Using the Discrete Fourier Transform · sample frequency is 11,025 samples/second. Listen to the...

6.003: Signal Processing

Using the Discrete Fourier Transform

March 14, 2019

Frequency Resolution

How many Fourier components are in a signal?

What are their amplitudes and frequencies?

Example: cello

Listen to the wav file DEb3.wav (fs = 44,100 Hz).

How can we characterize the frequency components?


Extract 1024 samples and calculate DFT (using FFT algorithm).

x[n]∣∣X1[k]

∣∣

Does it matter which part of the wav file we take the samples from?

Why use 1024 samples?

Would more be better? Would fewer be better?


Most of structure is in low frequencies. Zoom in on that part.

x[n]∣∣X1[k]

∣∣

What value of k corresponds to the big peak?

What is corresponding value of frequency (in Hz)?


Most of structure is in low frequencies. Zoom in on that part.

x[n]∣∣X1[k]

∣∣

What value of k corresponds to the big peak? 7What is corresponding value of frequency (in Hz)?

koN/2 = fo

fs/2

fo = koN fs = 7

1024 × 44100 ≈ 301.46 Hz:

between D (293.66 Hz) and E flat (311.13 Hz)

Improving Frequency Resolution

The DFT provides integer resolution in k. Therefore, the peak at

k = 7 could be off by as much as ±12 .

To improve frequency resolution, increase length of analysis window.

Two methods

• Zero padding

− better assessment of peak frequency.

− does not improve resolution of multiple components

• Longer sequence

− improves resolution of multiple components

− averages frequency content over longer period (bad if signal

changes with time)

Zero Padding

Original (N=1024).

x1[n]∣∣X1[k]

∣∣

Zero Padding

Lengthen by a factor of 2 (N=2048).

x2[n]∣∣X2[k]

∣∣

What (if any) relations exist between |X2[k]| (this slide) and |X1[k]|(previous slide)?

Zero Padding


x2[n]∣∣X2[k]

∣∣

What (if any) relations exist between |X2[k]| (this slide) and |X1[k]|(previous slide)?

X2[2k] = 12X1[k]

Zero Padding


x3[n]∣∣X3[k]

∣∣

Zero Padding


x4[n]∣∣X4[k]

∣∣

The stem plots can be distracting when they are close together.

(They also take a long time to compute!)

Replot using lines (but remember that the signals are DT).

Zero Padding

Original (N=1024).∣∣X1[k]∣∣ ∣∣X1[k]

∣∣

Zero Padding

Lengthen by a factor of 2 (N=2048).∣∣X2[k]∣∣ ∣∣X2[k]

∣∣

Zero Padding


∣∣

Zero Padding


∣∣

Peak is now at k = 55 suggesting that analysis for N = 1024 (i.e.,

k = 7) was only approximate. Here we would get k = 558 = 6.875.

Conclusion

Padding with zeros allowed us to compute more samples in fre-

quency.

The greater number of samples better illustrate the shape of the

window function that governs the between-frequency interpolation.

Regardless of the analysis width, resolution of the signal is deter-

mined by a square window of length 1024, since there are only 1024

samples of data available in any of the preceding analyses.

In order to increase frequency resolution, we need to include more

data.

Longer Sequence

Original (N=1024).

x5[n]∣∣X5[k]

∣∣

Longer Sequence


x6[n]∣∣X6[k]

∣∣

Longer Sequence


x7[n]∣∣X7[k]

∣∣

Longer Sequence


x8[n]∣∣X8[k]

∣∣

Switching again to line plots ...

Longer Sequence

Original (N=1024).∣∣X5[k]∣∣ ∣∣X5[k]

∣∣

Longer Sequence


∣∣

Longer Sequence

Lengthen by a factor of 16 (N=16,384).∣∣X9[k]∣∣ ∣∣X9[k]

∣∣

Longer Sequence

Lengthen by a factor of 32 (N=32,768).∣∣X10[k]∣∣ ∣∣X10[k]

∣∣

Clear peaks at k = 217 and k = 228 (f = 292.04 Hz and f = 306.85 Hz).

Notice that these are the second harmonics of lower frequencies.

The fundamental components were not clearly resolved with

N=1024 but are clear with N=32,768.

Summary: Frequency Resolution

Increasing the length of the analysis increases the number of fre-

quencies that result.

Increasing the length of the analysis by zero padding can better

illustrate the shape of the window function that governs the between-

frequency interpolation.

However, it does not increase frequency resolution.

To increase frequency resolution we must include more data.

Upsampling Time

We can similarly increase time resolution.

Example: Create a wav file to generate a tone at 441 Hz when the

sample frequency is 11,025 samples/second.

Listen to the wav file, and check that it sounds right.

Our goal is to scale the sound so that the new sample frequency

is 44,100 samples/second while preserving all of the same audible

frequency content.

Upsampling Time

Let x[n] represent the sampled signal.

fs = 11025f = 441Omega = 2*pi*f/fsN = 2**14x441 = [cos(Omega*n) for n in range(N)]wav_write(x441,fs,’x441.wav’)

Plot the first 25 samples of the signal.

Plot the magnitude of the DFT of these samples.

Upsampling Time

x441[n]∣∣X441[k]

∣∣

Upsampling Time

To increase frequency resolution, we increased the length of the

time signal (either by padding or including more samples).

To increase time resolution, stretch the time signal (inserting zeros).

Stretch x441[n] by a factor of 4:

y411 = []for n in range(len(x411)):

y411.append(x411[n])y411.append(0)y411.append(0)y411.append(0)

Upsampling Time

Plot the DFT of y[n].

y441[n]∣∣Y 441[k]

∣∣

Now there are 8 components instead of 2. Why?

What are the frequencies of other components?

Upsampling Time

Zero out the unwanted components and take the idft.

z441[n]∣∣Z441[k]

∣∣

Write to wav file and listen. It should sound exactly like the original

but it’s sampled 4 times faster.

Upsampling Time

Try it on an actual piece of audio.

x,fs = wav_read(’flights.wav’)x += (2**15-len(x))*[0]y = []for n in range(len(x)):

y.append(x[n])y.append(0)y.append(0)y.append(0)

Y = fft(y)N = len(x)for k in range(N//2,4*N-N//2+1):

Y[k] = 0z = ifft(Y)wav_write([4*s.real for s in z],4*fs,’fast_flights.wav’)

Using the Discrete Fourier Transform · sample frequency is 11,025 samples/second. Listen to the...

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