From Fourier Series to Analysis of Non-stationary Signals …euler.fd.cvut.cz/predmety/mathtools/files/lecture_07.pdf · · 2010-11-22Short Time Fourier Transform Spectrograms From

Short Time Fourier TransformSpectrograms

From Fourier Series to Analysis ofNon-stationary Signals - VII

prof. Miroslav Vlcek

November 23, 2010

prof. Miroslav Vlcek Lecture 7


Contents

1 Short Time Fourier Transform

2 Spectrograms



Contents

1 Short Time Fourier Transform

2 Spectrograms



An overview of Fourier transforms

continuous in time discrete in timeperiodic in frequency

infr

eque

ncy

x(t) =1

2π

∞∫

−∞

X(jω)ejωt dω x(n) =T

2π

+π/T∫

−π/T

X(ejωT)ejkωT dω

cont

inuo

us

X(jω) =

∞∫

−∞

e−jωt x(t)dt X(ejωT ) =∞∑

n=−∞

x(n)e−jkωT

Fourier transform Fourier transform t = nT

infr

eque

ncy

intim

e

x(t) =∞∑

k=−∞

X(k)ejkω0t x(n) =1

N

N−1∑k=0

X(k)e(j2π/N)kn

disc

rete

peri

odic

X(k) =ω0

2π

π/ω0∫

−π/ω0

x(t)e−jnω0t dt X(k) =N−1∑n=0

x(n)e−(j2π/N)kn

Fourier series Discrete Fourier transform



What is wrong with Fourier Transform?

Lets consider basis functions: sin(ωt), cos(ωt) and δ(t)

support region in time in frequency

sin(ωt) (−∞,∞) 0cos(ωt) (−∞,∞) 0δ(t) 0 (−∞,∞)

• The basis functions sin(ωt) and cos(ωt) are not localizedin time!

• The δ(t) is not localized in frequency!

We have learnt to localize a signal in time domain by windowing⇒ short time Fourier transform



Short Time Fourier Transform

• When sampling an audio signal at a sampling rate44.1 kHz

• ... 1 hour of stereophonic music would be44100 × 2 × 60 × 60 = 317520000 samples !

• If we want to compute DFT, the closest power-of-two FFTis 228 = 268435456 per channel.

• A better approach is to break the long signal into smallsegments and analyze each one with FFT ⇒ short timeFourier transform




• Assume that x(m) is an infinitely long sequence

• in order to localize energy in time and frequency wesegment the signal into short-time pieces and calculateDFT of each one

• sampled STFT for a window defined in the region0 ≤ m ≤ M − 1 is given by

X (k , ℓL) =M−1∑

m=0

x(ℓL − m)w(m)exp(

−j2πkmN

)









Spectrograms

In MATLAB the command

spectrogram(x,window,noverlap,nfft,fs,’yaxis’)

performs short-time Fourier transform and plots a 2Dfrequency-time diagram, where

• x is the signal specified by vector x• window is a Hamming window of length nfft

• noverlap is the number of overlapping segments thatproduces 50% overlap between segments



Spectrograms

• nfft is the FFT length and is the maximum of 256 or thenext power of 2 greater than the length of each segment ofx

• fs is the sampling frequency, which defaults to normalizedfrequency

• using ’yaxis’ displays frequency on the y-axis and timeon the x-axis

We also use command colorbar which appends a color scaleto the current axes.



Chirp signal analysis sin(2π(f0 + αt))



MATLAB project Spectrogram I

1 Start MATLAB. Load in the audio signal with commandsfilename = ’flute-C4.wav’;[x1 sr1] = wavread(filename);

2 The sampling rate is 11025 Herz, and the signal contains36 250 samples. If we consider this signal as sampled onan interval (0,T ), what is the time duration of the flutesound ?

3 Use command soundsc(x1,sr1) to obtain flute soundclick to play

4 Make resampling of the audio signal by fr = 4000[Hz] andwrite the sound file to disk usingwavwrite(x2, sr2, ’flute-resampled.wav’);




0 0.5 1 1.5 2 2.5 3 3.5

−0.05

0

0.05

Flute

Time [s]

0 0.5 1 1.5 2 2.5 3 3.5

−0.05

0

0.05

Resampled flute

Time [s])




5 Compute the DFT of the signal withX1 = fft(x1(1:1024)); and X2 =fft(x2(1:1024));

6 DFT of real-valued signals is always symmetric aroundsr/2 so we only need to plot the first half. Display themagnitude of the Fourier transform usingplot(f1(1:end/2+1), abs(X1(1:end/2+1)))

7 Q: What is the approximate fundamental frequency of theflute note C4?

8 A: Find the bin corresponding to the first peak in themagnitude spectrum.




9 You can use a systematic way to find the frequency of thepeaks in spectrum abs(X2)using following commands:% find local maximamag = abs(X2);mag = mag(1:end/2+1);peaks = (mag(1:end-2) < mag(2:end-1)) &mag(2:end-1) > mag(3:end);

10 Then evaluate the peaks at corresponding frequenciesabove a threshold:peaks = peaks & mag(2:end-1) > 0.5;fmax=f2(peaks)




0 1000 2000 3000 4000 5000 60000

0.2

0.4

0.6

0.8

1

1.2

1.4

Mag

nitu

de

Frequency (Hz)

Flute

0 500 1000 1500 2000 25000

1

2

3

4

5

6

7

Mag

nitu

de

Frequency (Hz)

Resampled flute




11 Finally we will use Spectrogram with folowingspecifications:nwin = 512; % samples of a windownoverlap = 256; % samples of overlapsnfft = 512; % samples of fast Fouriertransformspectrogram(x1, nwin, noverlap, nfft, sr2,’yaxis’);colorbarprint -djpeg90 figure-spect

12 Carefully study the options for the spectrogram using help!




0.5 1 1.5 2 2.5 30

200

400

600

800

1000

1200

1400

1600

1800

2000

Time

Fre

quen

cy (

Hz)

−140

−130

−120

−110

−100

−90

−80

−70

−60

−50



MATLAB project Spectrogram II

1 Construct a signal f of two sinusoids on intervalt ∈< 0,2 > sampled at 1 kHzt=[0:1999]/1000;f=cos(2*pi*137*t)+cos(2*pi*147*t);

2 Compute the DFT of f and display its magnitude in timeand frequency domain.

3 Use spectrogram routinespectrogram(f,512,510,512,1000,’yaxis’);colorbar







1 Construct a signal f with varying frequency within intervalt ∈< 0,2 > sampled at 1 kHzt=[0:1999]/1000;arg=100+20*cos(2*pi*t);f=cos(2*pi*arg.*t);

2 Compute the DFT of f and display its magnitude in timeand frequency domain.

3 Use spectrogram routinespectrogram(f,128,120,512,1000,’yaxis’);colorbar





From Fourier Series to Analysis of Non-stationary Signals …euler.fd.cvut.cz/predmety/mathtools/files/lecture_07.pdf · · 2010-11-22Short Time Fourier Transform Spectrograms From

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