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.