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I've been trying for a few days to convert an experimentally obtained (frequency against magnitude and frequency against phase) graphs to a bode plot and then construct a transfer function from the bode plot enter image description here

I constructed a table:

1) Converted each frequency from Herts to Rad/s by multiplying it by 2*Pi
2) Converted each magnitude to db by multiplying it by 20log(magnitude)

Note: As seen in the original graph, all the magnitudes are negative. When converting to db, you cannot take the log of a negative number so I assumed it to be positive. Am I correct? will this affect my bode plot?

The table can be seen below; enter image description here

Then I plotted the bode plot below: enter image description here

I understand that my transfer function will be a first order because the phase shifts 90 degrees meaning the highest power of the denominator will be 1. in other words, there will be 1 pole in the denominator.

In summery, my questions:

1) Was I correct in converting the negative magnitude to db by ignoring the negative value? 2) What will be the final transfer function of the constructed bode plot?

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  • \$\begingroup\$ Have you considered covering a wider range of frequency? over just 15:1, are you sure about those asymptotes? this looks like a pole (roll off, or drop in gain), followed by a zero (a rise in gain) used to flatted out the pole. Here the flattening is done at a higher frequency. \$\endgroup\$ – analogsystemsrf Jan 13 at 15:56
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    \$\begingroup\$ You already have it plotted in dB, so there was no need for conversion. By the looks of it, since you are starting your phase at around 0, then go to -90, you are missing a decade of critical information. \$\endgroup\$ – a concerned citizen Jan 13 at 15:59
  • \$\begingroup\$ @analogsystemsrf Thanks for the reply, the frequency range measured (from 0.1-1.6hz) was a requirement given from the lab. From my knowledge, I understand that we have 1 pole and 1 zero like you mentioned. But they need to rise by 20db/dec but our experimentally measured data takes more than a decade to rise up to 20db. What do you recommend I do? Shall I make assumptions? \$\endgroup\$ – Meti Jan 13 at 16:12
  • \$\begingroup\$ The pole interacts with the ZERO. You are using just 2 resistors and 2 capacitors, right? \$\endgroup\$ – analogsystemsrf Jan 13 at 16:39
  • \$\begingroup\$ Hi there. I realised I made a slight mistake in my question and corrected it. The system is indeed a first order system hence why the phase shifts to -90. May I ask if there is a possible way to make an asymptotic bode diagram from the accurate one I have constructed? Also you were correct. I need to inverse all my gain values to make the graph like how u have stated in the answer. May I also ask what makes the phase go to -90 and then back to zero? or is that because I previously stated that the system was a second order? thanks \$\endgroup\$ – Meti Jan 13 at 17:28
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I would draw a combined POLE_ZERO response like this, which assumes only 2 resistors and 2 capacitors. The interaction prevents reaching the 10:1 attenuation effect.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Can you explain why you chose this representation? It doesn't appear to be consistent with the data provided by the OP. Can you add values to the axes? As it is, this answer doesn't seem to be helpful, particularly to someone coming along later with a similar, but not identical, problem. \$\endgroup\$ – Elliot Alderson Jan 13 at 16:37
  • \$\begingroup\$ This is the response of a pole interacting with a zero. I've asked the OP for explanation. Is this being helpful to you? Then upvote my comment. \$\endgroup\$ – analogsystemsrf Jan 13 at 16:46
  • \$\begingroup\$ Can you explain how you determined that there was one pole and one zero, based on the information provided by the OP? The OP appears to be unclear about why there must be a zero and what affect that has on the Bode plot. Can you explain how the locations of the pole and zero are related to the shape of the graphs? Can you label the axes? \$\endgroup\$ – Elliot Alderson Jan 13 at 18:39
  • \$\begingroup\$ I see NO phase curves in the plots, on my computer. I just examined the rising gain response, noticing the leveling off. That behavior comes from a zero followed by a pole. However, the OP wrote of negating some values, meaning the gain (dB) response would be a falling response; if a falling response, this would be pole-followed-by-a-zero, Again, I saw no phase plot in the original question. \$\endgroup\$ – analogsystemsrf Jan 13 at 21:53

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