What is an "origin pole"? - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-09-15T18:11:08Z https://electronics.stackexchange.com/feeds/question/194374 https://creativecommons.org/licenses/by-sa/4.0/rdf https://electronics.stackexchange.com/q/194374 2 What is an "origin pole"? scanny https://electronics.stackexchange.com/users/70923 2015-10-09T07:04:28Z 2015-10-09T20:38:01Z <p>I'm studying Christophe Basso's book <em>Designing Control Loops for Linear and Switching Power Supplies</em>.</p> <p>In the book, he often uses the term "origin pole". This is what I think I understand about it so far:</p> <ul> <li>When a transfer function contains an "integrating" element, that element represents an <em>origin pole</em>. 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:</li> </ul> <p>$$\frac{1}{sR_2C_2}\text{ from, say, }\frac{(1+sR_1C_1)}{sR_2C_2(1+sR_3C_3)}$$</p> <ul> <li><p>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.</p></li> <li><p><em>(Not completely sure about this bit):</em> 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<br> \$\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 <em>breakpoint</em>.</p></li> </ul> <p>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 :) :</p> <p><a href="https://i.stack.imgur.com/sXhdY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sXhdY.png" alt="enter image description here"></a></p> <p>so this idea kind of threw me for a loop (no pun intended :) while I've been working to make sense of the book.</p> <p>So here's my question:</p> <ul> <li>Am I understanding this correctly so far?</li> <li>Do other folks use the term <em>origin pole</em> or is it something Christophe has introduced? The term doesn't seem to pull up too much on search.</li> <li>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? :)</li> </ul> https://electronics.stackexchange.com/questions/194374/-/194388#194388 6 Answer by Bimpelrekkie for What is an "origin pole"? Bimpelrekkie https://electronics.stackexchange.com/users/77865 2015-10-09T09:09:25Z 2015-10-09T20:38:01Z <p>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.</p> <p>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.</p> <p>\$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!</p> <p>I call this \$1/s\$ an <strong>integrator</strong> or <strong>pole at zero</strong>, they are useful in feedback loops to eliminate static errors. Almost every <a href="https://en.wikipedia.org/wiki/Phase-locked_loop" rel="nofollow">PLL</a> 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.</p>