Question

10. At 2 MHz a cable has an attenuation constant of alpha=0.05 Nepers/meter. At the transmitting end, I connect the line to a 5 volt, 2 MHz, sine wave. Answer all of the following questions A. What is the attenuation constant of the cable in dB/km? B. How far from the source do I have to go, before I notice the amplitude of the signal has dropped to 1 volt? C. At what velocity does the signal move down the cable? D. At what length does the cable change from being electrically short, to electrically long?

Answer 1

a Giren x = .6 Nepers / meter frequency af 22 MHZ L iput voltage = 6 vells gg I Neper z 8.6858 -- dB so the in .05 Neperz .05X 8.6858dB = .439 dB 05 Neper/mate a .434 dB Z 434 dB meter Km

so we have attenuation of 434 dB Capproximate] - Kn (6) on moving re meter the voltage drop from 5 to 1 Voltage drop in dB dropse 20 lag (5] = 20 ley 5 z 13:42

drop ? 13.97 dB we know our lire has a drop of .434 dB meter 0 4 34x nz 13.97 dB ez 13.97 e 32021 meters • 437 On mening 32.21 meter voltage drops from or to ir

C Taking 1x1 = 18) 2 405 We know the velocity of wave is given by 3 е 2T + = 21 2x100 rad sec vzw 229182X100 = 2.51x100 meter dec a(d) To find out lenght at which the calele change from electrically long to short wel first calculute the Wardlength of da vy zvls = 2.518108 a 125.66 meter 28166 when the lenght of cable is higher than 10% the the cable is electrically long and short when its lenght is less than 10% of a so

at looXo of de 125.66 X 10% 12.56 meters So at 12.56 meters short into electrically calele changes from electrically the long
in the answers I used approximate values for easier calculations