differentiation in matlab of two functions

Let

theta=atan(p/q)+pi

a(theta)=exp(2*pi*j*A*[cos(theta);sin(theta)]) where A ,is row vector and p,q,A are known

I need to find the in matlab.Please help me on this

Answer 1

[Alan Stevens]

a = @(theta) exp(2j*pi*A*[cos(theta);sin(theta)]);

dadtheta = @(theta) 2j*pi*A*[-sin(theta);cos(theta)]*a(theta);

Answer 2

[Stephan]

syms a(theta) p q A
eq1 = a(theta)==exp(2*pi*1i*A*[cos(theta);sin(theta)])    
eq2 = theta==atan(p/q)+pi
Deq1 = diff(eq1,theta)
sol = subs(rhs(Deq1),theta,rhs(eq2))
pretty(sol)

[Murali Krishna AG]

Hi Stephan,I wouId like to thank for ur answer ,

In my code

syms a(theta) p q A s(a)

eq1 = a(theta)==atan(p/q)+pi;

eq2 = s(a)==exp(2*pi*1i*A*[cos(a(theta));sin(a(theta))]);

Deq1 = diff(eq2,a(theta));

sol = subs(rhs(Deq1),a(theta),rhs(eq1));

Here a is function theta and s is function of a

Is this Correct?

And I have two doubts,

1. In my case A is 1*2 matrix,but here it is taking as 1*1 matrix finally am getting size (Deq1) is 2*1 since [cos(a(theta));sin(a(theta))] is 2*1 size..How to make size(Deq1) is 1*1?
2. After getting this sol,i have to multiply this with 0.68 but the result am getting here is zero .how to make it finite ??

[Stephan]

I dont think, that is really a function of theta - if we derive this w.r.t. theta the answer is zero - theta is just determined by and i think this is not a functoin of theta.

Regarding the size of the result, in case of A is of size 1x2 the result is 1x1:

syms s(theta) theta p q
A = sym('A', [1 2])
eq1 =s(theta)==exp(2*pi*1i*A*[cos(theta);sin(theta)])
eq2 = theta==atan(p/q)+pi;    
Deq1 = diff(eq1,theta)
sol = subs(rhs(Deq1),theta,rhs(eq2))

This results in:

A =
 
[ A1, A2]
 
 
eq1 =
 
s(theta) == exp(pi*A1*cos(theta)*2i + pi*A2*sin(theta)*2i)
 
 
Deq1 =
 
diff(s(theta), theta) == exp(pi*A1*cos(theta)*2i + pi*A2*sin(theta)*2i)*(pi*A2*cos(theta)*2i - pi*A1*sin(theta)*2i)
 
 
sol =
 
-exp(- (A1*pi*2i)/(p^2/q^2 + 1)^(1/2) - (A2*p*pi*2i)/(q*(p^2/q^2 + 1)^(1/2)))*((A2*pi*2i)/(p^2/q^2 + 1)^(1/2) - (A1*p*pi*2i)/(q*(p^2/q^2 + 1)^(1/2)))
 

For the multiplication:

>> sol = 0.68*sol
 
sol =
 
-(17*exp(- (A1*pi*2i)/(p^2/q^2 + 1)^(1/2) - (A2*p*pi*2i)/(q*(p^2/q^2 + 1)^(1/2)))*((A2*pi*2i)/(p^2/q^2 + 1)^(1/2) - (A1*p*pi*2i)/(q*(p^2/q^2 + 1)^(1/2))))/25

You could now use subs on this result and substitue A1 with the value of A(1), A2 with the value of A(2)... to get an symbolic result.

[Murali Krishna AG]

After getting sol from above

A=Amat;p=porg;q=qorg;

sol = -exp(- (A1*pi*2i)/(p^2/q^2 + 1)^(1/2) - (A2*p*pi*2i)/(q*(p^2/q^2 + 1)^(1/2)))*((A2*pi*2i)/(p^2/q^2 + 1)^(1/2) - (A1*p*pi*2i)/(q*(p^2/q^2 + 1)^(1/2)));

sola = subs(sol,A,Amat);

solb = subs(sola,p,porg);

solc = subs(solb,q,qorg);

Is this correct?

[Stephan]

syms A1 A2 p q
sol = -exp(- (A1*pi*2i)/(p^2/q^2 + 1)^(1/2) - (A2*p*pi*2i)/...
    (q*(p^2/q^2 + 1)^(1/2)))*((A2*pi*2i)/(p^2/q^2 + 1)^(1/2) -...
    (A1*p*pi*2i)/(q*(p^2/q^2 + 1)^(1/2)));

Amat = [0.5 2.5];
porg = 3;
qorg = -1;

sol_new = subs(sol, [A, p, q], [Amat, porg, qorg])
num_result = vpa(sol_new)

gives:

sol_new =
 
-(10^(1/2)*pi*exp((pi*10^(1/2)*7i)/5)*4i)/5
 
 
num_result =
 
7.7406505609803157366313776809726 - 1.8021645484713704266844120210115i

[Murali Krishna AG]

Thank u so much,It helped me a lot

[Stephan]

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