plotting a derivative of a function using surf command

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Wajahat
Wajahat 2019-3-8
编辑： Wajahat 2019-3-9
I am trying to plot a derivative of a function using surf command, but when I evaluate it an error occur "Data dimensions must agree".
This is what I typed:
close all;
X=-10:.1:10;
T=-10:.1:10;
mu1=-.01+1*1i;
a=(.1-2*1i);
b=(.1-.1*1i);
[x,t]=meshgrid(X,T);
x1=exp(-1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
x2=exp(1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y1=exp(-1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y2=exp(1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
A=2*1i*(1)*a*conj(b)*x1.*conj(x2).*(x2.*conj(y1)-y2.*conj(x1));
B=(a.*conj(a).*x1.*conj(x1).*y2.*conj(y2)+b.*conj(b).*x1.*conj(y1).*x2.*conj(y2)+a.*conj(a).*x2.*conj(x2).*y1.*conj(y1)+b.*conj(b).*y1.*conj(x2).*y2.*conj(x1));
r1=-2.*1i.*((A./B));
dr1=diff(r1);
dt=diff(t);
dr1dt=dr1./dt;
td=t(2:end);
surf(x,td,abs(dr1dt));
  1 个评论
KSSV
KSSV 2019-3-8
X=-10:.1:10;
T=-10:.1:10;
mu1=-.01+1*1i;
a=(.1-2*1i);
b=(.1-.1*1i);
[x,t]=meshgrid(X,T);
x1=exp(-1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
x2=exp(1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y1=exp(-1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y2=exp(1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
A=2*1i*(1)*a*conj(b)*x1.*conj(x2).*(x2.*conj(y1)-y2.*conj(x1));
B=(a.*conj(a).*x1.*conj(x1).*y2.*conj(y2)+b.*conj(b).*x1.*conj(y1).*x2.*conj(y2)+a.*conj(a).*x2.*conj(x2).*y1.*conj(y1)+b.*conj(b).*y1.*conj(x2).*y2.*conj(x1));
r1=-2.*1i.*((A./B));
dr1=gradient(r1);
dt=gradient(t);
dr1dt=dr1./dt;
td=t(2:end);
surf(x,t,abs(dr1dt)');
But you need re think on your code.

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回答（1 个）

KSSV
KSSV 2019-3-8
X=-10:.1:10;
T=-10:.1:10;
mu1=-.01+1*1i;
a=(.1-2*1i);
b=(.1-.1*1i);
[x,t]=meshgrid(X,T);
x1=exp(-1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
x2=exp(1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y1=exp(-1i*mu1.*x).*exp((1+(1i./(2*mu1)).*t));
y2=exp(1i*mu1.*x).*exp((1-(1i./(2*mu1)).*t));
A=2*1i*(1)*a*conj(b)*x1.*conj(x2).*(x2.*conj(y1)-y2.*conj(x1));
B=(a.*conj(a).*x1.*conj(x1).*y2.*conj(y2)+b.*conj(b).*x1.*conj(y1).*x2.*conj(y2)+a.*conj(a).*x2.*conj(x2).*y1.*conj(y1)+b.*conj(b).*y1.*conj(x2).*y2.*conj(x1));
r1=-2.*1i.*((A./B));
dr1=gradient(r1);
dt=gradient(t);
dr1dt=dr1./min(diff(T));
td=t(2:end);
surf(x,t,abs(dr1dt)');
To get dt you can use difference in T. You need not to take a matrix. ANote that dt is same i.e 0.01. If you use a matrix..it is coming out to be zero matrix and makes dr1dt a nan or inf matrix. I advice you to still rethink on your code.
  5 个评论
显示 3更早的评论
KSSV
KSSV 2019-3-8
diff reduces the dimension by one....ad you are subtracting the consecutive elements. Gradient will not reduce the dimensions.
Wajahat
Wajahat 2019-3-8
编辑：Wajahat 2019-3-9
@ KSSV,
To verify, I considered a simple example such as
X=-1:.05:1;
T=-1:.05:1;
mu1=1+1*1i;
[x,t]=meshgrid(X,T);
r1=mu1.*sin(x+4.*t);
dr1=gradient(r1);
dt=gradient(t);
dr1dt=dr1./min(diff(T));
td=t(2:end);
surf(x,t,abs(dr1dt));
I have plotted it.
Then I take the derivative of 'r1' w.r.t 't' and then plot the function. i..e,
X=-1:.05:1;
T=-1:.05:1;
mu1=1+1*1i;
[x,t]=meshgrid(X,T);
r1=4.*mu1.*cos(x+4.*t);
surf(x,t,abs(r1)');
You can see there is difference of amplitude in these plots. Why the two plots are not identical?

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