fourier analysis, part ii
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Fourier Analysis, part II. March 1, 2013. The Lay of the Land. I finally graded the second TOBI production exercise! Let’s check out the all-star team… Also: the first mystery spectrogram has been posted! A Fourier Analysis homework will be handed out after the weekend… - PowerPoint PPT PresentationTRANSCRIPT
Fourier Analysis, part II
March 1, 2013
The Lay of the Land• I finally graded the second TOBI production exercise!
• Let’s check out the all-star team…
• Also: the first mystery spectrogram has been posted!
• A Fourier Analysis homework will be handed out after the weekend…
• For now, you probably have enough to work on.
• Today’s goal: further down the rabbit hole of Fourier Analysis.
• Any questions so far?
Quick and Dirty Review• The three main elements of Fourier analysis:
1. A complex wave
2. Its component waves (the “harmonics”)
3. Its potential component waves
• = the “reference waves” from the last lecture
• A complex wave can be created by simply adding together the component waves.
• Normally when we do Fourier analysis, though, we want to find out what the component waves are.
• The last element is our primary tool: the dot product.
DFT, so far1. “Window” the signal
• = break it into smaller chunks
2. Smooth the window reduce the edges to 0.
• With the algorithm of your choice!
3. Determine the components of the smoothed chunk
• Calculate the dot product of the chunk with sine and cosine waves of likely component frequencies
• Non-components dot product = 0
4. Determine the amplitude of each component
• = dot product / power of the component
• Power = dot product of a wave with itself
Let’s Try Another• Let’s construct another example: 1 Hz sinewave + a 4 Hz cosine wave with half the amplitude.
1 2 3 4 5 6 7 8
A 1 Hz 0 .707 1 .707 0 -.707 -1 -.707
.5*B 4 Hz .5 -.5 .5 -.5 .5 -.5 .5 -.5
E Sum: .5 .207 1.5 .207 .5 -1.207 -.5 -1.207
• Let’s check the 1 Hz wave first:
E Sum: .5 .207 1.5 .207 .5 -1.207 -.5 -1.207
A 1 Hz 0 .707 1 .707 0 -.707 -1 -.707
E*A Dot: 0 .146 1.5 .146 0 .854 .5 .854
• Sum = 4
Yet More Dots• Another example: 1 Hz sinewave + a 4 Hz cosine wave with half the amplitude.
• Now let’s check the 4 Hz wave:
E Sum: .5 .207 1.5 .207 .5 -1.207 -.5 -1.207
B 4 Hz 1 -1 1 -1 1 -1 1 -1
E*B Dot: .5 -.207 1.5 -.207 .5 1.207 -.5 1.207
• The sum of these products is also 4.
• = half of the power of the 4 Hz cosine wave.
• The 4 Hz component has half the amplitude of the 4 Hz cosine reference wave.
• (we know the reference wave has amplitude 1)
Mopping Up, Part 2• Our component analysis gave us the following dot products:
• E*A = 4 (A = 1 Hz sinewave)
• E*B = 4 (B = 4 Hz cosine wave)
• Let’s once again normalize these products by dividing them by the power of the “reference” waves:
• power (A) = A*A = 4 E*A/A*A = 4/4 = 1
• power (B) = B*B = 8 E*B/B*B = 4/8 = .5
• These ratios are the amplitudes of the component waves.
• The 1 Hz sinewave component has amplitude 1
• The 4 Hz cosine wave component has amplitude .5
Footnote• Sinewaves and cosine waves are orthogonal to each other.
• The dot product of a sinewave and a cosine wave of the same frequency is 0.
1 2 3 4 5 6 7 8
A sin 0 .707 1 .707 0 -.707 -1 -.707
F cos 1 .707 0 -.707 -1 -.707 0 .707
A*F Dot: 0 .5 0 -.5 0 .5 0 -.5
• However, adding cosine and sine waves together simply shifts the phase of the complex wave.
• Check out different combos in Praat.
Problem #1• For any given window, we don’t know what the phase
shift of each frequency component will be.
• Solution:
1. Calculate the correlation with the sinewave
2. Calculate the correlation with the cosine wave
3. Combine the resulting amplitudes with the pythagorean theorem:
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At = Asin2 + Acos
2
• Take a look at the java applet online:
• http://www.phy.ntnu.edu/tw/ntnujava/index.php?topic=148
Sine + Cosine Example• Let’s add a 1 Hz cosine wave, of amplitude .5, to our previous combination of 1 Hz sine and 4 Hz cosine waves.
1 2 3 4 5 6 7 8
C 1+4: 1 -.293 2 -.293 1 -1.707 0 -1.707
.5*F cos .5 .353 0 -.353 -.5 -.353 0 .353
G Sum: 1.5 .06 2 -.646 .5 -2.06 0 -1.353
• Let’s check the 1 Hz sine wave again:
G Sum: 1.5 .06 2 -.646 .5 -2.06 0 -1.353
A 1 Hz 0 .707 1 .707 0 -.707 -1 -.707
G*A Dot: 0 .043 2 -.457 0 1.457 0 .957
• Sum = 4
Sine + Cosine Example• Now check the 1 Hz cosine wave:
G Sum: 1.5 .06 2 -.646 .5 -2.06 0 -1.353
F 1 Hz 1 .707 0 -.707 -1 -.707 0 .707
G*F Dot: 1.5 .043 0 .457 -.5 1.457 0 -.957
• Sum = 2
• Sinewave component amplitude = 4/4 = 1
• Cosine wave component amplitude = 2/4 = .5
• Total amplitude =
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(1*1) + (.5* .5) =1.118
• Check out the amplitude of the combo in Praat.
In Sum• To perform a Fourier analysis on each (smoothed) chunk of
the waveform:
1. Determine the components of each chunk using the dot product—
• Components yield a dot product that is not 0
• Non-components yield a dot product that is 0
2. Normalize the amplitude values of the components
• Divide the dot products by the power of the reference wave at that frequency
3. If there are both sine and cosine wave components at a particular frequency:
• Combine their amplitudes using the Pythagorean theorem.
Hold On A Second...• What would happen if our window length was 7 samples long, instead of 8?
• Back to the 1 Hz and 4 Hz wave combo:
1 2 3 4 5 6 7
Sum: 1 -.293 2 -.293 1 -1.707 0
2 Hz 0 1 0 -1 0 1 0
Dot: 0 -.293 0 .293 0 -1.707 0
• The sum of these products is -1.707, not 0. (!?!)
• The Fourier approach can only identify component sinewaves that can fit an integer number of cycles into the window.
Frequency Range• Q: What frequencies can we consider in the Fourier analysis?
• One possible (but unrealistic) setup:
• A window length of .25 seconds
• A sampling rate of 20,000 Hz
• (Note: 5,000 samples fit into a window)
• Longest possible period in window = .25 seconds, so:
• Lowest frequency component = 1 / 0.25 = 4 Hz
• Nyquist frequency = 10,000 Hz.
• A: We can check all frequencies from 4 to 10,000, in steps of 4 Hz.
• (10,000 / 4 = 250 possible frequencies)
Frequency Range, Part 2• Q: What frequencies can we consider in the Fourier analysis?
• Another, more realistic possible setup:
• A window length of .005 seconds
• A sampling rate of 20,000 Hz
• (Note: 100 samples fit into a window)
• Longest period = .005 seconds, so:
• Lowest frequency component = 1 / .005 = 200 Hz!
• Nyquist frequency = 10,000 Hz.
• A: from 200 to 10,000, in steps of 200 Hz.
• (10,000 / 200 = 50 possible frequencies)
Zero Padding• With short window lengths, we miss out on a lot of interesting frequencies…
• The solution is to “pad” the window with zeroes, until it’s long enough to enable us to look at an interesting frequency range.
• Example:
1 2 3 4 5 6 7 8
Sum: 1 -.293 2 -.293 1 -1.707 0 0
• Q: What effect do you think this would have on the power spectrum?
• Component frequencies have a reduced amplitude.
• Non-component frequencies have a non-zero amplitude.
Industrial Smoothing• Zero-padding “smooths” the spectrum.
• Spectral analysis of complex wave formed by 1 Hz and 4 Hz waves, with an 8 Hz sampling rate:
8 sample window 7 sample window, with zero padding
0
0.2
0.4
0.6
0.8
1
1 2 3 4
Frequency (Hz)
Amplitude
0
0.2
0.4
0.6
0.8
1
1 2 3 4
Frequency (Hz)
Amplitude
Another Example• Q: What would happen if we padded the window out to 16 samples?
• A: More frequencies we can check (resolution = .5 Hz)
• Also: even more smoothing
• What would happen if we increased the sampling rate?
• Upper end of analyzable frequency range increases
• ( higher Nyquist frequency) 7 sample window, with zero-
padding, 16 Hz sampling rate
0
0.1
0.2
0.3
0.4
0.5
0.6
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8
Frequency (Hz)
Amplitude
Trade-Offs• What happens if we increase the window length?
• (independent of zero padding)
• A: Increase the maximum analyzable period, so:
• Better frequency resolution
• ...without the smoothing.
• However:
• Temporal resolution is worse.
• (because the window length is less precise)
• Check it out in Praat.
Morals of the Fourier Story• Shorter windows give us:
• Better temporal resolution
• Worse frequency resolution
• = wide-band spectrograms
• Longer windows give us:
• Better frequency resolution
• Worse temporal resolution
• = narrow-band spectrograms
• Higher sampling rates give us...
• A higher limit on frequencies to consider.