how to use freqz function correctly

Hello,

Im trying to visualize the frequency response using the freqz function for a filter. My sampling frequency is 15 MHz, while the signal frequency is only 1 kHz. Due to this large difference between the sampling frequency and the signal frequency, the resulting frequency response plot is not clear. Is there anything i can do ?

Im using freqz(x), in the command window.

Answer 1

[Hassaan]

% Filter design
N = 10; % Filter order
Fcut = 2000; % Cut-off frequency of 2 kHz
Fs = 15000000; % Sampling frequency of 15 MHz

% Design a low-pass FIR filter
b = fir1(N, Fcut/(Fs/2));
a = 1; % For FIR filters, the denominator is 1

% Frequency range of interest
f_min = 0;     % Minimum frequency in Hz
f_max = 2000;  % Maximum frequency in Hz (around your signal frequency)

% Number of points to evaluate
n_points = 4096;

% Frequency vector (w radians/sample = f cycles/sec * (1/Fs) sec/sample * 2*pi rad/cycle)
% which is inverse of f = w * Fs / (2*pi) --- Thank you 'paul' for the
% sugguestion

w = linspace(0, f_max*2*pi/Fs, n_points);

% Calculate the frequency response
[h, w] = freqz(b, a, w);

% Convert rad/sample to Hz for plotting
f = w * Fs / (2*pi);

% Plot the magnitude response
figure;
plot(f, abs(h));
title('Frequency Response of the Designed FIR Filter');
xlabel('Frequency (Hz)');
ylabel('Magnitude');
grid on;

%Optional: If you want to plot phase response, uncomment the following lines
figure;
plot(f, angle(h));
title('Phase Response of the Designed FIR Filter');
xlabel('Frequency (Hz)');
ylabel('Phase (Radians)');
grid on;

In this code, the fir1 function designs a 10th-order low-pass FIR filter with a cutoff frequency of 2 kHz, which should be suitable for your 1 kHz signal. The frequency response is plotted for frequencies up to 2 kHz, which provides a more focused view on the relevant part of the spectrum. Feel free to adjust the filter order (N) and the cutoff frequency (Fcut) as per your specific requirements.

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[Paul]

Is this line correct?

% Frequency vector (in radians/sample)
w = linspace(0, f_max/(Fs/2), n_points);

Seems like it should be

% Frequency vector (w radians/sample = f cycles/sec * (1/Fs) sec/sample * 2*pi rad/cycle)
% which is inverse of f = w * Fs / (2*pi)
w = linspace(0, f_max*2*pi/Fs, n_points);

% Calculate the frequency response
[h, w] = freqz(b, a, w);
% Convert rad/sample to Hz for plotting
f = w * Fs / (2*pi);

[Hassaan]

@Paul you are right. I have fixed as per your suggestion. Thank you.

Answer 2

[Star Strider]

You can set the 'XLim ' range for the subplot axes, however that may be a bit of a stretch even then. Another option is to signifficantly increase the size of the fft that is calculated, although that could take a shile to calculate.

Try this —

Fs = 15E+6;
Fn = Fs/2;

Wp = 1/Fn;                                                          % Passband Frequency (Normalised)
Ws = 2/Fn;                                                          % Stopband Frequency (Normalised)
Rp =  1;                                                            % Passband Ripple
Rs = 60;                                                            % Passband Ripple (Attenuation)
[n,Wp] = ellipord(Wp,Ws,Rp,Rs);                                     % Elliptic Order Calculation
[z,p,k] = ellip(n,Rp,Rs,Wp);                                        % Elliptic Filter Design: Zero-Pole-Gain 

[sos2,g2] = zp2sos(z,p,k);                                          % Second-Order Section For Stability

figure
freqz(sos2, 2^16, Fs)                                               % Filter Bode Plot

figure
freqz(sos2, 2^20, Fs)                                               % Filter Bode Plot
set(subplot(2,1,1), 'XLim',[0 10])
set(subplot(2,1,2), 'XLim',[0 10])

figure
freqz(sos2, 2^24, Fs)                                               % Filter Bode Plot
set(subplot(2,1,1), 'XLim',[0 5])
set(subplot(2,1,2), 'XLim',[0 5])

This uses a lowpass elliptic filter as the example. Narrow the 'XLim' values to your area-of-interest. Longer fft lengths than this are primarily a problem of available memory.

.

Answer 3

[Paul]

If x is a digitalFilter object, then freqz can be called as

freqz(x,f,fs);

where f is vector of frequencies of interest to visualize the frequency response and fs is the sampiling frequency (15e6).

If x is not a digitalFilter object, then freqz(x) is not a valid, documented, call to freqz and you'll have to provide more information on what x is.