15.1 WAVES, TRANSMISSION LINES, WAVEGUIDES, AND ANTENNA FUNDAMENTALS 675
(a) For a rectangular air-filled waveguide witha=4.8 cm andb=2.4 cm, compute the
cutoff frequency. If the operating frequency is 4 GHz, find the waveguide’s characteristic
impedance.
(b) Calculate the diameter of an air-filled circular waveguide that will have a lower cutoff
frequency of 10 GHz.Solution(a)fc=c
2 a=3 × 108
2 × 4. 8 × 10 −^2= 3 .125 GHzZ ̄ 0 =R 0 =√^377
1 −( 3. 125 / 4 )^2=377
√
1 − 0. 8839=377
0. 3407= 1106. 5 (b)a=0. 293 c
fc=0. 293 × 3 × 108
10 × 109= 0 .88 cmDiameter= 2 a=1.76 cmNow referring to Figure 15.1.3, assuming matched and distortionless conditions, with signal
voltagex(t) across the line input, the resulting output voltagey(t) is given by
y(t)=Kx(t−td) (15.1.10)
whereKis the attenuation factor (less than unity) due to ohmic heating in the line dissipating part
of the input signal energy andtdis the delay time. Working with the average signal powersPin
andPout, thetransmission loss Lis defined as the power ratio
L=Pin/Pout (15.1.11)
whereL= 1 /K^2 >1 andPout<Pin. Regardless of the type of transmission line, the transmission
loss increases exponentially with distancelsuch that
L= 10 αl/^10 ) (15.1.12)
whereαis theattenuation coefficientof the line in decibels per unit length. ExpressingLin dB,
similar to power gain, we obtainLdB=10 logPin
Pout=αl (15.1.13)Typical values ofαrange from 0.05 to 100 dB/km, depending on the type of transmission line
and the frequency. Rewriting Equation (15.1.13) as
Pout
Pin= 10 −LdB/^10 = 10 −αl/^10 (15.1.14)Equation (15.1.14) reveals thatPoutwill be one-tenth ofPinwhenl= 10 /α, showing thereby
how rapidly the output power falls off as the distance increases.
The transmission loss can be overcome with the aid of one or more amplifiers connected
in cascade with the transmission line. Figure 15.1.4 shows such a system with a preamplifier
(transmitting amplifier) at the source, a receiving amplifier at the destination, and a repeater