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Unformatted text preview: PROBLEMS 751 SEBTIDN 14.1 FIELD PLDTTING 14.1 Use the program developed in Example 14.1 or your own equivalent code to plot the elec
tric ﬁeld lines and equipotential lines for the following cases: (a) Three point charges —1 C, 2 C, and 1 C placed at (—l, 0), (0,2), and (1, 0),
respectively. (b) Five identical point charges of 1 C located at (— 1, 1), (1, 1), (1, —1), (1, 1), and
(0, 0), respectively. SECTION 14.2 THE FINITE DIFFERENCE METHOD 14.2 Consider the onedimensional differential equation dzy
3+4y=0 subject to y(0) = 0, y(1) = 10. Use the ﬁnite difference (iterative) method to ﬁnd
y(0.25). You may take A = 0.25 and perform ﬁve iterations. , dV dZV .
14.3 (3) Obtain :1— and d—i— at x = 0.15 from the followmg table.
x x
x 0.1 0.15 0.2 0.25 0.3 V 1.0017 1.5056 2.0134 2.5261 3.0452 M (b) The data in the table are obtained from V = 10 sinh x. Compare your result in
part (a) with the exact values. 14.4 Show that the ﬁnite difference equation for Laplace’s equation in cylindrical coordinates,
V = V(p, z), is 1. h
V(po,zo) = Z [V000, z. + h) + V(10.,,z0  h) + (1 + 2p ) h
V(po + h,Zo) + — 2 Po ) V(po —' ha zo)] whereh = A2 = Ap. 14.5 Using the ﬁnite difference representation in cylindrical coordinates (p, (b) at a grid point P shown in Figure 14.42, let p = m Ap and d) = n AqS so that V(p, (Mp =
V(mAp, nA¢) = V2. Show that 2 1K 1>' ( ‘)
=—~—— ——— _— + ~—
VV'mm Apz 1 2m V’ml 1+ m an+1 1
+ 0» AW (V; 752 CHAPTER 14 NUMERICAL METHODS Figure 14.42 Finite difference grid in cylindri
cal coordinates; for Problem 14.5. 14.6 The four sides of a square trough are maintained at potentials 10 V, ~40 V, 50 V, and
80 V. Determine the potential at the center of the trough. 14.7 Use the ﬁnite difference method to calculate the potentials at nodes 1 and 2 in the poten
tial system shown in Figure 14.43. 1
14.8 Rework Problem 14.7 if pv = —%9 nC/m3, h = 0.1 m, and e = 80, where h is the mesh size. 14.9 Use the ﬁnite difference technique to ﬁnd the potentials at nodes 1 to 4 in the potential
system shown in Figure 14.44. Five iterations are sufﬁcient. 14.10 For the potential system shown in Figure 14.45 (divided into a square grid), write down
the ﬁnite differential equations of nodes 1 to 3. Solve for potential for nodes 1, 2, and 3. 14.11 For the rectangular region shown in Figure 14.46, the electric potential is zero on the
boundaries and the charge distribution pv is 50 nC/m3. Although there are six free nodes,
there are only four unknown potentials (V1 — V4) because of symmetry. Solve for the un
known potentials. Figure 14.43 For Problem 14.7. PROBLEMS 753 Figure 14.44 For Problem 14.9. 100V Figure 14.45 For Problem 14.10. Figure 14.46 For Problem 14.11. 754 CHAPTER 14 NUMERICAL METHDDS 100 V Figure 14.47 For Problem 14.12. 14.12 Consider the potential system shown in Figure 14.47. (a) Set the initial values at the free
nodes equal to zero and calculate the potential at the free nodes for ﬁve iterations.
(b) Solve the problem by the band matrix method and compare result with part (a). 14.13 Apply the band matrix technique to set up a system of simultaneous difference equations
for each of the problems in Figure 14.48. Obtain matrices [A] and [B]. 14.14 (a) How would you modify matrices [A] and [B] of Example 14.3 if the solution region
had charge density pv? (b) Write a program to solve for the potentials at the grid points shown in Figure 14.49
assuming a charge density pv = x(y  l) nC/m3. Use the iterative ﬁnite difference
method and take a, = 1.0.
14.15 The two—dimensional wave equation is given by
i 6245 _ 3% + 8245
CZ at2 6x2 6Z2 15V 100V (b) Figure 14.48 For Problem 14.13. PROBLEMS 755 Figure 14.49 For Problem 14.14. By letting 455,1,” denote the ﬁnite difference approximation of 45(xm, z", tj), show that the
ﬁnite difference scheme for the wave equation is . l . ._l . . .
(pin—tn = 2 4,711”: — (ping: + 0‘ (q)rjrz+1,n + ¢{71—1,}: ' 2 (prjnm) +
a ((pljn,n+l + ¢{u,n—l — 2 d’rjnm) whereh = Ax = A2 anda = (cAt/h)2. 14.16 Write a program that uses the ﬁnite difference scheme to solve the onedimensional wave
equation 62V ~ 62V 6x2 7972—, Ost1, t>0 given boundary conditions V(0, t) = 0, V(1, t) = 0, t> 0 and the initial condition
BV/at (x, 0) = 0, V(x, 0) = sin me, 0 < x < 1. Take Ax = At = 0.1. Compare your
solution with the exact solution V(x, t) = sin 7rx cos M for 0 < t < 4. 14.17 (a) Show that the ﬁnite difference representation of Laplace’s equation using the nine
node molecule of Figure 14.50 is V0: (b) Using this scheme, rework Example 14.4. Figure 14.50 For Problem 14.17. 756 CHAPTER 14 NUMERICAL METHODS Figure 14.51 For Problem 14.18. EEBTIDN 14.4 THE MDMENT METHDD 14.18 A transmission line consists of two identical wires of radius a, separated by distance d as
shown in Figure 14.51. Maintain one wire at 1 V and the other at 1 V and use the
method of moments to ﬁnd the capacitance per unit length. Compare your result with
exact formula for C in Table 11.1. Take a = 5 mm, d = 5 cm, 6 = 5 m, and e = 80. 14.19 Determine the potential and electric ﬁeld at point ( 1, 4, 5) due to the thin conducting
wire of Figure 14.18. Take Vo = l V, L = 1 m, a = 1 mm. 14.20 Two conducting wires of equal length L and radius a are separated by a small gap and in
clined at an angle 0 as shown in Figure 14.52. Find the capacitance between the wires by
using the method of moments for cases 0 = 10°, 20°, . . . , 180°. Take the gap as 2 mm,
a =1mm,L= 2m,s,= 1. 14.21 Given an inﬁnitely long thin strip transmission line shown in Figure 14.53(a), use the moment method to determine the characteristic impedance of the line. We divide each
strip into N subareas as in Figure 14.53(b) so that on subarea i, 2N
Vi = 2 AU Pi
j=l where
_ M In R,, i 9e j
_ 2’1r80 1
AU “MunAe 15] i=‘
27reo ' ’ J
Figure 14.52 For Problem 14.20.
L
~ 6
W Gap PROBLEMS 757 (b) Figure 14.53 Analysis of strip transmission line using moment
method; for Problem 14.21. R; is the distance between the ith and jth subareas, and V, = l or —1 depending on
whether the ith subarea is on strip 1 or 2, respectively. Write a program to ﬁnd the char
acteristic impedance of the line using the fact that V #030 C where C is the capacitance per unit length and N
iAlf
Q=i=zlp Zo= and Vd = 2 V is the potential difference between strips. Take H = 2 m, W = 5 m, and
N = 20. 14.22 Consider an Lshaped thin wire of radius 1 mm as shown in Figure 14.54. If the wire is
held at a potential V = 10 V, use the method of moments to ﬁnd the charge distribution
on the wire. Take A = 0.1. y Figure 14.54 For Problem 14.22. 758 CHAPTER 14 NUMERICAL METHODS (b) Figure 14.55 For Problem 14.23; coaxial line of (a) arbitrary cross
section, (b) elliptical cylindrical cross section. 14.23 Consider the coaxial line of arbitrary cross section shown in Figure l4.55(a). Using the moment method to ﬁnd the capacitance C per length involves dividing each conductor
into N strips so that the potential on the jth strip is given by 2N
VJ" : 2 Pi Aij
i=1
where
__ R..
M In J, i ¢ j
A” = 21m r0
‘1 — .
M [In[Lei 1.5], i =j
21w ro
and VJ = —l or 1 depending on whether A€i lies on the inner or outer conductor respec— tively. Write a Matlab program to determine the total charge per length on a coaxial cable
of elliptical cylindrical cross section shown in Figure l4.55(b) by using 1v
Q=§pi and the capacitance per unit length with C = Q/2. (a) As a way of checking your program, take A = B = 2 cm and a = b = 1 cm (coaxial line with circular cross section), and compare your result with the exact
value of C = 27re/ln (A/a). (b) TakeA = 2cm,B = 4cm,a = lcm,andb = 2cm.
Him: For the inner ellipse of Figure l4.55(b), for example, a
V sin2 (75 + vzcos2 d) where v = a/b, d6 = rd¢>. Take r0 = 1 cm. r: PROBLEMS 759 Figure 1456» For Problem 14.24. 14.24 A conducting bar of rectangular cross section is shown in Figure 14.56. By dividing the
bar into N equal segments, we obtain the potential at the jth segment as N
Vj = inij
where
1 . .
— z #1
47r8 R,~’
AU z 01] . .
2—“. r——’ 1:]
‘90 7rhA and A is the length of the segment. If we maintain the bar at 10 V, we obtain [AHq] = IOU]
Where[1] =[111. . .,1]Tandq,= pvt/1A. (a) Write a program to ﬁnd the charge distribution pv on the bar and take 6 = 2 m,
h = 2cm,t= lcm,andN= 20.
(b) Compute the capacitance of the isolated conductor by using C=%=(q1 +612+"~+qN)/10 $EE3TEUN 14.353 THE FlNETE ELEMENT METHDD 14.25 Another way of deﬁning the shape functions at an arbitrary point (x, y) in a ﬁnite element
is using the areas A1, A2, and A3 shown in Figure 14.57. Show that A
=, k=1,2,3
ozk A where A = A1 + A2 + A3 is the total area of the triangular element. 14.26 For each of the triangular elements of Figure 14.58, (a) Calculate the shape functions.
(b) Determine the coefﬁcient matrix. 760 CHAPTER 14 NUMERICAL METHODS l
, (x1, Y1) Figure 14.57 For Problem 14.25. 2 _
(x2! .YZ) 3’ Figure 14.58 Triangular elements
of Problem 14.26.
(I. 2) (2.5. 2) (l, 0.25) (0,0)
(3) (b) 14.27 The nodal potential values for the triangular element of Figure 14.59 are VI = 100 V,
' V2 = 50 V, and V3 = 30 V. (a) Determine where the 80 V equipotential line intersects
the boundaries of the element. (b) Calculate the potential at (2, 1). 14.28 The triangular element shown in Figure 14.60 is part of a ﬁnite element mesh. If
V1 = 8 V, V2 = 12 V, and V3 = 10 V, ﬁnd the potential at (a) (1, 2) and (b) the center of the element. 14.29 Determine the global coefﬁcient matrix for the twoelement region shown in Figure 14.61. 14.30 Calculate the global coefﬁcient matrix for the twoelement region shown in Fig
ure 14.62. Figure 14.59 For Problem 14.27. PROBLEMS 761 (1,4) Figure 14.60 For Problem 14.28. Figure 14.61 For Problem 14.29. y 3 Figure 14.62 For Problem 14.30. (0, 2) (0. 0) (4, 0) 14.31 Find the global coefﬁcient matrix of the two—element mesh of Figure 14.63. 14.32 For the twoelement mesh of Figure 14.63, let V1 = 10 V and V3 = 30 V. Find V2
and V4. 14.33 The mesh in Figure 14.64 is part of a large mesh. The shaded region is conducting and
has no elements. Find C5; and C5,]. 762 Figure 14.64 For Problem 14.33. 14.34 Use the program in Figure 14.33 to solve L aplace’s equation in the problem shown in
Figure 14.65, where V0 = 100 V. Comparet he ﬁnite element solution to the exact solu tion in Example 6.5; that is, W9C,” = 4Vo °° sinmnxsinhmry’ n = 2k+1
1r k=0 insmh mr 14.35 Repeat Problem 14.34 for V0 = 100 sin 1rx. Com
the theoretical solution [simil pare the ﬁnite element solution with
ar to Example 6.6(a)]; that is, PROBLEMS 763 Figure rams" For Problem 14.34. 14.38 The cross section of a transmission line is shown in Figure 14.67. Use the ﬁnite differ—
ence method to compute the characteristic impedance of the line. 14.39 Half a solution region is shown in Figure 14.68 so that the y—axis is a line of symmetry.
Use ﬁnite difference to ﬁnd the potential at nodes 1 to 9. Five iterations are sufﬁcient if
you use an iterative method. Figure 142.66 For Problem 14.37. 764 CHAPTER 14 NUMERICAL METHODS Figure 14.67 For Problem 14.38. 3cm { Figure 14.68 For Problem 14.39.
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This note was uploaded on 02/04/2012 for the course ECE 311 taught by Professor Peroulis during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
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