Calculation in trasmission lines

Discussion in 'Homework Help' started by Kayne, May 3, 2010.

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    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
     
  2. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    I'll give you a couple of clues:

    What's 250 - 183?

    Velocity factor is independent of frequency.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    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?
     
  4. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    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
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    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.
     
    Kayne likes this.
  6. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    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
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Correct

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

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

    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.

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

    Correct
     
  8. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    When you say 'free space velocity' does this mean 3x10^8.

    Thanks for your help
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That's right.
     
    Kayne likes this.
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