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Characteristics of FIR digital filters

The equations of the three-term moving average filter are shown below

2085_Characteristics of FIR digital filters.png

This is a crude low pass filter with simple linear phase, ∠H (ω) = -ω.

%Magnitude and phase response of 3-coefficient moving average filter

%Filter coefficients: h(n) = {1/3, 1/3, 1/3}

b3=[1/3, 1/3, 1/3], a=[1]

w=-pi: pi/256: pi; Hw3=freqz(b3, a, w);

subplot(2, 1, 1), plot(w, abs(Hw3)); legend ('Magnitude');

xlabel('Frequency \omega, rad/sample'), ylabel('Magnitude of H(\omega)'); grid

subplot(2, 1, 2), plot(w, angle(Hw3)); legend ('Phase');

xlabel('Frequency \omega, rad/sample'), ylabel('Phase of H(\omega)'); grid

612_Characteristics of FIR digital filters1.png

Normalized frequency We describe the r = ω/π. As ω goes from -p to p the value of  r goes from-1 to 1. That relates to a frequency range of -Fs/2 to Fs/2 Hz. In terms of the normalized frequency the frequency response of the three-term going average filter goes

938_Characteristics of FIR digital filters2.png

%Magnitude and phase response of 3-coefficient moving average filter

%Filter coefficients: h(n) = {1/3, 1/3, 1/3}

subplot(2,1,1); fplot('abs((1/3)*(1+exp(-j*pi*r)+exp(-j*2*pi*r)))', [-1, 1], 'k');

legend ('Magnitude');

xlabel('Normalized frequency, r'); ylabel('Magnitude of H(r)'); grid subplot(2,1,2);fplot('angle((1/3)*(1+exp(-j*pi*r)+exp(- *2*pi*r)))', [-1, 1], 'k'); legend ('Phase');

xlabel('Normalized frequency');ylabel('Phase of H(r)'); grid

203_Characteristics of FIR digital filters3.png

We explain below the characteristics of various kinds of FIR filter. The filter length N can be an odd (preferred) or an even number. Further, we are simply interested in linear part of phase. That needs the impulse response to have either odd or even symmetry about its center.

 

Example: Find the frequency response of the given FIR filters

A.  h(n) = {0.25, 0.5, 0.25}          Even symmetry

B.  h(n) = {0.5, 0.3, 0.2}              No symmetry

C.  h(n) = {0.25, 0.5, -0.25}         No symmetry

D.  h(n) = {0.25, 0, -0.25}            Odd symmetry

Solution

(A) The sequence h(n) = {0.25, 0.5, 0.25} has even symmetry.

 

%Magnitude and phase response of h(n) = {0.25, 0.5, 0.25}

%Filter coefficients - Even symmetry b3=[0.25, 0.5, 0.25],

a=[1]

w=-pi: pi/256: pi; Hw3=freqz(b3, a, w);

subplot(2, 1, 1), plot(w, abs(Hw3)); legend ('Magnitude');

xlabel('Frequency \omega, rad/sample'), ylabel('Magnitude of H(\omega)'); grid subplot(2, 1, 2), plot(w, angle(Hw3)); legend ('Phase = -\omega'); xlabel('Frequency \omega, rad/sample'), ylabel('Phase of H(\omega)'); grid

2121_Characteristics of FIR digital filters4.png

(B) The sequence h(n) = {0.5, 0.3, 0.2}is not symmetric.

 

%Magnitude and phase response of h(n) = {0.5, 0.3, 0.2}

%Filter coefficients - No symmetry b3=[0.5, 0.3, 0.2],

a=[1]

w=-pi: pi/256: pi; Hw3=freqz(b3, a, w);

subplot(2, 1, 1), plot(w, abs(Hw3)); legend ('Magnitude');

xlabel('Frequency \omega, rad/sample'), ylabel('Magnitude of H(\omega)'); grid subplot(2, 1, 2), plot(w, angle(Hw3)); legend ('Nonlinear Phase'); xlabel('Frequency \omega, rad/sample'), ylabel('Phase of H(\omega)'); grid

1283_Characteristics of FIR digital filters5.png

(C) The sequence h(n) = {0.25, 0.5, -0.25} is not symmetric.

 

%Magnitude and phase response of h(n) = {0.25, 0.5, -0.25}

%Filter coefficients - This is not odd symmetry b3=[0.25, 0.5, -0.25],

a=[1]

w=-pi: pi/256: pi; Hw3=freqz(b3, a, w);

subplot(2, 1, 1), plot(w, abs(Hw3)); legend ('Magnitude');

xlabel('Frequency \omega, rad/sample'), ylabel('Magnitude of H(\omega)'); grid subplot(2, 1, 2), plot(w, angle(Hw3)); legend ('Nonlinear Phase'); xlabel('Frequency \omega, rad/sample'), ylabel('Phase of H(\omega)'); grid

910_Characteristics of FIR digital filters7.png

(D) The sequence h(n) = {0.25, 0, -0.25} has odd symmetry.

 

%Magnitude and phase response of h(n) = {0.25, 0, -0.25}

%Filter coefficients - This is odd symmetry b3=[0.25, 0, -0.25],

a=[1]

w=-pi: pi/256: pi; Hw3=freqz(b3, a, w);

subplot(2, 1, 1), plot(w, abs(Hw3)); legend ('Magnitude');

xlabel('Frequency \omega, rad/sample'), ylabel('Magnitude of H(\omega)'); grid subplot(2, 1, 2), plot(w, angle(Hw3)); legend ('Phase = -\omega + \pi/2 '); xlabel('Frequency \omega, rad/sample'), ylabel('Phase of H(\omega)'); grid

1688_Characteristics of FIR digital filters6.png

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