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I have a question about drawing a magnitude characteristics of H(s).

So my transmittance function is for example:

enter image description here

and in the book the magnitude characteristics is shown as:

enter image description here

And my question is - how to count the value of abs(H(jΏ) ?

I understand that s = 1 is a zeros (local minima), and this function has two poles: -1-j and -1 +1 (local maxima), but I can't count the values highlighted in yellow.

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The most direct way is to turn your transfer function into frequency response by setting $$s = j\Omega$$ and then $$H(j\Omega) = \frac{j\Omega - 1}{-\Omega^2 + 2j\Omega + 2}$$ So for $\Omega = 0$, you get $$H(0) = -\frac{1}{2}$$ which gives $|H(0)| = \frac{1}{2}$ and is in agreement with your figure. Similarly, for $\Omega = \pm 1$, you get $$H(\pm j1) = \frac{\pm j - 1}{-1 \pm 2j + 2} = \frac{\pm j -1}{1 \pm 2j}$$ You can show that $|H(\pm j1)| = \sqrt{2/5}$, which also agrees with your magnitude response plot.

However, these are only some interesting values of the magnitude response. If you need to draw the magnitude response in greater detail without using some software to do this for you, you have to go for the geometric approach. Here is a good place to start.

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  • $\begingroup$ Taking the modulus of each complex number: it's $|H(\pm j1)| = \frac{|\pm j - 1|}{|1 \pm 2j|} = \frac{\sqrt{2}}{\sqrt{5}} = \sqrt{2/5}$ $\endgroup$ – GKH Jan 28 at 23:19
  • $\begingroup$ I am grateful, tanks a lot! $\endgroup$ – Krzysztof Bolek Jan 28 at 23:22

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