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TE2-SE-2009-2S

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Segunda Evaluación de Teoría Electromagnética II - 2009 - 2S …

Segunda Evaluación de Teoría Electromagnética II - 2009 - 2S
FIEC - ESPOL

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  • 1. ESCUELA SUPERIOR POLITÉCNICA DEL LITORAL POLIT CNICA TEORÍA ELECTROMAGNÉTICA II ELECTROMAGNÉTICA Profesor: Ing. Alberto Tama FrancoSEGUNDA EVALUACIÓN EVALUACIÓN viernes 05 de febrero de 2010 sAlumnos: ______________________________________________________________________Alumno ______________________________________________________________________PRIMER TEMA (20 puntos) puntos):(Engineering Electromagnetics, 6th Edition, William H. Hayt Fr. & John A. Buck) EngineeringUna línea ranurada en el aire cuya impedancia característica es de 50 [ Ω] se aplica a lamedición de una impedancia de carga. El patrón de voltaje obtenido directamente deldetector, tanto de los mínimos adyacentes, cuando la carga desconocida está conectada, los mínimos cargacomo de los mínimos cuando la carga es reemplazada por un cortocircuito, se muestra enla siguiente figura. Si la frecuencia de operación es de 400 [ MHz ] , determin eterminar: s, Γ y Z L . Ing. Alberto Tama Franco F FIEC-ESPOL – 2009 – 2S FIEC
  • 2. Black Magic Design 0.12 0.13 0.11 0.14 0.38 0.37 0.15 0.1 0.39 0.36 90 0.4 100 80 0.35 0.1 0 .09 6 45 70 50 0.4 1 110 40 0.3 1.0 4 0.9 1.2 0.1 55 .08 0.8 0 7 35 1.4 0.3 2 0 60 0.7 0.4 12 3 0.6 60 ) /Yo 1.6 0.1 0.0 7 E (+jB 30 8 3 NC 0.3 0.4 TA 0.2 1.8 2 0 EP 50 SC 65 13 SU 2.0 E 0.5 IV 06 0. IT 25 19 0. AC 44 0. AP 31 0. C 70 R ,O 0.4 o) 0 40 14 4 5 0. 0.2 0.0 /Z 5 20 0.3 jX 0.4 (+ 3.0 T 75 EN 0.6 N PO 4 0.2 0.0 0.3 OM 0 6 0.2 1 30 15 0.4 9 0.8 EC 15 > 4.0 R— 80 NC TO TA 1.0 0.22 AC ERA 0.47 0.28 5.0 RE 1.0 GEN 0.2 160 IVE 20 85 10 UCT ARD 8 0. 0.23 IND S TOW 0.48 0.27 ANG 90 0.6 ANG LE OF NGTH 10 LE OF 170 0.1 0.4 TRANSM 0.0 —> WAVELE 0.24 0.49 0.26 REFLECTION COEFFICIENT IN DE 20 0.2 ISSION COEFFICIENT IN 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 0.25 0.25 ± 180 0.0 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) D <— 0.2 20 RD LOA 0.24 0.49 0.26 0.4 70 0.1 DEGR −1 A GREE 10 S TOW 0.6 EES -90 S 0.23 ) 0.48 0.27 H /Yo NGT 8 (-jB 0. -10 ELE 160 CE −20 -85 0.2 V N − 1.0 WA 5.0 TA 0.22 7 0.28 0.4 EP 1.0 <— SC SU -80 4.0 0.8 -15 VE 04 0.2 50 TI 0.3 −3 UC −1 0. .46 0.2 1 0 D 0 9 IN 0.6 -75 R 3.0 ,O o) 5 -20 0.2 0.0 /Z 4 5 0. 0.3 X 0.4 40 (-j −4 −1 0 T 0.4 EN -70 N PO 06 0. 19 0. M CO -25 44 0. 0.5 31 0. CE 2.0 30 −5 AN 0 −1 -65 CT 0.1 7 EA 0.2 1.8 0.0 VE R 8 0.6 ITI 0.3 3 0.4 AC -30 2 1.6 CAP -60 20 .08 −6 0.1 0 0.7 −1 7 1.4 0 2 -35 0.3 0.8 0.4 3 1.2 -55 0.9 0.1 1.0 9 −70 0.0 0 −11 0 6 -4 0 -5 5 0.3 -4 1 0.4 0.1 −100 −80 0.15 4 −90 0.11 0.14 0.35 0.4 0.12 0.13 0.39 0.36 0.38 0.37 ] F RADIALLY SCALED PARAMETERS dB F . [ OE RT SW TOWARD LOAD —> <— TOWARD GENERATOR EN S C ] . P) N R TT S B T . ∞ 100 40 20 10 5 4 3 2.5 2 1.8 1.6 1.4 1.2 1.1 1 15 10 7 5 4 3 2 1 A . LO [d NS R L d SRF FL OSS BS W S O P S. LO (C F, L. . C [d ∞ 40 30 20 15 10 8 6 5 4 3 2 1 1 1 1.1 1.2 1.3 1.4 1.6 1.8 2 3 4 5 10 20 ∞ . K F C OE B] O 0 1 2 3 4 5 6 7 8 9 10 12 14 20 30 ∞ 0 0.1 0.2 0.4 0.6 0.8 1 1.5 2 3 4 5 6 10 15 ∞ FL EA OE R .P C I EF FF . or F, , P .W SM ,E E 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.01 0 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 3 4 5 10 ∞ S N FF or A E I 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.99 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 TR O .C SM CENTER N A 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 TR ORIGIN
  • 3. SEGUNDO TEMA (30 puntos):(Field and Wave lectromagnetics, 2nd Edition, David K. Cheng)Un estudiante de la materia Teoría Electromagnética II, determina que para acoplar unacarga Z L = 100 + j50 [ Ω] , se requiere la conexión de un sintonizador reactivo doble. Porlas condiciones físicas, se imposibilita conectar el primer stub directamente en losterminales de la carga, debiendo ser conectado a una cierta distancia d , tal como semuestra en la figura.Determinar el valor límite mínimo de dicha distancia, así como las longitudes mínimasrequeridas que deberán tener los sintonizadores (en corto circuito) con la finalidad delograr un acoplamiento perfecto. 3λ /8 d • • Z o = 300 [ Ω ] Z L = 100 + j 50 [ Ω ] • • l2 l1d= l1 = l2 = Ing. Alberto Tama Franco FIEC-ESPOL – 2009 – 2S
  • 4. Black Magic Design 0.12 0.13 0.11 0.14 0.38 0.37 0.15 0.1 0.39 0.36 90 0.4 100 80 0.35 0.1 0 .09 6 45 70 50 0.4 1 110 40 0.3 1.0 4 0.9 1.2 0.1 55 .08 0.8 0 7 35 1.4 0.3 2 0 60 0.7 0.4 12 3 0.6 60 ) /Yo 1.6 0.1 0.0 7 E (+jB 30 8 3 NC 0.3 0.4 TA 0.2 1.8 2 0 EP 50 SC 65 13 SU 2.0 E 0.5 IV 06 0. IT 25 19 0. AC 44 0. AP 31 0. C 70 R ,O 0.4 o) 0 40 14 4 5 0. 0.2 0.0 /Z 5 20 0.3 jX 0.4 (+ 3.0 T 75 EN 0.6 N PO 4 0.2 0.0 0.3 OM 0 6 0.2 1 30 15 0.4 9 0.8 EC 15 > 4.0 R— 80 NC TO TA 1.0 0.22 AC ERA 0.47 0.28 5.0 RE 1.0 GEN 0.2 160 IVE 20 85 10 UCT ARD 8 0. 0.23 IND S TOW 0.48 0.27 ANG 90 0.6 ANG LE OF NGTH 10 LE OF 170 0.1 0.4 TRANSM 0.0 —> WAVELE 0.24 0.49 0.26 REFLECTION COEFFICIENT IN DE 20 0.2 ISSION COEFFICIENT IN 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 0.25 0.25 ± 180 0.0 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) D <— 0.2 20 RD LOA 0.24 0.49 0.26 0.4 70 0.1 DEGR −1 A GREE 10 S TOW 0.6 EES -90 S 0.23 ) 0.48 0.27 H /Yo NGT 8 (-jB 0. -10 ELE 160 CE −20 -85 0.2 V N − 1.0 WA 5.0 TA 0.22 7 0.28 0.4 EP 1.0 <— SC SU -80 4.0 0.8 -15 VE 04 0.2 50 TI 0.3 −3 UC −1 0. .46 0.2 1 0 D 0 9 IN 0.6 -75 R 3.0 ,O o) 5 -20 0.2 0.0 /Z 4 5 0. 0.3 X 0.4 40 (-j −4 −1 0 T 0.4 EN -70 N PO 06 0. 19 0. M CO -25 44 0. 0.5 31 0. CE 2.0 30 −5 AN 0 −1 -65 CT 0.1 7 EA 0.2 1.8 0.0 VE R 8 0.6 ITI 0.3 3 0.4 AC -30 2 1.6 CAP -60 20 .08 −6 0.1 0 0.7 −1 7 1.4 0 2 -35 0.3 0.8 0.4 3 1.2 -55 0.9 0.1 1.0 9 −70 0.0 0 −11 0 6 -4 0 -5 5 0.3 -4 1 0.4 0.1 −100 −80 0.15 4 −90 0.11 0.14 0.35 0.4 0.12 0.13 0.39 0.36 0.38 0.37 ] F RADIALLY SCALED PARAMETERS dB F . [ OE RT SW TOWARD LOAD —> <— TOWARD GENERATOR EN S C ] . P) N R TT S B T . ∞ 100 40 20 10 5 4 3 2.5 2 1.8 1.6 1.4 1.2 1.1 1 15 10 7 5 4 3 2 1 A . LO [d NS R L d SRF FL OSS BS W S O P S. LO (C F, L. . C [d ∞ 40 30 20 15 10 8 6 5 4 3 2 1 1 1 1.1 1.2 1.3 1.4 1.6 1.8 2 3 4 5 10 20 ∞ . K F C OE B] O 0 1 2 3 4 5 6 7 8 9 10 12 14 20 30 ∞ 0 0.1 0.2 0.4 0.6 0.8 1 1.5 2 3 4 5 6 10 15 ∞ FL EA OE R .P C I EF FF . or F, , P .W SM ,E E 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.01 0 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 3 4 5 10 ∞ S N FF or A E I 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.99 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 TR O .C SM CENTER N A 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 TR ORIGIN
  • 5. TERCER TEMA (25 puntos):(Theory and Problems of Electromagnetics, 2nd Edition, Joseph A. Edminister)Un circuito ideal LC , de parámetros concentrados, tal como se muestra en la figura, esutilizado para acoplarse a una línea de transmisión sin pérdidas cuya impedanciacaracterística es de 100 [ Ω] , la misma que se encuentra a la entrada de un transistor deradio frecuencia cuya frecuencia de operación es de 1 [GHz ] . Si la carga de dichotransistor es de Z L = 35 − j 50 [ Ω ] . Determine los valores de L y C que cumplen con lacondición de acoplamiento perfecto. ¿Es posible obtener la condición de acoplamientoperfecto, intercambiando L por C y viceversa? Si su respuesta es afirmativa, determineentonces los nuevos valores de L y C . L r to Z o = 100 [ Ω ] sis C an Tr Ing. Alberto Tama Franco FIEC-ESPOL – 2009 – 2S
  • 6. Black Magic Design 0.12 0.13 0.11 0.14 0.38 0.37 0.15 0.1 0.39 0.36 90 0.4 100 80 0.35 0.1 0 .09 6 45 70 50 0.4 1 110 40 0.3 1.0 4 0.9 1.2 0.1 55 .08 0.8 0 7 35 1.4 0.3 2 0 60 0.7 0.4 12 3 0.6 60 ) /Yo 1.6 0.1 0.0 7 E (+jB 30 8 3 NC 0.3 0.4 TA 0.2 1.8 2 0 EP 50 SC 65 13 SU 2.0 E 0.5 IV 06 0. IT 25 19 0. AC 44 0. AP 31 0. C 70 R ,O 0.4 o) 0 40 14 4 5 0. 0.2 0.0 /Z 5 20 0.3 jX 0.4 (+ 3.0 T 75 EN 0.6 N PO 4 0.2 0.0 0.3 OM 0 6 0.2 1 30 15 0.4 9 0.8 EC 15 > 4.0 R— 80 NC TO TA 1.0 0.22 AC ERA 0.47 0.28 5.0 RE 1.0 GEN 0.2 160 IVE 20 85 10 UCT ARD 8 0. 0.23 IND S TOW 0.48 0.27 ANG 90 0.6 ANG LE OF NGTH 10 LE OF 170 0.1 0.4 TRANSM 0.0 —> WAVELE 0.24 0.49 0.26 REFLECTION COEFFICIENT IN DE 20 0.2 ISSION COEFFICIENT IN 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0 4.0 5.0 10 20 50 0.25 0.25 ± 180 0.0 50 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) D <— 0.2 20 RD LOA 0.24 0.49 0.26 0.4 70 0.1 DEGR −1 A GREE 10 S TOW 0.6 EES -90 S 0.23 ) 0.48 0.27 H /Yo NGT 8 (-jB 0. -10 ELE 160 CE −20 -85 0.2 V N − 1.0 WA 5.0 TA 0.22 7 0.28 0.4 EP 1.0 <— SC SU -80 4.0 0.8 -15 VE 04 0.2 50 TI 0.3 −3 UC −1 0. .46 0.2 1 0 D 0 9 IN 0.6 -75 R 3.0 ,O o) 5 -20 0.2 0.0 /Z 4 5 0. 0.3 X 0.4 40 (-j −4 −1 0 T 0.4 EN -70 N PO 06 0. 19 0. M CO -25 44 0. 0.5 31 0. CE 2.0 30 −5 AN 0 −1 -65 CT 0.1 7 EA 0.2 1.8 0.0 VE R 8 0.6 ITI 0.3 3 0.4 AC -30 2 1.6 CAP -60 20 .08 −6 0.1 0 0.7 −1 7 1.4 0 2 -35 0.3 0.8 0.4 3 1.2 -55 0.9 0.1 1.0 9 −70 0.0 0 −11 0 6 -4 0 -5 5 0.3 -4 1 0.4 0.1 −100 −80 0.15 4 −90 0.11 0.14 0.35 0.4 0.12 0.13 0.39 0.36 0.38 0.37 ] F RADIALLY SCALED PARAMETERS dB F . [ OE RT SW TOWARD LOAD —> <— TOWARD GENERATOR EN S C ] . P) N R TT S B T . ∞ 100 40 20 10 5 4 3 2.5 2 1.8 1.6 1.4 1.2 1.1 1 15 10 7 5 4 3 2 1 A . LO [d NS R L d SRF FL OSS BS W S O P S. LO (C F, L. . C [d ∞ 40 30 20 15 10 8 6 5 4 3 2 1 1 1 1.1 1.2 1.3 1.4 1.6 1.8 2 3 4 5 10 20 ∞ . K F C OE B] O 0 1 2 3 4 5 6 7 8 9 10 12 14 20 30 ∞ 0 0.1 0.2 0.4 0.6 0.8 1 1.5 2 3 4 5 6 10 15 ∞ FL EA OE R .P C I EF FF . or F, , P .W SM ,E E 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.01 0 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.5 3 4 5 10 ∞ S N FF or A E I 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.99 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 TR O .C SM CENTER N A 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 TR ORIGIN
  • 7. CUARTO TEMA (25 puntos):(Fundamentals of Applied Electromagnetics, 5th Edition, Fawwaz T. Ulaby)Una guía de ondas rectangular cuyas dimensiones son a = 5 [ cm] y b = 3 [ cm] , se llenacon un dieléctrico de permitividad desconocida. La expresión espacio temporal de lacomponente eléctrica de la onda que se propaga en su interior, en el modo transversaleléctrico, está dada por: Ex = −36 cos 40π x sen 100π y sen ( 2.4π x1010 t − 52.9π z ) [V /m ]Determinar:a) El modo de propagación.b) La frecuencia de corte.c) La frecuencia de operación.d) El número de onda de la guía.e) La permitividad relativa del material dieléctrico que ocupa la guía de ondas.f) La expresión espacio temporal de H y Ing. Alberto Tama Franco FIEC-ESPOL – 2009 – 2S