I have a Nyquist Diagram that looks like this:
The angular frequency at point A is $\sqrt 2$ rad/s. Now I would like to derive the transfer function based on the diagram. Now, I am given the choices of the open-loop transfer function as below:
\begin{align}
(a)L(s) = \dfrac{s+k}{s(s^2+as+b)}\\
(b)L(s) = \dfrac{k}{s(s^2+as+b)}\\
(c)L(s) = \dfrac{s+k}{s^2+as+b}\\
(d)L(s) = \dfrac{k}{s^2+as+b}
\end{align}
Currently, as far as I know, the place on the real-axis with the number 3 should have $\omega$ = 0. If I replace $L(s)$ with $L(j\omega)$ and put in $\omega$ with 0, then options (a) and (b) will not satisfy because the results of them will be infinity. However, I am not certain how to derive the full transfer function in a standard way and find k, a, b. I have tried to use (c) and (d) and replace $\omega$ with $\sqrt 2$, but it failed either.
This is how I did when trying to get the transfer function based on option (c) and (d):
\begin{equation}
L(j\sqrt 2) = \dfrac{k}{(b-2)+ja\sqrt 2} * \dfrac{(b-2)-ja\sqrt 2}{(b-2)-ja\sqrt 2} = \dfrac{k[(b-2)-ja\sqrt 2]}{(b-2)^2 + 2a^2} = 0 – j\dfrac{3}{4}\sqrt 2
\end{equation}
\begin{equation}
L(j\sqrt 2) = \dfrac{k + j\sqrt 2}{(b-2)+ja\sqrt 2} * \dfrac{(b-2)-ja\sqrt 2}{(b-2)-ja\sqrt 2} = \dfrac{[k + j\sqrt 2][(b-2)-ja\sqrt 2]}{(b-2)^2 + 2a^2} = \dfrac{k(b-2) + 2a – j(k-b+2)\sqrt 2}{(b-2)^2 + 2a^2} = 0 – j\dfrac{3}{4}\sqrt 2
\end{equation}
From option (d), it looks like k, a, and b should satisfy the following equations:
\begin{equation}
\begin{cases}
k(b-2) + 2a = 0\\
(b-2)^2 + 2a^2 = 4\\
k-b = 1
\end{cases}
\end{equation}
With option (c), this seems not to make sense because I think a could not be an imaginary number in this case. As for option (d), the equation, there seems no value pairs meet at the same time…, What did I miss?
Please help me with this question. Thank you for any advice.
Best Answer
With options (a) and (b),
$$ \begin{align*} L_a(jω) & = \dfrac{jω + k}{jω(-ω^2 + jaω + b)}, L_b(jω) = \dfrac{k}{jω(-ω^2 + jaω + b)} \\ L_a(j0) & = ∞, L_b(jω) = ∞ \\ \end{align*} $$
Therefore, options (a) and (b) do not correspond to the graph.
With option (c) and (d), $$ \begin{align*} L_c(jω) & = \dfrac{jω + k}{(b-ω^2) + jaω} * \dfrac{(b-ω^2) - jaω}{(b-ω^2) - jaω} = \dfrac{(jω + k)(b-ω^2-jaω)}{(b-ω^2)^2 + (aω)^2} \\ L_d(jω) & = \dfrac{k}{(b-ω^2) + jaω} * \dfrac{(b-ω^2) - jaω}{(b-ω^2) - jaω} = \dfrac{k(b-ω^2-jaω)}{(b-ω^2)^2 + (aω)^2} \\ \end{align*} $$
With $ω = \sqrt 2, L_c(j\sqrt 2)$ has a real part, hence the possible solution is (d)
Based on option (d) $$ \begin{align*} L_d(jω) & = \dfrac{k}{(b-ω^2) + jaω} * \dfrac{(b-ω^2) - jaω}{(b-ω^2) - jaω} = \dfrac{k(b-ω^2-jaω)}{(b-ω^2)^2 + (aω)^2} \\ L_d(j0) & = 3 = \dfrac{k}{b} \\ L_d(j\sqrt 2) & = 0 - j\dfrac{3}{4}\sqrt 2 = \dfrac{k(b-2-ja\sqrt 2)}{(b-2)^2 + 2a^2} \end{align*} $$
Therefore,
$$ \begin{align*} \begin{cases} k = 3b \\ b - 2 = 0 \\ \dfrac{-jka\sqrt 2}{2a^2} = -j\dfrac{3}{4}\sqrt 2 \end{cases} \end{align*} $$
$$ b = 2, k = 6, a = 4 $$