I'm learning about transfer functions, and I'm trying to understand the convention for getting poles and zeros from the transfer function. Let's say I have a transfer function:

\$H(s) = \frac{1}{(s+3)(s+2)} = 1/6*\frac{1}{(s/3+1)(s/2+1)}\$.

Are the poles of this transfer function -3 and -2, or +3 and +2?

Looking at links like http://en.wikipedia.org/wiki/Pole%E2%80%93zero_plot, it would seem that poles are supposed to make the denominator zero, so that would suggest the former.

But looking at links like http://www.onmyphd.com/?p=bode.plot, the poles would +3 and +2.

Are there two different conventions? Thanks


The poles in your \$H(s)\$ are \$s = -3\$ and \$s = -2\$ because they make the denominator zero. I'm not sure why you think the Bode plots suggest the poles are positive, but perhaps your confusion has to do with the fact that a Bode plot uses \$j\omega\$ as the \$x\$-axis where \$\omega\$ is the angular frequency. The poles are on the real (\$x\$) axis in the \$s\$-plane so they are symmetric about the imaginary (\$y\$) axis, meaning the Bode plot is the same whether the frequency \$\omega\$ is positive or negative.

The source of the confusion may also be due to the fact that there is an error in the second link you posted. The author uses the form


for the transfer function but claims that the poles are at \$s = p_0\$, etc. This is incorrect in general because at \$s = p_0\$ the relevant term of the denominator is \$p_0/p_0 + 1 = 1 + 1 = 2 \neq 0\$. The pole is actually at \$s = -p_0\$ so that the relevant term of the denominator is \$-p_0/p_0 + 1 = -1 + 1 = 0\$. The author either meant to say that the poles were at \$s = -p_0\$, etc., or use the form \$s/p_0 - 1\$ for each term.

  • \$\begingroup\$ Hmm, But according to the second link, if a transfer function is of the form: \$H(s) = A\frac{(s/z_0 + 1)(s/z_1 + 1)\cdots(s/z_n + 1)}{(s/p_0 + 1)(s/p_1 + 1)\cdots(s/p_n + 1)}\$, then the poles are \$p_0\$, etc. Wouldn't that make the poles +3, +2? \$\endgroup\$ Aug 25 '14 at 3:45
  • \$\begingroup\$ I managed to find a document which might help you visualize the s-plane vs. the Bode plot. See p. 3. The LHP case is the closest to your example except it has one pole instead of two. The key is to look where the w is pointing in the two plots. Also, in the s-plane the magnitude of the frequency is inversely proportional to the distance to the pole -- so you can see that it drops off as w --> infinity or w --> -infinity. \$\endgroup\$
    – Null
    Aug 25 '14 at 3:47
  • \$\begingroup\$ @AbubakarAbid In your new H(s) the poles are -p_0, etc. You need s/p_0 = -1 for the denominator to be 0. \$\endgroup\$
    – Null
    Aug 25 '14 at 3:49
  • \$\begingroup\$ I see. What you're saying makes a lot of sense, and thank you for the explanation. But if you look at the second link I put (onmyphd.com/?p=bode.plot), it says very clearly \$p_0\$, \$p_1\$ are the poles. Is this wrong? (I've also seen this in other places..) \$\endgroup\$ Aug 25 '14 at 3:50
  • 1
    \$\begingroup\$ Ah, I see that now. Yes, that is a mistake on the author's part. At \$s = p_0\$, etc. the denominator has a term of 2 instead of 0 and there is not a pole there. The author either meant to say \$-p_0\$, etc., or use the form \$(s/p_0 - 1)\$ for each term. \$\endgroup\$
    – Null
    Aug 25 '14 at 3:55

Poles are the values of "s" that make the denominator zero. In both expressions you get that.

  1. s+3=0; s=-3; s+2=0; s=-2;

  2. s/3+1=0; s/3=-1; s=-3; s/2+1=0; s/2=-1; s=-2;

In the second link I can't find anything that suggests the opposite. What it does is to show how to draw an asymptotic bode plot.

In the expression |H(s)|=A ... it does say that p0, p1, p2... are the pole values, but pole values would be -p0, -p1, -p2 and so on.


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