7/31/2023 0 Comments Matlab help freqzH = polyval(fliplr(b),exp(-j*w))./polyval(fliplr(a),exp(-j*w)) Īlso, this is really evaluating the transfer function at discrete equally spaced angular frequencies w = 2*pi*k/N which corresponds to the Discrete Fourier Transform (DFT). This would be equivalent to the following which makes use of the builtin polyval: N = 513 % number of points at which to evaluate the transfer function h,w freqz (sysobj,'Arithmetic',arithType) analyzes the filter System object, based on the arithmetic specified in arithType, using either of the previous syntaxes. N = 513 % number of points at which to evaluate the transfer function freqz uses the transfer function associated with the filter to calculate the frequency response of the filter with the current coefficient values. For the purpose of illustration, I will be assuming bs are the coefficients of the x sequence and as are the coefficients of the y sequence, such that the b are in the numerator and the as are in the denominator:Ī direct evaluation with Matlab could be done with: b = 1 Have you tried freqz () It returns the frequency response vector, h, and the corresponding angular frequency vector, w, for the digital filter with numerator and denominator polynomial coefficients stored in b and a, respectively. There are of course different ways to implement this with Matlab. abs(h) is the magnitude responseĮdit : Does wn=0.34 corresponds to 42.I would indeed be as simple as substituting exp(j*w) in your transfer function. Plot(f,20*log10(abs(h))) % we will use 20log10() for ploting the mag. X=fir1(N,wn,ftype,kaiser(N+1,beta)) % designing the corresponding fir filter ( note that fir1 takes an even filter order so we wrote N+1) , order of the filter, filter type and betta according to the design specs. This function will return the cuttoff freq. =kaiserord(fcuts,mags,devs,fs) % using kaiser window for designing the fir filter. Mags= % The required filter amplitude in these bands (Absolute value not in dB)ĭevs= % Passband and stopband ripples (Absolute value not in dB) matlab freqs freqz 1.freqs h freqs (b,a,w) h,w freqs (b,a) h,w freqs (b,a,f) freqs (b,a) 1. The complex-valued frequency response is calculated. So what is wrong with my code? Here is the code (with the required filter specifications): fs=250 įcuts= % passband and stopband frequencies The freqz function calculates the frequency response for a filter from the filter transfer function Hq(z). My question is that when I plot the amplitude response, the -3dB point occurs at frequency above 100 Hz, which means that the cutoff frequency is not equal to 42.5 Hz. According to the study material, the frequency response can be plotted in MATLAB by using ones (1, L) / L for filter coefficients: bb ones (1, L) / L ww -pi: (pi / 100):pi HH freqz (bb, 1, ww) subplot (2, 1, 1), plot (ww, abs (HH)) subplot (2, 1, 2), plot (ww, angle (HH)) Now I'm starting to get confused. Doing this, the returned wn is equal to 0.34 (returned by kaiserord()), which when I convert it to Hertz it gives me 42.5 Hz as required. I need to plot the filter's magnitude and phase responses with the x-axis as frequency in Hertz instead of plotting them with the normalized angular frequency. I am designing an FIR filter in Matlab using a Kaiser window.
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