How do I correctly use vpasolve?

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Goncalo Costa
Goncalo Costa 2022-11-17
评论： Walter Roberson 2022-11-18
I am trying to numerically solve the following set of equations
M_mag = 1 ; %just some constant simplified here as 1
om = 2*pi*new_freq ; %new_freq is an array of values
C1 = exp(-1i*L.*om./c); % L and c are constant
% phi = exp(1i*om*L/c);
nume = exp(1i.*om.*n*L/c)*4.*n./((1+n).^2); % terms depending on n
denom = 1 + exp(2*1i.*om.*n*L/c).*((n-1)./(n+1)).^2; % terms depending on n
syms n
S = vpasolve(M_mag./C1 == nume./denom , n , guess) %guess is a numerical approach done via another equation, in this case it is 1.7
Whenever I try to use this, I get the following text:
Error using mupadengine/feval_internal
More equations than variables is only supported for polynomial
systems.
Error in sym/vpasolve (line 172)
sol = eng.feval_internal('symobj::vpasolve',eqns,vars,X0);
I have seen similar things being explain in the following link, but I don't understand the explanation. Same for this explanation.
Why can I not solve the equation this way? Is it the exponentials terms?
Thank you very much for your help.
  1 个评论
Star Strider
Star Strider 2022-11-17
I usually use solve, and then vpa the result. It is generally more reliable, especially with functions with multiple roots.
Too much of this code is ‘over the horizon’ and out of sight to provide a specific response.

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回答（2 个）

Torsten
Torsten 2022-11-17
编辑：Torsten 2022-11-17
You have 2 equations for 1 unknown. MATLAB's symbolic toolbox does not solve such systems.
Is n supposed to be real or complex ?
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显示 4更早的评论
Goncalo Costa
Goncalo Costa 2022-11-18
n is a complex number. Can I change my code for a complex result?
Thank you very much for your help.
Goncalo Costa
Goncalo Costa 2022-11-18
@Walter Roberson that makes sense, I thought that it would solve one-by-one. I didn't understand the limitations of this tool. How can I go around this to have this equation solved for an array? Would a for loop be needed?

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Walter Roberson
Walter Roberson 2022-11-17
M_mag = 1 ; %just some constant simplified here as 1
om = 2*pi*new_freq ; %new_freq is an array of values
C1 = exp(-1i*L.*om./c); % L and c are constant
% phi = exp(1i*om*L/c);
syms n
nume = exp(1i.*om.*n*L/c)*4.*n./((1+n).^2); % terms depending on n
denom = 1 + exp(2*1i.*om.*n*L/c).*((n-1)./(n+1)).^2; % terms depending on n
S = arrayfun(@(EQN) vpasolve(EQN,n,guess), M_mag./C1 == nume./denom)
  2 个评论
Goncalo Costa
Goncalo Costa 2022-11-18
Dear @Walter Roberson, I have never understood how arrafun works, but it seems to be giving me the right answer. Have you got any link that explain the function written above?
Thank you for your help.
PS: For some reason, the first cell in the cell array separates the complex from the real part of the number.But that can be easily sorted.
Walter Roberson
Walter Roberson 2022-11-18
result = arrayfun(FUNCTION, ARRAY)
is effectively the same as
if isempty(ARRAY)
result = [];
else
out1 = FUNCTION(ARRAY(1));
result = zeros(size(ARRAY), class(out1));
result(1) = out1;
for K = 2 : numel(ARRAY)
result(K) = FUNCTION(ARRAY(K));
end
end
and
result = arrayfun(FUNCTION, ARRAY, 'uniform', 0)
is effectively the same as
result = cell(size(ARRAY));
for K = 1 : numel(ARRAY)
result{K} = FUNCTION(ARRAY(K));
end
and
result = arrayfun(FUNCTION, ARRAY1, ARRAY2)
is effectively the same as
assert(isequal(size(ARRAY1), size(ARRAY2)), 'input arrays must be the same size');
if isempty(ARRAY1)
result = [];
else
out1 = FUNCTION(ARRAY1(1), ARRAY2(1));
result = zeros(size(ARRAY1), class(out1));
result(1) = out1;
for K = 2 : numel(ARRAY1)
result(K) = FUNCTION(ARRAY1(K), ARRAY2(K));
end
end
So you input a function handle (typically), and one or more arrays that must be exactly the same size. The output is the same size as the array. When 'uniform', 0 is specified, the output is a cell array containing the result of executing the function on each corresponding sets of values from the array in turn. If 'uniform', 0 is not specified, the output is a regular array containing the result of executing the function on each corresponding set of values from the array in turn.

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