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
    Draw the bode plot of the following circuit: Opamp Ideal


    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)

    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.
    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
    I got the same transfer function.
    Wikkez likes this.
  3. RBR1317

    Active Member

    Nov 13, 2010
    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:
    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
    Wikkez likes this.
  4. Wikkez

    Thread Starter New Member

    Jan 26, 2013
    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

    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:

    [2] The transfer function for the non inverting amp (as shown in the schematics) is:

    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.