How I explicit one function in terms of another function using symbolic?

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Rubem Pacelli
Rubem Pacelli 2019-9-3
评论： Walter Roberson 2020-1-4
syms t T
syms a(t) c0(t) d(t)
I(t) = a(0)*c0(t) + a(2)*c0(t - 2*T) + a(4)*c0(t - 4*T)
Q(t) = a(1)*c0(t - T) + a(3)*c0(t - 3*T) + a(5)*c0(t - 5*T)
s(t) = I(t)*cos(t) - Q(t)*sin(t)
Suppose I wish to express s(t) in terms of I(t) and Q(t). That is, I wish to see
s(t) = I(t)*cos(t) - Q(t)*sin(t)
in the command window. How i do that?
OBS: I don't wanna substitute the awser by I(t) and Q(t). I just wanna explicit I(t) and Q(t) in my expression.
(I know, it's don't seems productive. But in longs expressions, it do).
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Walter Roberson
Walter Roberson 2019-11-7
You would need to evalin(symengine) or feval(symengine) using hold() constructs on the symbols I(t) and Q(t) to prevent them from being evaluated to the definitions you gave for the functions. However, most of the time at least one mupad eval() operation is done on your behalf, often at least three of them, and that would replace the symbols with the contents. You would need to experiment at the feval() or evalin() level to find the perfect number of nested hold() operations to put around the symbols so that the mupad print operation applied to the expression removed all of hold objects without execution of the symbols to their content while still permitting the expression to be evaluated more fully to get the expanded form when needed.
It would be a heck of a lot easier if you were willing to live with a printed representation that did not use exactly the same symbol names, and then used subs() when you wanted to expand.
Rubem Pacelli
Rubem Pacelli 2020-1-4
Walter Roberson , Thanks very much. It's seems pretty complicated, actually. I expected to do it easily, because it's so basic (my though). However, if I will do it, I take a look in this topic to refresh my mind. Thks!

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Walter Roberson
Walter Roberson 2020-1-4
编辑：Walter Roberson 2020-1-4
You define functions I(t) and Q(t) and then use the expression
s(t) = I(t)*cos(t) - Q(t)*sin(t)
and hope to see I(t) and Q(t) in the result even though those have been defined as functions.
In order to do that, you would somehow have to prevent I(t) and Q(t) from executing, and yet not lose their definitions. That involves hacking the way that MATLAB evaluates expressions, which is not going to end well.
You would find it easier to do
syms t T
syms a(t) c0(t) d(t)
syms I_(t) Q(t)
s(t) = I_(t)*cos(t) - Q(t)*sin(t)
I_(t) = a(0)*c0(t) + a(2)*c0(t - 2*T) + a(4)*c0(t - 4*T)
Q(t) = a(1)*c0(t - T) + a(3)*c0(t - 3*T) + a(5)*c0(t - 5*T)
Now if you ask to display s, you would get
I_(t)*cos(t) - Q(t)*sin(t)
and you could then do
subs(s)
to have I_ and Q evaluated to get
cos(t)*(a(0)*c0(t) + a(2)*c0(t - 2*T) + a(4)*c0(t - 4*T)) - sin(t)*(a(1)*c0(t - T) + a(3)*c0(t - 3*T) + a(5)*c0(t - 5*T))
Note the change I had to make to change I(t) to I_(t) . This change is because the symbolic engine treats I specially because inside the symbolic engine, I is used to refer to the imaginary constant sqrt(-1) .
If you ***really** want s to be in terms of I(t) then you can instead use
syms t T
syms a(t) c0(t) d(t)
syms I(t) Q(t)
s(t) = I(t)*cos(t) - Q(t)*sin(t)
I(t) = a(0)*c0(t) + a(2)*c0(t - 2*T) + a(4)*c0(t - 4*T)
Q(t) = a(1)*c0(t - T) + a(3)*c0(t - 3*T) + a(5)*c0(t - 5*T)
s
subs(s, {str2sym('I(t)'), str2sym('Q(t)')}, {I, Q})
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Rubem Pacelli
Rubem Pacelli 2020-1-4
Fantastic Walter Roberson, you solved my problem in a simple way!
You mean I need declare s first. Then i declare Q and I_, right? I swapped the these lines just out for curiosity, like that:
syms t T
syms a(t) c0(t) d(t)
syms I_(t) Q(t)
I_(t) = a(0)*c0(t) + a(2)*c0(t - 2*T) + a(4)*c0(t - 4*T)
Q(t) = a(1)*c0(t - T) + a(3)*c0(t - 3*T) + a(5)*c0(t - 5*T)
s(t) = I_(t)*cos(t) - Q(t)*sin(t)
I this mode, if I ask to display s, I would get:
s(t) =
cos(t)*(a(0)*c0(t) + a(2)*c0(t - 2*T) + a(4)*c0(t - 4*T)) - sin(t)*(a(1)*c0(t - T) + a(3)*c0(t - 3*T) + a(5)*c0(t - 5*T))
But If I follow your order, I achieve my goal, this is, I get:
s(t) =
I_(t)*cos(t) - Q(t)*sin(t)
Walter Roberson
Walter Roberson 2020-1-4
Yes, exactly. If you assign to s(t) first in terms of the unresolved symbolic functions I_(t) and Q(t) then if you just display s(t) then you get it in terms of the unresolved names, and when you then want to do the full expansion, subs() will do the expansion.

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