Chances are you have all 4 powers on the right. But you really need to write everything in terms of the common denominator to see it.1eqinfinity wrote:Z1202 wrote:You can't have an absolutely desired transfer function, because you don't have a choice in terms of the denominator (aside from being able to cancel some of its factors). But you can choose the numerator freely.

Hmm...

As far as I understand the desired transfer function absolutely must be expressable by the powers of s available in stages' transfer functions on the right. If the desired H has some power that is not present on the right, it is impossible to derive it from the current topology, right?

E.g. for transistor ladder the numerators are

s^4+4s^3+6s^2+4s+1

s^3+3s^2+3s+1

s^2+2s+1

s+1

1

By combining them together you can express any polynomial of s.

Just from the general intuitive considerations the same should be possible for the diode ladder, although I didn't check it.

It's just that for transistor ladder the numerators have a nicer form (s+1)^n. For diode ladder you should have less regularity.

If you want to have s^5, that of course wouldn't be possible. You can't have a larger power in the numerator than in the denominator in an integrator-based filter.