Vectors must be the same length error

Hello everyone, im facing a problem that i cant solve it. Im new to MatLab and im having a , Vectors must be the same length, error. I dont know that well about grafics in matlab so if someone could help me it would be apreciated

heres my code

bny = input('Insira o codigo binario a codificar: ' ,'s');
idx = ismember(bny,'01');
assert(all(idx),'O codigo binario só pode conter 0s e 1s, mas contem o(s) número(s) %s',bny(~idx))
fprintf('O seu codigo é: %s\n',bny)
V = [-3,3];
n = V(bny-'/');

i=1;
a=0;
b=0.5;
t=0:0.01:length(bny);
for j=1:length(bny)
    if t(j)>=a && t(j)<=b
        y(j)=V(i);
    elseif t(j)>b && t(j)<=i
        y(j)=0;
    else
        i= i+1;
        a=a+1;
        b=b+1; 
    end
end
plot(t,y,'k');axis([0 length(bny) -5 5]);title('Rz Polar');
xlabel('time-->');
ylabel('Amplitude-->');

[Adam Danz]

What input to 'bny' produces the error?

When I input 0 or 1 there is no error.

Also, copy-paste the full error message.

[António Pereira]

rz

Insira o codigo binario a codificar: 100110

O seu codigo é: 100110

Error using plot

Vectors must be the same length.

Error in rz (line 23)

plot(t,y,'k');axis([0 length(bny) -5 5]);title('Rz Polar');

[Adam Danz]

For input '100110', the variable 't' (below) will be a vector with 601 elements.

t=0:0.01:length(bny);  %length(bny) = 6

The for-loop (below) will have 6 iterations so the variable 'y' created within the for-loop will be a vector of 6 elements.

for j=1:length(bny)    %length(bny) = 6
    ...
end

Then you're trying to plot y as a function of t (below) and to do this, 'y' and 't' need to have the same ammount of data. But your code produces 601 data points for 't' and only 6 data points for 'y' (for the inputs 100110)

plot(t,y,'k')

I haven't gone through your code to understand what you're trying to do but this explains the source of the error.

[António Pereira]

I think i got it, but im new to matlab so i dont know that well how to do that

[Adam Danz]

Can you explain the purpose of the code, the inputs and expected outputs?

[António Pereira]

Sure.The purpose is to make a simulation of a RZ coded digital signal. The input is to add the 1´s and 0´s (binary) that will be coded. The expected output would be a grafic that when value is "1" it would go y=3 (and then halfway point it would go down to the middle again) i will show a image of that down below.

Answer 1

[Adam Danz]

After understanding the purpose of the code in the comments under the question, I rewrote the code to produce a plot that was described above. If this isn't what you were looking for or if you have specific quesitons about your original code, I'd be glad to help more.

%bny = input('Insira o codigo binario a codificar: ' ,'s');
bny = '100110'; 
bvec =  bny-'0'; %convert to numerical vector

% create x,y values of step function
stepHeight = 3; 
stepIdx = 0 : 0.5 : length(bvec)-0.5; 
stepX = repelem(stepIdx, 2); 
stepY = repmat([0, stepHeight, stepHeight, 0], 1, length(bvec)); 

% now we have a step function that's all positive 
%     figure
%     plot(stepX, stepY, 'r-')
%     ylim([-stepHeight, stepHeight]*2)

% flip sign of steps associated with bny=0
zeroIdx = repelem(bvec == 0, 4);   %4 because there are 4 values in stepX/Y for each step
tallIdx = stepY > 0; 
stepY(zeroIdx & tallIdx) = -1 * stepHeight; 

% plot results
figure
plot([min(stepX), max(stepX)], [0,0], 'k-', 'LineWidth', 4)     %reference line at y = 0 
hold on
plot(stepX, stepY, 'r-', 'LineWidth', 3)                        %step function
ylim([-stepHeight, stepHeight]*2)

[António Pereira]

Thank you so much. It works great. But my question if it is possible to be de user to chosse the 0's and 1's insted of doing it on the code.

[Adam Danz]

Absolutely. Just replace the line #2 of my code with line #1.

[António Pereira]

Wow, this is working really nice. Thank you so much!!! Priciate your time.

Just one last thing. If i wanted to block the user from using other numbers (not allow him to use numbers besides 0 and 1) how would i do it?

[Adam Danz]

Glad it helped!

Answer 2

[KSSV]

YOu need to rethink on your code.

bny = input('Insira o codigo binario a codificar: ' ,'s');
idx = ismember(bny,'01');
assert(all(idx),'O codigo binario só pode conter 0s e 1s, mas contem o(s) número(s) %s',bny(~idx))
fprintf('O seu codigo é: %s\n',bny)
V = [-3,3];
n = V(bny-'/');
i=1;
a=0;
b=0.5;
% t=0:0.01:length(bny);
t = linspace(0,length(bny),length(bny)) ;
for j=1:length(bny)
    if t(j)>=a && t(j)<=b
        y(j)=V(i);
    elseif t(j)>b && t(j)<=i
        y(j)=0;
    else
        i= i+1;
        a=a+1;
        b=b+1; 
    end
end
plot(t,y,'k');axis([0 length(bny) -5 5]);title('Rz Polar');
xlabel('time-->');
ylabel('Amplitude-->');

[António Pereira]

Hey, with that correction it still does give the same error