Is this correct?

1 visualización (últimos 30 días)
Allison Sims
Allison Sims el 18 de Jul. de 2022
Comentada: Chunru el 18 de Jul. de 2022
Use loops to compute and plot (not animated)the following piecewise function for -15<=x<=15 .
F(x)={5x, x<0
{x^2, 0<=x<2
{ 2lnx, x>=2
My code:
x=-15:15;
if x<0;
5*x;
elseif x>=0 & x<2
y=x^2;
else x>=2;
y=2*log(x);
end
plot(x,y)
Warning: Imaginary parts of complex X and/or Y arguments ignored.
  4 comentarios
M.B
M.B el 18 de Jul. de 2022
Check your variable names and indexation.
Chunru
Chunru el 18 de Jul. de 2022
Although the code is not doing what it intends to, the code can be run and MATLAB only gives a warning message (orange color).

Iniciar sesión para comentar.

Respuesta aceptada

Chunru
Chunru el 18 de Jul. de 2022
x=-15:15;
y = zeros(size(x));
% This is more MATLAB way
idx = x<0;
y(idx) = 5 * x(idx);
idx = x>=0 & x<2;
y(idx) =x(idx).^2;
idx = x>=2;
y(idx) = 2*log(x(idx));
plot(x,y)
% This is more conventional way (some othter programming language)
x=-15:15;
y = zeros(size(x));
for i=1:length(x)
if x(i)<0
y(i) = 5 * x(i);
elseif x(i)>=0 & x(i)<2
y(i) =x(i).^2;
elseif x(i)>=2;
y(i) = 2*log(x(i));
end
end
figure
plot(x,y)
  1 comentario
Allison Sims
Allison Sims el 18 de Jul. de 2022
That makes more sense thank you!

Iniciar sesión para comentar.

Más respuestas (1)

M.B
M.B el 18 de Jul. de 2022
x = -15:1:15;% init x
y = nan*x;% init y to be of same size as x
for index = -15:15;
if index<0;
y(index+16) = 5*index;
elseif index>=0 & index<2
y(index+16) = index^2;
else index>=2;
y(index+16) = 2*log(index);
end
end
plot(x,y)
  2 comentarios
Allison Sims
Allison Sims el 18 de Jul. de 2022
what does the y(index+16) mean?
Chunru
Chunru el 18 de Jul. de 2022
The "index" is the loop variable the code uses and it need to be added with 16 in order to become array index (which is 1,2,...31). It's working code, but it may be further improved.

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by