Bode plot of Magnitude and Phase Angle, Transfer function Simple circuit

Discussion in 'Homework Help' started by Wikkez, Jul 12, 2016.

  1. Wikkez

    Thread Starter New Member

    Jan 26, 2013
    6
    0
    Draw the bode plot of the following circuit: Opamp Ideal


    [​IMG]

    I'll start with my transfer function for this circuit, because I think it's wrong. After that I'll try to draw the bode plot

    1) The input voltage appears at the node between the impedances(Ideal opamp)
    [​IMG]

    So, I combine the resistor and the 10 C capacitor to get Z1.
    I see a voltage divider
    I get the gain (Vout/Vin) and write it out.
    [​IMG]
    Can this transfer function be correct? I always get this and I dont think it can be.

    So I get a zero at 1/ 11 RC and a pole at 1/10RC ?

    Can someone also give me a better way to draw in MS Paint. :)

    Thanks for anyone who will help me.
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,392
    497
    I got the same transfer function.
     
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  3. RBR1317

    Active Member

    Nov 13, 2010
    232
    48
    While I have seen some very complex schematics rendered quite well in MS Paint, it remains a tool whose main redeeming quality is simplicity. The best results will be obtained by creating a symbol library of schematic components so they only need to be drawn once, and then copied and pasted thereafter leaving only the interconnecting lines to be drawn.

    A much better tool for schematics is a vector drawing program; however, a commercial CAD program might be overkill. What I use is the LibreOffice Draw free-software program. I'm still working on making a good symbol library, with particular attention to aligning the symbol handles and connection points so that when placing the symbol and it snaps to a grid point, the connection points will also be located at a grid point. That makes it easy to attach a connection line to a connection point. Libre Draw also has extensive annotation capabilities and an equation editor. Here is an example of something drawn up in less than 5 minutes:
    Node-Equation-Example-A.png
    Note the solution via node equation. Whenever you write & solve a node equation for a circuit to obtain the answer, versus some clever approach like a voltage divider, one can always trust the results from the node equation whereas the clever solution leads to self doubt. Here is an extract from a Maple worksheet solving the node equation (one types in black, the computer responds in blue):
    Screenshot from 2016-07-12 18-56-20.png
     
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  4. Wikkez

    Thread Starter New Member

    Jan 26, 2013
    6
    0
    Thank you both very much!

    @ RBR1317 I will look into it!

    I'm now unavailable, but if I get home, I'm gonna try to sketch the bode plot
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hi,

    When you have simple negative feedback the solution is ultra simple.
    You have the feedback impedance Z1 (the R and the 10C) and the impedance Z2 (the lone C) from the inverting input to ground Z2 (or Z2 is from inverting input to Vin).

    [1] The transfer function for the inverting amp configuration is:
    Vout/Vin=Z1/Z2

    [2] The transfer function for the non inverting amp (as shown in the schematics) is:
    Vout/Vin=Z1/Z2+1

    You only have to reduce the result to one Numerator over one Denominator to finish up.

    This comes up so much that it is very useful to remember.
     
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