HW4_solutions

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Homework 4 EECS 230 - Fall 2023 Solutions 1. (15pts, 5pts each) Given: a lossless transmission line with Z 0 =100Ω, Z L =50+j25Ω, and frequency, f=1GHz. Using a Smith Chart, match the load to the line, using a lumped-element in parallel with the load. Find: (a) the lumped-element distance nearest to load : d, in wavelengths (b) element-type, (capacitor or inductor) (c) element-value Solution: Step1: z L = Z L /Z 0 = (50+j25)/100 = 0.5+j0.25 Step2: plot on smith chart: pointA Step3: draw SWR circle through pointA Step4: y L on smith chart is 180 o away, on same SWR circle: pointB Step5: Note that y L is at _0.3015___ wavelengths-toward-generator Step6: identify g L =1 circle, choose intersection with SWR circle as nearest to load: is pointC y-at-pointC= _1.0____ +j__-0.7906____ Step7: note that pointC is at _0.345___ wavelengths-toward-generator Step8: distance along xline between pointB and pointC is: _0.345____ - _0.3____ = _ _0.045___ wavelengths = d Step9: need to cancel the imag-part of y C , which = __-0.7906______ so need shunt_y = +j_0.7906______ so shunt_Z = 1/(shunt_y*Y 0 ) = Z 0 /(j_0.7906____) = j_-126.5____Ω Step10: 𝜔 = 2pi f = 2pi 1.X10 9 Hz = 6.28X10 9 rad/sec if shunt_Z<0: capacitor: Z C =1/(j 𝜔 C), C=1/(j 𝜔 Z C ) = 1/(j2pi(1e9Hz)(-j126.5)) , so C=1.26 pF if shunt_Z>0: inductor: Z L = j 𝜔 L,

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2. (10pts, 5pts each) Given: a lossless transmission line with Z 0 =150Ω, Z L =35+j80 Ω, and frequency, f=15GHz. Using a Smith Chart, match the load to the line, using a shorted stub (with the same Z 0 ) in parallel with the load. Find: (a) the stub distance ( nearest to the load ), in wavelengths, (b) the stub length ( shortest ), in wavelengths. Solution: Step1: z L = Z L /Z 0 = (35+j80)/150 = 0.23+j0.533 Step2: plot on smith chart: pointA Step3: draw SWR circle through pointA Step4: y L on smith chart is 180 o away, on same SWR circle: pointB Step5: Note that y L is at __0.332__ wavelengths-toward-generator Step6: identify g L =1 circle, intersection with SWR circle is pointC y-at-pointC= ___1__ +j_1.9397_____ Step7: note that pointC is at _0.187___ wavelengths-toward-load Step8: distance along xline between pointB and pointC is: (0.5-.332) + 0.187 = _ _0.3556___ wavelengths =d Step9: need to cancel the imag-part of y C , which = __1.9397______ so need shunt_y = -j_1.9397______ Step10: using a new smith chart: short is at left-most point along real axis. y of the short is at right-most point along real axis, pointE Step11: we must stay on the outermost circle of the smith chart, and move towards the load until y is equal to the admittance needed from step9. note that this is at WTG=_0.32576____lambda Step12: subtract 0.25lambda to get the stub length: _0.32576____ - 0.25 = _0.07576_ lambda

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3. (10pts, 5pts each) TDR: Given: step excitation of an originally-matched transmission line, but now with a fault, located a distance d from the sending end, that can be modeled as a parallel resistance, R f , and given the measured V(t) as shown in the figure below. Z 0 =135Ω. Note that the insulation has a dielectric constant of 3.6. Find: (a) d, (b) R f Solution: Step1: Since the line was matched, we know: R g =R L =Z 0 Step2: Get Vg: V 1 + = V g Z 0 / (R g +Z 0 ) = V g /2 From the figure V 1 + =11.6V, so: V g =23.2V Step3: Phase velocity on the line: u p =c/ = 3X10 8 m/sec / = 1.58X10 8 m/sec ε 𝑟 3. 6 Step4: Round-trip time delay of echo from the fault is from the figure: 54 msec Step5: And this time is distance/velocity = 2d/u p = 54 msec so: d = (0.5) (54 X 10 -6 sec) (1.58X10 8 m/sec) d = 4266 m (3a) Step6: from the plot: V 1 + + V 1 - = 4.2V, so V 1 - = -7.4V = Ⲅ f V 1 + so Ⲅ f =-7.4V/11.6V = -0.638 Step7: since Ⲅ f = (R lf -Z 0 )/(R lf +Z 0 )

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get: R lf Ⲅ f + Z0 Ⲅ f = R lf -Z 0 R lf ( Ⲅ f -1) = -Z 0 (1+ Ⲅ f ) R lf = Z0 (1+ Ⲅ f )/(1- Ⲅ f ) R lf = 135Ω (1-0.638)(1+0.638) R lf = 135Ω (0.362)/1.638 R lf = 29.84Ω Step8: Since R lf is the parallel combination of R f and Z 0 , we get: R lf = R f Z 0 / (R f + Z 0 ) R lf (R f + Z 0 ) = R f Z 0 R lf R f + R lf Z 0 - R f Z 0 =0 R f (R lf - Z 0 ) = -R lf Z 0 R f = R lf Z 0 / (Z 0 -R lf ) R f = (29.84Ω) (135Ω)/(135Ω - 29.84Ω) R f = 4028.4/105.2 Ω R f = 38.3Ω (3b)

4. (4pts, 1 each) Convert from cartesian (x,y,z) to cylindrical coords (r, 𝜙 ,z) (a) (1.,2.,3.) (b) (4., 10., -3.) (c) (18.5, -23., 14.) (d) (-1.,2.,3.) Solution: r= ? 2 + ? 2 𝜙 = arctan(y/x) (but in the appropriate quadrant) z=z (a) r= = = = 2.24 1 2 + 2 2 1 + 4 5 𝜙 =atan(2/1) = 63.4 o . z=3. in cylindrical coords: (2.24, 63.4 o , 3.) (4a) (or 1.107rad) (b) r= = = = 10.77 4 2 + 10 2 16 + 100 116 𝜙 =atan(10/4) = 68.2 o . z=-3. in cylindrical coords: (10.77, 68.2 o , -3.) (4b) (or 1.19rad) (c) r= = = = 29.52 18. 5 2 + (− 23) 2 342. 25 + 529 871. 25 𝜙 =atan(-23./18.5) = -51.2 o . z=14. in cylindrical coords: (29.52, -51.2 o , 14.) (4c) (or -0.894rad) (d) r= = = = 2.24 (− 1) 2 + 2 2 1 + 4 5 𝜙 =atan(2/-1) = -63.4 o . must be in 2nd quadrant: add 180 o , get: 116.6 o , z=3. in cylindrical coords: (2,24, 116.6 o , 3.) (4d) (or 2.035rad)

5. (4pts, 1 each) Convert from cartesian (x,y,z) to spherical coords (R, 𝜃 , 𝜙 ). (a) (1.,2.,3.) (b) (4., 10., -3.) (c) (18.5, -23., 14.) (d) (-1.,2.,3.) Solution: r= ? 2 + ? 2 + ? 2 𝜃 = arctan ( (but range is 0 o to 180 o ) ? 2 + ? 2 /?) 𝜙 = arctan(y/x) (but in the appropriate quadrant) (a) r = = = = 3.74 1 2 + 2 2 + 3 2 1 + 4 + 9 14 𝜃 = arctan ( / 3) = arctan( / 3) = arctan( / 3) = arctan( 2.24/3) = 36.7 o , 1 2 + 2 2 1 + 4 5 𝜙 = arctan (2./1.) = 63.4 o , in spherical coords: (3.74, 36.7 o , 63.4 o ) (5a) or (3.74, 0.641rad, 1.107rad) (b) r= = = = 11.2 4 2 + 10 2 + (− 3) 2 16 + 100 + 9 125 𝜃 = arctan ( / (-3)) = arctan ( / (-3)) = arctan ( / (-3) ) 4 2 + 10 2 16 + 100 116 = arctan( 10.77/(-3)) = -74.4 o , put in right range: add 180 o : 105.6 o , 𝜙 = arctan (10./4.) = 68.2 o , in spherical coords: (11.2, 105.6 o , 68.2 o ) (5b) or (11.2, 1.84rad, 1.19rad) (c) r= = = = 32.7 18. 5 2 + (− 23) 2 + 14 2 342. 25 + 529 + 196 1067. 25 𝜃 = arctan ( / 14)= arcan( / 14) 18. 5 2 + (− 23) 2 342. 25 + 529 = arctan( / 14) = arctan( 29.52 / 14) = arctan(2.11) = 64.6 o , 871. 25 𝜙 = arctan (-23./18.5) = arctan(-1.24) = -51.2 o , this is in the correct quadrant in spherical coords: (32.7, 64.6 o , -51.2 o ) (5c) or (32.7, 1.13rad, -0.894rad) (d) r= = = = 3.74 (− 1) 2 + 2 2 + 3 2 1 + 4 + 9 14 𝜃 = arctan ( / 3) = arctan( / 3) = arctan( / 3) (− 1) 2 + 2 2 1 + 4 5 = arctan( 2.24/3) = 36.7 o , 𝜙 = arctan (2./-1.) = -63.4 o , must be in 2nd quadrant: add 180 o , get: 116.6 o , in spherical coords: (3.74, 36.7 o , 116.6 o ) (5d) or (3.74, 0.641rad, 2.035rad)

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6. (4pts, 1 each) Convert from cylindrical coords (r, 𝜙 ,z) to spherical coords (R, 𝜃 , 𝜙 ). (a) (3.,0.23rad,12.) (b) (7.5, 156 o ,-34.) (c) (51.,-112 o ,89.) (d) (12.,-0.45rad,-12.) Solution: R = 𝑟 2 + ? 2 𝜃 = arctan(r/z) 𝜙 = 𝜙 (a) R = = = = 12.4 3 2 + 12 2 9 + 144 153 𝜃 = arctan(3./12.) = arctan(0.25) = 14.03 o , 𝜙 = 0.23rad = 13.2 o , in cylindrical coords: (12.4, 14.03 o , 13.2 o ) (6a) or (12.4, 0.245rad, 0.23rad) (b) R = = = = 34.82 7. 5 2 + (− 34) 2 56. 25 + 1156 1212. 25 𝜃 = arctan(7.5/-34.) = arctan(-0.22) = -12.4 o , convert to right range: add 180 o : 167.6 o , 𝜙 = 156 o , in cylindrical coords: (34.8, 167.6 o , 156 o ) (6b) or (34.8, 2.93rad, 2.72rad) (c) R = = = = 102.6 51 2 + 89 2 2601 + 7921 10522 𝜃 = arctan(51/89) = arctan(0.57) = 29.8 o , 𝜙 = -112 o , in cylindrical coords: (102.6, 29.8 o , -112 o ) (6c) or (102.6, 0.52rad, -1.95rad) (d) R = = = = 17. 12 2 + (− 12) 2 144 + 144 288 𝜃 = arctan(12/-12) = arctan(-1) = -45 o , convert to right range: add 180 o : 135 o , 𝜙 = -0.45rad = -25.8 o , in cylindrical coords: (17., 135 o , -25.8 o ) (6d) or (17., 2.36rad, -0.45rad)

7. (12pts) Given the range of coords, sketch (2pts) and determine the surface area (2pts). (a) cartesian coords: x from 1. to 2.5, y from -10. to -5. (b) cylindrical coords: r from 3. to 6, 𝜙 from 12 o to 35 o . (c) spherical coords: R=10., 𝜃 from 3 o to 10 o , 𝜙 from 5 o to 10 o . Solution: (a) green shows the area in the sketch. area=dx X dy= |2.5-1.| X |-10-(-5)| X= 1.5 X 5, so area=7.5 (7a) (b) green shows the area in the sketch. area = 𝑟=3 6 ∫ ϕ=12 𝑜 35 𝑜 ∫ 𝑟 𝑑𝑟 𝑑ϕ to do the integral, need to convert degrees to radians:

12 o = 12*pi/180= 0.209rad 35 o = 35*pi/180= 0.611rad area = 0.5 = 0.5(36-9)(0.611-0.209) 𝑟 2 | 3 6 𝜙| 0.209𝑟𝑎𝑑 0.611𝑟𝑎𝑑 area = 5.43 (7b) (c) green shows the area in the sketch. area = R 2 sin 𝜽 d 𝜽 d 𝜙 θ=3 𝑜 10 𝑜 ∫ ϕ=5 𝑜 10 𝑜 ∫ convert angles to radians: 3 o (pi/180 o ) = 0.0524rad 10 o (pi/180 o )= 0.175rad 5 o (pi/180 o ) = 0.0873rad area = R 2 (-cos 𝜽 ) 𝜙 | 0.0524𝑟𝑎𝑑 0.175𝑟𝑎𝑑 | 0.0873𝑟𝑎𝑑 0.175𝑟𝑎𝑑 area = 10 2 (-0.985 + 0.999) (0.175-0.0873) area = 100 (0.014) (0.0877) area = 0.123 (7c)

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8. (12pts) Given the range of coords, sketch(2pts) and determine the volume (2pts). (a) cartesian coords: x from 1.5 to 10., y from -10. to -7., z from 2. to 5. (b) cylindrical coords: r from 13. to 16, 𝜙 from 112 o to 125 o , z from 100. to 125. (c) spherical coords: R from 10. to 15.5, 𝜃 from 13 o to 25 o , 𝜙 from 3 o to 12 o . Solution: (a) green shows the volume in the sketch. volume=dx X dy X dz = |10.-1.5| X |-10-(-7)| X |5-2| = 8.5 X 3 X 3, so volume = 76.5 (8a) (b) green shows the volume in the sketch.

volume = 𝑟=13 16 ∫ ϕ=112 𝑜 125 𝑜 ∫ ?=100 125 ∫ 𝑟 𝑑𝑟 𝑑ϕ 𝑑? convert angles to radians: 112 o ( 𝜋 /180 o )= 1.95rad 125 o ( 𝜋 /180 o )= 2.18rad volume = 0.5 r 2 = (0.5) (16 2 -13 2 )(2.18rad-1.95rad)(125-100) | 13 16 ϕ| 1.95𝑟𝑎𝑑 2.18𝑟𝑎𝑑 ?| 100 125 volume = (0.5) (87)(0.23rad)(25) volume = 250.125 (8b) (c) green shows the volume in the sketch. volume= R 2 sin 𝜽 dR d 𝜽 d 𝜙 𝑅=10 15.5 ∫ θ=13 𝑜 25 𝑜 ∫ ϕ=3 𝑜 12 𝑜 ∫ convert angles to radians: 13 o (pi/180 o ) = 0.227rad 25 o (pi/180 o )= 0.436rad 3 o (pi/180 o ) = 0.0524rad 12 o (pi/180 o ) = 0.209rad volume = (0.3333) R 3 (-cos 𝜽 ) 𝜙 | 10 15.5 | 0.227𝑟𝑎𝑑 0.436𝑟𝑎𝑑 | 0.0524𝑟𝑎𝑑 0.209𝑟𝑎𝑑 volume = (0.3333)(15.5 3 -10 3 ) (-0.9064 + 0.974) (0.209-0.0524) volume = (0.3333)(3723.9-1000) (0.0676) (0.157) volume = (0.3333)(2723.9) (0.0676) (0.157) volume = 9.636 (8c)

9. (10pts, 5pts each) Given the following vector field: = (a/r) + (bz) + (3) 𝑉 𝑟 ϕ ? and the cylindrical surface: r=3 Find: (a) the vector component of normal to the surface 𝑉 (b) the vector component of tangential to the surface 𝑉 Solution: (a) Find an expression for the outward surface normal: for a cylindrical surface, outward normal is 𝑟 To find the scalar normal component, use the dot-product: normal component of V = ᐧ = (a/r) 𝑉 𝑟 as a vector: normal = (a/r) (9a) 𝑉 𝑟 (b) Tangential component: tangential = - normal 𝑉 𝑉 𝑉 so: tangential = (bz) + (3) (9b) 𝑉 ϕ ?

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10. (10pts, 5pts each) Given the following vector field: = (a/R) + (cz) + (bR) 𝑉 𝑅 θ ϕ and the spherical surface: R=6.5 Find: (a) the vector component of normal to the surface 𝑉 (b) the vector component of tangential to the surface 𝑉 Solution: (a) Find an expression for the outward surface normal: for a cylindrical surface, outward normal is 𝑅 To find the scalar normal component, use the dot-product: normal component of V = ᐧ = (a/r) 𝑉 𝑅 as a vector: normal = (a/R) (10a) 𝑉 𝑟 (b) Tangential component: tangential = - normal 𝑉 𝑉 𝑉 so: tangential = (cz) + (bR) (10b) 𝑉 θ ϕ

11. (9pts, 3 pts each) Given each of the vectors, Find: (a) A vector perpendicular to it, in the same coordinate system (2pts). (b) Verify using the dot product (1pt). (a) cartesian: (3., -5, 12.) (b) cylindrical: (21., 0.35rad, -12.) (c) spherical: (2., 25 o , 50 o ) Solution: (a) There are many vectors that are perpendicular to another. (a1) One way to get such a vector is to pick another vector, not parallel to the one in question, and then take the cross-product of the 2. Pick v2=(1,0,0) v2Xv1 = (1,0,0) X (3,-5, 12) = determinant of: ? ? ? 1 0 0 3 -5 12 cross-product = ( (0)(12) - (0)(-5)) - ( (1)(12) - (0)(3)) + ( (1)(-5) - (0)(3)) ? ? ? = (0) + (-12) + (-5) ? ? ? perpendicular = (0, -12, -5) (11a1) (a2) verify using dot-product: expect the dot-prod to be zero: (3,-5,12) ᐧ (0,-12, -5) = (3)(0) + (-5)(-12) + (12)(-5) =0. checks out. (11a2) (b) pick v2 = (1,0,0) v2Xv1 = (1,0,0) X (21,0.35, -12) = determinant of: 𝑟 ϕ ? 1 0 0 21 0.35 -12 cross-product = ( (0)(-12) - (0)(0.35)) - ( (1)(-12) - (0)(21)) + ( (1)(0.35) - (0)(21)) 𝑟 ϕ ? = (0) - (-12) + (0.35) 𝑟 ϕ ? = (0) + (12) + (0.35) 𝑟 ϕ ? perpendicular = (0, 12, 0.35) (11b1)

(b2) verify using dot-product: expect the dot-prod to be zero: (21,0.35,-12) ᐧ (0,12, 0.35) = (21)(0) + (0.35)(12) + (-12)(0.35) =0. checks out. (11b2) (c) pick v2 = (1,0,0) convert angles to radians: 25 o (pi/180) = 0.436 50 o (pi/180) = 0.873 v2Xv1 = (1,0,0) X (21,0.35, -12) = determinant of: 𝑅 θ ϕ 1 0 0 2 0.436 0.873 cross-product = ( (0)(0.873) - (0)(0.436)) - ( (1)(0.873) - (0)(2)) + ( (1)(0.436) - (0)(2)) 𝑅 θ ϕ = (0) - (0.873) + (0.436) 𝑅 θ ϕ = (0) + (-0.873) + (0.436) 𝑅 θ ϕ perpendicular = (0, -0.873, 0.436) (11c1) (c2) verify using dot-product: expect the dot-prod to be zero: (2,0.436,0.873) ᐧ (0,-0.873, 0.436) = (2)(0) + (0.436)(-0.873) + (0.873)(0.436) =0. checks out. (11c2)

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