# Butterworth filter higher order

I am new (very beginner) to signal processing and I am trying to understand how the Butterworth filter operates.

According to my current understanding of filters, there seems to be a tradeoff between the the passband ripples and the transition band. Like if the former decreases, the latter increases and vice-versa. Now this makes sense from the definition of the Butterworth filter - a maximally flat frequency response at the expense of a wider transition band.

What I don't understand is that how is it possible that by increasing the order of the Butterworth filter, the transition band will decrease but the passband remains flat??

Also what does an unstable filter mean?

• Welcome to SE.SP! Butterworth filters do not have "ripples" in the passband. That's the point of maximal flatness: the first and higher order derivatives are all zero at a given frequency. Can you explain what you mean by "ripples" ?
– Peter K.
Commented Jun 19, 2021 at 21:15
• by ripples i mean fluctuations in the passband. So even if you keep increasing the order, the passband will remain flat? but then why don't we just keep increasing the order of the BW until we reach the ideal 'brickwall' filter in practice?
– Dumb
Commented Jun 19, 2021 at 21:22
• The the delay, computational cost, and sensitivity to numerical noise (quantization and rounding causing the calculation to "blow up") become greater with a higher order filter. So the ideal brickwall would take forever, and require infinite precision arithmetic thus requiring more petabytes than exist in any computer. Commented Jun 20, 2021 at 0:14

According to my current understanding of filters, there seems to be a tradeoff between the the passband ripples and the transition band.

No. Butterworth filters have neither passband nor stopband ripples.

What I don't understand is that how is it possible that by increasing the order of the Butterworth filter, the transition band will decrease but the passband remains flat??

The magnitude squared of the transfer function of an analog Butterworth lowpass filter of order $$N$$ with a cutoff frequency of $$f_0$$ is

$$|H(f)|^2 = \frac{1}{1+(f/f_0)^{2N}}$$

The passband is determined $$f/f_0 \ll 1$$ in which case we simply get $$|H(f)| \approx 1$$, so there is no ripple .

For $$f/f_0 \gg 1$$ we get $$|H(f)| \approx (f_0/f)^{N}$$ so it falls with the frequency raised to the order of the filter. The higher the order, the quicker it falls.

Please note that is $$|H(f)|^2$$ falls monotonic over all frequencies. There are no ripples anywhere.

Also what does an unstable filter mean?

Certain types of filters (like a Butterworth) are recursive, i.e. they feed part of their output back into the input. This can create a "run-away" condition. Let's look a simple difference equation

$$y[n] = x[n] + 0.5\cdot y[n-1]$$

If we feed a unit impulse into this system the output will be 1, 0.5, 0.25, 0.125 ... It keeps going for a while but will die down over time.

Now let's look at an unstable example.

$$y[n] = x[n] + 2\cdot y[n-1]$$

For the same unit impulse input the output will be 1, 2, 4, 8, 16, 32, ... so there is unbounded growth. That's unstable.

Answering part #2 of your question: In a well conditioned numerical computation, the quantization and rounding errors stay small with respect to the size of the desire result. An unstable filter can do the opposite, where tiny numerical noise incurred during doing the DSP arithmetic can rapidly increase until it becomes vastly larger than any desired result. Since real computers have finite memory, an unstable filter might make a garbage result impossible to avoid.