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From the Bode Diagram which is also a minimum phase system,

enter image description here

I have tried to figure out the transfer function it could have been like this,

$$ \begin{align*} L(s) = \dfrac{1}{s}\cdot\dfrac{2}{s + \sqrt{2}} \end{align*} $$

Which, I see that there are two different slopes in the graph, I guess there are two poles in the transfer function. Besides, the frequency at 1 and root 2 are also the points to identify the coefficients of the transfer function. Thus, I guess this should be the approximated function. However, I use Wolfram to plot the bode diagram of this function, it does not correspond with the graph shown here.

enter image description here

What could be the incorrect guess about the derived transfer function?

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    \$\begingroup\$ The pole locations are incorrect. Correct that and then you can adjust the magnitude to match the question \$\endgroup\$
    – sai
    Commented Jul 10, 2023 at 5:58
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    \$\begingroup\$ There seems to be a pole at the origin meaning you should have a division by \$s\$ scaled by a coefficient in your expression so that the gain goes infinite when \$s=0\$. Also, when you write a pole \$\frac{1}{1+b_1s}\$, \$b_1\$ has the dimension of time which is inverse of the pole frequency. \$\endgroup\$ Commented Jul 10, 2023 at 7:14
  • \$\begingroup\$ Thanks, I have adjusted the transfer function which is derived from the bode diagram. Besides, I have tried this on Wolfram, and the diagram looks quite similar to the one in the OP. \$\endgroup\$
    – Sonamu
    Commented Jul 28, 2023 at 4:46

2 Answers 2

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Without also having the phase response, it is impossible to determine if the poles lie in the LHP or the RHP of the s-plane. Assuming that the system is stable one can deduce the following:

  • There exists a single pole at \$s=0\$ because the magnitude slope is -20 dB/dec right from the beginning.

  • There exists a single pole at \$s = -\sqrt{2}\$ because the magnitude slope starts decreasing by -40 dB/dec after \$\omega = 2 \: \text{rad/s}\$.

So currently, the transfer function is

$$H(s) = K\frac{1}{(s-0)(s+\sqrt{2})} = K\frac{1}{s(s+\sqrt{2})} $$

where \$K\$ is some gain.

If we set \$K=1\$ the transfer function has the magnitude \$|H(j\omega)| = -4.82 \: \text{dB}\$ at \$\omega = 1 \: \text{rad/s}\$. But we know from OP's graph that the magnitude should be \$20 \log_{10}(2) \approx 6 \: \text{dB}\$, so this means that K must be \$ K = \pm 10^\frac{6+4.82}{20} = \pm 10^\frac{10.82}{20} \approx \pm 3.4754\$.

Again, because we don't know the phase, we cannot determine the sign of \$K\$. However, assuming that \$K>0\$ we end up with the transfer function

$$H(s) = \frac{3.4754}{s(s+\sqrt{2})} $$

which yields the following (satisfying) Bode Plot: -

enter image description here

Again, I emphasize that we cannot uniquely identify the transfer function because the phase characteristic is left out of the question. Hence, there are multiple transfer functions that realize the magnitude Bode Plot, one example being:

$$ H(s) = \frac{3.4754}{s(s-\sqrt{2})}$$

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  • \$\begingroup\$ Yes, the system is a minimum phase system, I forgot to mention this at the beginning and I think the poles are correct in your post except for the final transfer function there is a typo in the sign. \$\endgroup\$
    – Sonamu
    Commented Jul 28, 2023 at 5:01
  • \$\begingroup\$ The gain K seems like the sum of the initial gain plus the reduction of the pole at root 2 which is quite an interesting thing. I did not understand how to derive this when I first tried this. Yet, this is quite a task to derive the phase margin with this not well-looking gain. \$\endgroup\$
    – Sonamu
    Commented Jul 28, 2023 at 5:06
  • \$\begingroup\$ From the graph at gain equal to 1, I can see that the phase margin is approximately 41 degrees, yet, if I want to calculate it, besides the traditional way, in which I found the cross gain frequency, and then, bring it to the H(s) with jω replace s. Is there any better way to do this? \$\endgroup\$
    – Sonamu
    Commented Jul 28, 2023 at 5:11
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It is not possible to derive the transfer function from such a simplified diagram. The only way is to find an approximate expression. The function starts with idealized integrating block (because only one pole is shown). The first guess for the integrating time constant is Ti=1/1.258 (derived from the gain at w=1 rad/s).

The next portion can be modelled with a first-order lowpass with a time constant T=1/wp=1/SQRT(2).

This gives the transfer function

H(s)=(1.258/s)[1/(1+s/1,414)]=1.258/[s+0.7071s²]

Of course, the magnitude at w=1 is NOT 2dB (1.258) due to the lowpass influence. Hence the numerator of the function should be increased correspondingly-

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