abs(h) is the magnitude responseĮdit : Does wn=0.34 corresponds to 42. 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.
#Matlab help freqz how to#
I have figured out how to plot the entire transform of my frequency response, but I only need half of it, and I need to normalize it from pi to 1 (where 0:pi represents my x axis, and I want that to go to 0:1). This function will return the cuttoff freq. I have a function which is basically recreating the freqz command in matlab.
=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 help freqz code#
So what is wrong with my code? Here is the code (with the required filter specifications): fs=250 įcuts= % passband and stopband frequencies 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. 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. Note Some help text displays the names of functions in uppercase characters to make them stand out from the rest of the text. H = 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).I am designing an FIR filter in Matlab using a Kaiser window. 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. Syntax help name help Description example help name displays the help text for the functionality specified by name, such as a function, method, class, toolbox, or variable. 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 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.
N = 513 % number of points at which to evaluate the transfer function The frequency response of a digital filter can be interpreted as the transfer function evaluated at z e j. 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 =
There are of course different ways to implement this with Matlab. I would indeed be as simple as substituting exp(j*w) in your transfer function. The Matlab function freqz can be used to easily compute the frequency response of a DT system that is specified in terms of its a and b coefficients (LCCDE).
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