1. Speech Signal Processing
Problem 1
Speech Signal Processing

2. Solution Filter length estimation: from transition band width
∆ 𝑓 1 = 𝑓 𝑠𝑡𝑜𝑝 − 𝑓 𝑝𝑎𝑠𝑠 𝑓 𝑠 𝑎𝑛𝑑 𝑁= 3.3 ∆𝑓
We select N = 25 for the number of filter coefficients and using the Hamming window method.
we next obtain from the transition band central frequency
𝑓 𝑐 = 𝑓 𝑝𝑎𝑠𝑠 +𝑠𝑡𝑜𝑝 2
The normalized lower and upper cutoff frequencies are calculated as:
and
𝑁=2 𝑀+1=25
𝑀=12

3. Design using MATLAB

5. w : Window coefficients
h: filter transfer function coefficients
b: filter coefficients

7. Filter Coefficients

9. Original speech and processed speech using the band-pass filter.
The band-pass frequency components contain a small portion of speech energy

10. Spectral comparison of the original speech and processed speech using the ban-dpass filter.
low and high frequencies are removed by the band-pass filter.
Low scale amplitude

11. Problem 2
1. Calculate the filter coefficients for a 5-tap FIR band-pass filter with a lower cutoff frequency of 1,000 Hz and an
upper cutoff frequency of 1,500 Hz and a sampling rate of 5,000 Hz.
2. Determine the transfer function and plot the frequency responses with MATLAB.
3. apply the hamming window function to obtain windowed coefficients;
4. plot the impulse response ℎ(𝑛).and windowed impulse response ℎ 𝑤 (𝑛).
Hamming window

12. Solution Calculating the normalized cutoff frequencies leads to
Ω 𝐿 = 2𝜋 𝑓 𝐿 𝑓 𝑠 = 2𝜋 =0.4 𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
Ω 𝐻 = 2𝜋 𝑓 𝐻 𝑓 𝑠 = 2𝜋 =0.6 𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
Since 2M +1= 5 in this case, using the equation in Table
ℎ 𝑛 = Ω 𝐻 − Ω 𝐿 𝜋 𝑛=0 sin⁡(Ω 𝐻 𝑛) 𝑛𝜋 − sin⁡(Ω 𝐿 𝑛) 𝑛𝜋 𝑛≠0 𝑎𝑛𝑑−2≤𝑛≤2
Calculations for non-causal FIR coefficients are listed as
ℎ 0 = Ω 𝐻 − Ω 𝐿 𝜋 = 0.6𝜋−0.4𝜋 𝜋 =0.2
ℎ 1 = sin⁡(Ω 𝐻 𝑛) 𝑛𝜋 − sin⁡(Ω 𝐿 𝑛) 𝑛𝜋 = sin 0.6𝜋 ×1 1×𝜋 − sin 0.4𝜋 ×1 1×𝜋 =0.0035
ℎ 2 = sin⁡(Ω 𝐻 𝑛) 𝑛𝜋 − sin⁡(Ω 𝐿 𝑛) 𝑛𝜋 = sin 0.6𝜋 ×2 2×𝜋 − sin 0.4𝜋 ×2 2×𝜋 =0.0035

15. Problem 3
Let us design a first-order digital low-pass filter with a 3-dB cutoff frequency of
𝜔 𝑐 =0.25 𝜋 by applying the bilinear transformation to the analog Butterworth
filter:
𝐻 𝑎𝑝 (𝑠)= 1 1+ 𝑠 Ω 𝑐

16. 1. Frequency Warping (mapping digital frequency to analog frequency)
Solution
1. Frequency Warping (mapping digital frequency to analog frequency)
Because the 3-dB cutoff frequency of the Butterworth filter is Ω 𝑐 for a cutoff frequency 𝜔 𝑐 =0.25 𝜋 in the digital filter, we must have:
Ω 𝑐 = 2 𝑇 𝑠 𝑡𝑎𝑛 0.25𝜋 2 = 𝑇 𝑠
2. Prototype Transformation (From Prototype analog LPF to desired analog filter)
Therefore, the system function of the analog filter is (we use directly the given analog Butterworth filter ):
𝐻 𝑎 𝑠 = 1 1+ 𝑠 Ω 𝑐 = 𝑠𝑇 𝑠
3. Bilinear Transformation (From analog filter transfer function to digital filter Transfer Function)
Applying the bilinear transformation to the analog filter gives
𝐻 𝑎 𝑧 = 𝐻 𝑎 𝑠 𝑠= 2 𝑇 𝑠 1− 𝑧 −1 1+ 𝑧 −1 = − 𝑧 −1 1+ 𝑧 −1 = 𝑧 −1 1− 𝑧 −1
Note that the parameter 𝑇 𝑠 , does not enter into the design.

17. Matlab
%- freqz returns the frequency response based on the current filter coefficients
b=[ ]; a=[ ];
[hz,f]=freqz(b,a); % 512 nb of data plot
%- angle(hz)returns the phase angles in radians,
phi = 180*unwrap(angle(hz))/pi; %- unwrap Correct phase angles to produce smoother phase plots
subplot(2,1,1), plot(f, abs(hz)),grid;
%axis([0 fs/2 0 1]);
%xlabel('Frequency (Hz)'); ylabel('Magnitude Response')
subplot(2,1,2), plot(f, phi); grid;
%axis([0 fs/ ]);
%xlabel('Frequency (Hz)'); ylabel('Phase (degrees)')

18. Problem 4
Illustrating the simple pole-zero method of calculating filter coefficients a band-pass digital filter is required to meet the following specifications:
Assuming a sampling frequency of 500 Hz, obtain
1. The transfer function of the filter by suitably placing z-plane poles and zeros.
2. its difference equations.
3. Block diagram representation of filter.

19. Solution
Since a complete rejection is required at zero and 250Hz we need to: place zeros at the corresponding points on the z-plane. These are at angles of 0° and 360×250/500=180° on the unit circle.
place zeros at :
𝜃= 𝑓 𝑓 𝑠 ° = ° = 0 °
and
𝜃= ° = 18 0 °
𝑧 1 =1 𝑎𝑛𝑑 𝑧 2 =−1
To have the pass-band centered at 125Hz requires us to place poles at ±360×125/500=±90°.
place poles at :
𝑝 1 =𝑟 𝑒 𝑗 𝜋 2 𝑎𝑛𝑑 𝑝 2 =𝑟 𝑒 −𝑗 𝜋 2
To ensure that the coefficients are real , it is necessary to have complex conjugate pole pair.
The radius of the poles is determined by the desired bandwidth. An approximate relationship between r, for r > 0.9 , and bandwidth bw is:
𝑟~1− =0.937
for our specifications bw =10Hz and Fs = 500Hz . This leads to an r value of r = 0.937
The difference equation is

20. 𝑓 𝑠 /2
Pole-zero dıagram
Block diagram representation of filter

21. Matlab
f0 = 125 ; bw = 10; fs= 500;
r=1-(bw/fs)*pi; % magnitude of the pole
theta=(f0/fs)*360; % angle of the pole
k=((1-r)*sqrt(1-2*r*cos(2*theta*(pi/180))+r^2))/(2*abs(sin(theta*(pi/180))));
% the filter transfer function
b=[k 0 -k];a=[1 -2*r*cos(theta*(pi/180)) r^2];
%- freqz returns the frequency response based on the current filter coefficients
[hz,f]=freqz(b,a,512,fs); % 512 nb of data plot
%- angle(hz)returns the phase angles in radians,
phi = 180*unwrap(angle(hz))/pi; %- unwrap Correct phase angles to produce smoother phase plots
subplot(2,1,1), plot(f, abs(hz)),grid;
axis([0 fs/ ]);
xlabel('Frequency (Hz)'); ylabel('Magnitude Response')
subplot(2,1,2), plot(f, phi); grid;
axis([0 fs/ ]);
xlabel('Frequency (Hz)'); ylabel('Phase (degrees)')