written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Electronics Engineering > Sem 5 > Electromagnetic Engineering
Marks: 10 Marks
Year: May 2015, May 2016
written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Electronics Engineering > Sem 5 > Electromagnetic Engineering
Marks: 10 Marks
Year: May 2015, May 2016
written 7.8 years ago by |
FDM method is used for finite geometric electromagnetic computation.
In this complete element is divided into form of meshes.
From Poisson’s equation
$∇^2 v=0$
Since it’s a 2-D figure
$∇^2 v=\dfrac{d^2 v}{dx^2} ×\dfrac{d^2 v}{dy^2} \\ \dfrac{d^2 v}{dx^2}=\dfrac{v(i+1)- 2v(i)+ v (i-1)}{∆i^2}……..(1) \\ \dfrac{d^2 v}{dy^2}=\dfrac{v(j+1)- 2v (j)+ v(j-1)}{∆j^2}………..(2) \\ ∆i= ∆j=1 \\ \dfrac{d^(2 ) v}{dx^2} + \dfrac{d^2 v}{dy^2}=v(i+1,j)+ v(i,j+1)+ v(i-1 ,j)+ v(i,j-1)- ∆v(i,j)$
From the above diagram,
$4v_0=v_1+ v_2+ v_3+ v_4 \\ v_0=\dfrac{1}{4} [v_1+ v_2+ v_3+ v_4 ]$
For example –
Find out voltage on following object
For voltage $v_1$
$v_1=\dfrac{100+100+0+v_2}{4} \\ 4v_1- v_2=200…………(a)$
For voltage $v_2$,
$v_2={100+v_1+ v_3+ 100}{4} \\ -v_1+ 4v_2- v_3=200……(b)$
For voltage $v_3$,
$v_3=\dfrac{v_2+ 0+0+100}{4} \\ -v_2+ 4v_3=100…………(c)$
Solving equation (a)(b) & (c).
$v_1=69.64v \\ v_2=78.57v \\ v_3=44.64v$
Advantage of FDM:
Disadvantage of FDM: