# Bode Plots and Transfer Function

How would you find generic form of this transfer function given these bode plots? I am having a hard time since the phase angle "dips" instead of going up.

• I don't follow you. What do you mean by "dips" while also saying "instead of going up?" You toss out two curves with no marks on them at all to help guide us to look at what you are looking at. Why not mark them up a bit, put numbers or letters near each point or area you care about and then talk about each one? I get that you are having a hard time. But I can't read your mind. Sep 16, 2023 at 3:14
• Allow me to make this starting suggestion. I've added two comments and pointed at some areas of interest. What can you work out from what I wrote down? Anything? Sep 16, 2023 at 3:47
• Well, if you don't have anything to say, neither do I. I'll be happy to discuss this further (as should others here) should you decide to write something more. Until then... Sep 16, 2023 at 5:14
• Hi, thanks for the comment, what I meant to say is that there is a first order pole near 10^7 rad/s, which, I think, should make the phase angle go up to +90 degree instead of "dipping" to -90 degree. Sep 17, 2023 at 0:32
• No, a pole causes the angle to turn -90 degrees, so if starting from 0 then to -90. Not +90. A zero causes the angle to turn +90 degrees, so if starting from -90 then ending at 0. So your picture says you have a single zero and a single pole and that the zero is at a higher frequency than the pole. Sep 17, 2023 at 1:56

One nice thing about your question is that it isn't all that complicated. Just barely enough so that the answer isn't downright trivial. Just almost so.

Let's look at what I posted in comments, where I added a couple of notes to your plots:

It starts out at $$\0\:\text{dB}\$$, makes a corner somewhere near $$\\omega=1\times 10^7\:\frac{\text{rad}}{\text{s}}\$$, and at first goes into a straight line descent at about $$\20\:\text{dB}\$$ per decade. This is typical for a low-pass filter with a single pole (1st order.) So the denominator will be a first order expression.

(Note, there's more to come because there's another corner to be found. A simple low-pass filter would not exhibit like this. So there is more yet to see. As it turns out this will be an all-pass filter -- both a low-pass and a high-pass.)

Somewhere near $$\\omega=1\times10^{10}\:\frac{\text{rad}}{\text{s}}\$$ it makes another corner, but this one opposes the previous one. So the prior descent slope is killed and returned to a straight horizontal line (no decline remaining.) This behavior is represented by a zero. But we know it's only one zero because there was only one pole, earlier, and this one is just exactly and only enough to counter it. So the numerator will be a first order expression, too.

Usually, when I see something like that for the magnitude plot, I expect to see an inflection point right at $$\-45^\circ\$$ around the point where the magnitude corners downward. There should be some range over which the phase remains more negative than $$\-45^\circ\$$ but then turns around at some point more to the right, where I should then find another inflection point right at $$\-45^\circ\$$ as it rises back to $$\0^\circ\$$.

A pole does not cause the phase to go from $$\0^\circ\$$ to $$\+90^\circ\$$ as you seem to have imagined when writing in comments, "... should make the phase angle go up to +90 degree instead of "dipping" to -90 degree.". However, in that same comment you did get this right: "what I meant to say is that there is a first order pole near 10^7 rad/s". Your recognition of that fact tells me that you have a few clues. Good!

The inflection points for phase are important. These won't only occur at $$\-45^\circ\$$ on their way towards $$\-90^\circ\$$, though. They might occur at $$\-90^\circ\$$ on their way towards $$\-180^\circ\$$ if its a pole pair, and not one pole. So it's not the momentary phase's value, so much as the phase's infection, that tells you where the corner is.

In this case, you've two corners: one at $$\\omega=1\times 10^7\:\frac{\text{rad}}{\text{s}}\$$ and another at $$\\omega=1\times 10^{10}\:\frac{\text{rad}}{\text{s}}\$$. But one of these corners is for the pole and the other one is for the zero.

Note also that it appears likely that behavior at DC will be a gain of 1 ($$\0\:\text{dB}\$$.) So this means the Laplace equation isn't multiplied by some other gain value. That keeps it simple enough.

Each pole or zero looks about like this: $$\\left(\frac{s}{\omega_c}+1\right)\$$. So at DC when $$\s\$$ is exactly zero, the value is just 1. And at daylight (as $$\s\to\infty\$$) the added 1 no longer matters and the expression just becomes $$\\frac{s}{\omega_c}\$$ for all purposes.

Since there is a zero divided by a pole, or $$\\frac{\frac{s}{\omega_1}+1}{\frac{s}{\omega_0}+1}\$$, at DC when both of these expressions are 1, the result is 1 when dividing one by the other. So expect to see $$\0\:\text{dB}\$$ near DC. But when at daylight ($$\s\to\infty\$$), then the result becomes $$\\frac{\omega_0}{\omega_1}\$$. Since here you have $$\\omega_0=1\times 10^7\:\frac{\text{rad}}{\text{s}}\$$ and $$\\omega_1=1\times 10^{10}\:\frac{\text{rad}}{\text{s}}\$$ you should find that this is $$\20\cdot\log_{10}\left(\frac{10^{7}\:\frac{\text{rad}}{\text{s}}}{10^{10}\:\frac{\text{rad}}{\text{s}}}\right)=-60\:\text{dB}\$$.

And that is, in fact, where the magnitude curve bottoms out at. So this is yet another confirmation that the use of the two inflection points to spot the corner frequencies worked correctly for us.

Or, putting it another way, this final value of $$\-60\:\text{dB}\$$ tells us the ratio of the two corner frequencies. So if we saw, instead, that it bottomed out at $$\-40\:\text{dB}\$$ then we'd know that the ratio was instead 1:100 and not 1:1000, as is the case here. So if you needed a way to help you spot the corner separation, this value (when compared with the DC value) would help do that.

So the y-axis separation of magnitude, from DC to daylight, is related to the x-axis separation of the two corners in phase. In this case, anyway.

Here's an example, using LTspice to illustrate:

The above shows that the final plateau varies depending upon $$\\omega_1\$$'s value. Just as I mentioned.

Expanding on Periblepsis's hint in the comments: You could create a circuit that looks like the following which will match the plots. The high & low pass sections are independent of each other which allows you to easily set the high & low pass corner frequencies. For the frequencies in the plot, you would need an impossibly wide bandwidth op amp (i.e., ideal op amp).
C1 & R1 form a high-pass, C2 & R2 form a low-pass.
I'll leave it up to you to sort out the values and/or the equations.

simulate this circuit – Schematic created using CircuitLab

• Thank you so much! Sep 17, 2023 at 2:56
• The product R1*C1 must be extremely tiny and I wonder about opamp output current limits and/or its internally compensated pole. Sep 17, 2023 at 3:29

Hint

So, how would you achieve this with a simple circuit? Try and evolve a circuit made from resistors and capacitors that can achieve this.

I am having a hard time since the phase angle "dips" instead of going up.

That's because it isn't a simple low-pass filter. In a simple low-pass filter the phase response at higher frequencies remains at 90°. It's sometimes called a shelving filter.

You should also notice that the basic filtering is "1st order" due to the 20 dB/decade slope of the roll-off of the gain curve. And, if you are still struggling, note that it took me two minutes and 3 components to get this response: -

I'm not saying it has the same break-points as your curve because I only spent two minutes simulating it.

• Thanks for the comment, I am trying to simulate this only using 2 capacitors and 1 resistor, and I completely forgot to mention that in the main post. I am not too sure how I would have to add the second capacitor to the RC low pass filter to simulate the "shelving" or the second inflection point for the phase plot. Sep 17, 2023 at 1:02
• Well, without the extra capacitor, you have three circuit nodes namely; input, output and ground so, there are only 3 places it can go and, two of those places make no sense. @RoyHwang if we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. Sep 17, 2023 at 9:06

The response shown in those Bode plots is the response of a phase lag compensator.

The standardised form of the transfer function is :-