Numerical Integration. CSE245 Lecture Notes. Content. Introduction Linear Multistep Formulae Local Error and The Order of Integration Time Domain Solution of Linear Networks. Transient analysis is to obtain the transient response of the circuits.
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CSE245 Lecture Notes
Transient analysis is to obtain the transient response of the circuits.
Equations for transient analysis are usually differential equations.
Numerical integration: calculate the approximate solutions Xn.
Linear multistep formulae are the primary numerical integration method.
Introductionk the circuits.
k
iXni + h iXni = 0
i=0
i=0
Linear Multistep FormulaeX = F(X)
 0 = 0
 Xn is the only unknown variable
 0 0
 Xn, Xn are all unknown variables.
X(t) the circuits.
Xn
X(tn)
Xn1
tn
tn1
t
Linear Multistep FormulaeXn– Xn1– h Xn1 = 0
where 0 = 1, 1 = 1, 0 = 0, 1 = 1
Xn– Xn1– h Xn = 0
where 0 = 1, 1 = 1, 0 = 1, 1 = 0
Xn– Xn1– h (Xn + Xn1 )/2= 0
where 0 = 1, 1 = 1, 0 = 1/2, 1 = 1/2
k the circuits.
k
iX(tni) + h iX(tni)
i=1
i=0
Local Error and Order of IntegrationEk = X(tn) +
k the circuits.
k
i((tntni)/h)l + h (l/h)i((tntni)/h)l1
i=0
i=0
k
i = 0
i=0
k
k
(ii  i) = 0
[(ii  pi)ip1] = 0
i=0
i=0
Order of Integration0 = 1, 1 = 1, 0 = 0, 1 = 1
Sol = 00 + 1 = 1 + (1) = 0;
l = 100 + 11  0  1 = 10 + (1)1  0 – (1) = 0;
l = 2(11  21)1 = ((1)1  2(1))1 = 1 0;
The forward Euler is 1th order.
0 = 1, 1 = 1, 0 = 1, 1 = 0
Sol = 00 + 1 = 1 + (1) = 0;
l = 100 + 11  0  1 = 10 + (1)1  (1)  0 = 0;
l = 2(11  21)1 = ((1)1  20)1 = 1 0;
The backward Euler is 1th order.
0 = 1, 1 = 1, 0 = 1/2, 1 = 1/2
Sol = 00 + 1 = 1 + (1) = 0;
l = 100 + 11  0  1 = 10 + (1)1  (1/2) – (1/2) = 0;
l = 2(11  21)1 = ((1)1  2(1/2))1 = 0;
l = 3(11  31)12 = ((1)1  3(1/2))1 = 1/2 0;
The trapezoidal method is 2th order
 Choose p, the order of the numerical integration method needed;
 Choose k, the number of previous values needed;
 Write down the (p+1) equations of pth order accuracy;
 Choose other (2kp) constrains of the coefficients and ;
 Combine and solve above (2k+1) equations;
 Get the result coefficients and .
k the circuits.
k
iXni + h iXni = 0
i=0
i=0
Solution of Linear NetworksMX = GX + Pu
(1)
(2)
k the circuits.
k
iXni + h iXni = 0
i=1
i=1
Solution of Linear Networks(1) Xn + h0Xn +
Xn + h0Xn + b = 0
Xn = (1/h0)( Xn + b)
(2)+(3)M[(1/h0)( Xn + b)] = GXn + Pu
(1/h0) Xn = GXn + Pu + (M/h0)b
(3)
i the circuits.c
(C/h0)
– (C/h0) bc
vc
ic
vc
Solution of Linear NetworksC vc = ic
C [(1/h0)( vc + bc)] = ic
(C/h0) vc– (C/h0) bc = ic
i the circuits.l
– (L/h0) bl
+

vl
(L/h0)
il
vl
Solution of Linear NetworksL il = vl
L [(1/h0)( il + bl)] = vl
(L/h0) il– (L/h0) bl = vl
“Interconnect Analysis and Synthesis”, Wiley and Sons, 2000
“Computer Methods for Circuit Analysis and Design”, 1983