andy.zhou.nuaa

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12 years, 25 days

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These are answers submitted by andy.zhou.nuaa

You answer is very useful to me. Although there is a little dismatch on the condition of vds>300, the precision is enough for my appliction. 

tks again for ur help.

id=gfs*(vgs-Vth), other condition keep the same!

tks very much,Robert Israel !

best regards!

andy

   forgive my poor knowledge, Some conclusion you made such as "Provided w1/w2 is not a rational number, Vcs_max = V1 + V2" i don't undertstand very well, i need further help if you would like to help me. could u post a workseet showing how to find the max value of voltage on Cs.

     Note: is there any books or articles focus on high-order resonant circuit , could recommend one for me?

    because the presence of  Cp, i think status which Cs stores all the energy is impossilbe occurs in this circuit. 

  i know it can be solved by substitue constants with values. i want to know does the value has generality, i worry some problems come up:

 Does the following problem will be exsit:

 when Lp=1mH, the function is  the monotone function.

 when Lp=10mH, the function is the monotonous fuction.

that is what i worry!

  with ur help. the problem has been solved!

 

  assmue A,B, theta is real and there is a function Tran() in maple

 i write these command in maple

 Tran(A*sin(t)+B*cos(t));

 then the result will be  sqrt(A^2+B^2)*sin(t+arctan(B/A));

My problem is how to build this function ?

   let me describe the question again!

 how to translate A*sin(t) + B*cos(t) to C*sin(t+theta) by maple?  where theta = arctan(B/A)
C = sqrt(A^2+B^2)
 

 For acer:

    my expression is  A*sin(t)+B*cos(t) not A*cos(t)+B*sin(t) , so when A=1,B=1, t=Pi, the equation is also  establish. 
 

Thanks for jakubi's help, but your answer is not i want. maybe something wrong with description of my problem, so rewrite it:

  assuming:

     v := A*sin(t)+B*cos(t); 

  we both know v can be simplified to following expression:

  v := C*sin(t+theta)   where  C := sqrt(A^2+B^2) , theta := arctan(B/A).

So i want to know how this could be done by maple?  Can the results be simplified in ODE equations directly?

 Hi:

       you mean this equations doesn't have an solution because there
is an identity not involving those variables. Can u write in detail, are there any math theory prove this conclusion.

   Tanks very much !

Equations:

eqns := {La+Lp/nlpa^2+Lm/(nsp^2*npa^2), Lm = LEF1, Lp+Lm/nsp^2 = LAB1, Lp+Rs^2*Lm/((omega^2*Lm^2+Rs^2)*nsp^2) = LAB2, La+Lp/nlpa^2+Rs^2*Lm/((omega^2*Lm^2+Rs^2)*nsp^2*npa^2) = LCD2, (nsp^2*Rp^2+omega^2*nsp^2*Lp^2+omega^2*Lp*Lm)*nsp^2*Lm/(nsp^4*Rp^2+2*omega^2*Lp*Lm*nsp^2+omega^2*Lm^2+omega^2*nsp^4*Lp^2) = LEF2};

vars:

 vars := {Lm, Lp, nsp, La, nlpa, npa};

solve(eqns, vars);  #  "nothing display in maple"

 

 This is the right Equation:

 eqns := a*(x/sqrt(x^2-1)+2*x/(Pi*(x^2-1))-1-2/(Pi*x)) = 1

 it have the same problem with the first equation!

 i am very sorry for waste someone's time!!

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