Graphing Transfer Function

[MrAl]

Hi again,


On the graphs make sure to find the pass band gains and the -3db points as that is what this is all about.

 

[EN0]

I apologize for not coming on sooner! I fixed my error in the other post, so it should look ok now.

I’ve found some math software’s, which I’m looking into. I’ll let you know when the graphs are produced.

 

[EN0]

Hey MrAl,

I am very sorry this took so long to generate, but I had a lot of school work and then I needed to learn the syntax of Freemat. Anyway, here it is:

-3dB Point of Ordinary LPF: 15915.49 Hz
-3dB Point of LPF With Load: 15915.49 Hz

Given:

R1 = 1kΩ
C = 10nF
R2 = 1kΩ
Vin = 5V

Is that correct?

 

[MrAl]

Hi again,


Something doesnt look right in the graphs. How does the amplitude get as high as 5v in the top trace for example?

 

[EN0]

Well the top graph is without a load, so it should start from 5V, shouldn't it? The lower graph is with the load so it begins at 2.5V. I should have told you beforehand which was which, sorry.

 

[MrAl]

Hi again,

So you are using an input voltage of 5v then? Usually you would use 1v to graph it. Of course it helps to graph in db gain too rather than voltage so you can easily spot the -3db point.

 

[EN0]

After all that, I need to change the graphs again?

Alright, I'll fix it up and put it in terms of dB. One volt for the input voltage is a good idea, I'll implement that as well.

Till Then,

Austin

 

[EN0]

Hey MrAl,

Apparently there is a mistake in our mathematics? Hayato provided the graph that is attached and his equations are different.

 

[MrAl]

Hello,

I'll double check and get back here.

 

[EN0]

Hi MrAl,

I believe our first transfer function is correct, but the second one isn't. I went over the math myself and will show you my results.

Conventional Low-Pass Filter

Final Equation

[LATEX]Amplitude = \frac{V_{IN}}{\sqrt{1 + Z_R^2 \omega^2 C^2[/LATEX]

Low-Pass Filter With Load

[LATEX]i_1 = i_2 + i_3[/LATEX]

[LATEX]i_1 = \frac{V_{IN} - V_O}{R}[/LATEX]

[LATEX]i_2 = V_O j \omega C[/LATEX]

[LATEX]i_3 = \frac{V_O}{R_L}[/LATEX]

[LATEX]\frac{V_{IN} - V_O}{R} = V_O j \omega C + \frac{V_O}{R_L}[/LATEX]

[LATEX]V_{IN} = V_O + V_O R j \omega C + \frac{V_O R}{R_L}[/LATEX]

[LATEX]V_{IN} R_L = V_O + V_O R j \omega C + V_O R[/LATEX]

[LATEX]V_{IN} = V_O(1 + R j \omega C + R)[/LATEX]

[LATEX]V_O = \frac{V_{IN} R_L}{1 + R j \omega C + R}[/LATEX]

[LATEX]\frac{(R_N^2 + I_N^2)}{(R_D^2 + I_D^2)}[/LATEX]

[LATEX]\frac{(V_{IN} R_L)^2}{(1 + R)^2 + (R \omega C)^2}[/LATEX]

[LATEX]Amplitude = \sqrt{\frac{V_{IN}^2 R_L^2}{1 + 2R + R^2 + R^2 \omega ^2 C^2}}[/LATEX]

Final Equation

[LATEX]Amplitude = \frac{V_{IN} R_L}{\sqrt{1 + 2R + R^2 + R^2 \omega^2 C^2}}[/LATEX]

I believe that's correct?

 

[Hayato]

Plot the graph again, if they are similar to the ones I've plotted, they are ok.

 

[MrAl]

Hi MrAl,

I believe our first transfer function is correct, but the second one isn't. I went over the math myself and will show you my results.

Conventional Low-Pass Filter

Final Equation

[LATEX]Amplitude = \frac{V_{IN}}{\sqrt{1 + Z_R^2 \omega^2 C^2[/LATEX]

Low-Pass Filter With Load

[LATEX]i_1 = i_2 + i_3[/LATEX]

[LATEX]i_1 = \frac{V_{IN} - V_O}{R}[/LATEX]

[LATEX]i_2 = V_O j \omega C[/LATEX]

[LATEX]i_3 = \frac{V_O}{R_L}[/LATEX]

[LATEX]\frac{V_{IN} - V_O}{R} = V_O j \omega C + \frac{V_O}{R_L}[/LATEX]

[LATEX]V_{IN} = V_O + V_O R j \omega C + \frac{V_O R}{R_L}[/LATEX]

[LATEX]V_{IN} R_L = V_O + V_O R j \omega C + V_O R[/LATEX]

[LATEX]V_{IN} = V_O(1 + R j \omega C + R)[/LATEX]

[LATEX]V_O = \frac{V_{IN} R_L}{1 + R j \omega C + R}[/LATEX]

[LATEX]\frac{(R_N^2 + I_N^2)}{(R_D^2 + I_D^2)}[/LATEX]

[LATEX]\frac{(V_{IN} R_L)^2}{(1 + R)^2 + (R \omega C)^2}[/LATEX]

[LATEX]Amplitude = \sqrt{\frac{V_{IN}^2 R_L^2}{1 + 2R + R^2 + R^2 \omega ^2 C^2}}[/LATEX]

Final Equation

[LATEX]Amplitude = \frac{V_{IN} R_L}{\sqrt{1 + 2R + R^2 + R^2 \omega^2 C^2}}[/LATEX]

I believe that's correct?

Hello again,


Ok since you only have a problem with the circuit with the load resistor we'll concentrate on that one.

With R1 being the input resistor and R2 being the load resistor and C the only capacitor, the transfer function is:
Vout=(Vin*R2)/(s*C*R1*R2+R1+R2)

and the amplitude (after subst s=j*w) is:
VoutAmpl=(Vin*R2)/sqrt(w^2*C^2*R1^2*R2^2+R2^2+2*R1*R2+R1^2)

Now this last result has been verified in a circuit simulator, which by the way you should be doing too. You can compare your result to that VoutAmpl and if you dont get that or an equivalent expression then it can not be correct so you'll have to see what went wrong.
You might note that both R1 and R2 appear in the denominator too. I also think that using Latex just complicates matters rather than making them simpler. You should probably avoid that until you get the final correct solution.