What is wrong?Help with converting to matlab

> restart; with(PDEtools); > i(t) := -a*v(t)+b*v(t)^3 ; i[D] := i[S]+i(t);
i:= t -> -a v(t) + b v(t)^3
i[D] := i[S] - a v(t) + b v(t)
> eq[1] := i[C]+i[D]+i[R]-i[L](t) = 0;
eq1:=i[C] + i[S] - a v(t) + b v(t) + i[R] - i[L](t) = 0
> V[D] := V[S]+v(t); V[C] := V[D]; V[R] := V[D]; V[D]:=V[S] + v(t) V[C}:=V[S] + v(t) V[R]:=V[S] + v(t) >i[C] := C*(diff(V[C], t)); i[R] := V[R]/R;
/ d \
i[C]:= |--- v(t)|
\ dt /
i[R]:= V[S] + v(t)
-----------
R
>eq[1];
/ d \ 3 V[S] + v(t)
C |--- v(t)| + i[S] - a v(t) + b v(t) + ----------- - i[L](t) = 0
\ dt / R
>eq[2] := diff(eq[1], t);
/ d / d \\ / d \ 2 / d \
eq2:=C |--- |--- v(t)|| - a |--- v(t)| + 3 b v(t) |--- v(t)|
\ dt \ dt // \ dt / \ dt /
d
--- v(t)
dt / d \
+ -------- - |--- i[L](t)| = 0
R \ dt /
>V[L] := V[S]-V[D];V[b]=V[D] V[L]:= -v(t)
>de := subs(diff(i[L](t), t) = V[L]/L, eq[2]); / d / d \\ / d \ 2 / d \ C --- |--- v(t)| - a --- v(t) + 3 b v(t) --- v(t) \ dt \ dt // \ dt / \ dt /
d
--- v(t)
dt v(t)
+ -------- + ---- = 0
R L
>expand(de/C); / d \ 2 / d \ d a --- v(t) 3 b v(t) --- v(t) --- v(t) / d^2 \ \ dt / \ dt / dt --- v(t) - ------------ + -------------------- + -------- \ dr^2 / C C C R
v(t)
+ ---- = 0
C L
>collect(%, diff(v(t), t));
/ 2 \ 1 a 3 b v(t) | / d \ / v(t) / d^2 \\ |-- - - + --------- |* -- v(t) + | --- + | --- v(t)|| *v(t) = 0 \CR C C / \ dt / \ CL \ dt^2 //
>de[2] := subs(v(t)/(C*L) = w^2*v(t), %); / 2 \ | a 3 b v(t) 1 | / d \ / d ^2 \ de2:= - - + --------- + ---* --- v(t) + w^2 v(t)+ ------- v(t) = 0 \ C C C R/ \ dt / \ dt^2 /
2
>tr := {t = tau/w, v(t) = x(tau)*sqrt(a-1/R)/sqrt(3*b)};
\
/
| / \ 1
| x(tau sqrt (|a - -|sqrt(3))
tr:= tau 1 \ / R
|t = -----, v(t) = - ----------------------
| omega 3 sqrt (3*b)
\ /
>sqrt(3*b)*dchange(tr, de[2], [x(tau), tau], simplify)/(omega^2*sqrt(a-1/R));
expand(%); / d \ x(tau) ----- x(tau) a \ dtau / / d / d \\ ------------------------ + x(tau) + ----- |----- x(tau)| omega C R \ dtau \ dtau //
2 / d \ / d \ d
x(tau) |----- x(tau)| |----- x(tau)| a ----- x(tau)
\ dtau / \ dtau / dtau
- ---------------------- - ---------------- + ------------ = 0
omega C R omega C omega C R
>collect(%, diff(x(tau), tau), factor);
/ d \
(x(tau) - 1) (x(tau) + 1) (R a - 1) |----- x(tau)|
\ dtau /
-------------------------------------------------- + x(tau)
omega C R
/ d / d \\
+ |----- |----- x(tau)|| = 0
\ dtau \ dtau //
>de[3] := subs(x(tau)-1 = -(1-x(tau)^2)/(x(tau)+1), %); / 2\ / d \ \1 - x(tau) / (R a - 1) ----- x(tau) \ dtau / - -------------------------------------- + x(tau) omega C R
/ d / d \\
+ |----- |----- x(tau)|| = 0
\ dtau \ dtau //
>subs(R*a-1 = epsilon*omega*C*R, %); / 2\ / d \ -\1 - x(tau) / epsilon ----- x(tau) + x(tau) \ dtau /
/ d / d \\
+ |----- |----- x(tau)|| = 0
\ dtau \ dtau //
provided that epsilon is positive. Since a,w(omega),and C are positive, this condition requires that the resistance R>1/a.

1 Kommentar

Rik
Rik am 1 Mai 2017
Have a read here and here. It will greatly improve your chances of getting an answer.

Melden Sie sich an, um zu kommentieren.

Antworten (0)

Kategorien

Mehr zu Functions finden Sie in Hilfe-Center und File Exchange

Tags

Gefragt:

am 1 Mai 2017

Kommentiert:

Rik
am 1 Mai 2017

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by