Spring 2018 - University of Arizona

1 Problem At an operating frequency of 300 MHz, a lossless 50- air-spaced transmission line m in length is terminated with an impedance ZL = (40 + j20) . Find the input impedance. Solution: Given a lossless transmission line, Z0 = 50 , f = 300 MHz, l = m, and ZL = (40 + j20) . Since the line is air filled, up = c and therefore, from Eq. ( ), 2 300 106. = = = 2 rad/m. up 3 108. Since the line is lossless, Eq. ( ) is valid: . ZL + jZ0 tan l (40 + j20) + j50 tan (2 rad/m m). Zin = Z0 = 50. Z0 + jZL tan l 50 + j(40 + j20) tan (2 rad/m m). = 50 [(40 + j20) + j50 0] 50 + j(40 + j20) 0. = (40 + j20) . Problem A 6-m section of 150- lossless line is driven by a source with vg (t) = 5 cos(8 107t 30 ) (V). and Zg = 150 . If the line, which has a relative permittivity r = , is terminated in a load ZL = (150 j50) , determine: (a) on the line. (b) The reflection coefficient at the load. (c) The input impedance. (d) The input voltage Vei . (e) The time-domain input voltage vi (t).

2 (f) Quantities in (a) to (d) using CD Modules or Solution: vg (t) = 5 cos(8 107t 30 ) V, . Veg = 5e j30 V. 150 I~. i Transmission line Zg +. + ~. IL. +. ~. Vg ~. Vi Zin Z0 = 150 . ~. VL ZL (150-j50) . - - - Generator l=6m Load z = -l z=0. ~.. Ii Zg +. +. ~ ~ Zin Vg Vi - - Figure : Circuit for Problem (a). c 3 108. up = = = 2 108 (m/s), r up 2 up 2 2 108. = = = = 5 m, f 8 107. 8 107. = = = (rad/m), up 2 108. l = 6 = (rad). Since this exceeds 2 (rad), we can subtract 2 , which leaves a remainder l = . (rad). ZL Z0 150 j50 150 j50 . (b) = = = = e . ZL + Z0 150 j50 + 150 300 j50. (c).. ZL + jZ0 tan l Zin = Z0. Z0 + jZL tan l . (150 j50) + j150 tan( ). = 150 = ( + ) . 150 + j(150 j50) tan( ). (d).. e Veg Zin 5e j30 ( + ). Vi = =. Zg + Zin 150 + + . j30 + = 5e + . = 5e j30 e = e (V). (e). vi (t) = Re[Vei e j t ] = Re[ e e j t ] = cos(8 107t ) V.. Problem A generator with Veg = 300 V and Zg = 50 is connected to a load ZL = 75 through a 50- lossless line of length l =.

3 (a) Compute Zin , the input impedance of the line at the generator end. (b) Compute Iei and Vei . (c) Compute the time-average power delivered to the line, Pin = 21 Re[Vei Iei ]. (d) Compute VeL , IeL , and the time-average power delivered to the load, PL = 21 Re[VeL IeL ]. How does Pin compare to PL ? Explain. (e) Compute the time-average power delivered by the generator, Pg , and the time- average power dissipated in Zg . Is conservation of power satisfied? Solution: 50 . Transmission line +. ~ 75 . Vg Zin Z0 = 50 . - Generator l = Load z = -l z=0. ~.. Ii Zg +. +. ~ ~ Zin Vg Vi - - Figure : Circuit for Problem (a). 2 . l = = 54 , .. ZL + jZ0 tan l 75 + j50 tan 54 . Zin = Z0 = 50 = ( ) . Z0 + jZL tan l 50 + j75 tan 54 . (b). Veg 300 . Iei = = = e (A), Zg + Zin 50 + ( ).. Vei = Iei Zin = e ( ) = e (V). (c). 1 1 . Pin = Re[Vei Iei ] = Re[ e e ]. 2 2. = cos( ) = 216 (W). 2. (d). ZL Z0 75 50. = = = , ZL + Z0 75 + 50.. 1 e j54 . V0+ = Vei.

4 = = 150e (V), e + e j l j l j54 j54. e + e . VeL = V (1 + ) = 150e +. 0. j54. (1 + ) = 180e j54 (V), . V0+ 150e j54 . IeL = (1 ) = (1 ) = e j54 (A), Z0 50. 1 1 . PL = Re[VeL IeL ] = Re[180e j54 e j54 ] = 216 (W). 2 2. PL = Pin , which is as expected because the line is lossless; power input to the line ends up in the load. (e). Power delivered by generator: 1 1 . Pg = Re[Veg Iei ] = Re[300 e ] = 486 cos( ) = (W). 2 2. Power dissipated in Zg : 1 1 1 1. PZg = Re[IeiVeZg ] = Re[Iei Iei Zg ] = |Iei |2 Zg = ( )2 50 = (W). 2 2 2 2. Note 1: Pg = PZg + Pin = W. Problem The circuit shown in Fig. consists of a 100- lossless transmission line terminated in a load with ZL = (50 + j100) . If the peak value of the load voltage was measured to be |VeL | = 12 V, determine: (a) the time-average power dissipated in the load, (b) the time-average power incident on the line, (c) the time-average power reflected by the load. Rg +. ~ Z0 = 100 ZL = (50 + j100).

5 Vg . Figure : Circuit for Problem Solution: (a). ZL Z0 50 + j100 100 50 + j100 . = = = = . ZL + Z0 50 + j100 + 100 150 + j100. The time average power dissipated in the load is: 1. Pav = |IeL |2 RL. 2. 2. 1 VeL . = RL. 2 ZL . 1 |VeL |2 1 50. = 2. RL = 122 2 = W. 2 |ZL | 2 50 + 1002. (b). i Pav = Pav (1 | |2 ). Hence, i Pav Pav = 2. = = W. 1 | | 1 (c). r Pav = | |2 Pav i = ( )2 = W. Problem A 200- transmission line is to be matched to a computer terminal with ZL = (50 j25) by inserting an appropriate reactance in parallel with the line. If f = 800 MHz and r = 4, determine the location nearest to the load at which inserting: (a) A capacitor can achieve the required matching, and the value of the capacitor. (b) An inductor can achieve the required matching, and the value of the inductor. Solution: (a) After entering the specified values for ZL and Z0 into Module , we have zL. represented by the red dot in Fig. (a), and yL represented by the blue dot.

6 By moving the cursor a distance d = , the blue dot arrives at the intersection point between the SWR circle and the S = 1 circle. At that point y(d) = To cancel the imaginary part, we need to add a reactive element whose admittance is positive, such as a capacitor. That is: C = ( ) Y0. = = = 10 3 , Z0 200. which leads to 10 3. C= = 10 12 F. 2 8 108. Figure (a). (b) Repeating the procedure for the second intersection point [Fig. (b)] leads to y(d) = + , at d2 = . To cancel the imaginary part, we add an inductor in parallel such that 1 = , L 200. from which we obtain 200. L= = 10 8 H. 2 8 108. Figure (b). Problem A voltage generator with vg (t) = 5 cos(2 109t) V. and internal impedance Zg = 50 is connected to a 50- lossless air-spaced transmission line. The line length is 5 cm and the line is terminated in a load with impedance ZL = (100 j100) . Determine: (a) at the load. (b) Zin at the input to the transmission line. (c) The input voltage Vei and input current I i.

7 (d) The quantities in (a) (c) using CD Modules or Solution: (a) From Eq. ( ), ZL Z0 (100 j100) 50 . = = = . ZL + Z0 (100 j100) + 50. (b) All formulae for Zin require knowledge of = /up . Since the line is an air line, up = c, and from the expression for vg (t) we conclude = 2 109 rad/s. Therefore 2 109 rad/s 20 . = = rad/m. 3 108 m/s 3. Then, using Eq. ( ), . ZL + jZ0 tan l Zin = Z0. Z0 + jZL tan l " #. (100 j100) + j50 tan 203 rad/m 5 cm = 50 . 50 + j(100 j100) tan 203 rad/m 5 cm " #. (100 j100) + j50 tan 3 rad = 50 = ( ) . 50 + j(100 j100) tan 3 rad . (c) In phasor domain, Veg = 5 Ve j0 . From Eq. ( ), Veg Zin 5 ( ) . Vei = = = (V), Zg + Zin 50 + ( ). and also from Eq. ( ), . Vei . Iei = = = (mA). Zin ( ). Problem Two half-wave dipole antennas, each with an impedance of 75 , are connected in parallel through a pair of transmission lines, and the combination is connected to a feed transmission line, as shown in Fig.. 75 . (Antenna).

8 Zin1. Zin Zin2. 75 .. (Antenna). Figure : Circuit for Problem All lines are 50 and lossless. (a) Calculate Zin1 , the input impedance of the antenna-terminated line, at the parallel juncture. (b) Combine Zin1 and Zin2 in parallel to obtain ZL , the effective load impedance of the feedline. (c) Calculate Zin of the feedline. Solution: (a).. ZL1 + jZ0 tan l1. Zin1 = Z0. Z0 + jZL1 tan l1.. 75 + j50 tan[(2 / )( )]. = 50 = ( ) . 50 + j75 tan[(2 / )( )]. (b). Zin1 Zin2 ( )2. ZL = = = ( ) . Zin1 + Zin2 2( ). (c). l = . Zin ZL'. Figure : (b) Equivalent circuit.. ( ) + j50 tan[(2 / )( )]. Zin = 50 = ( ) . 50 + j( ) tan[(2 / )( )]. Problem A 75- resistive load is preceded by a /4 section of a 50- lossless line, which itself is preceded by another /4 section of a 100- line. What is the input impedance? Compare your result with that obtained through two successive applications of CD Module Solution: The input impedance of the /4 section of line closest to the load is found from Eq.

9 ( ): Z2 502. Zin = 0 = = . ZL 75. The input impedance of the line section closest to the load can be considered as the load impedance of the next section of the line. By reapplying Eq. ( ), the next section of /4 line is taken into account: Z02 1002. Zin = = = 300 . ZL Problem If the two-antenna configuration shown in Fig. is connected to a generator with Veg = 250 V and Zg = 50 , how much average power is delivered to each antenna? /2 ZL1 = 75 . (Antenna 1). 50 /2 e2. Lin + A C. 250 V Zin Line 1.. B D Lin e3. Generator /2 ZL2 = 75 . (Antenna 2). Figure : Antenna configuration for Problem Solution: Since line 2 is /2 in length, the input impedance is the same as ZL1 = 75 . The same is true for line 3. At junction C D, we now have two 75- . impedances in parallel, whose combination is 75/2 = . Line 1 is /2 long. Hence at A C, input impedance of line 1 is , and Veg 250. Iei = = = (A), Zg + Zin 50 + 1 1 ( )2 Pin = Re[IeiVei ] = Re[Iei Iei Zein.]

10 ]= = (W). 2 2 2. This is divided equally between the two antennas. Hence, each antenna receives 2 = (W). Problem A 25- antenna is connected to a 75- lossless transmission line. Reflections back toward the generator can be eliminated by placing a shunt impedance Z at a distance l from the load (Fig. ). Determine the values of Z. and l. l=? B A. Z0 = 75 Z=? ZL = 25 . Figure : Circuit for Problem Solution: . 90. 100 80 9. 6. 70 1 110 4. 8. 7. ) .42 /Yo 0 60 3. 0 12 (+jB. CE 7 AN 8. PT 3 SCE SU 2. 0 VE 50. 13 TI. CI. 06. 0. PA. 19. 0. CA. 44. 0. 31. 0. R. ,O 4. 0. o). /Z. 0. 40. 5. 14. jX. (+. 5. T. EN. N. PO. SWR Circle 4. M. CO. 6. 0. 1. 30. 15. 9. CE. >. R . AN. TO. CT. ERA. EA. ER. GEN. 160. TIV. 20. 8. 0. ARD. UC. IND. TOW. ANG. GTHS. 10. LE OF. 170. > WAVELEN. REFLECTION. 20. 50.. 10. 20. 50. A B. 180. COEFFICIENT IN. 50. RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo). D LOAD <. 20. 0. OWAR. DEGR. -17. 10. EES. HS T.