How can I specify the dimension of a differentiated function?

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Ambali Odebowale
Ambali Odebowale 2023-1-27
移动：Walter Roberson 2023-1-27
I am getting error when I tried to find h_w_net and the error is "Matrix dimensions must agree" at the last line.
The dimension of k_r=500 by 1, PTC=500 by 500, theta=1 by 500. After I differentiated theta (rep by theta_diff) its dimension is 1 by 200.
How can I make dimension of theta_diff 1 by 500?
Thanks
%Here is the code
T=300;
theta = h_bar*w ./ ( exp( (h_bar*w)/(k_B*T) ) - 1 ) ;
theta_diff=diff(theta,T);
%Thermal conductance calculation
h_w_net= (1/4*pi^2) *theta_diff.* k_r .* PTC ;
  2 个评论
Rik
Rik 2023-1-27
Ran in:
Did you really intend to take the 300th derivative? I get the impression you don't really understand what the diff function is doing.
When you put in an array, if will calculate the difference between subsequent values. So it doesn't really find the derivative, as it find the numerical approximation of the derivative on the midpoints.
x = [1 2 3];
y = [7 3 6];
diff(y),corresponding_x=mean([x(1:(end-1)) ; x(2:end)])
ans = 1×2
-4 3
corresponding_x = 1×2
1.5000 2.5000
Ambali Odebowale
Ambali Odebowale 2023-1-27
编辑：Ambali Odebowale 2023-1-27
No, I want to differentiate the function theta wrt T. Kindly ignore the T value. It's an error on my part

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回答（2 个）

Walter Roberson
Walter Roberson 2023-1-27
diff() is generally successive differences. It is only derivative if the first parameter is sym or symfun, both from the Symbolic Toolbox
Ambali Odebowale
Ambali Odebowale 2023-1-27
移动：Walter Roberson 2023-1-27
I was able to solve it by using syms
Thanks for your time

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