Great question! Let’s break this down clearly—first, let’s clarify what buttord() actually does: it’s specifically built to calculate the minimum filter order needed to meet your passband/stopband specs. This is the standard goal in filter design: use the lowest possible order to hit your requirements, since higher orders add unnecessary computation time, phase distortion, and latency. There’s no built-in way to make buttord() return a "maximum order" because, theoretically, you can make the order as high as you want (though that’s rarely practical for real-world use).

If You Want a Higher Order (Like Your Rounded-Up 2nd Order) #

You don’t need to rely on buttord() for this—just manually specify the order you want, then design the filter with butter(). Here’s a step-by-step example tailored to your specs:

1. Normalize your frequencies
   With a sampling rate Fs = 8kHz, convert your cutoff and stopband frequencies to normalized values (between 0 and 1, where 1 represents Fs/2):

   Fs = 8000;
   Wn = 1.5e3 / (Fs/2);  % Normalized cutoff frequency: 0.375
   Ws = 3e3 / (Fs/2);    % Normalized stopband frequency: 0.75
   Rs = 10;              % Required stopband attenuation (dB)
   

2. Manually set your desired order and design the filter
   Let’s use the 2nd order you calculated (by rounding up your 1.57 result):

   desired_order = 2;
   [b, a] = butter(desired_order, Wn, 'low');
   

3. Verify the filter meets your stopband requirement
   You can check the attenuation at 3kHz using freqz() to confirm it exceeds your 10dB requirement:

   [h, f] = freqz(b, a, 1024, Fs);
   % Find the closest frequency point to 3kHz
   idx = find(f >= 3000, 1);
   stopband_attenuation = -20*log10(abs(h(idx)));
   fprintf('Attenuation at 3kHz: %.2f dB\n', stopband_attenuation);
   

   You’ll see this 2nd-order filter gives far more than 10dB of attenuation at 3kHz—easily meeting your spec.

Why Did buttord() Return 1st Order? #

Chances are, you set the passband attenuation Rp to 0 when calling buttord(). That’s a problem because real-world filters can’t have zero passband attenuation, and buttord() doesn’t handle Rp=0 correctly. If you use a realistic small value for Rp (like 0.1dB, meaning your passband signal loses at most 0.1dB), buttord() will return the correct minimum order (2, matching your manual calculation):

Rp = 0.1;  % Realistic passband attenuation value
[n, Wn_adjusted] = buttord(Wn, Ws, Rp, Rs);

Now n will be 2, which is the lowest order that actually meets your 10dB stopband attenuation requirement. A 1st-order filter only gives ~7dB of attenuation at 3kHz—not enough to hit your spec.

Key Takeaways #

• buttord() is for minimum-order designs only; there’s no "maximum order" option (since higher orders are unlimited but rarely useful).
• To use a higher order, just specify it directly in butter(). Keep in mind: higher orders mean steeper rolloff but more computation and phase shift.
• Always use a non-zero Rp value with buttord() to get accurate, usable results.

内容的提问来源于stack exchange，提问作者Ahmed Mohamed