how to replace variable by another variable

2023 2 8
3 个回答
更新时间：2023 4 29
9 次查看（30 天）

0 个投票

Hn=-2*P/(Mu*exp(n^2*pi^2*T/Mu))
I want to replace T by t and want to do calculation

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Venkat Siddarth
Venkat Siddarth 2023-2-8
Hi @vikas singh,
Can you elaborate the query,on whats the result you are getting when you replace "T" with "t" in the equation itself i.e
Hn=-2*P/(Mu*exp(n^2*pi^2*t/Mu))?
vikas singh
vikas singh 2023-2-10
actually I have to replace T by t/L and again i have to give different value to t for calculation. and some for loop is also involve in the equation for n

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回答（3 个）

Walter Roberson
Walter Roberson 2023-2-8

0 个投票

if you are using the symbolic toolbox then subs()

6 个评论
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vikas singh
vikas singh 2023-2-10
if i want to give some numeric value to new variable then how to give
Walter Roberson
Walter Roberson 2023-2-11
Ran in:
Ran in:
syms L Mu n P T t
Pi = sym(pi);
Hn=-2*P/(Mu*exp(n^2*Pi^2*T/Mu))
Hn = 
Hn_new = subs(Hn, T, t/L)
Hn_new = 
subs(Hn_new, t, (1:5).')
ans = 
vikas singh
vikas singh 2023-2-11
it is a great help for me.
vikas singh
vikas singh 2023-2-13
can you please tell me if I want to give value of t as 5, 10, 15, 20,25,.......,1000 then how to give here in the code. thanks in advance.

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vikas singh
vikas singh 2023-2-15

0 个投票

when I'm using the following code to evaluate z value . I'm getting the Q value and all the value after Q is zero. I m not getting where I'm wrong.
clc;
clear all;
close all;
l=10000;
x=0:100:l;
t=[0;1;5;10;15;20;25];
H=30;
Mu=2.8e-1;
D=31;
k=0.01132;
k1=l/H;
for n=1:100
a=2*H;
b=(1-(-1)^n*exp(-k*l))/((n*pi+(k^2*l^2)/(n*pi)));
c=1/(n*pi);
Cn(n)=a*(b-c);
end
s=0;
for n=1:100
d= sin(n*pi*x/l);
q= exp((-n^2*pi^2*t)/(Mu^2*l));
aa=Cn(n).*d.*q;
s=s+aa;
end
syms X T
for n=1:100
H0=@(X,T)Cn(n).*sin(n*pi*X).*exp(-n^2*pi^2*T/Mu)+H*X/(D-H)+H/(D-H);
y=diff(H0*diff(H0,X),X);
Q=int(y*sin(n*pi*X),X,0,1);
P=int(Q*exp(n^2*pi^2*T/Mu),T,0,T);
Hn=-2*P/(Mu*exp(n^2*pi^2*T/Mu));
H1=Hn.*sin(n*pi*X);
for i=1:size(t,1)
for j=1:length(x)
H=(subs(H1,{X,T},{x(j)/l,t(i,1)/(Mu*l)}));
ZZ(j,1)=double(H);
end
XX(i,1)=sum(ZZ);
clear ZZ
end
s1=((D-H)^2).*XX/l;
end
j=(H*x)/l;
z=s-j-s1
plot(x,z)

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Walter Roberson
Walter Roberson 2023-2-16
I think you are getting some underflow, exp(-number) terms that evaluate in floating point to zero even though mathematically they are not zero.
vikas singh
vikas singh 2023-3-6
suppose I have a variable z and it has 100 values for each time variable t (suppose t= 0,1,5, 10,15). suppose one variable cell V{} contains 100 value for different time, then how to minus z and V{} values
vikas singh
vikas singh 2023-4-25
I want to multiply r and f value in the following code. how to do it.
clc;
clear all;
close all;
L=600;
T=365;
delt=0.025;
k=8.64;
aa=delt*k*(1+delt);
epc=0.3;
H=50;
mu=epc/(k*delt*H);
w1=86400*1*10^-8;
w2=8*86400*10^-8;
dx=1;
dt=1;
d=(mu*dt)/(dx^2);
N=L/dx +1;
M=T/dt +1;
for i=1:N
x(i)=0+(i-1)*dx;
end
for n=1:M
t(n)=0+(n-1)*dt;
end
for n=1:M
for i=1:N
if x(i)>=250 && x(i)<=350
r(i)=w2;
else
r(i)=w1;
end
end
if t(n)<=182.5
f(n)=sin(2*pi*t(n)/365).^2;
else
f(n)=0;
end
end

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vikas singh
vikas singh 2023-4-25
编辑：Walter Roberson 2023-4-25
Ran in:

0 个投票

Ran in:
I'm writing the following but it is giving me zero and showing matrix dimension is not agree
clc;
clear all;
close all;
L=600;
T=365;
delt=0.025;
k=8.64;
aa=delt*k*(1+delt);
epc=0.3;
H=50;
mu=epc/(k*delt*H);
w1=86400*1*10^-8;
w2=8*86400*10^-8;
dx=1;
dt=1;
d=(mu*dt)/(dx^2);
N=L/dx +1;
M=T/dt +1;
for i=1:N
x(i)=0+(i-1)*dx;
end
for n=1:M
t(n)=0+(n-1)*dt;
end
for n=1:M
for i=1:N
if x(i)>=250 && x(i)<=350
r(i)=w2;
else
r(i)=w1;
end
end
if t(n)<=182.5
f(n)=sin(2*pi*t(n)/365).^2;
else
f(n)=0;
end
f2=r.*f
end
f2 = 1×601
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Arrays have incompatible sizes for this operation.

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Walter Roberson
Walter Roberson 2023-4-25
for n=1:M
for i=1:N
if x(i)>=250 && x(i)<=350
r(i)=w2;
else
r(i)=w1;
end
end
r is a vector of length N after that for i loop.
if t(n)<=182.5
f(n)=sin(2*pi*t(n)/365).^2;
else
f(n)=0;
end
f is a vector of length n after that if test -- it is being expanded as do more interations of for n
f2=r.*f
r is 1 x N vector. f is a 1 x n vector. They can only be .* together if the changing variable n == N or n == 1 (in which case 1 x N .* 1 x 1 would be scalar expansion.)
vikas singh
vikas singh 2023-4-28
编辑：Walter Roberson 2023-4-28
I want to plot between x and -(z(n+1,i)).^0.5. how to get it?
clc;
clear all;
close all;
L=600;
T=365;
delt=0.025;
k=8.64;
aa=delt*k*(1+delt);
epc=0.3;
H=50;
mu=(k*delt*H)/epc;
w1=1*86400*10^-8;
w2=2*86400*10^-8;
dx=1;
dt=1;
d=(mu*dt)/(dx.^2);
N=L/dx +1;
M=T/dt +1;
for i=1:N
x(i)=0+(i-1)*dx;
end
for n=1:M
t(n)=0+(n-1)*dt;
end
for n=1:M
for i=1:N
if x(i)>=250 && x(i)<=350
r(i)=w2;
else
r(i)=w1;
end
if t(n)<=182.5
f(n)=sin(2*pi*t(n)/365).^2;
else
f(n)=0;
end
Q(n,i)=(2*r(i)*f(n)*mu)/aa;
end
end
for n=1:M
z(n,1)=0;
z(n,N)=0;
end
for i=1:N
z(1,i)=0;
end
for n=1:M-1
for i=2:N-1
z(n+1,i)=z(n,i)+d*(z(n,i+1)-2*z(n,i)+z(n,i-1))+Q(n,i)*dt;
end
end
Walter Roberson
Walter Roberson 2023-4-28
I want to plot between x and -(z(n+1,i)).^0.5
Do you mean that you have an independent variable 1:N on the x axis, and you want to treat the variable x and that particular expression as dependent variables to be drawn and you want to fill the area between the two lines?
Or is your x variable to be treated as the x axis and you want to draw -(z(n+1,i)).^0.5 even though that appears to be independent of x?
vikas singh
vikas singh 2023-4-29
tell me for both the conditons as well

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