Script not producing the correct plot

Hi...

I need to simulate a signal in MatLab but I can't make my simple script to work properly.

The function is a Transfer Function of a given circuit.

The expression is the following

The script I'm trying to use to simulate this expression is:

    if true
R1=100;
R2=1e3;
C=1e-6; 
% fc=1/(2*pi*R*C); 
f=1:1:1e4; 
ter1=R2^2;
ter2=(2*pi.*f*C*R1*R2).^2;
ter3=(R1+R2).^2;
ter4=(2*2*pi.*f*C*R1*R2).^2;
mhf=(ter1+ter4)/(ter2+ter3); 
mhf=power(mhf, 0.5);
% mod_hf=abs(hf); 
% fase_hf=angle(hf); 
figure(1); 
subplot(2,1,1); plot(f,mhf); grid on; 
title('Funcao de Transferencia do Filtro Passa-baixo RC') 
xlabel('Frequencia (Hz)') 
ylabel('|H(f)|') 
% subplot(2,1,2); plot(f,fase_hf);grid on; 
% xlabel('Frequency (Hz)') 
% ylabel('arg[H(f)] (rad)') 
    end

I have calculated the result for a few frequencies in calculator and I fot the following values:

f=0Hz--> |H(f)| = 0.9091
f=50Hz--> |H(f)| = 0.9081
f=200Hz--> |H(f)| = 0.8932
f=500Hz--> |H(f)| = 0.8274
f=1000Hz--> |H(f)| = 0.7072
f=5000Hz--> |H(f)| = 0.5169
f=5MHz--> |H(f)| = 0.5

which matches an High-Pass Filter but I can't get the correct plot!!! Any help?

Answer 1

[Star Strider]

Change:

mhf=(ter1+ter4)/(ter2+ter3);

to:

mhf=(ter1+ter4)./(ter2+ter3);

and it works. You need to do element-by-element division as well.

[Star Strider]

José Carlos Ferreira’s ‘Answer’ moved here:

Ok, it's working but it's not consistent with results from calculator...

It should start around 0.9091 for f=0Hz and stabilize around 0.5 for high frequencies!!! And the plot is starting around 0.9 and goes up until 2...

[Star Strider]

I would have to see your circuit and analyse it myself to see if there is a problem. You coded your equation correctly with respect to the equation you posted as far as I can determine.

[José Carlos Ferreira]

Ok, I'll try to represent it here:

           ____R1____
  ________|          |______________________________o
          |_____C1___|        |
                            __|__
                           |     |
                           |     |
                           R2    C2
                           |     |
                           |_____|
                              |
  ____________________________|_____________________o

Where C1 = C2 = 1uF and R1 = 100 ohms and R2 = 1000ohms

[Star Strider]

No matter how I analyse that circuit, I get a lowpass characteristic out of it. I can’t reproduce your result.

Stopping here.

[José Carlos Ferreira]

If you try to calculate a few values for H(f) and for H(f) phase in a calculator, are your calcs consistent with MatLAb?

Answer 2

[José Carlos Ferreira]

I've changed a little bit the script to use the original Transfer Function and let MatLab compute the Abosulte and Angle values for it but I can't understand why the results from MatLab are not consistent with my calculator results...

R=2.2e3; 
C=0.33e-6; 
fc=1/(2*pi*R*C); 
f=1:1:1e5; 
hf=(R2+j*2*pi.*f*C*R1*R2)./((R1+R2)+j*2*pi.*f*C*R1*R2); 
mod_hf=abs(hf); 
fase_hf=angle(hf); 
figure(1); 
subplot(2,1,1); plot(f,mod_hf); grid on; 
title('Funcao de Transferencia do Filtro Passa-baixo RC') 
xlabel('Frequencia (Hz)') 
ylabel('|H(f)|') 
subplot(2,1,2); plot(f,fase_hf);grid on; 
xlabel('Frequency (Hz)') 
ylabel('arg[H(f)] (rad)')

[José-Luis]

Please don't place a comment as a new answer.

[José Carlos Ferreira]

This was not a comment. Was actually a new reply... I'm not very used to this kind of forum! Sorry!