31 views (last 30 days)

Basically i want to normalize and display the maximum and minimum values between y and y2 for the following code so that they both display on the same plot

t=0:0.001:0.05;

y= 11.18*cos(60*pi*t+26.565);

y2= -60*pi*11.18*sin(60*pi*t+26.565);

title('Phasor Waveforms')

f1=max(y);

f2=max(y2);

hold on

plot(t,y)

plot(t,y2)

hold off

Star Strider
on 11 Sep 2019

Try this:

t=0:0.001:0.05;

y= 11.18*cos(60*pi*t+26.565);

y2= -60*pi*11.18*sin(60*pi*t+26.565);

title('Phasor Waveforms')

f1=max(y);

f2=max(y2);

figure

yyaxis left

plot(t,y)

yyaxis right

plot(t,y2)

Star Strider
on 11 Sep 2019

As always, my pleasure!

I’m not sure what you want to do.

Try this:

t=0:0.001:0.05;

y= 11.18*cos(60*pi*t+26.565);

y2= -60*pi*11.18*sin(60*pi*t+26.565);

title('Phasor Waveforms')

f1=max(y);

f2=max(y2);

idymax = find(y == max(y));

figure

yyaxis left

plot(t,y)

hold on

plot(t(idymax),y(idymax),'o','MarkerFaceColor','red','MarkerSize',15)

hold off

yyaxis right

plot(t,y2)

If you want to plot a point or a series of values, the subscripts for those values need to be the same for all coordinates.

The find function will return the indices of all the values that are equal to ‘max(y)’ here. If there is only one ‘y’ maximum, an alternative could be:

[f1,idymax]=max(y);

since the max function will return the index to the first maximum it discovers.

Star Strider
on 11 Sep 2019

If you assign that logical operation to a variable it returns a logical vector of false (0) values, except for the maximum (true,1) to the variable:

q = y == max(y)

Otherwise, without the assignment, it does not appear to do the logical operation. (I did that experiment.)

Sign in to comment.

KSSV
on 11 Sep 2019

t=0:0.001:0.05;

y= 11.18*cos(60*pi*t+26.565);

y2= -60*pi*11.18*sin(60*pi*t+26.565);

title('Phasor Waveforms')

y=y/max(y) ;

y2=y2/max(y2) ;

[f1,idx1]=max(y);

[f2,idx2]=max(y2);

hold on

plot(t,y)

plot(t,y2)

plot(t(idx1),f1,'*r')

plot(t(idx2),f2,'*b')

hold off

Sign in to comment.

Sign in to answer this question.

Opportunities for recent engineering grads.

Apply Today
## 0 Comments

Sign in to comment.