Bode Plots

[schmidty99]

Can anyone explain Bode plots/diagrams to me? I'm working specifically on NCEES 114. I've been trying to figure out how to solve these problems but I can't seem to wrap my head around it. Thanks in advance!

 

[CincinnatiControlsGuy]

The first step is to find the transfer function. In this case, using KCL at the op-amp's negative lead (virtual short means that this node is ground) we find that v1/R + v2*jwC = 0. This leads to v2/v1 = -1/(jwRC) = j/(wRC). The magnitude is the square root of v2/v1 times its conjugate, which is 1/wRC.

The best way to look at this is in terms of poles and zeroes. Take the log of this magnitude (log(1/wRC)). Remember that with logarithms multiplication = addition and division = subtraction. This can be rewritten as log(1) - log(wRC). log(1) = 0, so we're left with -log(wRC). Think about the limits. What happens to -log(wRC) as w -> infinity? -log(infinity) = -infinity. Now what happens as w -> 0? -log(wRC) = -log(0) = - * (-infinity) = infinity. How about at w = 1/RC? Here -log(1) = 0dB. What about w = 10/RC? -log(10) = -1. Using this logic you can see that it's a straight line with a negative slope. Therefore, the answer is D.

The best way to learn Bode plots is by practicing with transfer functions in the s-domain and with multiple situations: double poles/zeroes, complex conjugate poles/zeroes, etc. Simple poles and zeroes correspond to +/-20dB/decade changes, respectively. I hope this helps.

 

[schmidty99]

Yeah, transfer functions aren't really a strong point for me. But finding the transfer function was the first step I didn't know how to do. I'll study your answer and see if I can get it to make sense. Any chance you would know of a resource that has more problems like this one? Thanks!!

Ok, after studying your answer my first questions comes from the KCL. I understand it except the part about v2*jwC. Since we're finding currents (I), I don't understand how we can multiply v2 by jwC. In my mind, that should be division from I=V/R. I understand that you're correct, but i don't quite follow the second part fo the KCL equation. From your equation, I was able to mostly understand the math to get to v2/v1=1/wRC. Although if you could show me the math step by step, I'd feel better about it.

The part about thinking about it in logs is very helpful. My problem is I have never been trained to look at them that way. So this is baptism by fire. :smileyballs:

 

[benbo]

I don't have the problem, but I can visualize it and I suspect the reason is because the impedance of a Cap Zc = 1/jwC (the reciprocal). So you are using ohms law but dividing by the reciprocal - -

I = V/Zc = V/(1/jwC) = V*jwC.

THe impedance of a coil is jwL.

Is this a low pass filter? Does it have a cap in the feedback loop? Just in case I am completely off base.

Assuming it is a LPF - with a resistor from V1 to the v- input, and a cap in the FB the KCL at the inverting input is-

Voltage V- = 0 (as GB said, virtual ground)

(V- - v1)/R = (V- - V2)/(1/jwC)

V- = 0

(0-V1)/R = (0-V2)/(1/jwC)

-V1/R = -V2*jwC

V2/V1 = 1/jwRC

 

[CincinnatiControlsGuy]

It's a low pass (sort of) and is the same as the circuit attached with no resistor in the feedback loop.

Benbo is correct about the impedance of C, Zc = 1/jwC.

Schmidty99,

When studying I went through Engineering Circuit Analysis, 6th edition, by Hayt, Kemmerly and Durbin instead of the EERM for circuit analysis. I felt that I needed to be as strong as I was in school as far as circuits were concerned. It was my college circuits book and contains material on DC/AC circuit analysis, op-amps, filters, Bode Plots, Laplace/Fourier Transform analysis, Magnetically coupled circuits, Power and Two-Port Networks. That book is worth its weight in gold if you know it well. Either this weekend or early this week at lunch at work, I'll outline 114 along with another transfer function (s-domain) to show you how to tackle these problems. In fact, a detailed, worked, problem can be printed out and taken in with you to the exam for reference. As for transfer functions, teach yourself to be comfortable in that area as they are very important. The best practice, IMO, is to know op-amps inside and out. Gain is always given as a transfer function. Anyway, I have to warn you ahead of time that I have TERRIBLE handwriting. I'll do my best to make it readable.

 

[schmidty99]

Thanks guys. As soon as benbo said that, I knew what I missed. I just couldn't see it that way for some reason. GB: I'll be looking forward to whatever you send me. I'll keep hammering away at it. I appreciate it!

 

[CincinnatiControlsGuy]

Ok, here is the long explanation of 114 and a transfer function to Bode plot from my engineering circuit analysis book. The latter is more complex than I remember it, so you'll have to bear with me there, though the ideas are sound. When you download 114, let me know, and I'll post the other problem/solution so as to not exceed attachment space limits.

 

[schmidty99]

Thanks GB! That's awesome! I appreciate the work and I have it downloaded.

 

[CincinnatiControlsGuy]

Ok, here is the other example. I didn't do the phase plot for it because that would be pretty lengthy as well and, as you can see, there is no way that the PE board would ask you to quickly sketch a Bode Plot like this as it takes too much time. The key is to always find the transfer function, put it into suitable form, convert to decibels, sketch each component and then graphically add each component. Anyway, I hope this helps in some fashion.

 

[schmidty99]

Ok, I got your example. I'll look it over when I get a chance and hopefully understand it. Just glancing at it, it seems to make sense. But boy, it sure does get complicated. I appreciate your help with this. I have a much better understanding of transfer functions and bode plots than I did 3 days ago, still a long way to go though.

Be forwarned, I will most likely be hammering the Boards with questions more and more as the test approaches. Any help you and everyone else is willing give will be most greatly appreciated. I pretty much know that I won't pass this test without help from you folks!