Produce the same result but without using a for-loop.
Mostrar comentarios más antiguos
% Produce the same result but without using a for-loop.
% Increment
dx = 35/300;
% Independent variable
x = -5:dx:30; % x = linspace(-5,30,300)
for n = 1:length(x)
if x(n) >= 9
y(n) = 15*sqrt(4*x(n)) + 10;
elseif (x(n) >= 0) && (x(n) <= 9)
y(n) = 10*x(n) + 10;
else
y(n) = 10;
end
end
plot(x,y), xlabel('x'), ylabel('y'), grid on
1 comentario
Respuestas (2)
Cris LaPierre
el 6 de Abr. de 2021
Editada: Cris LaPierre
el 6 de Abr. de 2021
0 votos
If your assignment doesn't allow that, then create 3 vectors, one for each equation, and solve each for all values of x. Then multiply each by a logical array created from x and the corresponding conditional statement(s). Finally, add all three vectors together.
See Ch 12 of MATLAB Onramp if you need help creating the logical array, and Ch 6 if you need help on elementwise multiplication.
8 comentarios
laith awwad
el 6 de Abr. de 2021
Cris LaPierre
el 6 de Abr. de 2021
Editada: Cris LaPierre
el 6 de Abr. de 2021
Both techniques worked for me. Please share your code, as well as the error message (all the red text) that you are getting.
laith awwad
el 6 de Abr. de 2021
Editada: Rik
el 6 de Abr. de 2021
Cris LaPierre
el 6 de Abr. de 2021
Editada: Cris LaPierre
el 6 de Abr. de 2021
I can't reproduce your error, but here's one suggestion.
You have not written your second condition in valid MATLAB syntax. You need two separate conditions joined with an '&', similar to what you did in the elseif line previously.
Also be aware of your endpoints of your different conditions. It is best if they are not shared. For example, if you have x>=9 in one, then the next range should be x<9 and not x<=9.
laith awwad
el 6 de Abr. de 2021
Walter Roberson
el 6 de Abr. de 2021
Editada: Walter Roberson
el 6 de Abr. de 2021
You have not written your second condition in valid MATLAB syntax. You need two separate conditions joined with an '&', similar to what you did in the elseif line previously.
That turns out not to be true for piecewise():
syms x
A = piecewise(2 <= x & x <= 3, 1, 0)
B = piecewise(2 <= x <= 3, 1, 0)
A - B
Also be aware of your endpoints of your different conditions. It is best if they are not shared.
The rule for piecewise() is that the left-most case is the one that will be picked; which is to say that the first matching condition is the one that will be used. People overlap all the time, and it is often fine:
syms x
A = piecewise(x == 0, 2, -1 <= x & x <= 1, 1/x, 0)
However, when people have two matching boundary conditions, it is sometimes (but not always) a sign that they have no paid attention to be sure that the value matches at the boundary. When equations with bounds are extracted from textbooks or research papers, the authors will generally have ensured that the bounds match if the bounds are equal... but authors are known to make mistakes.
Cris LaPierre
el 6 de Abr. de 2021
Editada: Cris LaPierre
el 6 de Abr. de 2021
You have two competing conditions in your second case.
- 0>=x means x less than or equal to 0.
- x>=9 is the same as your condition for your first condition and means x greater than or equal to 9.
Can you see how the two conditions cannot be met? Think through how to define your ranges.
laith awwad
el 6 de Abr. de 2021
Walter Roberson
el 6 de Abr. de 2021
syms y(x)
y(x) = piecewise(x>=9, 15*sqrt(4*x)+10, 0>=x=<9, 10*x+10,10)
^^
MATLAB does not have an =< operator; it has a <= operator.
dx = 35/300;
xvalue = -5:dx:30;
fplot(y)
fplot() by default plots from -5 to +5 . Your assignment to xvalue does not affect that. To have it plot over a different interval, the interval would have to be passed as the second parameter, such as
fplot(y, [-5, 30])
Only the endpoints of the interval would be passed.
fplot() always chooses its own points to plot at. If you only want to plot at particular points, you should not use fplot(); you should use plot() instead.
plot() does not accept formulas, so you would need to
plot(xvalue, y(xvalue))
1 comentario
laith awwad
el 6 de Abr. de 2021
Categorías
Más información sobre Mathematics en Centro de ayuda y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
