A 0 dB TEM wave is to be transmitted through a loss coaxial

A 0 dB TEM wave is to be transmitted through a loss coaxial cable. The attenuation per meter is 0.3 dB. If the cable is 10 m long, determine the following: Power (in W) at the input of the cable. Total attenuation in dB due to the 10 m long cable. Power (in dBW) at the end of the cable. Power (in W) at the end of the cable. Attenuation constant in Np/m.

Solution

a) Given : 0dBW TEM wave which is power of input signal.

10logP =0 dBW

P_in=10⁰=1 Watt [answer]

b)

For 10 m long cable ,total attenuation will be A=0.3dB/meter*10 meter = 3 dB [answer]

c)

Power at the end of cable P_out = P_in(dB) - A(dB) =0 dBW- 3 dB = -3 dBW [ANSWER]

d)

Power at the end of cable P_out=-3dBW

10logP_out= - 3

P_out=10^-3 W= 1 mW [ANSWER]

e)

Attenuatino constant in dB/m :A= 0.3 dB /m

as we know 1Np = 8.68 dB

A= 0.3 dB /m =0.3/8.68 = 0.0346 Np/m [answer]