# Bode plot and poles

When we draw a bode plot: is the dominant pole the biggest one or the smallest one?

What happens if there are positive poles or zeros?

Is it possible to implement a filter with positive poles and zeros?

• Poles in the right half of the plane means you have an oscillator... exponentially growing function. I think the dominate pole is the lowest frequency one. – George Herold Feb 8 '17 at 20:03

1) There are no "small" and "large" poles. A pole is at a certain frequency and at that frequency the bode plot takes an extra 20dB/decade roll-off over frequency.

2) When there are 2 poles are at the same frequency then each adds that same 20 dB/decade roll-off so they add up to 40 dB/decade roll-off in total (for frequencies higher than the frequency of those poles.

3) The same is valid for more than 2 poles.

4) Often for low-pass behaviour the lowest frequency pole is called the "dominant pole" if the next pole or poles are at a considerably higher frequency than the lowest frequency pole so that the system can be treated as having only one pole. This is useful for systems with feedback as such a one-dominant pole system is often (but not always !) inherently stable.

5) Systems with poles in the right half plane have signals which amplify themselves. These we call oscillators. Filters with poles in the right half plane generate their own signals so they're not usable as a proper filter.

Is it possible to implement a filter with positive poles and zeros?

It is possible to build a system like that but it would not behave like a filter since it would be oscillating by itself. For a filter that makes it useless.

(1) Zeros: We must distinguish between (a) "Allpole filters" (e.g. Butterworth, Chebyshev, Bessel characteristics) which have only zeros in the left half of the s-plane (LHP) and (b) "Real-zero filters" (Chebyshev-inverse, elliptical Cauer characteristics) which have zeros on the imag. axis of the s-plane. Furthermore, we have (c) "allpass filters" with a conjugate-complex pair of zeros in the right half of the s-plane (RHP).

(2) Poles: Stable systems have poles in the left half (LHP) only. Any complex pole pair in the RHP will cause rising oscillations. A pair of poles which are (idealized case) positioned directly on the imag. axis will cause continuous oscillations with a constant amplitude. A real pole in the RHP will drive the amplifier immediately into saturation (latch-up).

Explanation/Comment: The above explanations can be easily understood if we switch to the time domain. From system theory we know that the roots of the transfer functions denominator (poles) are identical to the solutions of the corresponding differential equation (characteristic equation) in the time domain. That means: A pole s=a+jw in the frequency domain will cause an exponential expression exp(a+jw)=exp(a)*exp(jw) in the time domain.

As we can see, the real part a must be negative for a decaying function (whereas the expression exp(jw) gives the corresponding frequency). For pos. values of a the amplitudes of the corresponding frequency will rise until it is limited by external restrictions (supply voltages).