Convention for Transfer Function Poles and Zeros

Question

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

Answer 1

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.

Answer 2

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.