how do I extend my function on the time axis by the function value of 0 ?

Hey guys i m new to matlab and i need help,

i generated a acceleration function for a simulink model. to make it work i have to increase the time axes to tFinal =10 sec but the acceleration must be 0 . the original function goes to t= 3.677s. Here the acceleration is alomost 0.

I thought i ll use a for loop to increase the time axis but it ist not working. can someone help me ?

%Fourierkoeffizienten eigenes Model
sk=[0.0 -0.00850314 0.0 0.00377249 0.0 -0.000182763 0.0 0.000029564 0.0000652721 0.000052017];
ck=[0.0240577 -0.0241291 0.0 -0.00210694 0.00208514 -0.00044844 0.000515647 -0.0000502304 0.000126237 -0.0000402195];
%Taktzeit in s
tT = 3/5*2*pi;

%Übertragungsfaktor phi = omega * t mit Omega in rad/s
omega = (1/tT)*2*pi;

%Zeitvektor für neues Signal
t_mHsl = linspace(0,tT,1000);

%Bestimmung des Beschleunigungsvorgabe
s_mhsl = zeros(length(1),length(t_mHsl));
v_mhsl = zeros(length(1),length(t_mHsl));
a_mhsl = zeros(length(1),length(t_mHsl));

for k = 0:length(sk)-1
    %Wegvorgabe
    p = 0;
    s_mhsl = s_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));
    
    %Beschleunigungsvorgabe
    p = 1;
    v_mhsl = v_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));
    
    %Beschleunigungsvorgabe
    p = 2;
    a_mhsl = a_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));
end
% plot Beschl.Verl nach Erzeugung der Bewegungsvorgabe mit mHsl
mhsl.time = t_mHsl;
mhsl.a = a_mhsl;
% Beschl.
figure()
plot(mhsl.time,mhsl.a)
% Beschl.verlauf Errerfunktion verlängert von tT auf tFinal 
t_mHsl_tFinal = linspace(0,tFinal,1000);
BeschlmHsl = zeros(length(1),length(t_mHsl_tFinal));
for i = 1:length(t_Final)
    if t_mHsl_tFinal(i)<=t_mHsl(end) % endwert der Erregerfunktion für tT = 3/5 *2pi
        BeschlmHsl(i) =  a_mhsl(1,:);
    else
        BschlmHsl(i) =0;
        
    end
end
figure()
plot(t_Final,BschlmHsl);

[Jan]

I've formatted your code. Please use the tools on top of the editing field. Thanks.

length(1) determines the number of elements of a scalar. It is easier to write "1" directly:

s_mhsl = zeros(1, numel(t_mHsl));

The command length is fragile, because it determines the longest dimension. This does, what you expect, for vectors. But for arrays it is a frequent cause of bugs. Use size(X, dim) or numel(X) instead.

What does this mean: "I thought i ll use a for loop to increase the time axis"?

The statement "but it ist not working" is less clear than a description of what you observe.

[enrico Rothe]

thanks a lot for the help!

sk=[0.0 -0.00850314 0.0 0.00377249 0.0 -0.000182763 0.0 0.000029564 0.0000652721 0.000052017];

ck=[0.0240577 -0.0241291 0.0 -0.00210694 0.00208514 -0.00044844 0.000515647 -0.0000502304 0.000126237 -0.0000402195];

%Taktzeit in s

tT = 3/5*2*pi;

%Übertragungsfaktor phi = omega * t mit Omega in rad/s

omega = (1/tT)*2*pi;

%Zeitvektor für neues Signal

t_mHsl = linspace(0,tT,1000);

%Bestimmung des Beschleunigungsvorgabe

s_mhsl = zeros(length(1),length(t_mHsl));

v_mhsl = zeros(length(1),length(t_mHsl));

a_mhsl = zeros(length(1),length(t_mHsl));

for k = 0:length(sk)-1

%Wegvorgabe

p = 0;

s_mhsl = s_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

%Beschleunigungsvorgabe

p = 1;

v_mhsl = v_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

%Beschleunigungsvorgabe

p = 2;

a_mhsl = a_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

end

% plot Beschl.Verl nach Erzeugung der Bewegungsvorgabe mit mHsl

mhsl.time = t_mHsl;

mhsl.a = a_mhsl;

% Beschl.

figure()

plot(mhsl.time,mhsl.a)

% Beschl.verlauf Errerfunktion verlängert von tT auf tFinal

t_mHsl_tFinal = linspace(0,tFinal,1000);

BeschlmHsl = zeros(1,numel(t_mHsl_tFinal));

for i = 1:numel(t_mHsl_tFinal)

if t_mHsl_tFinal(i)<=t_mHsl(end) % endwert der Erregerfunktion für tT = 3/5 *2pi

BeschlmHsl(1,i) = a_mhsl(1,:);

else

BeschlmHsl(1,i) =a_mhsl(1,:);

end

end

figure()

plot(t_Final,BeschlmHsl);

What does this mean: "I thought i ll use a for loop to increase the time axis"?

i m able to plot my acceleration function for the timevector t_mHsl. (t_mHsl final value = 3.66..sec) to use it in my simulink model i have to expent the acceleration function to t=10 sec but the acceleration values must be 0 after t = 3.66 sec.. and i thought i can solve it with a for loop ?

if i run the code now i still get the error Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-1000.

for BeschlmHsl(1,i) = a_mhsl(1,:);

Answer 1

[Jan]

I'm not sure, what the purpose of the loop is, but this is not a loop at all:

for i = 1:length(t_Final)    

If t_Final is 10, as you explain in the text, length(t_Final) is 1. Then the loop runs from 1 to 1.

Maybe you mean:

for i = 1:numel(t_mHsl_tFinal)    

You have pre-allocated BeschlmHsl to zeros already. Then there is no reason to write 0 into the values again. Omit:

     else
        BschlmHsl(i) =0;

This cannot work:

        BeschlmHsl(i) =  a_mhsl(1,:);        

There is a scalar on the left and a vector on the right.

[enrico Rothe]

thanks for the quick help.

i changed the part of the code but it is still not working

%Fourierkoeffizienten eigenes Model

sk=[0.0 -0.00850314 0.0 0.00377249 0.0 -0.000182763 0.0 0.000029564 0.0000652721 0.000052017];

ck=[0.0240577 -0.0241291 0.0 -0.00210694 0.00208514 -0.00044844 0.000515647 -0.0000502304 0.000126237 -0.0000402195];

%Taktzeit in s

tT = 3/5*2*pi;

%Übertragungsfaktor phi = omega * t mit Omega in rad/s

omega = (1/tT)*2*pi;

%Zeitvektor für neues Signal

t_mHsl = linspace(0,tT,1000);

%Bestimmung des Beschleunigungsvorgabe

s_mhsl = zeros(length(1),length(t_mHsl));

v_mhsl = zeros(length(1),length(t_mHsl));

a_mhsl = zeros(length(1),length(t_mHsl));

for k = 0:length(sk)-1

%Wegvorgabe

p = 0;

s_mhsl = s_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

%Beschleunigungsvorgabe

p = 1;

v_mhsl = v_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

%Beschleunigungsvorgabe

p = 2;

a_mhsl = a_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

end

% plot Beschl.Verl nach Erzeugung der Bewegungsvorgabe mit mHsl

mhsl.time = t_mHsl;

mhsl.a = a_mhsl;

% Beschl.

figure()

plot(mhsl.time,mhsl.a)

% Beschl.verlauf Errerfunktion verlängert von tT auf tFinal

t_mHsl_tFinal = linspace(0,tFinal,1000);

BeschlmHsl = zeros(1,length(t_mHsl_tFinal));

for i = 1:length(t_mHsl_tFinal)

if t_mHsl_tFinal(i)<=t_mHsl(end) % endwert der Erregerfunktion für tT = 3/5 *2pi

BeschlmHsl(1,i) = a_mhsl(1,:);

else

BeschlmHsl(1,i) =a_mhsl(1,:);

end

end

figure()

plot(t_Final,BeschlmHsl);

i still get the size error for BeschlmHsl(1,i) = a_mhsl(1,:);

[enrico Rothe]

i got i working now but the acceleration curve doesent look like i need it :D i m not sure if i can solve my problem with one for loop. i might have to put my two for loops together the get the result i want ?

i you copy the code to matlab u get 2 figures. i need the curve of the first figure and after t => 3.766 the acceleration curve must be 0 ^^^

sk=[0.0 -0.00850314 0.0 0.00377249 0.0 -0.000182763 0.0 0.000029564 0.0000652721 0.000052017];

ck=[0.0240577 -0.0241291 0.0 -0.00210694 0.00208514 -0.00044844 0.000515647 -0.0000502304 0.000126237 -0.0000402195];

%Taktzeit in s

tT = 3/5*2*pi;

%Übertragungsfaktor phi = omega * t mit Omega in rad/s

omega = (1/tT)*2*pi;

%Zeitvektor für neues Signal

t_mHsl = linspace(0,tT,1000);

%Bestimmung des Beschleunigungsvorgabe

s_mhsl = zeros(1,length(t_mHsl));

v_mhsl = zeros(1,length(t_mHsl));

a_mhsl = zeros(1,length(t_mHsl));

for k = 0:length(sk)-1

%Wegvorgabe

p = 0;

s_mhsl = s_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

%Beschleunigungsvorgabe

p = 1;

v_mhsl = v_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

%Beschleunigungsvorgabe

p = 2;

a_mhsl = a_mhsl + (k^p*omega^p*(ck(k+1)*cos(p*pi/2+k*omega*t_mHsl)+sk(k+1)*sin(p*pi/2+k*omega*t_mHsl)));

end

% plot Beschl.Verl nach Erzeugung der Bewegungsvorgabe mit mHsl

mhsl.time = t_mHsl;

mhsl.a = a_mhsl;

% Beschl.

figure()

plot(mhsl.time,mhsl.a)

% Beschl.verlauf Errerfunktion verlängert von tT auf tFinal

t_mHsl_tFinal = linspace(0,tFinal,1000);

BeschlmHsl = zeros(1,numel(t_mHsl_tFinal));

for i = 1:numel(t_mHsl_tFinal)

if t_mHsl_tFinal(i) <= t_mHsl(end) % endwert der Erregerfunktion für tT = 3/5 *2pi

BeschlmHsl(i) = a_mhsl(1,1000);

else

BeschlmHsl(i) =0;

end

end

figure()

plot(t_Final,BeschlmHsl);

[Jan]

I have mentioned, that I have formatted the code for you in the question. Please apply a proper formatting by your own to improve the readaility of the code. Thanks.

"i need the curve of the first figure and after t => 3.766 the acceleration curve must be 0" - please consider, that the reader cannot guess, which variable is meant. While this detail is obvious for you, it is a puzzle for the readers.

Maybe you mean:

t_mHsl_tFinal = linspace(0, tFinal, 1000);
BeschlmHsl    = zeros(1, numel(t_mHsl_tFinal));
index         = t_mHsl_tFinal < t_mHsl(end);
BeschlmHsl(index) = a_mhsl(1, 1000);

No loop needed.