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pik1432

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6 years, 122 days
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Thank you for the clear answer....

pik1432 245
December 06 2020
0 0

@acer Thank you!

Thank you for the clear answer. Yes, th...

pik1432 245
December 06 2020
0 0

Thank you Acer and Kitonum for your answers. Yes, it was my mistake to add an unnecessary 's' into the denominator of the expression. Here is the corrected question + Kitonum's answer + Acer's answer. 


 

> 

restart;
> 

eq9_13_m3 := 2*R__R*s*omega__s*L__sigma_S/(s^2*L__sigma_S^2*omega__s^2 + R__R^2);

2*R__R*s*omega__s*L__sigma_S/(s^2*L__sigma_S^2*omega__s^2+R__R^2)

 (1)
> 

aux2 := s_hat = R__R/(L__sigma_S*omega__s);

s_hat = R__R/(L__sigma_S*omega__s)

 (2)
> 

desired := 2*(s/s_hat) / (1+(s/s_hat)^2);

2*s/(s_hat*(1+s^2/s_hat^2))

 (3)
> 

# 20201204 Kitonum
> 

restart;
> 

eq9_13_m3 := 2*R__R*s*omega__s*L__sigma_S/(s^2*L__sigma_S^2*omega__s^2 + R__R^2);

2*R__R*s*omega__s*L__sigma_S/(s^2*L__sigma_S^2*omega__s^2+R__R^2)

 (4)
> 

simplify(eq9_13_m3, {R__R/(L__sigma_S*omega__s)=s_hat});

2*s*s_hat/(s^2+s_hat^2)

 (5)
> 

# 20201204 Acer
> 

restart;
> 

eq9_13_m3 := 2*R__R*s*omega__s*L__sigma_S/(s^2*L__sigma_S^2*omega__s^2 + R__R^2);

2*R__R*s*omega__s*L__sigma_S/(s^2*L__sigma_S^2*omega__s^2+R__R^2)

 (6)
> 

aux2 := s_hat = R__R/(L__sigma_S*omega__s);

s_hat = R__R/(L__sigma_S*omega__s)

 (7)
> 

temp := simplify(eq9_13_m3, {(rhs=lhs)(aux2)});

2*s*s_hat/(s^2+s_hat^2)

 (8)
> 

expand(numer(temp)/s_hat)/(s_hat*expand(denom(temp)/s_hat^2));

2*s/(s_hat*(s^2/s_hat^2+1))

 (9)
> 

 

> 

 


Merry Christmas!

Download Q20201204_w_answers.mw

Thank you for your attention!...

pik1432 245
November 18 2020
0 0

@rlopez It is appreciated. 

Thank you for the clear answer....

pik1432 245
November 18 2020
0 0

@Kitonum Your answer gave me this alternative way:


 

> 

restart;
> 

expression := exp(-I/2*(-2*rho__m + Pi))*I;

I*exp(-((1/2)*I)*(-2*rho__m+Pi))

 (1)
> 

evalc(%);

cos(rho__m)+I*sin(rho__m)

 (2)
> 

convert(%, exp);

exp(I*rho__m)

 (3)
> 

 


 

Download Q20201118_kitonum.mw

Thank you for the clear answer....

pik1432 245
November 18 2020
0 0

@acer Thank you for this answer and your comments below.

Thank you for the clear answer+further c...

pik1432 245
November 09 2020
0 0

@Kitonum Thank you for the succinct answer and further clarification!

Thank you for the clear answer....

pik1432 245
November 09 2020
0 0

@nm : Thank you for your answer. It worked!

Thank you for the clear answer....

pik1432 245
October 28 2020
0 0

@Kitonum Your answer worked as well! Thank you!

Thank you for the clear answer....

pik1432 245
October 28 2020
0 0

@acer : Your answer worked! It is appreciated. 

Thank you for the clear answer....

pik1432 245
October 15 2020
0 0

@acer It worked! Thank you for your answer!

Thank you for the clear answer....

pik1432 245
October 07 2020
0 0

@acer Thank you!. Now the identity goes both ways. This was what I looking for, the identity relationship between one transformation (what EE people call as Park's transformation) and another transformation (Clarke's transformati
 

> 

restart;
> 

Clark_mag_invariant := 2/3*<1, -1/2, -1/2; 0, sqrt(3)/2, -sqrt(3)/2; 1/2, 1/2, 1/2>;

Matrix(3, 3, {(1, 1) = 2/3, (1, 2) = -1/3, (1, 3) = -1/3, (2, 1) = 0, (2, 2) = (1/3)*sqrt(3), (2, 3) = -(1/3)*sqrt(3), (3, 1) = 1/3, (3, 2) = 1/3, (3, 3) = 1/3})

 (1)
> 

Clark_mag_invariant_rotate := <cos(theta), sin(theta), 0; -sin(theta), cos(theta), 0; 0, 0, 1>;

Matrix(3, 3, {(1, 1) = cos(theta), (1, 2) = sin(theta), (1, 3) = 0, (2, 1) = -sin(theta), (2, 2) = cos(theta), (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1})

 (2)
> 

Ts_sksul:=2/3*<cos(theta), cos(theta-2/3*Pi), cos(theta+2/3*Pi);
            -sin(theta), -sin(theta-2/3*Pi), -sin(theta+2/3*Pi);
            1/2, 1/2, 1/2>;

Matrix(3, 3, {(1, 1) = (2/3)*cos(theta), (1, 2) = -(2/3)*cos(theta+(1/3)*Pi), (1, 3) = -(2/3)*sin(theta+(1/6)*Pi), (2, 1) = -(2/3)*sin(theta), (2, 2) = (2/3)*sin(theta+(1/3)*Pi), (2, 3) = -(2/3)*cos(theta+(1/6)*Pi), (3, 1) = 1/3, (3, 2) = 1/3, (3, 3) = 1/3})

 (3)
> 

Clark_mag_invariant_rotate2 := Clark_mag_invariant_rotate.Clark_mag_invariant;

Matrix(3, 3, {(1, 1) = (2/3)*cos(theta), (1, 2) = -(1/3)*cos(theta)+(1/3)*sin(theta)*sqrt(3), (1, 3) = -(1/3)*cos(theta)-(1/3)*sin(theta)*sqrt(3), (2, 1) = -(2/3)*sin(theta), (2, 2) = (1/3)*sin(theta)+(1/3)*cos(theta)*sqrt(3), (2, 3) = (1/3)*sin(theta)-(1/3)*cos(theta)*sqrt(3), (3, 1) = 1/3, (3, 2) = 1/3, (3, 3) = 1/3})

 (4)
> 

 

> 

expand~(Ts_sksul);

Matrix(3, 3, {(1, 1) = (2/3)*cos(theta), (1, 2) = -(1/3)*cos(theta)+(1/3)*sin(theta)*sqrt(3), (1, 3) = -(1/3)*cos(theta)-(1/3)*sin(theta)*sqrt(3), (2, 1) = -(2/3)*sin(theta), (2, 2) = (1/3)*sin(theta)+(1/3)*cos(theta)*sqrt(3), (2, 3) = (1/3)*sin(theta)-(1/3)*cos(theta)*sqrt(3), (3, 1) = 1/3, (3, 2) = 1/3, (3, 3) = 1/3})

 (5)
> 

convert~(Clark_mag_invariant_rotate2, phaseamp, theta);

Matrix(3, 3, {(1, 1) = (2/3)*cos(theta), (1, 2) = -(2/3)*cos(theta+(1/3)*Pi), (1, 3) = -(2/3)*sin(theta+(1/6)*Pi), (2, 1) = -(2/3)*sin(theta), (2, 2) = (2/3)*sin(theta+(1/3)*Pi), (2, 3) = -(2/3)*cos(theta+(1/6)*Pi), (3, 1) = 1/3, (3, 2) = 1/3, (3, 3) = 1/3})

 (6)
> 

 


 

Download DQ_Clark.mw

on)+rotation with a given angle. 

Thank you for the clear answer....

pik1432 245
October 05 2020
0 0

@Kitonum Your answer worked, thank you!

Thank you for the answer!...

pik1432 245
October 02 2020
0 0

@nm Thank you for introducing a different way!

Thank you for the clear answer....

pik1432 245
October 01 2020
0 0

@Kitonum Thank you for the answer! It worked. 

Thank you for the amazing answers!...

pik1432 245
September 02 2020
0 0

@acer Your step by step explanation is a great tour regarding how those individual commands work! Thank you for the amount of time and effort to unfold the explanation!

 

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