: a Full-Tone Scale

  13. New Topic #1: Logarithms in General; Sound Pressure Level • Here’s how logarithms are defined in general: y=logbx if and only if x=by • The “natural logarithm” is just the “log base e” (e=2.718282…), because 3≈ln(20)=loge(20) because 20≈e3 • For other calculations, it sometimes makes sense to define the log base 10, e.g., 3=log10(1000) because 1000=103

  14. Loudness p(t) = P0 + A cos(wt) • Perceived pitch is mostly determined by w. • Perceived loudness is mostly determined by the pressure amplitude, A (in Pascals).

  15. What Loudness Differences Can People Hear? • As of 1900, the answer was: • If p1(t)=A1cos(wt), and p2(t)=A2cos(wt) (both in Pascals), most people can hear the difference in loudness if and only if A2/A1 > 1.12 (about a 12% increase) log10(A2/A1) > log10(1.12), which is 0.05 20 log10(A2/A1) > 1

  16. Signal level comparisons • The relative levels of two acoustic signals are compared in decibels. • Level can be computed either • From pressure amplitude, L = 20 log10( A2/A1), • or from the squared amplitude L = 10 log10( (A2/A1)2 ) = 10 log10( A22/A12 ) • Teaser: We will find out later that intensity (W/m2) is A2/z0 for some constant z0. What are the units of z0?

  17. Sound Pressure Level • Choose a reference amplitude, PREF=0.00002 Pa = 20mPa. This is almost, but not quite, the softest audible sound. • By convention, we compute the SPL of a sound using its RMS pressure (PRMS=A/√2 in the case of a cosine), instead of using “A” directly. • Then any sound p(t)=Acos(wt) is L dB louder than PREF, where L = 20 log10(PRMS/PREF) = 10 log10(PRMS2/PREF2) • L is the C-weighted sound pressure level (SPL) of p(t). • A-weighted SPL is almost the same formula, but sounds with less audible frequencies are discounted.

  18. Examples of Sound Pressure Level(see e.g. http://www.makeitlouder.com/Decibel%20Level%20Chart.txt)All numbers without distances are suspicious – level drops 6dB per doubling of distance.

  19. Review, Topic #1: Sound Pressure Level Humans can hear a difference in amplitude of about A2/A1 = 1.12 (12 percent, or 1dB) The relative levels of two signals are L2-L1 = 20 log10(A2/A1) = 10 log10(A22/A12) Sound Pressure Level = 20 log10(PRMS/PREF) = 10 log10(PRMS2/PREF2)

  20. Topic #2: Stiffness and Mass (and Newton’s Law) • Newton’s Second Law: f(t) = m dv/dt m = mass of an object v(t) = velocity dv/dt = acceleration • Definition of Linear Stiffness: f(t) = k(x0 – x(t)) x0 = “resting” position of the spring x(t) = current position of the spring k = spring constant / stiffness / Young’s modulus

  21. Spring-Mass System • Newton’s Second Law --- Mass on the end of a Spring (bouncing…): m dv/dt = k(x0 – x(t)) • If we can choose to set x0=0, then m dv/dt = -k x(t) • Since v(t)=dx/dt, we can write d2x/dt2 = - (k/m) x(t)

  22. Solution of the Spring-Mass System • Newton’s second law for the spring-mass system is: d2x/dt2 = –(k/m) x(t) • Is this equation true for any value of x(t)? No, it isn’t. The spring constant “k” and the mass “m” put some constraints on the possible positions x(t). But what kind of constraints? • Suppose somebody tells us that the solution is x(t) = A cos(wt-f), where A is in meters • Do we believe them? Can we tell what are the values of A, w, or f?

  23. Solution of the Spring-Mass System • Newton’s second law for the spring-mass system is: d2x/dt2 = –(k/m) x(t) • Let’s start by working out the derivatives: x(t) = A cos(wt-f) dx/dt = –w A sin(wt-f) = wA cos(wt–f+p/2) d2x/dt2 = (–w)(wA) sin(wt–f+p/2) = –w2A cos(wt–f)

  24. Testing the Solution of the Spring-Mass System • Newton’s second law for the spring-mass system is: d2x/dt2 = –(k/m) x(t) • Plug in x(t)=A cos(wt-f), and we get –w2A cos(wt–f) = –(k/m) A cos(wt-f) • So x(t)=A cos(wt-f) can only be the solution if w=√(k/m).

  25. Natural Frequency and Initial Conditions • The physical system determines a “natural” value of w. • w0=√(k/m) is called the natural frequency of the spring-mass system. • If there’s no friction, the spring can bounce forever at frequency w=w0 • (notice: Acos(wt-f) goes on forever…) • Everlasting bounces at any other frequency are impossible • If you start a bounce at any other frequency, it will quickly die away. • A and f are determined by the initial conditions. • Where did you start the spring? x(0)=Acos(–f). • How hard a push did you give? v(0)= –wAsin(–f)

  26. Review, Topic #2 • Newton’s Second Law: f(t) = m dv/dt • Definition of Linear Stiffness: f(t) = –k x(t) • Putting them together: d2x/dt2 = –(k/m) x(t) • The spring can bounce forever, as: x(t) = A cos(w0t-f), w0=√(k/m) Acos(–f) = x(0), the starting position in meters –wAsin(–f) = v(0), the starting velocity in m/s

  27. Topic #3: Circular Motion • Consider a girl going around a merry-go-round. • Her left-to-right position is x(t)=Acos(wt-f) • Her back-to-front position is y(t)=Asin(wt-f) • f = starting position (radians) • A = radius of the circle (meters) • w = angular velocity ( x(t),y(t) ) = (Acos(wt-f), Asin(wt-f)) y A f x

  28. y x(t) = sin(wt) x x(t) = cos(wt) Define the boldface notation: x(t) = (x(t),y(t)) For example, on this slide, x(t) = (cos(wt), sin(wt))

  29. y y(t) = sin(wt-p/2) (which equals –cos(wt)) x x(t) = cos(wt-p/2) (which equals sin(wt)) x(t) = (cos(wt-p/2), sin(wt-p/2))

  30. y y(t) = sin(wt+p/2) (which equals cos(wt)) x x(t) = cos(wt+p/2) (which equals -sin(wt)) x(t) = (cos(wt+p/2), sin(wt+p/2))

  31. Imagining a Circle • Remember that sound is an air pressure fluctuation: p(t) = A cos(wt-f) • It is sometimes convenient to imagine a circle: p(t) = ( Acos(wt-f), Asin(wt-f) )

  32. Imagining a Circle • A cos(wt-f) is the “real part” of p(t) Re{p(t)} = A cos(wt-f) Usually, the “real part” is the measurable pressure signal – the bit that exists in the real world. • A sin(wt-f) is the “imaginary part” of p(t) Im{p(t)} = A sin(wt-f) Usually, the “imaginary part” is a convenient fiction.

  33. Review, Topic #3: Circular Motion • A little girl going in a circle, starting at position f, moves according to x(t) = ( Acos(wt-f), Asin(wt-f) ) • Often it will be convenient for us to imagine that the sound wave, p(t)=Acos(wt-f), is the “real part” of a circular motion: p(t) = ( Acos(wt-f), Asin(wt-f) ) p(t) = Re{p(t)}

  34. Topic #4: Euler’s Identity • There is a standard way to write circular motion (called “Euler’s Identity”): Aej(wt-f) = ( Acos(wt-f), Asin(wt-f) ) • … where we define the new variable to be j=√-1

  35. OK, Here’s what we’re about to do. • The only thing you need to know about Euler’s identity, this week, is the definition: • Aej(wt-f) = (Acos(wt-f), Asin(wt-f)). • But eventually (future weeks), you’ll want to know why anyone would ever propose such a bizarre-looking definition. • The next three slides will tell you why. • Summary: this definition makes the spring-mass system very easy to solve. • Warning: proving that it works is kind of messy. • Reason you shouldn’t panic: you don’t need to be able to prove it yourself. So just sit back and watch the show.

  36. The Meaning of j=√-1:The second derivative of ejwt is negative,just like the second derivative of cos(wt). • Remember the derivatives of cosine: d cos(wt)/dt = -w sin(wt) d2 cos(wt)/dt2 = -w2 cos(wt) • Here are the derivatives of eat: d eat/dt = aeat d2 eat/dt2 = a2 eat • Here’s what happens if a=jw: D ejwt/dt = jw ejwt d2 ejwt/dt2 = j2w2 ejwt = -w2 ejwt Just like a cosine!!!

  37. … so the derivatives of Euler’s identity work d2 ( Acos(wt-f), Asin(wt-f) )/dt2 = -w2 ( Acos(wt-f), Asin(wt-f) ) d2 Aej(wt-f) /dt2 = -w2 Aej(wt-f) d2x/dt2 = -w2x(t) Isn’t that the simplest second derivative you’ve ever seen? That is the whole, total, entire purpose of Euler’s Identity.

  38. Newton’s Law for a Spring, Redux d2x/dt2 = –(k/m) x(t) if x(t)=Aej(wt-f), then –w2 x(t) = –(k/m) x(t) w=√(k/m)

  39. Review, Topic #4 • ejwt is a shorthand notation for the path a little girl takes while going around a merry-go-round: Aej(wt-f)= ( Acos(wt-f), Asin(wt-f) ) • Sound is often written as p(t) = Re{p(t)}, p(t)=Aej(wt-f) • It seems wasteful to invent an “imaginary part” of the acoustic wave (isn’t the real part already complicated enough?), but … • The spring-mass system (and the acoustic wave equation!!) are easier to solve using p(t) than using p(t). • When we’re done solving using p(t), we take the real part to get p(t)=Re{p(t)} back again. It works!

  40. Review: This Week • Decibels: • Humans can hear level differences of 1dB • SPL = 20log10(PRMS/0.00002 Pa). • Stiffness and Mass • Newton’s Law for a spring-mass system: • d2x/dt2= – k x(t) • The solution: x(t)=Acos(w0t-f), for w0=√(k/m) • Circular Motion • x(t)=(x(t),y(t)) = (Acos(wt-f), Asin(wt-f)) • A compact way to write that: x(t)=Aej(wt-f)

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