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Wolfram plot of $\frac{s}{1-s}$ is $\pm\infty$ at $s=1$. But, bode plot of $\frac{s}{1-s}$ results in $1$ at $s=1$. Obviously, this is wrong. Why?

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    $\begingroup$ You could look at what a bode plot is. $\endgroup$ – Peter Shor May 27 '13 at 18:43
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    $\begingroup$ @PeterShor I think the asker's question is valid. Even from reading the wiki article, it's not entirely clear what's happening here if you're new to Bode plots and frequency domain. $\endgroup$ – Ataraxia May 27 '13 at 18:47
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The x-axis in the Bode plot isn't $s$, but $\omega$. Remember that $s=j\omega$, so what you're seeing in the Bode plot is as follows:

$$20\log|H(j1)|=20\log|\frac{j1}{1-j1}|=20\log|\frac{j(1+j)}{2}|$$

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  • $\begingroup$ Wow. Thanks. I see that I can choose W=1 and get one plot or choose s=1 and, then, get w=-i and another plot. Which one should I choose? Can I understand the difference between Laplace and Fourier based on that? $\endgroup$ – Val May 27 '13 at 18:57
  • $\begingroup$ @Val Most likely $\omega=1$ will prove the most useful. See, the difference between $s$ and $j\omega$ is $s$ is a complex number, whereas $j\omega$ is only an imaginary number. The real component of $s$ is what is known as the damping frequency, whereas the imaginary part is the frequency that most of us are familiar with: the periodicity. The chances are, if you're looking at a Bode plot, you're more interested in the latter. $\endgroup$ – Ataraxia May 27 '13 at 19:09

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