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I have the following transfer function:

$$P(s) = \frac{3}{(s-1)(s+2)(s+3)}, s= j\omega$$

I got the starting and endpoints:
$$\omega_0 = -\frac{1}{2}, \omega_\infty = 0$$

When I split the equation up in real -and imaginary parts I get this:
$$ P(j\omega) =\frac{ -12\omega^2-18 }{ \omega^6+14\omega^4+50\omega^2+36 } - j\frac{ 3(\omega-\omega^3) }{\omega^6+14\omega^4+50\omega^2+36}$$

But there must be something wrong. There must be a faster way, because this is an example of an exam and I don't have the time to calculate through this big equation...

I also cannot detect the zero crossing at $-0.3$ like you can see in this plot.

So how can I evaluate the curve?
Is there something like a standard technique?

Edit: I made a mistake in my calculation, now its correct.
But I found out that it is enough to just separate the denominator in real and imaginary parts for getting the intersections on each axes.

$$ P(j\omega) = \frac{ 3 }{ -6 -4\omega^2 +j\omega(1-\omega^2) } $$

Now I just need to know how I can draw the curve. :)
I think you can read it from the phase of the bode plot, but I don't know.

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you should add a few words about the definitions of the terms you are using - people without degrees in EE are generally more familiar with terms like "Laplace Transform" then they are with "Transfer Function" . – nbubis Jan 14 '13 at 10:16
up vote 4 down vote accepted

Rewrite $P(j\omega)$ as $P(j\omega)=\text{Re}\{P(j\omega)\}+j\;\text{Im}\{P(j\omega)\}$:

$$ \begin{equation*} \frac{3}{-6-4\omega ^{2}+j\omega (1-\omega ^{2})}=3\frac{-6-4\omega ^{2}}{\left( -6-4\omega ^{2}\right) ^{2}+\left( \omega -\omega ^{3}\right) ^{2}}-3\frac{\omega -\omega ^{3}}{\left( -6-4\omega ^{2}\right) ^{2}+\left( \omega -\omega ^{3}\right) ^{2}}j \end{equation*} $$

The real and imaginary parts are

$$ \text{Re}\{P(j\omega)\}=3\frac{-6-4\omega ^{2}}{\left( -6-4\omega ^{2}\right) ^{2}+\left( \omega -\omega ^{3}\right) ^{2}}$$


$$ \text{Im}\{P(j\omega)\}=-3\frac{\omega -\omega ^{3}}{\left( -6-4\omega ^{2}\right) ^{2}+\left( \omega -\omega ^{3}\right) ^{2}}. $$

EDIT. The plot I got in SWP was this one (the horizontal axis is the real part and the vertical axis is the imaginary part) for $-500\le\omega\le 500$:

enter image description here

ADDED. Since the curve is parametrized in terms of the variable $\omega$, we can choose some values of $\omega$ and compute the corresponding values of $\text{Re}\{P(j\omega)\}$ and $\text{Im}\{P(j\omega)\}$.

ADDED2. Numerically I got

$$\begin{array}{ccc} \hline \omega & \text{Re}\left\{ P\left( j\omega \right) \right\} & \text{Im}\left\{ P\left( j\omega \right) \right\} \\ \hline -50 & -1.910\,4\times 10^{-6} & -2.385\,7\times 10^{-5} \\ -10 & -1.063\,8\times 10^{-3} & -2.594\times 10^{-3} \\ -5 & -1.240\,4\times 10^{-2} & -1.404\,3\times 10^{-2} \\ -1 & -0.3 & 0 \\ 0 & -0.5 & 0 \\ 1 & -0.3 & 0 \\ 5 & -1.240\,4\times 10^{-2} & 1.404\,3\times 10^{-2} \\ 10 & -1.063\,8\times 10^{-3} & 2.594\times 10^{-3} \\ 50 & -1.910\,4\times 10^{-6} & 2.385\,7\times 10^{-5} \\ \hline \end{array}$$ And some more values: $$\begin{array}{ccc} \hline \omega & \text{Re}\left\{ P\left( j\omega \right) \right\} & \text{Im}\left\{ P\left( j\omega \right) \right\} \\ \hline -4 & -2.470\,6\times 10^{-2} & -2.117\,6\times 10^{-2} \\ -3 & -5.384\,6\times 10^{-2} & -3.076\,9\times 10^{-2} \\ -2 & -0.126\,92 & -3.461\,5\times 10^{-2} \\ 0 & -0.5 & 0 \\ 2 & -0.126\,92 & 3.461\,5\times 10^{-2} \\ 3 & -5.384\,6\times 10^{-2} & 3.076\,9\times 10^{-2} \\ 4 & -2.470\,6\times 10^{-2} & 2.117\,6\times 10^{-2} \\ \hline \end{array}$$

Remark: to improve the efficiency of the computation notice that $$\text{Re}\left\{ P\left( j\omega \right) \right\}=\text{Re}\left\{ P\left( j(-\omega \right)) \right\}$$ and $$\text{Im}\left\{ P\left( j\omega \right) \right\}=-\text{Im}\left\{ P\left( j(-\omega \right)) \right\}.$$

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Thank you, but is there a way to get the vague form of the curve. You gave me the exact solution, but in my exam I am only allowed to use a simple calculator, and so it takes a lot of time just for getting one coordinate. – madmax Jan 14 '13 at 14:15
@madmax You are welcome. The curve is parametrized in terms of $\omega$. I would choose a few values of the variable $\omega$ and compute the corresponding Re(P(jω))- and Im(P(jω)-values. – Américo Tavares Jan 14 '13 at 14:24

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