Issue while plotting the equation

Vfb=1.2;
phit=0.026;
Nd=1e19;
ni=1.5e10;
q=1.6e-19;
Tsc=10^-6;
tox=0.8*10^-7
W=10^-4
L=10^-4
eox=25*8.8*1e-14
esi=8.8*11.7*1e-14
Cox=eox/tox
Ld=sqrt((esi*phit)/(q*Nd))
gamma=(sqrt(2*q*Nd*esi))/(Cox)
mu=100
mub=200
d=4*Cox^2*gamma^2*phit
p=q*Nd*Tsc
a=0.001
Vds=0.01
tf=12*10^-7;
Qs=2.7*10^-5
Qc=2*10^-5
Ec=10^6
f=2*esi*3.14*0.026*q*q*Nd*Nd*Tsc;

alpha_0=(((3*(Qs^4)*(Qc^2))-(5*(Qc^4)*(Qs^2)))/((4*(Qs^2)*(Qc^5))-(2*(Qs^4)*(Qc^3))-(2*Qc^7)))*Ec
beeta_0=(((5*Qc^4)-Qs^4)/((4*(Qs^2)*(Qc^5))-(2*(Qs^4)*(Qc^3))-(2*Qc^7)))*Ec
damma_0=(((Qs^2)-3*Qc^2)/((4*(Qs^2)*(Qc^5))-(2*(Qs^4)*(Qc^3))-(2*Qc^7)))*Ec


alpha=2*tf*alpha_0
beeta=4*tf*beeta_0
damma=6*tf*damma_0

N=500;
Vgd=linspace(0,1.2,N);
Vgs=linspace(0,1.2,N);
Qd=zeros(1,N);
Qs=zeros(1,N);
Ids=zeros(1,N);

for i=1:500
    syms x
    eqnLeft = Vgd(i)-Vfb-Vds+((Tsc/(8*esi))*(x+p));
    eqnRight =0.026*(log(2*sqrt(((x+p)*Tsc)/(8*3.14*0.026*esi))*(1-((x+p)/(p)))))+((alpha+(1/Cox))*((x+p)/2))+(beeta*(((x+p)/2).^3))+(damma*(((x+p)/2).^5));
    Qd(1,i)=vpasolve(eqnLeft == eqnRight,x,10^-11);
end


for i=1:500
    syms x
    eqnLeft = Vgs(i)-Vfb+((Tsc/(8*esi))*(x+p));
    eqnRight =0.026*(log(2*sqrt(((x+p)*Tsc)/(8*3.14*0.026*esi))*(1-((x+p)/(p)))))+((alpha+(1/Cox))*((x+p)/2))+(beeta*(((x+p)/2).^3))+(damma*(((x+p)/2).^5));
    Qs(1,i)=vpasolve(eqnLeft == eqnRight,x,10^-11);
end


hold on
plot(Vgd,abs(Qd))
plot(Vgs,abs(Qs))
hold off

While plotting Qd and Qs I am getting its complex values in workspace, how can i just plot the real values??

Answer 1

[KSSV]

z = 3+1i*2 

z = 3.0000 + 2.0000i

real(z)

ans = 3

imag(z)

ans = 2

[Mohd Ayaz]

Not like that i mean there will be 5 roots since it is 5th order eqtn then how can i just take real root amongst 5 of the root

[KSSV]

If A is your vector of both real and complex numbers, to extract only real numbers use:

A_real = A(imag(A) == 0)

Answer 2

[Mohd Ayaz]

Actually if you look at the code then x should have 5 roots but it's just storing 1 complex root in Qd or Qs but i want to store real root amongst all 5 roots is there any method with which i can do this?

[Walter Roberson]

KSSV showed you how: A_real = A(imag(A) == 0) where A is the name of the variable that has all of the roots.

[Mohd Ayaz]

But how to get variable A which has all roots ? I mean by my method I am only able to get 1 root

Answer 3

[Walter Roberson]

Vfb=1.2;
phit=0.026;
Nd=1e19;
ni=1.5e10;
q=1.6e-19;
Tsc=10^-6;
tox=0.8*10^-7;
W=10^-4;
L=10^-4;
eox=25*8.8*1e-14;
esi=8.8*11.7*1e-14;
Cox=eox/tox;
Ld=sqrt((esi*phit)/(q*Nd));
gamma=(sqrt(2*q*Nd*esi))/(Cox);
mu=100;
mub=200;
d=4*Cox^2*gamma^2*phit;
p=q*Nd*Tsc;
a=0.001;
Vds=0.01;
tf=12*10^-7;
Qs=2.7*10^-5;
Qc=2*10^-5;
Ec=10^6;
f=2*esi*3.14*0.026*q*q*Nd*Nd*Tsc;

alpha_0=(((3*(Qs^4)*(Qc^2))-(5*(Qc^4)*(Qs^2)))/((4*(Qs^2)*(Qc^5))-(2*(Qs^4)*(Qc^3))-(2*Qc^7)))*Ec;
beeta_0=(((5*Qc^4)-Qs^4)/((4*(Qs^2)*(Qc^5))-(2*(Qs^4)*(Qc^3))-(2*Qc^7)))*Ec;
damma_0=(((Qs^2)-3*Qc^2)/((4*(Qs^2)*(Qc^5))-(2*(Qs^4)*(Qc^3))-(2*Qc^7)))*Ec;


alpha=2*tf*alpha_0;
beeta=4*tf*beeta_0;
damma=6*tf*damma_0;

N=5;
Vgd=linspace(0,1.2,N);
Vgs=linspace(0,1.2,N);
%Qd=zeros(1,N);
%Qs=zeros(1,N);
Ids=zeros(1,N);

for i=1:N
    syms x
    eqnLeft = Vgd(i)-Vfb-Vds+((Tsc/(8*esi))*(x+p));
    eqnRight =0.026*(log(2*sqrt(((x+p)*Tsc)/(8*3.14*0.026*esi))*(1-((x+p)/(p)))))+((alpha+(1/Cox))*((x+p)/2))+(beeta*(((x+p)/2).^3))+(damma*(((x+p)/2).^5));
    Qd{i} = solve(eqnLeft == eqnRight, x, 'real', true);
end

Warning: Unable to find explicit solution. For options, see help.

Warning: Unable to find explicit solution. For options, see help.

Warning: Unable to find explicit solution. For options, see help.

Warning: Unable to find explicit solution. For options, see help.

Warning: Unable to find explicit solution. For options, see help.

for i=1:N
    syms x
    eqnLeft = Vgs(i)-Vfb+((Tsc/(8*esi))*(x+p));
    eqnRight =0.026*(log(2*sqrt(((x+p)*Tsc)/(8*3.14*0.026*esi))*(1-((x+p)/(p)))))+((alpha+(1/Cox))*((x+p)/2))+(beeta*(((x+p)/2).^3))+(damma*(((x+p)/2).^5));
    Qss{i} = solve(eqnLeft == eqnRight, x, 'real', true);
end

Warning: Unable to find explicit solution. For options, see help.

Warning: Unable to find explicit solution. For options, see help.

Warning: Unable to find explicit solution. For options, see help.

Warning: Unable to find explicit solution. For options, see help.

Warning: Unable to find explicit solution. For options, see help.

%{
hold on
plot(Vgd,abs(Qd))
plot(Vgs,abs(Qs))
hold off
%}

Qd{1}

 
ans =
 
Empty sym: 0-by-1
 

Qss{1}

 
ans =
 
Empty sym: 0-by-1
 

residue = eqnLeft - eqnRight

residue = 

string(residue)

ans = "(38759912637870261*x)/274877906944 - (13*log(-(9444732965739290427392*x*((38685626227668133590597632*x)/26014240345388328125 + 61897001964269002752/26014240345388328125)^(1/2))/7555786372591431))/500 + (5954689997320791*(x/2 + 7555786372591431/9444732965739290427392)^3)/8 - 1958130468122061235552256*(x/2 + 7555786372591431/9444732965739290427392)^5 + 292861619712054503059211338333491/1298074214633706907132624082305024"

%fplot([real(residue), imag(residue)], [-1 1])

[Walter Roberson]

There are NO real-roots of that residue expression, which is the 5th Qss output.

Look at the square root: that has to be positive or else te sqrt() will be imaginary. But you need a negative x so that the -x*sqrt() gives a positive number. So you need x to be between solve() of the expression under the square root, and 0. But if you graph residue over that range you can see that it never crosses zero there.

You are incorrect that there are 5 roots. There is an (x+c)^5 term, but there is also that log(sqrt()) term.

So, your problem cannot be solved.