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The platform’s transfer function

I am not able to understand what the frequency where the hydraulic pole occurs (as in the diagram) signifies? To give an overview, there is a vibrating platform and we are seeing its response motion at different input frequencies using an inductive position sensor (IPS). Now I know poles are points where the transfer function’s denominator is zero. But what does that translate into for a real life system? I was guessing that it might be the point where the amplitude of motion becomes maximum (in reality it can’t become infinity hence this thought). But in the diagram, I just can’t figure out what makes the pole point’s frequency (around 20 mHz) special? My reasoning might be awful and I apologise if that is so.

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  • \$\begingroup\$ What was the input magnitude? was it 10⁰=1? \$\endgroup\$ – jDAQ Aug 15 at 14:48
  • \$\begingroup\$ How do you know the hydraulic pole is at 20 mHz? \$\endgroup\$ – Andy aka Aug 15 at 15:11
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    \$\begingroup\$ This question (and my answer LOL) might give you an insight. \$\endgroup\$ – Andy aka Aug 15 at 15:12
  • \$\begingroup\$ @jDAQ Nothing is mentioned about input magnitude in the text, so I am not sure. \$\endgroup\$ – Niket Aug 15 at 16:16
  • \$\begingroup\$ @Andyaka The number is mentioned in the text containing this diagram. \$\endgroup\$ – Niket Aug 15 at 16:18
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Poles are the roots of the characteristic equation (the denominator of the transfer function), so if you have a transfer function such as: $$ G(s) = \frac{1}{s^2+a \cdot s+b} $$

That can also be rearranged to look like this, $$ G(s) = \frac{1}{(s-s_1)(s-s_2)} $$

where \$s_1,s_2\$ are the poles of it.

You are correct that,

Now I know poles are points where the transfer function’s denominator is zero.

But, when you are inputing a sine/cosine to the system you should look at the response for the Fourier transform $$ G(j\omega) = \frac{1}{(j\omega-s_1)(j\omega-s_2)} $$ where \$\omega = 2\pi \cdot f\$. Notice that, any of the poles can have the form \$ s_1 = x+ jy\$. So, a term of the form \$(j\omega-s_1)\$ will be able to "zero" the imaginary part of the pole, but not the real part, that will remain unchanged. That is why inputing the right frequency can result in a peak output but not a infinite one. Observe this, if the system can be represented as

$$ G(j\omega) = \frac{1}{(j\omega+a+j)(j\omega+a-j)} $$

and for, $$ G(j) = \frac{1}{(a+j2)(a)} $$.

For a very small \$a\$, let say,

$$ a=0.001 -> G(j) \approx \frac{1}{(j2)(0.001)} = \frac{1000}{j2} $$.

These are cases where you have a peak, when the pole is very close to the imaginary axis. Then, by inputing the right frequency the output will have a big gain. But, if \$a\$ is large, such as

$$ a=100 -> G(j) = \frac{1}{(100+2j)(100)} \approx \frac{1}{10000} $$

there is no peak. At peak at a pole is discussed the same confusion on poles leading to a very big output. The resonant frequency happens when you have a pole on the the imaginary axis or very close to it, then the system tends to oscillate increasingly.

At slide 14 you can see a nice graph on the different behaviors that can be observed at the pole frequency, also, note that it is also explained the approximations usually used on bode plots. In these approximations, it is said that after a pole (as the frequency increases) the amplitude of the signal decreases 20dB/decade, while after a zero it increases 20dB/decade. But that is just an approximation! After all, the two examples I gave show the amplitude of the output for a signal with the frequency of the pole can either cause a peak or not.

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  • \$\begingroup\$ Right. But in this diagram the hydraulic pole doesn't give peak amplitude. So then, what is special about it? I used to look at this like when you are at the pole, your transfer function 'blows up' (in real damped systems becomes very large) and so your output also becomes large. Hence I would think that amplitude should be the largest here and this should correspond to the resonant frequency of the system. But it clearly isn't the case. \$\endgroup\$ – Niket Aug 15 at 16:26
  • \$\begingroup\$ in your system the pole seems to be far from the imaginary axis, hence no peak. Did that help out? \$\endgroup\$ – jDAQ Aug 15 at 17:24
  • \$\begingroup\$ Yes it did. Looks like the pole frequency is not important when looking at amplitude response since it doesn't show any different amplitude response than other frequencies. \$\endgroup\$ – Niket Aug 15 at 17:28
  • \$\begingroup\$ it does, because before the pole (in a single pole case) the amplitude of the response remains about the same (some constant), after the pole (whether it has a peak or not) it starts to diminish. \$\endgroup\$ – jDAQ Aug 15 at 17:54
  • \$\begingroup\$ Ah right! This is what I wanted. Thank you. \$\endgroup\$ – Niket Aug 16 at 1:16
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As @jDAQ said, poles are the roots of the characteristic equation.

But a pole can take a real value or come as a complex pair, and your response plot is for a sine wave sweep -- and in the Laplace (s) domain the only frequencies in a sine wave are \$\pm j \omega\$.

A good physical interpretation of a transfer function such as \$H(s) = k/(s + a)\$ is that the system will have a strong response to sine waves at frequencies below \$\omega = a\$, medium at \$\omega = a\$ (strictly speaking, at half power), and diminishing response above. Another one is that the system has a time-domain response containing the exponential \$e^{-at}\$. Both of these are consistent with one another -- they're just two different ways of looking at the same problem.

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  • \$\begingroup\$ So does this physical interpretation mean that the 'strongest' response of the system is at $w=a$? If so, then this is worse because here the hydraulic pole frequency doesn't give peak amplitude. I am not able to understand what does this mean here then? It doesn't look like this is the resonant frequency, which I used to think the poles would give. \$\endgroup\$ – Niket Aug 15 at 16:27
  • \$\begingroup\$ No, not at all. For a single pole the response at the pole's frequency has a magnitude of \$1/\sqrt{2}\$ of the response to a constant input. \$\endgroup\$ – TimWescott Aug 15 at 16:38
  • \$\begingroup\$ By constant do you mean the zero frequency input? I am guessing not because here the response at pole's frequency is greater than zero frequency response, though it might be the case that this is not a single pole transfer function. \$\endgroup\$ – Niket Aug 15 at 17:12
  • \$\begingroup\$ he means constant amplitude, the input signal is a sine/cosine of constant amplitude \$\endgroup\$ – jDAQ Aug 15 at 17:27

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