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EE513 Audio Signals and Systems. Digital Signal Processing (Analysis) Kevin D. Donohue Electrical and Computer Engineering University of Kentucky. Spectra. The spectrum of the impulse response indicates the following filter/system properties:

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### EE513Audio Signals and Systems

Digital Signal Processing (Analysis)

Kevin D. DonohueElectrical and Computer EngineeringUniversity of Kentucky

The spectrum of the impulse response indicates the following filter/system properties:

• resonance and frequency sensitivities through the magnitude spectrum

• delay and dispersive properties through the phase spectrum.

The discrete Fourier transform (DFT) computes complex samples of the spectrum.

• For deterministic signals both the phase and magnitude are important for characterizing the signal or response.

• For stationary noise processes the square of the DFT magnitudes can be averaged from independent segments to create a power spectral density.

In order to take a DFT of a signal, a finite window (time interval) must be extracted from the original sampled time signal:

The process can be repeated with overlapping windows covering entire signal.

• If the signal is deterministic and changes over time, the DFTs for small consecutive intervals must be displayed to show the changes in spectrum over time (Spectrogram).

• If the signal is random with statistics that do not change over time, the DFT magnitudes can be averaged (Power Spectral Density (PSD)).

DFT Magnitude

Squared

DFT Phase

DFT points are evaluated at discrete frequency samples (frequency sampling). The DFT for an N point sequence is defined as:

This inverse DFT is a similar computation to the DFT except the complex conjugate of the kernel is used:

DFT with W notation

Let WN equal the complex sinusoid with the lowest nonzero frequency (k = 1):

WN is effectively a fundamental frequency in a harmonic expansion, since the frequency domain is sampled in equal increments. DFT and its inverse can now be written as:

Compute the DFT for the 4 point and 8 point signals given below:

What was the effect of extra zeros padded on the end of the signal in the second example?

120

60

3

W

= j

4

150

30

2

W

= -1

0

4

180

0

W

= 1

4

210

330

2

W

= -j

4

240

300

270

Computation Examples

IM

Compute the DFT for the 4 point

RE

W

= j

8

90

120

60

7

W

=

5

W

=

8

8

150

30

0

W

= 1

8

180

0

4

W

= -1

8

210

330

3

1

W

=

W

=

8

8

240

300

270

2

W

= -j

8

Computation Examples

Compute the DFT for the 8 point

• What symmetries exist in the WN factor in the DFT computation that suggest a more efficient computation than a direct multiply and summation? The implementation that takes advantage of these symmetries is called a Fast Fourier Transform (FFT).

• If the sequence is a power of 2, the FFT is most efficient. A direct implementation of the DFT requires N2 complex multiplications. An FFT requires Nlog2N complex multiplications.

• Compute the ratio between complex multiplications for the FFT and DFT for a 128 point sequence. Compute the same ratio for a 2048 point sequence.

• See http://www.cmlab.csie.ntu.edu.tw/cml/dsp/training/coding/transform/fft.htmlFor more details on the FFT derivation

Derive each of the symmetry relationships below for the WN factor:

Time sample the following FS equations to show it equivalence to the DFT.

Sample time axis with Ts where NTs = T

Distortion and error result from the truncated and frequency sampled signal because:

• Lost energy from truncated time portion will distribute error over the signal.

• Truncating signal at the endpoints can introduce a sharp transition that may not be part of the signal.

• Sampling signal in frequency produces periodicity in time, which results in circular time-domain convolution from multiplying spectra in frequency domain.

The DFT can be applied to a finite duration signal. If signal support is infinite, part of it must be truncated. The DFT definition is applied to the finite number of points to obtain a frequency sampled spectrum, which results in a periodic extension of the signal in the time window.

The DFT representation implies the time domain signal was periodic (sampling in time produces periodic spectra, likewise sampling in frequency produces periodic time segments) as would be the case for Fourier series.

Multiplication in frequency is convolution in time. For discrete signals, this can be performed with the fft command in Matlab on the filter coefficients h(n) and input signal x(n), then the ifft command on their frequency domain product:

Time

Domain

Frequency

Domain

Time

Domain

Linear convolution implemented in the time domain. Note the convolved signal length is greater than either of the originals. Why?

Convolution implemented in the frequency domain. Note the convolved signal length is the same as the originals. Why?

Why is there an apparent artifact at the beginning of the signal?

Convolution implemented in the frequency domain with zero padding. Why is the apparent artifact from the nonzero pad example no longer present?

Use the FFT to filter a 32 point square pulse s[n]:

(index n starts a 0)

where the FIR filter has impulse response h[n]:

• Use the fft command to take signals into frequency domain without padding with zeros. Take inverse fft to obtain time domain signal. Plot the filtered signals and explain what you observe.

• Use the fft command and pad with zeros to double the signal lengths and repeat part (a)

The FFT can be applied directly to finite duration energy signals to examine the spectral content of the signal.

For Matlab’s FFT:

• FFT(X) is the discrete Fourier transform (DFT) of vector X. For matrices, the FFT operation is applied to each column.

• FFT(X,N) is the N-point FFT, padded with zeros if X has less than N points and truncated if it has more.

Write a Matlab script to Generate a tone modulated by a Rayleigh envelope. Plot the signal, its spectra, and play the sound.

% Simulate a Rayleigh envelope ring, truncate it, take its FFT and examine

% its spectral content

fr = 250; % Frequency of the ring

sigma = .05; % effective duration TIMES 2 of ring

fs = 8000; % Sampling frequency

dur = .25; % signal duration in seconds

nfft = 4096;

t = [0:round(dur*fs)-1]/fs; % Create time axis

ring = (t/sigma).*exp((-t.^2)/(sigma^2)).*sin(2*pi*t*fr); % Generate Ring signal

figure(1)

plot(t, ring) % Plot ring

title('Rayleigh Envelope Ring Signal')

xlabel('Seconds')

ylabel('Amplitude')

soundsc(ring,fs) % Play sound

spec0 = fft(ring); % No Zero pad

spec1 = fft(ring,nfft); % Zero pad (or truncate) to NFFT points

spec2 = fft(ring,2*nfft); % Zero pad to 2*NFFT

% Create frequency axes for each of the spectra

faxis0 = fs*[0:length(spec0)-1]/length(spec0);

faxis1 = fs*[0:length(spec1)-1]/length(spec1);

faxis2 = fs*[0:length(spec2)-1]/length(spec2);

% Plot spectrum with zero padded version on the same graph and compare

figure(2) % Magnitude

plot(faxis0, abs(spec0),'k',faxis1, abs(spec1),'r--',faxis2, abs(spec2),'g.')

legend([int2str(length(t)) ' point FFT'], [int2str(nfft) ' point FFT'], [int2str(2*nfft) ' point FFT'])

xlabel('Hertz')

ylabel('Magnitude')

figure(3) % Phase

plot(faxis0, 180*phase(spec0)/pi,'k',faxis1, 180*phase(spec1)/pi,'r--',faxis2, 180*phase(spec2)/pi,'g.')

legend([int2str(length(t)) ' point FFT'], [int2str(nfft) ' point FFT'], [int2str(2*nfft) ' point FFT'])

xlabel('Hertz')

ylabel('Degrees')

Create a signal consisting of 2 sine waves with amplitude 1 and sampled at 4000 Hz. Set one with frequency 250 Hz and the other with 254 Hz.

a) Take the DFT using window length of 0.05, 0.25 and 0.5 seconds. Describe what you see and make generalizations about the impact of signal length on frequency resolution.

b) Repeat part (a) using zeros padding so that each signal length is effectively 1 seconds. Describe what you see and make generalization about the impact zero padding on frequency resolution.

• When the signal is long with changing dynamics, as would be the case for a speech or music signal, the FFT must be applied to short segments of these signals.

• To reduce truncation effects, a tapering is used to bring the ends of the segment close to zero a the points of truncation.

• To ensure that information is not lost in the truncation and tapering, an overlap between signal segments is used.

The windowed DFT is the product of the original signal and the windowing function:

where x[n] is a signal of infinite extend, n is the sample index, g[n] is a finite widow.

In the frequency domain this becomes convolution with the window DFT:

For a signal that deterministically changes spectral properties over time, the DFT over local time windows must be computed and indexed. This function is referred to as the spectrogram:

where m is the time index, T is the sample increment to the next window position, which effectively slides the N point analysis window over consecutive (often overlapping) segments of the signal x[n]. The DFTs are plotted along the y-axis, while the x-axis is time (usually corresponding to the short-time window centers. The result is a function that shows how the spectrum changes over time.

The following parameters must be set properly for a usable analysis:

window length N, window type, window increment (overlap between consecutive windows), number of FFT points.

The Matlab spectrogram command is appropriate for analyzing long deterministic or non-stationary signals.

For Matlab’s SPECTROGRAM:

>> S = SPECTROGRAM(X,WINDOW,NOVERLAP,NFFT,Fs)

calculates the spectrogram for the signal sampled at Fs in vector X. SPECTROGRAM splits the signal into overlapping segments, windows each with the WINDOW vector and forms the columns of S with their zero-padded, length NFFT discrete Fourier transforms. Thus each column of S contains an estimate of the short-term, time-localized frequency content of the signal X (positive frequencies only). Time increases linearly across the columns of S, from left to right.

If plotting, it is convenient to use the optional output arguments:

[S,F,T] = SPECTROGRAM(A, WINDOW, NOVERLAP, NFFT, Fs) returns a column of frequencies F for each time in T at which the spectrum was computed. F has length equal to the number of rows of S, and T has length equal to the number of columns. If you leave Fs unspecified, SPECTROGRAM assumes a default of 2 Hz. Optional output parameters F and T are the frequency and time axes useful for plotting the spectrum:

>> imagesc(T, F, abs(S))

Write a Matlab script to Generate a frequency sweep from 20 Hz to 1.9 kHz, where Fs is 4 kHz. Then plot the magnitude of the spectrogram and play the sound.

% Simulate a frequency sweep signal and take its spectrogram and examine

% its spectral content over time

% Signal Parameters

flow = 20; % Starting frequency of sweep in Hertz

fend = 1900; % Ending frequency of sweep in Hertz

dt = 4; % Time duration of sweep in seconds

fs = 4000; % Sampling frequency

% Spectrogram Parameters

wlen = 128; % Length of actual point extracted from signal segment

nfft = 1024; % number of FFT point (zero padding)

olap = floor(wlen/2); % Points of overlap between segments

wn = hamming(wlen); % Create tapering window (also try boxcar - square window)

% Generate signal

t = [0:round(dt*fs)-1]/fs; % Create time axis

fsw = flow + ((fend-flow)/2)*[0:length(t)-1]/length(t); % Create Frequency ramp - Why divid by 2?

swp = sin(2*pi*t.*fsw); % Generate sweep signal

soundsc(swp,fs) % Play sound

% Create Spectrogram

[b,faxis,taxis] = spectrogram(swp,wn,oplap,nfft,fs);

% Plot over time and frequency

figure(1); imagesc(taxis, faxis, abs(b)) % Plot spectrogram

axis('xy') % Flip y axis to put zero Hz on bottom

colorbar % Include colorbar to determine color coded magnitudes on graph

title(['Frequency Sweep from ' num2str(flow) ' Hz to ' num2str(fend) ' Hz']); xlabel('Seconds'); ylabel('Hz')

% Just plot at a single time instant

figure(2); tindex = find(taxis >= dt/2); %Find index of halfway point over the time axis

tindex = tindex(1);

plot(faxis,abs(b(:,tindex)))

title(['Single column from Spectrogram at ' num2str(taxis(tindex)) ' seconds']); xlabel('Hz'); ylabel('Magnitude')

Use the scale program created in a previous homework problem to generate a scale for 2 octaves starting at 256 Hz with tones of amplitude 1 and duration 0.25 seconds. Use a sampling rate of 4kHz.

A) Compute and plot the spectrogram magnitude in dB, labeling all axes correctly. Use the parameters you think best for estimating the spectrogram for this signal. Comment on the frequencies of the tones generated with your program and those identified through the spectrogram.

B) Quadruple the number of FFT points from that used in (A). Compute and plot the spectrogram, and explain difference observe from part (A).

C) Increase the window length by a factor of 4 and set the number of FFT points to twice the amount of the window length. Compute and plot the spectrogram, and explain the observed changes.

If a process is changing over time, while the statistics of the process do not change over time, the process is referred to as a stationary process. In this case averaging and moment computations result in useful characterization.

If the signal’s (or process’) probability density function never changes over time, then it is referred to as strict-sense stationary (SSS), if the first and second moments remain constant over time, the process is referred to as wide-sense stationary (WSS).

The Power Spectral Density (PSD) is the average power in a signal as a function of frequency. This can be estimated by averaging DFT magnitudes squared over segments of signals from the same noise process.

where E[] is the expected value, S(f) is the data spectrum plus noise (i.e. modeled as a random variable) and is the DFT or short-time FT estimated from independent signal segments. is the true or underlying spectrum.

A popular approach to spectral estimation is to is to window a long segment of data into a sequences of shorter windows (similar to the spectrogram) and average all the FFT magnitudes together. This is referred to as Welch's or the hopping window method. The Matlab command pwelch() implements this method.

Example: Filter white noise through a band-pass filter and compute the PSD. Compare PSD to the transfer function magnitude of the filter. Assume a sampling rate of 8kHz and a 6th order Butterworth filter with passband from 500 to 1500 kHz.

fs = 8000; % Sampling Frequency

dur = 20; % Sound duration in seconds

ord = 6;

plen = 32; % PSD segment length

% Bandlimit on the filter

f1 = [500]; % lower bandlimit

f2 = [1500]; % corresponding upper bandlimit

% colors for the plots

col = ['g', 'r', 'b', 'k', 'c', 'b'];

no = randn(1,round(fs*dur)); % Generate noise signal

[b,a] = butter(ord,2*[f1 f2]/fs); % Generate filter

% perform filter operation to get colored noise

cno = filter(b,a,no);

% Compute PSD of noise

[p, fax] = pwelch(cno,hamming(plen),fix(plen/2),2*plen, fs);

figure(1); lh = plot(fax,abs(fs*p/2),col(1)) % Plot PSD

set(lh,'LineWidth',2) % Make line thicker

hold on

% Find filter transfer function

[h,fq] = freqz(b,a,2*plen,fs);

plot(fq,abs(h).^2,col(2))

hold off

% Label figure

xlabel('Hertz', 'Fontsize', 14); ylabel('PSD', 'Fontsize', 14)

As a measure of how quickly a signal changes on average, or for the detection of periodicities, the autocorrelation can be used. It directly measures the correlation between samples separate in time by a particular interval or lag.

Define the autocorrelation of a sequence as:

Repeat previous example for signal generation and compute the autocorrelation function. Show autocorrelation for white noise (filter input) and colored noise (filter output).

% Compute auto correlation of input sequence

mxlag = fs*.02; % Only compute lags up to .1 seconds

[acwno, lagwno] = xcorr(no, mxlag, 'coef');

% Plot lags

figure(2); plot(1000*lagwno/(fs),acwno)

xlabel('milliseconds','Fontsize', 14); ylabel('Correlation coefficient','Fontsize', 14)

% Compute auto correlation of output sequence

mxlag = fs*.02; % Only compute lags up to .1 seconds

[accno, lagcno] = xcorr(cno, mxlag, 'coef');

% Plot lags

figure(3); plot(1000*lagcno/(fs),accno)

xlabel('milliseconds','Fontsize', 14); ylabel('Correlation coefficient','Fontsize', 14)

Repeat previous example for signal generation and compute the autocorrelation function. Show autocorrelation for white noise (filter input) and colored noise (filter output).

White noise AC

Colored noise AC

The AC and PSD are DFT pairs. The DFT of the AC is shown below:

Colored noise AC

DFT of AC

Generate 3 seconds of white noise at a sampling rate of 44.1 kHz. Use the following coefficients in an IIR filter to filter the white noise and generate pink noise (bnum are numerator and aden are denominator coefficients).

bnum = [ 0.04957526213389, -0.06305581334498, 0.01483220320740 ]

aden = [ 1.00000000000000, -1.80116083982126, 0.80257737639225]

Coefficients from http://music.columbia.edu/pipermail/music-dsp/2001-November/046099.html

A) Estimate the AC and PSD of the Pink noise and present their plots. Comment on the observed differences from white noise.

B) Add a sinusoid of frequency 220 Hz to the pink noise signal with power equal to that of the pink noise (i.e. p = std(pink noise signal) and sig = sqrt(2)*p*sin()) repeat part (A). Show the plots and explain the differences observed with the plots from part (A). Does this make sense since the sine function is not random?

a) Create a tone at 450 Hz with sampling rate 8000, amplitude 0.707, and duration 3 seconds. Add white noise (use the randn function) with a signal-to-noise ratio of 12 dB. Use the fir1 command to design a 30 and 120th order band-pass filter from 400 Hz to 506 Hz. Plot the magnitude response of the filters. Use filter to filter the signal and listen to the sound before and after filtering. Plot the signal spectral magnitudes before and after filtering (use the fft function). Describe the differences you hear between the signals for before and after filtering and compare the before and after filtering spectra.

b) Repeat part (a) comparing a 5th order elliptical filter with passband ripple of .5 dB and stopband ripple of 30 dB and a 5th order Butterworth filter.

Use Sinc function interpolation to upsample by a factor of 50 the band-limited signal is given in terms of its samples below. Assume signal is bandlimited to 0.5 Hz. The sampling rate (B) is 1 Hz.

F = [0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0]

Note:

a) Matlab has a sinc function (see help sinc)

b) A step edge is not bandlimited so the interpolated bandlimited function will have overshoot and undershoot between the given samples (i.e. it will not look like a sharper step).