I have a block diagram,

enter image description here

From the diagram, I think the two summing points at the middle of the diagram can be switched, therefore, I got open and close loop transfer function as below:

$$ \begin{align*} L(s) & = \dfrac{K(s+2)}{s(s+1)} \\\ G(s) & = \dfrac{L(s)}{1+L(s)} = \dfrac{K(s+2)}{s^2+(K+1)s+2K} \end{align*} $$

Now, I am being questioned about at what value K the under-damping coefficient(ratio) will be minimized. I have tried to solve this from Routh Criterion and the general formula of a second-order lag system,

The Routh Criterion simply gives me K > 0, this does not seem to be the answer. Meanwhile, if I wrote

$$ \begin{align*} & G(s) = \dfrac{L(s)}{1+L(s)} = \dfrac{K(s+2)}{s^2+(K+1)s+2K}, w_n = \sqrt{2K} \\\ & 2\zeta w_n = (K+1), 2\sqrt{2K} \zeta = (K+1), \zeta = \dfrac{K+1}{2\sqrt{2K}} \end{align*} $$

However, I am still not able to locate the value K, what did I miss?

Besides, when giving $$K = 1, \quad r(t) = 10\sin{2t}$$

I would like to find its steady-state output y(t), therefore, I tried

$$ \begin{align*} y(t) & = \lim_{t\to ∞}y(t) = \lim_{s\to 0}sG(s)R(s) \\\ & = \lim_{s\to 0}s\dfrac{s+2}{s^2+2s+2}\dfrac{10(2)}{s^2+4} = 0 \end{align*} $$

But, this seems quite weird that the steady-state output is 0. Personally speaking, I think the step response should not be 0. Did I get the wrong transfer function?

Based on the suggestion provided by Tim and RussellH, I have tried to use calculus on question 1.

$$ \begin{align*} \zeta & = \dfrac{K+1}{\sqrt{8K}}, \\ \space \zeta' & = \dfrac{(K+1)'\cdot\sqrt{8K} - (K+1)\sqrt{8K}'}{\sqrt{8K}^2} = \dfrac{8K - (K+1)}{2(8K)^{3/2}} = 0 \\ K & = \dfrac{1}{15} \end{align*} $$

  • 1
    \$\begingroup\$ I'm not familiar with an "under-damping coefficient". Do you mean "damping coefficient" ζ? As for your two questions: 1. you find an expression for ζ, and it seems you are tasked with finding the minima. How would you find the minima of an expression? 2. Under what conditions will a transfer function be "zero", what characteristic will it contain? Finally, you think about the step response, but do not explain why; your second question regards a sine input. You seem to have confused the two? (If these comments answer your questions, let me know and I will move them there.) \$\endgroup\$ Commented Jul 10, 2023 at 1:49
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    \$\begingroup\$ Up until the "However " line is correct. 1. Apply standard Calculus min-max to your expresion for \$\zeta\$. 2. The Final value theorem applies only if the limits exist. \$\endgroup\$
    – RussellH
    Commented Jul 10, 2023 at 2:21
  • \$\begingroup\$ @TimWilliams Yes, the "damping coefficient", 0 < ζ < 1 is called the "under-damping coefficient" from some textbooks I have. 1. I think I can express it as a function and use the derivative to find its extrema. 2. Yes, you are right, I was confused about the sin input with step response, they are different and the result is thus not a step response but the frequency response. Yet, I am still not understanding the conditions a transfer function will be "zero". \$\endgroup\$
    – Sonamu
    Commented Jul 10, 2023 at 3:14
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    \$\begingroup\$ You should re-do \$\frac{d\zeta}{dK}\$ \$\endgroup\$
    – RussellH
    Commented Jul 10, 2023 at 7:28
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    \$\begingroup\$ @Sonamu Note that in most litterature \$\zeta\$ is called the damping ratio. \$\endgroup\$
    – Carl
    Commented Jul 10, 2023 at 10:01

1 Answer 1

  1. Find K results in the minimum damping ratio

$$ \begin{align*} & ω_n = \sqrt{2}, \quad ζ = \dfrac{K+1}{\sqrt{8K}} \\ \dfrac{dζ}{dK} & = \dfrac{(K+1)'(\sqrt{8K}) - (K+1)(\sqrt{8K})'}{(\sqrt{8K})^2} = \dfrac{1(\sqrt{8K}) - (K+1)(\dfrac{8}{2\sqrt{8K}})}{8K} \\ & = \dfrac{8K - 4(K+1)}{(8K)^{3/2}} = \dfrac{4(K-1)}{(8K)^{3/2}} = 0, \quad K = 1 \end{align*} $$

  1. Find the steady state of the transfer function

$$ \begin{align*} & G(jω) = \dfrac{jω + 2}{2-ω^2+2jω} \\ & G(jω)_{ω=2} = \dfrac{2j + 2}{4j - 2} \\ & 10|G(jω)|_{ω=2} = 10\dfrac{|2j+2|}{|4j-2|} = 10\dfrac{\sqrt{8}}{\sqrt{20}} = 2\sqrt{10} = y(t) \end{align*} $$


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