# I want to plot an implicit equation

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mohd ayaz on 21 Apr 2022
Answered: Sam Chak on 21 Apr 2022 I have attached the equations and its all variable I don't know how to plot.
mohd ayaz on 21 Apr 2022
@David Goodmanson yes Qm is negative

David Goodmanson on 21 Apr 2022
Edited: David Goodmanson on 21 Apr 2022
Hi mohd,
This is not really an implicit equation since Vgs appears only once, by itself, on the lhs. So you can take all the other stuff to the rhs and use Qm as the independent variable and Vgs as the dependent variable. You can experiment around for the range of Qm that gives the correct range for Vgs, assuming there is one. When you fine the right range, don't be reluctant to use lots of points in the Qm array. Ater you find the corresponding Vgs array you can plot(Vgs,Qm). One catch is if Qm+qNdtsi goes negative, in which case you will run into problems with the log term.

Sam Chak on 21 Apr 2022
I tried to plot this way to first visualize how the curve looks like for the positive part of .
Qm = linspace(-1, 0, 1001);
Vgs = (1 + 1)*(Qm + 1)/2 + 1*((Qm + 1)/2).^3 + 1*((Qm + 1)/2).^5 + 1 + 1 - 1*(Qm + 1) + 1*log(- Qm.*sqrt((Qm + 1)/(1)));
plot(Qm, Vgs, 'linewidth', 1.5)
grid on
xlabel('Q_{m}')
ylabel('V_{gs}') So, you want to invert the axes to plot(Vgs, Qm). You can check this documentation:
f = @(Vgs, Qm) Vgs - ((1 + 1)*(Qm + 1)/2 + 1*((Qm + 1)/2).^3 + 1*((Qm + 1)/2).^5 + 1 + 1 - 1*(Qm + 1) + 1*log(- Qm.*sqrt((Qm + 1)/(1))));
fimplicit(f, [-5 2 -1 0])
grid on
xlabel('V_{gs}')
ylabel('Q_{m}') Now, make the substitution of the original constants.

R2020a

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