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.
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
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.