Picking up the correct index values

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Ganesh Kini
Ganesh Kini el 27 de Mayo de 2020
Comentada: Ganesh Kini el 28 de Mayo de 2020
I have 4 arrays, two combinations of Nwell and Pwell voltages of Transistor 1 and Transistor 2
For transistor 1
vnw_1 =
0.00000 0.90000
vpw_1 =
-1.8000 -1.2000
So we define as (Vn, Vp) as (0.00000, -1.8000) and another (0.90000, -1.2000) these are the combination respectively.
for transistor 2
vnw_2 =
0.90000 0.00000
vpw_2 =
-1.20000 0.20000
Similarly we have (Vn, Vp) as (0.90000, -1.2000) and another (0.00000, 0.20000)
we can see that (0.90000, -1.2000) is common in T1 and T2.
I want to get the common voltage as the output
Answer (0.90000, -1.2000)
How do i do that? There is nothing linked to the concepts of transistors, just i need that combination as the output. How do i do that? Please help me

Respuesta aceptada

Rik
Rik el 27 de Mayo de 2020
You should probably rethink the way you store these values, because this is much more difficult to index than it needs to be.
vnw_1 = [ 0.00000 0.90000];
vpw_1 = [-1.80000 -1.20000];
vnw_2 = [ 0.90000 0.00000];
vpw_2 = [-1.20000 0.20000];
Answer=[];
for T1_ind=1:2
T1=[vnw_1(T1_ind) vpw_1(T1_ind)];
for T2_ind=1:2
T2=[vnw_2(T2_ind) vpw_2(T2_ind)];
if isequal(T1,T2)
Answer=T1;
end
end
end
  9 comentarios
Rik
Rik el 28 de Mayo de 2020
First convert your four vectors to two arrays so you can compare them. Make sure the things you want to compare are on the rows of your array. Then you can use ismembertol. You could also try with ismember first, but if your decimal values can't be expressed as a binary fraction you may get rounding errors. (just like you can't store 1/3 in a decimal value, Matlab can't store every value without rounding)
VNW1=[-0.20000 -0.20000];
vpw1=[-0.90000 -1.80000];
VNW2=[1.20000 0.30000 0.90000 -0.20000 1.20000];
VPW2=[0.00000 -1.20000 -0.90000 -1.80000 -0.30000];
T1=[VNW1(:) vpw1(:)];
T2=[VNW2(:) VPW2(:)];
L=ismembertol(T1,T2,'rows');
matches=T1(L,:);
Ganesh Kini
Ganesh Kini el 28 de Mayo de 2020
Thanks for the help it works fine

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