Solucionario fundamentos de electromagnetismo con aplicaciones a la ingenieria stuart wentworth

http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS 2-1 Solutions for Chapter Problems Vectors in the Cartesian Coordinate System P2.1: Given P(4,2,1) and APQ=2ax +4ay +6az, find the point Q APQ = ax + ay + az = (Qx-Px)ax + (Qy-Py)ay+(Qz-Pz)az Qx-Px=Qx-4=2; Qx=6 Qy-Py=Qy-2=4; Qy=6 Qz-Pz=Qz-1=6; Qz=7 Ans: Q(6,6,7) P2.2: Given the points P(4,1,0)m and Q(1,3,0)m, the vectors found in (a) through (f) Vector a Find the vector A AOP = ax + ay from the origin to P b Find the vector B BOQ = ax + ay from the origin to Q c Find the vector C CPQ = -3 ax + ay from P to Q d Find A + B A + B = ax + a y e Find C – A C - A = -7 ax + ay f Find B - A B - A = -3 ax + ay fill in the table and make a sketch of Mag 4.12 Unit Vector AOP = 0.97 ax + 0.24 ay 3.16 aOQ = 0.32 ax + 0.95 ay 3.61 aPQ = -0.83 ax + 0.55 ay 6.4 7.07 3.6 a = 0.78 ax + 0.62 ay a = -0.99 ax + 0.14 ay a = -0.83 ax + 0.55 ay a AOP = (4-0)ax + (1-0)ay + (0-0)az = ax + ay A OP  42  12  17  4.12 ax  a y  0.97a x  0.24a y 17 17 (see Figure P2.2ab) aOP  b BOQ =(1-0)ax + (3-0)ay + (0-0)az = ax + ay BOQ  12  32  10  3.16 Fig P2.2ab ax  a y  0.32a x  0.95a y 10 10 (see Figure P2.2ab) aOQ  c CPQ = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + ay C PQ  32  2  13  3.61 3 ax  a y  0.83ax  0.55a y 13 13 (see Figure P2.2cd) a PQ  Fig P2.2cd 2-2 d A + B = (4+1)ax + (1+3)ay + (0-0)az = ax + ay A  B  52  42  41  6.4 ax  a y  0.78a x  0.62a y 41 41 (see Figure P2.2cd) a e C - A = (-3-4)ax + (2-1)ay + (0-0)az = -7 ax + ay C  A  72  12  50  7.07 7 ax  a y  0.99a x  0.14a y 50 50 (see Figure P2.2ef) a FigP2.2ef f B - A = (1-4)ax + (3-1)ay + (0-0)az = -3 ax + ay B  A  32  22  13  3.6 3 ax  a y  0.83a x  0.55a y 13 13 (see Figure P2.2ef) a P2.3: MATLAB: Write a program that will find the vector between a pair of arbitrary points in the Cartesian Coordinate System A program or function for this task is really overkill, as it is so easy to perform the task Enter points P and Q (for example, P=[1 3]; Q=[6 4]) Then, the vector from P toQ is simply given by Q-P As a function we could have: function PQ=vector(P,Q) % Given a pair of Cartesian points % P and Q, the program determines the % vector from P to Q PQ=Q-P; Running this function we have: >> P=[1 3]; >> Q=[6 4]; >> PQ=vector(P,Q) PQ = Alternatively, we could simply perform the math in the command line window: 2-3 >> PQ=Q-P PQ = >> Coulomb’s Law, Electric Field Intensity, and Field Lines P2.4: Suppose Q1(0.0, -3.0m, 0.0) = 4.0nC, Q2(0.0, 3.0m, 0.0) = 4.0nC, and Q3(4.0m, 0.0, 0.0) = 1.0nC (a) Find the total force acting on the charge Q3 (b) Repeat the problem after changing the charge of Q2 to –4.0nC (c) Find the electric field intensity for parts (a) and (b) (a) F13  so Q1Q2 a13 , where R13 = ax + ay =, R13 = 5m, a13 = 0.8 ax + 0.6 ay 4 o R132  4a  3a  x10 C 1x10 C    4 10 F 36 m   5m  9 F13 9 x y 9 FV NM C VC  1.15 x109 ax  0.86 x109 a y N Similarly, F23  1.15 x109 ax  0.86 x109 a y N , so FTOT  2.3a x nN (b) with Q2 = -4 nC, F13 is unchanged but F23  1.15 x109 ax  0.86 x109 a y N , so FTOT  1.7a y nN 9 FTOT  2.3x10 ax N  VC V (c) Ea    2.3a x -9 Q3 m 1x10 C  Nm Likewise, Eb  1.7a y V m Fig P2.4 P2.5: Find the force exerted by Q1(3.0m, 3.0m, 3.0m) = 1.0 C on Q2(6.0m, 9.0m, 3.0m) = 10 nC Q1Q2 a12 , where 4 o R122 R12 = (6-3)ax + (9-3)ay + (3-3)az = ax + ay m 3a  6a y R12  32  62  45m, a12  x ,and 45 F12  2-4 1x10 C 10 x10 C  3a  4 10 F 45m  36 m   6 F12 9 9 x  6a y FV NM , so F12  0.89a x  1.8a y  N C VC 45 Fig P2.5 P2.6: Suppose 10.0 nC point charges are located on the corners of a square of side 10.0 cm Locating the square in the x-y plane (at z = 0.00) with one corner at the origin and one corner at P(10.0, 10.0, 0.00) cm, find the total force acting at point P We arbitrarily label the charges as shown in Figure P2.6 Then ROP = 0.1 ax + 0.1 ay ROP = 0.141 m aOP = 0.707 ax + 0.707 ay FOP  10nC 10nC  0.707   ax  a y  9  4  10 F 36 m  0.141m     32  a x  a y   N FTP  FSP  10nC 10nC  ay 9  4  10 F 36 m  0.1m     90a y  N   90a x  N 10nC 10nC  a x 9  4  10 F 36 m  0.1m   Fig P2.6 and then the total (adjusting to significant digits) is: FTOT  120  a x  a y   N 2-5 P2.7: 1.00 nC point charges are located at (0.00, -2.00, 0.00)m, (0.00, 2.00, 0.00)m, (0.00, 0.00, -2.00)m and (0.00, 0.00, +2.00)m Find the total force acting on a 1.00 nC charge located at (2.00, 0.00, 0.00)m Figure P2.7a shows the situation, but we need only find the x-directed force from one of the charges on Qt (Figure P2.7b) and multiply this result by Because of the problem’s symmetry, the rest of the components cancel 2a  2a y QQt F1t  a R , R  2a x  2a y , R  m, a R  x , 4 o R 1x109 C 1x109 C   2a x  2a y  so F1t   796 x1012  a x  a y  N 9 4 10 F 8m  36 m  The force from all charges is then FTOT     796 x1012 a x  nN  3.2a x nN   Fig P2.7a Fig P2.7b P2.8: A 20.0 nC point charge exists at P(0.00,0.00,-3.00m) Where must a 10.0 nC charge be located such that the total field is zero at the origin? For zero field at the origin, we must cancel the +az directed field from QP by placing Q at the point Q(0,0,z) (see Figure P2.8) Then we have Etot = EP + EQ = 20 x109 C  a z  QP FV V So, E P  a   20 a z R 9 4 o R m 4 10 F 3m C 36 m     2-6 and EQ  Q 4 o R aR 10 x10 C  (a ) 9   9 4 10 F 36 m z   z ( m)   90 az z2 So then 90 20a z  a z  0, z 90 z  , z  2.12 20 Thus, Q(0,0,2.12m) Fig P2.8 The Spherical Coordinate System P2.9: Convert the following points from Cartesian to Spherical coordinates: a P(6.0, 2.0, 6.0) b P(0.0, -4.0, 3.0) c P(-5.0,-1.0, -4.0)   o 1   o (a) r  62  22  62  8.7,   cos 1    47 ,   tan    18  8.7  6 3  4  (b) r  02  42  32  5,   cos 1    53o ,   tan 1    90o 5    1  o 1  1  o (c) r  52  12  42  6.5,   cos 1    130 ,   tan    190  6.5   5  P2.10: a b c Convert the following points from Spherical to Cartesian coordinates: P(3.0, 30., 45.) P(5.0, /4, 3/2) P(10., 135, 180) (a) x  r sin  cos   3sin 30o cos 45o  1.06 y  r sin  sin   3sin 30o sin 45o  1.06 z  r cos   3cos 30o  2.6 so P (1.1,1.1, 2.6) (b) 2-7 x  r sin  cos   5sin 45o cos 270o  y  r sin  sin   5sin 45o sin 270o  3.5 z  r cos   5cos 45o  3.5 so P (0, 3.5,3.5) (c) x  r sin  cos   10sin135o cos180o  7.1 y  r sin  sin   10sin135o sin180o  z  r cos   10 cos135o  7.1 so P (7.1, 0, 7.1) P2.11: Given a volume defined by 1.0m ≤ r ≤ 3.0m, ≤ ≤ 0, 90 ≤ ≤ 90, (a) sketch the volume, (b) perform the integration to find the volume, and (c) perform the necessary integrations to find the total surface area (a) Fig P2.11 (b) V   r sin  drd d   r dr 2 So volume V = 14 m 90o  sin  d   d  13  13.6m3 (c) There are surfaces: an inner, an outer, and identical sides  S side   rdrd   rdr  d  2 m ; Souter   r sin  d d   90o  sin  d m ; STOT  11 m  34.6m 2 So Stotal = 35 m2 Sinner  S sides  6 m 2   d  9 m 2-8 Line Charges and the Cylindrical Coordinate System P2.12: Convert the following points from Cartesian to cylindrical coordinates: a P(0.0, 4.0, 3.0) b P(-2.0, 3.0, 2.0) c P(4.0, -3.0, -4.0) 4 (a)   02  42  4,   tan 1    90o , z  3, so P(4.0,90o ,3.0) 0   (b)   22  32  3.6,   tan 1    124o , z  2, so P(3.6,120o , 2.0)  2   3  (c)   42  32  5,   tan 1    37 o , z  4, so P(5.0, 37 o , 4.0)   P2.13: a b c Convert the following points from cylindrical to Cartesian coordinates: P(2.83, 45.0, 2.00) P(6.00, 120., -3.00) P(10.0, -90.0, 6.00) (a) x   cos   2.83cos 45o  2.00 y   sin   2.83sin 45o  2.00 z  z  2.00 so P (2.00, 2.00, 2.00) (b) x   cos   6.00 cos120o  3.00 y   sin   6.00sin120o  5.20 z  z  3.00 so P (3.00,5.20, 3.00) (c) x   cos   10.0 cos(90.0o )  y   sin   10.0sin( 90.0o )  10.0 z  z  6.00 so P (0, 10.0, 6.00) P2.14: A 20.0 cm long section of copper pipe has a 1.00 cm thick wall and outer diameter of 6.00 cm a Sketch the pipe conveniently overlaying the cylindrical coordinate system, lining up the length direction with the z-axis b Determine the total surface area (this could actually be useful if, say, you needed to an electroplating step on this piece of pipe) c Determine the weight of the pipe given the density of copper is 8.96 g/cm 2-9 (a) See Figure P2.14 (b) The top area, Stop, is equal to the bottom area We must also find the inner area, Sinner, and the outer area, Souter 2 Stop    d  d    d   d  5 cm Sbottom  Stop 2 20 0 Souter    d dz   d  dz  120 cm 2 20 0 Sinner    d dz   d  dz  80 cm The total area, then, is 210 cm2, or Stot = 660 cm2 (c) Determining the weight of the pipe requires the volume: V    d  d dz 2 20 0    d   d  dz  100 cm3 g   M pipe   8.96  100 cm3  cm    2815 g So Mpipe = 2820g Fig P2.14 P2.15: A line charge with charge density 2.00 nC/m exists at y = -2.00 m, x = 0.00 (a) A charge Q = 8.00 nC exists somewhere along the y-axis Where must you locate Q so that the total electric field is zero at the origin? (b) Suppose instead of the 8.00 nC charge of part (a) that you locate a charge Q at (0.00, 6.00m, 0.00) What value of Q will result in a total electric field intensity of zero at the origin? (a) The contributions to E from the line and point charge must cancel, or E  E L  EQ For the line: E L   2nC / m  L V a  a y  18 a y 9 2 o  m 2 10 F 2m 36 m     10-42 We then have: L' 101.2 L1 ''    40.2nH BW 2  400 x106  C1 ''  2  400 MHz  BW   0.110 pF 2 o L1 '  2   2392 MHz 2 101.2  L2 ''  2  400 MHz  BW   0.560nH 2 o C2 '  2   2392 MHz 2  0.01988  C2 ''  C2 ' 0.01988   7.93 pF BW 2  400 x106  To plot the insertion loss using ML1043, we use the impedances shown in Figure P10.42(b), given as j j Z1  j L1 '' , Z  j L2 '' C1 '' C2 '' We also define Z3 as the impedance seen by the voltage v1: Z  Z  R  Z1  Now we can find the voltage ratio relations:    R  Z3 vLF    vs  v1 , and v1    R  Z1   Z  R  Z1  Combining, and realizing vL = vs/2, we arrive at: vL  R  Z1  R  Z1  Z   vLf RZ The insertion loss is then:  v  IL  20 log  L   vLf    This is plotted by ML1042 in Figure P10.42(c) % MLP1042% % Design N=3 BPF, dB ripple % clc clear R=50; flo=2.2e9;fhi=2.6e9; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi); 10-43 Fig P10.42(c) g1=2.0236;g2=.9941; L=g1*R;C=g2/R; L1=L/BW; C1=BW/((wo^2)*L); L2=BW/((wo^2)*C); C2=C/BW; f=1:0.01:5; %f in GHz w=2*pi.*f*1e9; Z1=j.*(w.*L1-1./(w.*C1)); Z2=parallel(j.*w.*L2,-j./(w.*C2)); Z3=parallel(Z2,R+Z1); A=((R+Z1).*(R+Z1+Z3))./(2.*R.*Z3); IL=20*log10(abs(A)); plot(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('N=3 BPF: 900-1100 MHz') 10-44 P10.43: Starting with the Figure 10.28(b) circuit configuration, design a 3rd order Chebyshev bandstop filter for a 50  system The stopband is to be from 900 MHz to 1100 MHz with dB of ripple allowed Plot the insertion loss (a) Fig P10.43a&b (b) Assuming dB of ripple, from Table 10.3 we have: g1  g3  3.3487, and g  0.7117 (see Figure P10.28(b)) Now we perform the impedance transformation: g L1 '  L3 '  g1 R  167.4, and C2 '   0.01423 R The frequency transformation is performed using Table 10.4, where L1 ' is transformed into a parallel circuit containing L1 '' and C1 '' C2 ' is transformed into a series circuit containing L2 '' and C2 '' This is shown in Figure P10.43(a) To perform the transformation, we calculate BW and wo: BW  2  200 MHz  o  2  900MHz 1100MHz   2  995MHz  We then have: 1 C1 ''    4.75 pF BW L1 ' 2  200 x106  167.4  L1 ''  L2 ''  C2 ''  BW L1 '  o  2  200 MHz 167.4   2   995MHz  2  5.39nH 1   55.9nH BW C2 ' 2  200 MHz  0.01423 BW C2 ' o2  2  200 x106   0.01423  2   995MHz  2  0.458 pF 10-45 To plot the insertion loss using ML1043, we use the impedances shown in Figure P10.43(b), given as j j Z  j L2 '' , Z1  j L1 '' C2 '' C1 '' We also define Z3 as the impedance seen by the voltage v1: Z  Z  R  Z1  Now we can find the voltage ratio relations:    R  Z3 vLF    vs  v1 , and v1    R  Z1   Z  R  Z1  Combining, and realizing vL = vs/2, we arrive at: vL  R  Z1  R  Z1  Z   vLf RZ The insertion loss is then:  v  IL  20 log  L   vLf    This is plotted by ML1043 in Figure P10.43(c) % MLP1043 % % Design N=3 band-stop filter, dB ripple % clc;clear R=50; flo=900e6;fhi=1100e6; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi); g1=3.3487;g2=.7117; L=g1*R;C=g2/R; C1=1/(BW*L); L1=BW*L/((wo^2)); C2=BW*C/((wo^2)); L2=1/(BW*C); Fig P10.43c f=.8:0.001:1.2; %f in GHz w=2*pi.*f*1e9; Z2=j.*(w.*L2-1./(w.*C2)); Z1=parallel(j.*w.*L1,-j./(w.*C1)); 10-46 Z3=parallel(Z2,R+Z1); A=((R+Z1).*(R+Z1+Z3))./(2.*R.*Z3); IL=20*log10(abs(A)); plot(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('N=3 Band-Stop Filter: 900-1100 MHz') axis([0.8 1.2 50]) P10.44: Starting with the Figure 10.28(a) circuit configuration, design a 5th order Chebyshev bandstop filter for a 50  system The stopband is to be from 2.3 GHz to 2.5 GHz with dB of ripple allowed Plot the insertion loss From Table 10.3 for dB of ripple and N = we find: g1  g5  2.1349 g  g  1.0911 g3  3.0009 We first perform transformation: g C1 '   0.0427, Ro the impedance (a) L2 '  g Ro  54.56, C3 '  g3  0.0602 Ro Figure P10.44(b) shows the filter portion of the circuit after transforming based on Table 10.4 Fig P10.44a&b To perform the frequency transformation, we first calculate: BW  2  2.5GHz  2.3GHz   2  200 MHz  o  2 Then, L1 ''   2.3GHz  2.5GHz   2  2.398GHz  1   18.6nH BW C1 ' 2  200 x106   0.0427  (b) 10-47 C1 ''  C2 ''  L2 ''  L3 ''  C3 ''  BW C1 '  o  2  200 MHz  0.0427   2   2.398GHz  2  0.236 pF 1   14.6nH BW L2 ' 2  200MHz  54.56  BW L2 ' o2  2  200 x106   54.56   2   2.398GHz  2  0.302nH 1   13.3nH BW C3 ' 2  200MHz  0.06002  BW C3 ' o2  2  200 x106   0.06002   2   2.398GHz  2  0.332 pF The voltage ratios needed to determine insertion loss are found with the aid of Fig P10.44(c) Here we have j Z1  j L1 ' , C1 ' Z  j L2 ' j  C2 ' j C3 ' Fig P10.44(c) Some of these impedances are also lumped together: Z a  Z  Z  Z1 R  and Z b  Z1  Z  Z a  Z  j L3 ' The voltage ratios are then:  Z1 R   Za   Zb  vLf    v1 , v1    v2 , v2    vs ,  Za  Z2   R  Zb   Z1 R  Z  so vLf  Z1 R   Z a   Z b      vs  Z1 R  Z   Z a  Z   R  Z b  Now, since vL=vs/2, then after manipulation we have vL  Z1 R  Z   Z a  Z  R  Z b   vLf Z a Z b  Z1 R  and this is used to find IL  20 log vL using MLP1044: vLf 10-48 % MLP1044 % Design N=5 Band-Stop filter, dB ripple % clc;clear R=50; flo=2.3e9;fhi=2.5e9; BW=2*pi*(fhi-flo); wo=2*pi*sqrt(flo*fhi); g1=2.1349;g2=1.0911;g3=3.0009; Ca=g1/R;L=g2*R;Cb=g3/R; L1=1/(BW*Ca); C1=BW*Ca/((wo^2)); L2=BW*L/((wo^2)); C2=1/(BW*L); L3=1/(BW*Cb); C3=BW*Cb/((wo^2)); f=1.8:0.0001:3; w=2*pi.*f*1e9; %f in GHz Z1=j.*w.*L1-j./(w.*C1); Z2=parallel(j.*w.*L2,-j./(w.*C2)); Z3=j.*w.*L3-j./(w.*C3); Za=parallel(Z1,R); Zb=parallel(Z3,Z2+Za); Zc=parallel(Z1,Z2+Zb); A=(Za+Z2).*(Zb+Z2).*(Zc+R); B=2.*Za.*Zb.*Zc; IL=20*log10(abs(A./B)); plot(f,IL) grid on xlabel('frequency (GHz)') ylabel('IL (dB)') title('N=5 Band-Stop Filter: 2.3 - 2.5 GHz') axis([1.8 50]) 10-49 Fig P10.44(d) Amplifiers P10.45: (JustAsk): The following S-Parameters were measured at 2.0 GHz in a 50  system:  S11  0.68e j125 S12  S 21  3.6e j 40 S 22  0.86e  j 74 (a) Determine the gain, in dB, without any matching networks maximum gain, assuming optimized matching networks   (b) Determine the (a) With out matching networks, we have S = L = as indicated in Figure P10.45 So our transducer gain term is: GT   S 2 S 21 1 L 2  S 21  12.96   IN  S  S 22  L So GT (dB) = 10 log(12.96) = 11.1 dB (b) Now assuming optimized matching, the transducer gain term reduces to: Fig P10.45 10-50 GT  S 21  92.6  S 21  S 22 and GT (dB) = 10 log(92.6) = 19.7dB P10.46: For P10.45, (a) design open-ended shunt stub matching networks (b) You are to realize the matching networks in microstrip constructed on 25.0 mil thick Teflon Determine the required microstrip width, and provide a labeled sketch of your network similar to Figure 10.42 Referring to Figure P10.46(a) we find the open-ended shunt stub matching network to  * achieve  L  S 22  0.86e j 74 We move from the open-ended stub in the admittance chart (point a’) a distance 0.206 to reach the point b’ The normalized admittance at this point, added to the matched load admittance of y = 1+j0, puts us at point c’ From here, we move a distance (0.397– 0.209 = 0.188) along the constant || circle to point d’ Transforming this to the impedance chart, we reach our desired impedance point d The same procedure is used for the source matching network Referring to Figure P10.46(b) we find the open-ended shunt stub matching network to achieve   S  S11*  0.68e  j125 We move from the open-ended stub in the admittance chart (point a’) a distance 0.173 to reach the point b’ The normalized admittance at this point, added to the matched load admittance of y = 1+j0, puts us at point c’ From here, we move a distance (0.5000.174 – 0.185 = 0.489) along the constant || circle to point d’ Transforming this to the impedance chart, we reach our desired impedance point d The final stub matching network is indicated in Figure P10.46(c) To realize the stub matching network in microstrip, we employ ML0605 for microstrip design: Microstrip Design width & thickness will be in the same units enter the desired impedance: 50 enter the substrate thickness: 25 enter substrate rel permittivity: 2.1 w = 79.3268 eeff = 1.8015 up = 223363676.0028m/s >> The length of a guide wavelength is 10-51 (a) (b) (c) (d) Fig P10.46 G  c f  eff  x108 m / s  1mil     4400mils 6 x10 1.8015  25.4 x10 m  This is used to find the physical lengths required of the stubs, as shown in Figure 10.46(d) 10-52 P10.47: For P10.45, design a matching network using lumped elements In the sketch of your solution, indicate line lengths in terms of wavelength Fig P10.47(a) We first design the lumped element matching network to achieve  *  L  S 22  0.86e j 74 For this type of problem we actually work backwards, starting from point c and working back to point a, to identify the location of the points Then we work the matching network forwards to determine the component values From the matched condition at point a, we add a shunt inductance to point b’ (-j2.0) Then, transforming to the impedance chart, we add a series inductance to reach point c (j1.3 – j0.4 = j0.9) For the shunt inductance we have:  jZ o  j 2.0  , or L 50 L  2nH  2   x109  For the series inductance, j L j 0.9  , or Zo L  0.9  50  2  x109   3.6nH Now we design the lumped element matching network to achieve   S  S11*  0.68e  j125 From the matched condition at point a, we add a shunt capacitance to point b’ (j1.8) Then, transforming to the impedance chart, we add a series capacitance to reach point c (-j0.49 + j0.42 = -j0.07) For the shunt capacitance we have: j1.8  jCZ o , or Fig P10.47(b) C 1.8  2.9 pF 2  x109   50  For the series capacitance, 10-53 Fig P10.47(c)  j 0.07  C j , or CZ o  23 pF  0.07  2  x109   50  The overall circuit is given in Figure P10.47(c) P10.48: The following S-Parameters were measured at 10 GHz in a 50  system:  S11  0.72e j 76 S12  S 21  4.4e j125 S 22  0.58e  j 30 (a) Determine the gain, in dB, without any matching networks maximum gain, assuming optimized matching networks   (b) Determine the (a) With out matching networks, we have S = L = So our transducer gain term is: 2  S  L GT  S21  S21  19.36 2   IN  S  S22  L So GT (dB) = 10 log(19.36) = 12.9 dB (b) Now assuming optimized matching, the transducer gain term reduces to: 1 GT  S21  60.58 2  S21  S22 and GT (dB) = 10 log(60.59) = 17.8dB P10.49: For P10.48, design shorted shunt stub matching networks with the overall line lengths minimized In the sketch of your solution, indicate line lengths in terms of wavelength Referring to Figure P10.49(a) we find the shorted shunt stub matching network to achieve  *  L  S22  0.58e j 30 We move from the open-ended stub in the admittance chart (point a’) a distance 0.099 to reach the point b’ The normalized admittance at this point, added to the matched load admittance of y = 1+j0, puts us at point c’ From here, we 10-54 (a) (b) (c) Fig P10.49 move a distance (0.458– 0.326 = 0.132) along the constant || circle to point d’ Transforming this to the impedance chart, we reach our desired impedance point d The same procedure is used for the source matching network Referring to Figure P10.49(b) we find the shorted shunt stub matching network to achieve   S  S11*  0.72e  j 76 We move from the open-ended stub in the admittance chart (point a’) a distance 0.072 to reach the point b’ The normalized admittance at this point, added to the matched load admittance of y = 1+j0, puts us at point c’ From here, we move a distance (0.500+ 0.106– 0.311 = 0.295) along the constant || circle to point d’ Transforming this to the impedance chart, we reach our desired impedance point d Figure P10.49(c) shows the completed circuit Note that this pair of matching networks results in the smallest overall sum of stub lengths 10-55 P10.50: For P10.48, design the matching networks using lumped elements We first design the lumped element matching network to achieve  *  L  S22  0.58e j 30 For this type of problem we actually work backwards, starting from point c and working back to point a, to identify the location of the points Then we work the matching network forwards to determine the component values From the matched condition at point a, we add a series inductance to point b (+j1.6) Then, transforming to the admittance chart, we add a shunt capacitance to reach point c’ (-j.25 –- j.44 = j0.19) For the series inductance we have: 50 1.6   j L j1.6  , or L   1.27nH Zo  2  10 x109  For the shunt capacitance, j 0.19  j CZ o , or C  (a) Fig P10.50  0.19  2 10 x109   50   060 pF (b) 10-56 Now we design the lumped element matching network to achieve   S  S11*  0.72e  j 76 From the matched condition at point a, we add a shunt capacitance to point b’ (+j1.2) Then, transforming to the impedance chart, we add a series capacitance to reach point c (j1.2 j.48 = -j0.72) For the shunt capacitance we have: 1.25 j1.25  j CZ o , or C   0.40 pF  2  10 x109   50  For the series capacitance, j  j 0.72  , or C   44 pF  CZ o 2 10 x10   50  0.72  Receiver Design P10.51: Determine the IF power, in watts, exiting a mixer that has a 6.0 dB conversion loss if dBm of RF power and of LO power enters the mixer P  CL  10log  RF   6dB,  PIF  PRF  100.6  3.981 PIF PIF  PRF 1mW   0.25mW 3.981 3.981 P10.52: (JustAsk): Referring to Example 10.21 and Figure 10.48, suppose you require a 100 W output power level and the antenna receives -80 dBm If you have several of each amplifier available, design the receiver You are also allowed to insert a fixed value attenuator One design is shown in Figure P10.52 Fig P10.52 ... eo=8.854e-12; d=8 ;a= 2; 2-20 delta_r=40;delta_theta=72;delta_phi=144; % Perform calculation for k=(1:delta_phi) for j=(1:delta_theta) for i=(1:delta_r) r=i *a/ delta_r; theta=(pi/2)+j*pi/(2*delta_theta); phi=k*2*pi/delta_phi;...   r o A d ; C2  o A d ; C1 48  r   4.0 C2 12 P2.71: A parallel plate capacitor with a 1.0 m surface area for each plate, a 2.0 mm plate separation, and a dielectric with relative permittivity... LaPlace equation approach instead Laplace:  2Vcyl    V         ; here we see V only depends on   V  A; V  A ln   B ;  where A and B are constants Now apply boundary conditions