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I have seen quite many of this type of question and I want to find how to solve them but I haven't cracked it yet. For example this is one: enter image description here

My thoughts so far: Replace $$s \rightarrow j\omega$$, $$G(j\omega) = \frac{k(a-j\omega)^2(1-j\omega)^n}{(a^2+\omega^2)(1+\omega^2)^n}$$
For $$\omega \rightarrow \infty$$ it is: $$G(j\infty)=\frac{k\omega^2*(-1)^n\omega^n}{\omega^{4+2n}}$$ $$=\frac{k(-1)^n}{\omega^{2+n}}$$ which is zero for $$n \geq -2$$ and infinite otherwise. But we can see that there is no point at which the plot approaches infty, so it has to be $$G(j\infty)=0$$
We also can see that $$Re(G(j0.5) ) = 0 $$
My problem is that with that not know parameter n , I cannot split the transfer function into Real and Imaginary part so there must be some hint to find the parameter n, before anything else. Is there a way to identify how many poles there are by just looking at the plot?

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From the Nyquist plot we see that the maximum angle, $\angle H(s) = -360^{\circ}$. As the transfer function $$ H(s) = \frac{k}{(s+a)^2(s+1)^n} $$ has no zeros and all the poles are in the open left half-plane (LHP) as $a > 0$, the phase of $\angle H(s) = -360^{\circ}$ can only be achieved if the transfer function has 4 poles in the LHP, i.e. $n = 2$.

The other two variables can then be determined by substituting $s$ with $j\omega$ and solve the equation for the following two constraints $$ \begin{align} \mathrm{Re}\left\lbrace H(j\omega_0)\right\rbrace &= 0, \quad \omega_0 = 0.5\\ \mathrm{Im}\left\lbrace H(j \omega) \right\rbrace &= -2.\\ \end{align} $$

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