This is a problem from Paul Claytons's Introduction to Electromagnetics.

My question is why the cable gain or loss which is mentioned in dB gets added to input power which is dBm.

Do you need to convert cable loss 4.5dB/100ft to dBm/100ft and then subtract from the input power to calculate the output power.

 

The cable gain term is effectively a multiplier which translates to an addition when using logarithms. So no - you don't need to convert this multiplier to dBm.

Consider the situation without using logarithms.

The cable gain (really a loss) is -6.75dB which is an equivalent multiplying factor of 0.21135 (less than unity is a loss)

The cable input power is -30dBm which is 1uW.

The cable output power is 0.21135*1uW = .21135uW which translates to -36.75dBm.

 

Based on your reply I have following questions:

1. Do you mean even if cable loss (dB/ft) can be added directly to power either dBuV or dBmV or dB?

2. in general, if I have amplifier block with some gain lets say 32dB it can be straight away subtracted from output reading to get input reading, is it right?

 

Q1. Yes - however unless the input and output signals share a common impedance base the voltage and power gains differ.

Pgain(dB)=10log(Pout/Pin)
Vgain(dB)=20log(Vout/Vin)

Q2. Yes


Consider an amplifier with fixed Voltage Gain Av=100 fed by an ideal source voltage of Vin=1mV. Say the amp input resistance is 1kΩ and output resistance is zero. The load impedance connected to the output is 100Ω.

Av(dB)=20log(100)=40dB
Vin=1mV=20Log(1000) dBuV=60 dBuV

Vout(dBuV)=40+60=100 dBuV=10^5 uV = 100mV

Pin=1nW=-60dBm
Pout=100uW=-10dBm

Pgain(dB)=-10-(-60)=50dB