# Pole at infinity and zero at finite frequency

Is there any circuit that has a pole at an infinite frequency and zero at some finite frequency? A first-order RC Low pass filter has a finite pole and zero at infinity and it makes sense, I am just curious if another way around is possible in circuits?

• Theoretically -An ideal differentiator using op-amp but practically filter will be saturated after certain value of frequency and doesn't behave linearly . – user215805 2 days ago
• 'zero at finite frequency' makes no sense, you mean 'a finite zero' – Chu 2 days ago
• @user215805 Thank you! – sumita sahu 2 days ago

a pole at an infinite frequency ...

No.

RC Low pass filter has a finite pole and zero at infinity and it makes sense

Low pass will be $$\ H(s) = \frac{1}{1+s}\$$ and it has no zero. It is not the same as "zero at infinity" , which is not really a thing, and would be $$\ G(s) = \frac{s-\infty}{1+s}\$$ and for any $$\s \in \mathbb{C}\$$ it would be ill defined.

• This accepted answer says otherwise -electronics.stackexchange.com/questions/159939/… – user215805 Apr 8 at 16:48
• @user215805 the difference start with this OP asking if there is a circuit that does that, which is no. Second, I agree with the linked answer about the behavior as s->infinity. But I disagree with the analysis saying s=infinity then we will have a zero/pole, because $\infty \not \in \mathbb{C}$ – jDAQ 2 days ago
• Isn't in some context frequency used for imaginary part of S and pole at infinite frequency mean gain (magnitude of frequency response ) tend infinite when frequency tends to infinite ? – user215805 2 days ago
• @user215805 Transfer functions are s-space and include both frequency as well as exponential decay (or its opposite, rarely.) (Multiplication in s-space involves both rotation as well as scaling.) So I think jDAQ is accurate in stating that $\infty$ is ill-defined here. That said, transfer functions are often examined only for their frequency response, setting $\sigma=0$, so that may also be a fair question. But then it's changing the question from Laplace to Fourier, so to speak. I prefer jDAQ's rigorous approach, yet I see the OP specifying "frequency," too. So I'm on the fence about it. – jonk 2 days ago
• I'm very confused on what is the issue here. The poles and zeros are defined with respect to H(s) even though people plot its bode plot and use frequency responses ( H(jw) ) to analyze and design their circuits. One rough "definition" of poles and zeros using H(jw) would be looking at the bode plot and seeing if at a certain point the slope of the plot changes. But still, an infinity frequency would never "show up" in the bode plot, so it only makes sense to describe what happens as s->infinity (but not s=infinity) or how it behaves as you change one of the poles/zeros->infinity. – jDAQ 2 days ago