# 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

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
3,920
600
I got the same transfer function.

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3. ### RBR1317 Active Member

Nov 13, 2010
377
68
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):

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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 Distinguished Member

Jun 17, 2014
3,601
754
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.