Question

I need help in MATLAB. I'm working on a circuits lab report and I want to plot the derivative of an input signal. The circuit is a differentiator OpAmp. It is receiving a triangle wave as an input and should output a square wave. (I've included my existing code.)

The output formula is:

Vout = -(Rf)*C*(dVin/dt) Where Rf is feedback resistance: Rf = 1*10^6; and C = 1*10^-6.

EXISTING CODE:

%% This section is copied, and then modified from another problem for use as the input.

%%Does not need to be changed.

clc

clear all

close all

R1 = 1000;

C1 = (1*10^-3);

t = linspace(0,20);

In1 = square(t);

f=@(t) [2*[(-1/(R1*C1))*(integral(@(t) square(t),0,t))]/(3)]+1;

for i=1:numel(t)

Vo1(i) = f(t(i));

end

%% Setting up new circuit values.

Rf = 1*10^6;

C2 = 1*10^-6;

%% Now I need to use the outut of the modified previous problem as input.

% Vout = -(R_feedback)*C*[d(Vin)/dt] *Again Vin here, is Vo1 from

% the previous problem.*

%

%% I originally thought this was where my mistake was, however I believe its actually in the subplot.

% My answers for Vo2 in the array are oscillating the way a square wave does, but it doesn't plot properly.

Vo2 = -Rf*C2*diff(Vo1)

%% Plotting outputs.

subplot(2,1,1)

plot(t/pi,Vo1,'r.-')

title('Input Voltage')

xlabel('t')

ylabel('V(t)')

xlim([0 5])

ylim([-1.5, 1.5])

grid on

subplot(2,1,2)

plot(t/pi,Vo2,'g.-')

title('Inverted Output Voltage')

xlabel('t')

ylabel('V(t)')

xlim([0 5])

grid on

Answer 1

%the diff function returns a vector of length 1 less than the original , since it calculates the difference. Thats why it gave error because you tried plotting it against the same length t vector. I skipped the last t value, so now it plots.

%% This section is copied, and then modified from another problem for use as the input.

%%Does not need to be changed.

clc

clear all

close all

R1 = 1000;

C1 = (1*10^-3);

t = linspace(0,20);

In1 = square(t);

f=@(t) [2*[(-1/(R1*C1))*(integral(@(t) square(t),0,t))]/(3)]+1;

for i=1:numel(t)

Vo1(i) = f(t(i));

end

%% Setting up new circuit values.

Rf = 1*10^6;

C2 = 1*10^-6;

%% Now I need to use the outut of the modified previous problem as input.

% Vout = -(R_feedback)*C*[d(Vin)/dt] *Again Vin here, is Vo1 from

% the previous problem.*

%

%% I originally thought this was where my mistake was, however I believe its actually in the subplot.

% My answers for Vo2 in the array are oscillating the way a square wave does, but it doesn't plot properly.

Vo2 = -Rf*C2*diff(Vo1)

%% Plotting outputs.

subplot(2,1,1)

plot(t/pi,Vo1,'r.-')

title('Input Voltage')

xlabel('t')

ylabel('V(t)')

xlim([0 5])

ylim([-1.5, 1.5])

grid on

subplot(2,1,2)

plot(t(1:end-1)/pi,Vo2,'g.-')%change made, plotting against t with last value unused

title('Inverted Output Voltage')

xlabel('t')

ylabel('V(t)')

xlim([0 5])

grid on