I'm studying Christophe Basso's book Designing Control Loops for Linear and Switching Power Supplies.

In the book, he often uses the term "origin pole". This is what I think I understand about it so far:

  • When a transfer function contains an "integrating" element, that element represents an origin pole. An integrating element is a denominator element with an \$s\tau\$ factor on its own, one that is not part of a \$(1 + s\tau)\$ factor. This is consistent with the idea that the Laplace transform for an integral is \$1/s\$. \$\tau\$ is commonly an RC time constant. This would be an example of an integrating element:

$$\frac{1}{sR_2C_2}\text{ from, say, }\frac{(1+sR_1C_1)}{sR_2C_2(1+sR_3C_3)}$$

  • Mathematically, an origin pole has infinite gain at DC (\$s = 0\$), from which "point" the gain declines at 20dB/decade. In practice, this rise to infinity is halted at some point, such as when the available gain of the op amp is reached.

  • (Not completely sure about this bit): The gain curve of the origin pole, if unaffected by other poles or zeros, crosses 0dB at \$\omega_o\$, the frequency of the pole, \$\frac{1}{2\pi\tau}\$, which is perhaps typically
    \$\frac{1}{2\pi RC}\$. This is markedly different than a "regular" pole, whose \$\omega_p\$ is the point of a downward inflection in the gain, a so-called breakpoint.

Before starting the book, I thought that all poles were located at a 3dB breakpoint and looked like this, but maybe I slept through the day origin poles were mentioned in class :) :

enter image description here

so this idea kind of threw me for a loop (no pun intended :) while I've been working to make sense of the book.

So here's my question:

  • Am I understanding this correctly so far?
  • Do other folks use the term origin pole or is it something Christophe has introduced? The term doesn't seem to pull up too much on search.
  • Is there anything else interesting about origin poles that I and other curious readers yearning for knowledge might like to know, particularly in the realm of control loop transfer functions? :)
  • 2
    \$\begingroup\$ scanny-I could not detect any error or misunderstanding within your text. However, as you probably know such a pole at the origin (another term for "origin pole") is not possible in reality. \$\endgroup\$ – LvW Oct 9 '15 at 7:35
  • \$\begingroup\$ Chu-each real integrator has a lowpass response with a FINITE and very low cut-off frequency; only IDEAL integrators (not realizable) have a pole at the origin. \$\endgroup\$ – LvW Oct 9 '15 at 8:13
  • \$\begingroup\$ @LvW, going from velocity to displacement on a motor shaft is a pure integration. \$\endgroup\$ – Chu Oct 9 '15 at 8:18
  • 1
    \$\begingroup\$ I think it's just shorthand for "a pole at the origin on the graph" - i.e. at zero frequency as you thought.. \$\endgroup\$ – Brian Drummond Oct 9 '15 at 10:57
  • 1
    \$\begingroup\$ I've never heard the term origin pole, but the integrating response has always reminded me of the open loop gain of an (compensated) opamp. (I don't know if that adds any understanding.) \$\endgroup\$ – George Herold Oct 9 '15 at 13:37

The "origin pole" is indeed the \$1/s\$ term in the transfer function \$H(s)\$. In the bode plot it results in a first order transfer that does NOT flatten out for low frequencies.

Your Bode plot is that of a low pass filter $$H(s) = \frac{1}{1 + s}$$ Note how this \$H(s)\$ would result in \$H(0) = 1 = 0\text{ dB}\$ like in your Bode plot.

\$H(s) = 1/s\$ is different, \$H(0) = \infty\$! In theory at least. So the -20 dB /decade line in the Bode plot keeps going on forever to both sides. Note that a Bode plot has a logarithmic X-axis, where would that place the 0 Hz point ? At minus infinity!

I call this \$1/s\$ an integrator or pole at zero, they are useful in feedback loops to eliminate static errors. Almost every PLL has an integrator consisting of a charge-pump (switched current source) feeding current into a capacitor. What happens to the capacitor's voltage when you feed a current into it? Yes, it keeps rising forever. That's integrator behavior.

  • \$\begingroup\$ @Null Thanks once again for the nicer formatting :-) \$\endgroup\$ – Bimpelrekkie Oct 9 '15 at 14:02
  • \$\begingroup\$ You're helping me get the Copy Editor badge. :) BTW, a good reference for MathJax formatting can be found on Math.SE meta. There's a related thread on EE.SE meta. Note that this site uses \$ instead of $ for inline LaTeX/MathJax (see here). \$\endgroup\$ – Null Oct 9 '15 at 14:24
  • \$\begingroup\$ Should be -20 dB/decade rather than -10, shouldn't it? \$\endgroup\$ – scanny Oct 9 '15 at 17:56
  • 1
    \$\begingroup\$ @scanny You got me, it totally should be ! Just checking if you were paying attention ;-) No, just my mistake, I'll update the answer. \$\endgroup\$ – Bimpelrekkie Oct 9 '15 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.