Issue while plotting the equation

2021 12 12
3 个回答
更新时间：2021 12 12
7 次查看（30 天）

0 个投票

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??

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回答（3 个）

KSSV
KSSV 2021-12-12
Ran in:

0 个投票

Ran in:
z = 3+1i*2
z = 3.0000 + 2.0000i
real(z)
ans = 3
imag(z)
ans = 2

2 个评论
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Mohd Ayaz
Mohd Ayaz 2021-12-12
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
KSSV 2021-12-12
If A is your vector of both real and complex numbers, to extract only real numbers use:
A_real = A(imag(A) == 0)

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Mohd Ayaz
Mohd Ayaz 2021-12-12

0 个投票

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?

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Walter Roberson
Walter Roberson 2021-12-12
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
Mohd Ayaz 2021-12-12
But how to get variable A which has all roots ? I mean by my method I am only able to get 1 root

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Walter Roberson
Walter Roberson 2021-12-12
Ran in:

0 个投票

Ran in:
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])

1 个评论
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Walter Roberson
Walter Roberson 2021-12-12
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.

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