. -2 (ZL+jZotanf3l) Zo + jzl J31 _ (60+j20)+j50tan(21trad/mx2.5m))

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1 CHAPTER Section 8-6: Input Impedance Problem 8.17 At an operating frequency of 300 MHz, a loss less air-spaced transmission line in length is terminated with an impedance 4'. = (60+ j20) o.. Find the input impedance. Solution: Given a lossless transmission line, 2D= 50 0., f = 300 MHz, I = , and ZL = (60 + j20) o.. Since the line is air filled, lip = c and therefore, from Eq. (8.38), f3 = ro = 21t X 300 X 106 Up 3 x IOn = 21t rad/m. Since the line is lossless, Eq. (8.69) is valid: Zm (ZL+jZotanf3l) Zo + jzl J31 _ (60+j20)+j50tan(21trad/mx2.5m)) + j( + (21t x 2.5 m) = 50 ( j(60+ j20) + j20) j50 x 0) = (60+ }'20) Q. Problem 8.18 A loss less transmission line of electrical length I = 0.35A IS terminated in a load impedance as shown in Fig (P8.18). Find r,s, and Zn. Figure P8.18: Loaded transmission line. From Eq. (8A9a), Eq. (8.59), r= ZL -Zo = (60+ j30) -100 = 0.307ei132.5 ZL +Zo (60+ j30) S= I + WI = = WI

2 278 CHAPTER 8 From Eq. (8.63) Zin =Zo (ZL Zo+ + jzl jzotanf3l) f31 ~!0O( j30) j(60+ + j j30)!00tan ( 0.35A.») (l1t{ado.35a.) = (64.8- j38.3) Q. Problem 8.19 Show that the input impedance of a quarter-wavelength long loss less line terminated in a short circuit appears as an open circuit. Solution: Zin = Zo (ZL Zo + JZL ~Zotan(31) tanf31. For I =~, f31 = 2;:. ~ =. With ZL = 0, we have Zin = Zo ---- =)00 (jzotanrr./2). (open circuit). Problem 8.20 Show that at the position where the magnitude of the voltage on the line is a maximum the input impedance is purely real. Solution: From Eq. (8.56), lmax = (Sr+2111t)/2f3, so ITom Eq. (8.61), using polar representation for r, which is real, provided 20 is real. Problem 8.21 A voltage generator with vg(t) = 5cos(2rr. x 109t) V and internal impedance Zg = 50 Q is connected to a 50-Q lossless air-spaced transmission line. The line length is 5 cm and it is terminated in a load with impedance ZL = (100- jioo) Q. Find (a) r at the load. (b) Zin at the input to the transmission line. (c) the input voltage Vi and input current~.

3 CHAPTER Section 8-7: Special Cases Problem 8.24 At an operating frequency of200 MHz, it is desired to use a section of a lossless 50-Q transmission line terminated in a short circuit to construct an equivalent load with reactance X = 25 Q. If the phase velocity of the line is 0.75c, what is the shortest possible line length that would exhibit the desired reactance at its input? Solution: (21t rad/cycle) x (200 x 106cycle/s) = 5.59 rad/m. 13= rojup = 0.75 x (3 x 108 m/s) On a lossless short-circuited transmission line, the input impedance is always purely imaginary; Le., Zf; = jxfnc, Solving Eq. (8.68) for the line length, I = ~tan-l 13 (XinC) Zo = 5.59 rad/m I tan-1 (250 Q) Q = (0.464+n1t) 5.59 rad/m rad. ' for which the smallest positive solution is 8.3 em (with n = 0). Problem 8.25 A lossless transmission line is terminated in a short circuit. How long (in wavelengths) its input terminals? should the line be in order for it to appear as an open circuit at Solution: From Eq. (8.68), Zj~= jzotanl31. If 131= (1t/2+ n1t), then Zf~= joo (Q). Hence, This is evident from Figure 8.15(d). I = 21t ~ (~ 2 + n1t) = ~ 4 + n'a. 2. Problem 8.26 The input impedance of a 31-cm-long loss less transmission line of unknown characteristic impedance was measured at 1 MHz. With the line terminated in a short circuit, the measurement yielded an input impedance equivalent to an inductor with inductance of ph, and when the line was open circuited, the measurement yielded an input impedance equivalent to a capacitor with capacitance of 20 pf. Find Zo of the line, the phase velocity, and the relative permittivity of the insulating material. Solution: Now ro = 21tf = 6.28 x 106 rad/s, so Zf; =jrol =j21t x 106 x x 10-6 = jo.804 Q

4 284 CHAPTER 8 and Z~c = IjjoX: = IjU2rc x 106 x 20 X 10-12) = -j8000 Q. From Eq. (8.74), Zo = vzi~z~c = VUO.804 Q)( - j8000 Q) = 80 Q. Using Eq. (8.75), ro rof U = - = ---:---==== p 13 tan-) V/-Z~cjZ':Jc m m 6.28 X 106 x tan-) (±V- jo.804j( - j8000)) 1.95 x 106! (±0.01 +l1rc ) m s, where 11 2: 0 for the plus sign and 11 2: I for the minus sign. For 11 = 0, up = 1.94 X 108 mls = 0.65cand Of = (cjupf = IjO.652 = 2.4. For other values of 11, up is very slow and Er is unreasonably high. Problem 8.27 A 60-Q resistive load is preceded by a Aj 4 section of a 50-Q loss less line, which itself is preceded by another Aj 4 section of a IOO-Qline. What is the input impedance? Solution: The input impedance of the 1../4 section ofline closest to the load is found from Eq. (8.77): z. m--- _ Z~ 502 ZL - 60 = Q. 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. (8.77), the next section of 1../4 line is taken into account: Z2 1 2 Zin = ~ _ 00 ZL = 240 Q. Problem 8.28 A 100-MHz FM broadcast station uses a 300-Q transmission line between the transmitter and a tower-mounted half-wave dipole antenna. The antenna impedance is 73 Q. You are asked to design a quarter-wave transformer to match the antenna to the line. (a) Determine the electrical length and characteristic impedance of the quarterwave section. (b) If the quarter-wave section is a two-wire line with d = 2.5 em, and the spacing between the wires is made of polystyrene with &- = 2.6, determine the physical length of the quarter-wave section and the radius of the two wire conductors.

5 CHAPTER ZL,=75 Q (Antenna I) Generator B ZL,= 75 Q (Antenna 2) Figure P8.32: Antenna configuration for Problem This is divided equally between the two antennas. Hence, each antenna receives 15~.37 = (W). Problem 8.33 For the circuit shown in Fig (P8.33), calculate the average incident power, the average reflected power, and the average power transmitted into the infinite line. The A/2 line is loss less and the infinitely long line is slightly lossy. (Hint: The input impedance of an infinitely long line is equal to its characteristic impedance so long as a =I- 0.) Solution: Considering the semi-infinite transmission line as equivalent to a load (since all power sent down the line is lost to the rest of the circuit), L[ = ZI = Since the feed line is A./2 in length, Eq. (8.76) gives.4n = ZL = and (31 = (2n/A)(A./2) = n, so e±j(31 = -1. From Eq. (8.49a), r = ZL-Zo = =.!.. ZL+Zo

6 292 CHAPTER 8. +~~ W=1_. _ 2V~_ a 20=50n ~ Qf 1 21 = 100n P~v-!-p~v I pr_1 av! Figure P8.33: Line terminated in an infinite line. Also, converting the generator to a phasor gives Vg = 2ejOO (V). Plugging all these results into Eq. (8.66),. v.+ o = ( Zg+Zin V.Z;, ) ( ej\3/+re-jj31 I ) ~ C x ) ( (-I)+t(-I) I ) = 1~180o = -1 (V). From Eqs. (8.82), (8.83), and (8.84), i IVo+12 l1ejl80012 Pay = 2Z0 = ~ _n = 10.0 mw, r 2 i 1 Pay = -In Pay = -"3 X 10 mw = P~v= Pay= P~y+P~v = 10.0 mw -1.1 mw, mw = 8.9 mw. Problem 8.34 An antenna with a load impedance Li = (75 +j25) Q is connected to a transmitter through a 50-Q lossless transmission line. If under matched conditions (50-Q load), the transmitter can deliver low to the load, how much power does it deliver to the antenna? Assume Zg = Zoo