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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

Michael Karas
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Curious Student
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2 Answers2

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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

$$H(s)=A\frac{(s/z_0+1)(s/z_1+1)\cdots(s/z_n+1)}{(s/p_0+1)(s/p_1+1)⋯(s/p_n+1)}$$

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.

Null
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  • 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? – Curious Student Aug 25 '14 at 03:45
  • I managed to find a [document](http://ece.wpi.edu/~mcneill/handouts/Bode_plot_guide.pdf) 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`. – Null Aug 25 '14 at 03:47
  • @AbubakarAbid In your new `H(s)` the poles are `-p_0`, etc. You need `s/p_0 = -1` for the denominator to be 0. – Null Aug 25 '14 at 03:49
  • 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 (http://www.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..) – Curious Student Aug 25 '14 at 03:50
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    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. – Null Aug 25 '14 at 03:55
  • I am new to the topic. I understand that the poles and zeros are in the complex plane while the Bode plot only concerns one axis (the imaginary one for s). Now I saw that poles are marked.on the Bode plot, how is that possible? For.instance https://www.mathworks.com/matlabcentral/answers/223008-mark-poles-and-zeroes-on-bode-plot – gamebm Jun 18 '22 at 15:31
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    @gamebm The poles on a Bode plot are "marked" at the frequency at which they are closest to the \$j\omega\$ axis and thus cause the transfer function to reach its peak. However, the poles are on the real rather than the imaginary (i.e. \$j\omega\$) axis so they are not really on the Bode plot -- if they were the transfer function would be infinity at that frequency. – Null Jun 19 '22 at 03:11
  • Thanks, so if I understand correctly, the poles are indicated on the Bode plot using their real part (of $s$, namely, the distance to the origin along the real axis), the corresponding value is indicated on the frequency-axis of Bode plot which is actually the imaginary part (of $s$). As you said, if the pole or zero locates exactly where they are indicated on the plot, the magnitude must either diverge or vanish. This is a convention, right? I searched google for an entire day but cannot find any where this is mentioned.... – gamebm Jun 19 '22 at 15:17
  • @gamebm It's hard to explain in words -- try looking at the [image on this other answer](https://electronics.stackexchange.com/a/160144/51760) which shows a Bode plot along with a pole-zero plot and a 3D representation of the pole-zero plot with the Bode plot on it. – Null Jun 21 '22 at 14:00
  • Thinks, I have read those posts and am aware of how a pole or zero in the complex plane will affect the value of a function on the axis, in a similar way how that poles of the frequency-domain Green's function works, in the context of theoretical physics. I did need the convention (in some textbook or explicit written form) how those poles are marked on the Bode plot, so that I will be sure about my understanding, as they did not indicate the real location, which seems to be practice in some paper I read. – gamebm Jun 21 '22 at 15:25
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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.

ehilario
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