Hi there,



I have worked though the problem state below can you tell me if i am correct in the working i have done to find the answers. Also the red writing i am unsure how to do this.


Thanks for your help




A Transmitted with an output of 250W operates at a frequency of 16MHz. it feeds an antenna with an impedance of 50 ohm though a 50m length of coaxial cable of R213.

The characteristic impedance is 50 ohm and is solid polyethylene dialectic. Velocity factor is 0.67. Power rating of suitable but due to the attenuation produced by the cable, only 183W of signal reaches the antenna.

Attenuation in dB of the cable

dB = 10 Log P1/P2

=10 Log 250/185
=1.307dB

The Free Space wavelength of the signal radiatied by the antenna
f=frequency,λFS=Wavelenght FreeSpaceC=fλFS
λFS=C/f
=3x10^8/16x10^6
= 18.75m


Velocity of the signal as it propagates down the cable to the antenna
V=velocity, f=frequency,λFS=Wavelenght FreeSpaceV=f λFS
=16x10^6 x 18.75m

= 300000km/s
I dont think I have done this part correctly



The wavelength of the signal as is propagates down the cable

Unsure what to use here to find out the answer. The cable is 50m and freq is 16MHz do you have to divide this?


How long is the electrical wavelength
λp=Electrical wave lenght, V=Velocity,λFS=Wavelenght FreeSpace
λp = VfλFS
=0.67 x 18.75m
=12.56m

Time is takes for the signal to progate down this cable
t=time, d=distance Vp=velocity

t=d/Vp
=50/3x10^8
=0.166us

What is the rate at which heat is produced in the cable
P=power, J=Joules, s=seconds
P=J/s
= 250 x 0.000000166
=0.000041666J/s

 

I'll give you a couple of clues:

What's 250 - 183?

Velocity factor is independent of frequency.

 

It's not clear if the antenna power is 183 or 185W. You used 185W for the cable loss calculation.

If the cable velocity factor is less than 1 then the wave propagates along the cable at less than the speed of light.

Isn't the power dissipation in the cable just the power loss between the transmitter and the antenna? You seem to have calculated that all the transmitter output of 250W is lost in the cable.

What's the idea with the 0.000000166 value? Isn't 1 Watt= 1 Joule / second?

 

TNK you are corret I made a Typo and the antenna should be 185W. I didnt pick up on that before sending.



For the power/heat produced I used

Power = Joules/Second

I new the Power = 250W
and Time (from the above question) = 0.0166us

This is how I got the answer of 0.00004166J/s


I will redo the question with the help you have provided and post again soon.
Thanks for your help

 

TNK you are corret I made a Typo and the antenna should be 185W. I didnt pick up on that before sending.



For the power/heat produced I used

Power = Joules/Second

I new the Power = 250W
and Time (from the above question) = 0.0166us

This is how I got the answer of 0.00004166J/s


I will redo the question with the help you have provided and post again soon.
Thanks for your help

I think you've missed the point on a couple of matters.

The 250W input isn't dissipated in the cable - rather 250W-185W=65W.

I see you used the period of the 16MHz signal as the time base. There's no reason why you would do that - the definition of power is the work done (heat lost) per unit time - Hence 1 watt = 1 joule/sec.

 

Here is the question that i have redone, Hopefully have taken on everything you have said and this one is correct.



Thanks



Attenuation in dB of the cable

dB = 10 Log P1/P2

=10 Log 250/185
=1.307dB

The Free Space wavelength of the signal radiated by the antenna
f=frequency, λfs=Wavelength Free Space C=fλfs


λfs=C/f
=3x10^8/16x10^6
= 18.75m

Velocity of the signal as it propagates down the cable to the antenna
Vp=velocity, d= distance, t=time


Vp=d/t
=50/0.1666us
= 0.3km/s

The wavelength of the signal as is propagates down the cable
k=velocity factor, C=3x10^8, f=frequency

λ = kC/f
=0.67 x 3x10^8/16x10^6
=12.25m

How long is the cable in electrical wavelengths
λp=Electrical wave length, V=Velocity, λFS=Wavelength Free Space


λp = VfλFS
=0.67 x 50m
=33.7m

Time is takes for the signal to propagate down this cable
t=time, d=distance Vp=velocity

t=d/Vp
=50/3x10^8
=0.166us

What is the rate at which heat is produced in the cable
P=power, J=Joules, s=seconds


P=Ptransmitter-Pantenna
= 250 -186 = 65W
=65J/s




Thanks for your help

 

Attenuation in dB of the cable

dB = 10 Log P1/P2

=10 Log 250/185
=1.307dB

The Free Space wavelength of the signal radiated by the antenna
f=frequency, λfs=Wavelength Free Space C=fλfs


λfs=C/f
=3x10^8/16x10^6
= 18.75m

Correct

Velocity of the signal as it propagates down the cable to the antenna
Vp=velocity, d= distance, t=time

Vp=d/t
=50/0.1666us
= 0.3km/s

Incorrect - the cable velocity = free space velocity x cable velocity factor

The wavelength of the signal as is propagates down the cable
k=velocity factor, C=3x10^8, f=frequency

λ = kC/f
=0.67 x 3x10^8/16x10^6
=12.25m

Bit rusty on the calculator - based on your numbers your answer should be 12.56m - right idea.

How long is the cable in electrical wavelengths
λp=Electrical wave length, V=Velocity, λFS=Wavelength Free Space


λp = VfλFS
=0.67 x 50m
=33.7m

Incorrect - the electrical wavelength λ in meters is λ=cable_velocity/frequency

So how many of these electrical wavelengths make up 50m? You answer will be a dimensionless value.

Time is takes for the signal to propagate down this cable
t=time, d=distance Vp=velocity

t=d/Vp
=50/3x10^8
=0.166us

No - at reduced propagation velocity (the cable velocity) the wave takes longer to travel the 50m than it would at free space velocity.

What is the rate at which heat is produced in the cable
P=power, J=Joules, s=seconds

P=Ptransmitter-Pantenna
= 250 -186 = 65W
=65J/s

Correct

 

Cable Velocity = free space velocity x cable velocity factor

When you say 'free space velocity' does this mean 3x10^8.

Thanks for your help

 

That's right.