Confuction with cutoff frequency

Hello, I have this code that is from a chebyshev filter and I need to know if what is marked in red on the graph corresponds to the cutoff frequencies

Fs = 4400;

Fn = Fs/2;

n = 3; Rp = 0.5;

Wn = [250 2000]/Fn;

[b a]=cheby1(n,Rp,Wn,'bandpass')

[h,w]=freqz(b,a)

plot(w,abs(h))

Answer 1

[Star Strider]

The first red line (at 0.357) corresponds to the low cutoff frequency. The upper cutoff frequency should be at 2.856, so the second red line (at 2.000) does not correspond to it (or to anything else of significance).

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[Juan Chehin]

Hi, Thanks for the answer, but it contradicts the answer from @Yazan :/

[Star Strider]

My pleasure.

You will find that my Answer is correct:

Fs = 4400;          
Fn = Fs/2;         
Wn = [250 2000]/Fn * pi

Wn = 1×2

0.3570 2.8560

The frequency axis in that plot is normalised on the interval radians/time unit, and is not in units of Hz. The freqz call:

[h,w]=freqz(b,a)

forces that, although if you wanted the frequency axis to be in Hz, it would only be necessary to provide the sampling frequency as the last argument to force that.

EDIT — (5 Aug 2021 at 19:42)

I still have no idea what the second (higher-frequency) red line corresponds to.

The centre frequency of the passband is:

Fc = mean([250 2000]/Fn * pi)               % Radians/Time

Fc = 1.6065

or:

Fc = mean([250 2000])                       % Hz

Fc = 1125

Is the second red line supposed to actually mean something? If so, what?

.

[Yazan]

Well, now I am checking your graph in full size, I can say that the answer above is correct. I was assuming that the x-axis represents the frequency in Hz, but it does not seem so. Thus, the values of the red lines do not really represent anything of interest.

[Juan Chehin]

@Star Strider The second red line corresponds (according to me) to the cutoff frequency of 2000 Hz. But this statement may be wrong, that's why the question in the forum

[Star Strider]

The problem is the way you called freqz, since the default frequency axis (as I mentioned earlier) is .

If you want to plot in Hz:

Fs = 4400;          
Fn = Fs/2;         
 
n = 3; Rp = 0.5;
Wn = [250 2000]/Fn;
 
[b a]=cheby1(n,Rp,Wn,'bandpass');
[h,w]=freqz(b,a,2^16,Fs);

plot(w,abs(h))
hold on
plot(([1;1]*Wn*Fn), ([1;1]*ylim).', '-r')
hold off
legend('Passband','Cutof Frequencies', 'Location','S')

Note — The frequency axis is different from the frequency axis in the posted plot image.

.

[Juan Chehin]

This is the final graph I got

[Star Strider]

If the sampling frequency is not specified in the freqz call,

Fs = 4400;          
Fn = Fs/2;         
 
n = 3; Rp = 0.5;
Wn = [250 2000]/Fn;
 
[b a]=cheby1(n,Rp,Wn,'bandpass');
[h,w]=freqz(b,a,2^16);

plot(w,abs(h))
hold on
plot(([1;1]*Wn*pi), ([1;1]*ylim).', '-r')
hold off
legend('Passband','Cutof Frequencies', 'Location','S')

The frequency axis is different. The plot is otherwise the same.

.

[Juan Chehin]

Thanks very much!!!

[Juan Chehin]

One last question, the red line of the cutoff frequency that you put, I see that on the y axis cuts at 0.9. Shouldn't it cut at 0.707? Where the filter gain falls. Green lines

[Star Strider]

I had to refresh my memory with respect to Chebyshev Type 1 filters.

‘The common practice of defining the cutoff frequency at −3 dB is usually not applied to Chebyshev filters; instead the cutoff is taken as the point at which the gain falls to the value of the ripple for the final time.’

So with a passband ripple defined as 0.5 dB, the cutoff frequencies correspond to that amplitude (dotted black line).

Fs = 4400;          
Fn = Fs/2;         
 
n = 3; Rp = 0.5;
Wn = [250 2000]/Fn;
 
[b a]=cheby1(n,Rp,Wn,'bandpass');
[h,w]=freqz(b,a,2^16);


plot(w,mag2db(abs(h)))
hold on
plot(([1;1]*Wn*pi), ([1;1]*ylim).', '-r')
plot(xlim, -0.5*[1 1], ":k")
hold off
grid
ylim([-10, 0])
legend('Passband','Cutof Frequencies', 'Location','S')

.