CS 591 S1 Computational Audio -- Spring, 2017

Transcription

1 CS 591 S1 Computational Audio -- Spring, 2017 Wayne Snyder Department Boston University Lecture 11 (Tuesday) Discrete Sine Transform Complex Numbers Complex Phasors and Audio Signals Lecture 12 (Thursday) Discrete Fourier Transform in Complex Case Fast Fourier Transform Digital Audio Fundamentals: Multiplying/Squaring Signals Recall: For a window of length W samples, a window frequency is one whose period P is such that W = P * k for some integer k, i.e., an integral number of periods exactly fit within the window; alternately, it begins and ends at same instantaneous phase. We will use these signals as probe waves to analyze a musical signal and assume that all such probe waves (for now) start at phase

2 Digital Audio Fundamentals: Multiplying/Squaring Signals Recall: What happens when the signal is composite (not a simple sine wave)? Let s track the average sample value when multiplying a composite wave by a probe wave created from [ ( f, 1.0, 0.0 ) ] for various frequencies f X = makesignal( [ ( 3.0,0.5,0.0), (5.0,0.3,0.0), (10,0.2,0.0) ], 1.0) Z = multsignals(x,p) P = makesignal( [ ( 2.0,1.0,0.0)], 1.0) µ = dotproduct(x,p) / len(x) = avg sample so µ ~ Digital Audio Fundamentals: Multiplying/Squaring Signals Recall: What happens when the signal is composite (not a simple sine wave)? Let s track the average sample value when multiplying a composite wave by a probe wave created from [ ( f, 1.0, 0.0 ) ] for various frequencies f X = makesignal( [ ( 3.0,0.5,0.0), (5.0,0.3,0.0), (10,0.2,0.0) ], 1.0) Z = multsignals(x,p) P = makesignal( [ ( 3.0,1.0,0.0)], 1.0) µ = dotproduct(x,p) / len(x) = == 3.0 so µ >

3 Digital Audio Fundamentals: Multiplying/Squaring Signals If we graph the results of multiplying this composite signal by various probe waves, we get the following: [ ( 3.0,0.5,0.0), (5.0,0.3,0.0), (10,0.2,0.0) ] Avg Sample Frequency of probe wave Why are the amplitudes reported exactly half as big as the component amplitudes? 5 Digital Audio Fundamentals: Multiplying/Squaring Signals Well, let s look at the square of the probe wave: Component: (3.0, 0.5, 0.0) Probe: (3.0, 1.0, 0.0) Recall: sin(x)*sin(y) = ½ [ cos(x-y) cos(x+y) ] so A 1 *sin(x)*a 2 * sin(x) = (A 1 *A 2 ) / 2 * [ cos(x-x) cos(x+x) ] In this case: (0.5*1.0) / 2 * [ cos(0) cos(2x) ] = *cos(2x) 0.25 average sample of cos(2x) = 0.0 so average of product wave = center line of product sine wave = 0.25 Punchline: To get amplitudes of components, use probe waves of amplitude 1.0, and multiply the average sample value of the product wave by

4 Digital Audio Fundamentals: Discrete Sine Transform Doing this consistently for all window frequencies gives us X = makesignal( [ ( 3.0,0.5,0.0), (5.0,0.3,0.0), (10,0.2,0.0) ], 1.0) S = DST( X ) S[0]: 0.0 S[1]: e-12 S[2]: e-12 S[3]: S[4]: e-12 S[5]: S[10]: S[11]: e-12 S[22049]: e-12 Why len(s) = W // 2?? Last frequency is W // 2 1 = / MAX_AMP = / MAX_AMP = / MAX_AMP = Digital Audio Fundamentals: Discrete Sine Transform Doing this consistently for all window frequencies gives us Note: The transform can ONLY detect window frequencies = k * f for f = 1 / W (in secs) So a window of 1.0 seconds can detect 0, 1, 2,., of 0.1 seconds can detect 0, 10, 20, 30,., of 0.2 seconds can detect 0, 5, 10,., ONLY Another problem is that this took 25 minutes to run! Double for loop with W = * = 972,405,000 executions of inner loop! 8 4

5 Digital Audio Fundamentals: Discrete Sine Transform Doing this consistently for all window frequencies gives us X = makesignal( [ ( 30.0,0.5,0.0), (50.0,0.3,0.0), (100.0, 0.2,0.0) ], 0.1) S = DST( X ) Bin Absolute Amp Freq Relative Amp S[0]: S[1]: e e-16 S[3]: S[4]: e e-16 S[5]: S[10]: S[2204]: e e-17 This took about 15 seconds to run 9 Digital Audio Fundamentals: Discrete Sine Transform Doing this consistently for all window frequencies gives us X = makesignal( [ ( 30.0,0.5,0.0), (50.0,0.3,0.0), (100.0, 0.2,0.0) ], 0.2) S = DST( X ) Bin Absolute Amp Freq Relative Amp S[0]: S[1]: e e-17 S[6]: S[10]: S[20]: S[4409]: e e-16 This took about 1 minute to run 10 5

6 def dst( X ): W = len(x) S = [0] * (W//2) # spectrum for f in range(w//2): # for each probe wave f in [0..N//2] for i in range(w): # S[f] = sum of product of X and probe S[f] += X[i] * sin(2 * pi * f * i / N) S[f] = S[f] / (W/2) # normalize to get actual amplitude return S Returns a spectrum of absolute amplitudes (in range -32K to 32K ) S = [ A 0, A 1, A 2,., A N//2-1 ] assuming w is even for window frequencies W f = [ 0, 1, 2,., N//1 1 ] and actual frequencies F = [ 0, 1R, 2R,., R*(N//2 1) ] for R = SampleRate / W Spectrum: [ ( 880, 0.8, 0 ), (1760, 0.6, 0), (2640, 0.4, 0) ] the Discrete Sine DST( X ) => Spectrum S of length len(x)//2 S[ f ] = amplitude of frequency component of sine wave at window freq f. 88 * SR / W = 880 Hz / = * SR / W = 1760 Hz / = 0.6 Freq in Hz = f * SampleRate / W W = * SR / W = 2640 Hz / = 0.4 6

7 Spectrum: [ ( 880, 0.8, 0 ), (1760, 0.6, 0), (2640, 0.4, 0) ] the Discrete Sine 88*SR/W = 880 Hz /32767 = *SR/W = 1760 Hz /32767 = *SR/W = 2640 Hz /32767 = 0.4 Spectrum: [ ( 880, 0.8, 0 ), (1760, 0.6, 0), (2640, 0.4, 0) ] the Discrete Sine 88*SR/W = 880 Hz /32767 = *SR/W = 1760 Hz /32767 = *SR/W = 2640 Hz /32767 = 0.4 7

8 Spectrum: [ ( 880, -0.8, 0 ), (1760, -0.6, 0), (2640, 0.4, 0) ] Component sine waves may have a negative amplitude. Component sine waves may have a negative amplitude; they will produce the negative of a squared wave, and report negative amplitudes just as they report positive amplitudes. 8

9 The same effect can be gotten by delaying the phase by pi or by using a negative frequency: all will produce negative amplitudes. Spectrum: [ ( 880, 0.8, pi ), (1760, 0.6, 0), (2640, 0.4, pi) ] the Discrete Sine Delaying a component by phase pi produces negative amplitudes. 9

10 Spectrum: [ ( 880, 0.8, 0 ), (-1760, 0.6, 0), (-2640, 0.4, 0) ] Negative frequencies produce negative amplitudes. Spectrum: [ ( 880, -0.8, 0 ), (-1760, -0.6, 0), (-2640, -0.4, pi) ] Doing combinations of these will flip the amplitude back and forth: 10

11 Spectrum: [ ( 880, 0.8, 0 ), (1760, 0.6, 0), (2640, 0.4, 0) ] What happens when we look at the spectrum past the Nyquist Limit, up to Hz?? Spectrum: [ ( 880, 0.8, 0 ), (1760, -0.6, 0), (2640, 0.4, 0) ] What happens when we look at the spectrum past the Nyquist Limit, up to Hz?? 11

12 Spectrum: [ ( -880, 0.8, 0 ), (-1760, -0.6, 0), (2640, 0.4, pi) ] What happens when we look at the spectrum past the Nyquist Limit, up to Hz?? Spectrum: [ ( 880, 0.8, 0 ), (1760, 0.6, 0), (2640, 0.4, 0) ] Testing for frequencies above the Nyquist limit will find the aliases The frequencies above the Nyquist Limit appear in reverse order with half amplitude. They have half amplitude because we normalized by len(x) instead of len(x)/2. 12

13 Digital Audio Fundamentals: Discrete Sine Transform There are three problems (so far): (1) This is horribly inefficient: O( N 2 ) for N = len(x) Solution: There will be a fast version of the transform presented next time, based on a recursive algorithm O( N log(n) ). (2) The resolution is limited to multiples of f Hz = 1 / W (secs) No solution, unfortunately, can try different window sizes, but stuck with this! (3) All components and probe waves have to be at the same phase (e.g., 0.0) Solution: If we do all the work with complex numbers, we can avoid issues of phase So onto Complex Numbers 25 13