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\prob{18}
In transmitting high frequency signals on a coaxial cable, it is important that the cable be terminated at an end with its characteristic impedance in order to avoid

  1. leackage of the signal out of the cable
  2. overheating of the cable
  3. reflection of signal from the terminated end of the cable
  4. attenuation of the signal propagating in the cable
  5. production of image currents in the other conductor

Lab Methods}Coax Cable

Elimination time. The first-pass question to answer is why is it important that a coax cable be terminated at an end:
(A) Perhaps...

(B) Probably not. Terminating the cable at an end would not help heat dissipation and thus should not prevent overheating.

(C) Perhaps...

(D) Probably not, since termination should attenuate the signal rather than to prevent it.

(E) Probably not, since image currents should be canceled by the outer sheath.



Choices (A) and (C) remain. Now, use the second fact supplied by ETS. The cable should be terminated by its characteristic impedance. Characteristic impedance has to do with resonance. Thus, it should prevent reflection of the signal.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
heypete
2010-11-05 15:34:36
Maybe I'm just old, but I remember 10BASE2 coaxial ethernet from my middle school. They had 50-ohm resistors across the end of the cables to prevent packets from being reflected.

While I didn't understand the physics behind it back then ("How can data reflect from the end of the cable? There's nothing there to reflect off of!"), that little tidbit of information stuck with me to this day.

From the Wikipedia entry:

"As was the case with most other high-speed buses, Ethernet segments had to be terminated with a resistor at each end. Each end of the cable had a 50 ohm (Ω) resistor attached. Typically this resistor was built into a male BNC and attached to the last device on the bus. If termination was missing, or if there was a break in the cable, the AC signal on the bus was reflected, rather than dissipated, when it reached the end. This reflected signal was indistinguishable from a collision, and so no communication would be able to take place."
Alternate Solution - Unverified
Comments
heypete
2010-11-05 15:35:34
Oops. Sorry for the double-post.NEC
heypete
2010-11-05 15:34:36
Maybe I'm just old, but I remember 10BASE2 coaxial ethernet from my middle school. They had 50-ohm resistors across the end of the cables to prevent packets from being reflected.

While I didn't understand the physics behind it back then ("How can data reflect from the end of the cable? There's nothing there to reflect off of!"), that little tidbit of information stuck with me to this day.

From the Wikipedia entry:

"As was the case with most other high-speed buses, Ethernet segments had to be terminated with a resistor at each end. Each end of the cable had a 50 ohm (Ω) resistor attached. Typically this resistor was built into a male BNC and attached to the last device on the bus. If termination was missing, or if there was a break in the cable, the AC signal on the bus was reflected, rather than dissipated, when it reached the end. This reflected signal was indistinguishable from a collision, and so no communication would be able to take place."
Alternate Solution - Unverified
GREview
2009-08-26 15:44:27
If one thinks of the coaxial cable as a waveguide, then you can use the analogy that there is an E&M wave propagating through the cable. It's different that your usual E&M wave due to the boundary conditions (see Griffiths E&M Ch 12 or so), but nonetheless it is a wave. And what do all waves do? They reflect and refract in different media.

The last piece of the puzzle is the impedance, which in the context of a this E&M wave can be thought of as an index of refraction. Thus if the wave hits a barrier with index of refraction (impedance) much greater from the index of the material in which it was traveling, then it will mostly reflect. Thus the impedance needs to be as close as possible to the material's characteristic impedance.
NEC
Jeremy
2007-10-30 13:31:19
Interesting note: My EM professor once mentioned that trumpets (and other similar instruments) have flared ends (making the "inside-outside" boundary more gradual) to prevent similar types of reflections. This probably isn't the scenario you visualize when you think of impedance matching, but that's essentially what it is. Pretty cool.NEC

Post A Comment!
You are replying to:
If one thinks of the coaxial cable as a waveguide, then you can use the analogy that there is an E&M wave propagating through the cable. It's different that your usual E&M wave due to the boundary conditions (see Griffiths E&M Ch 12 or so), but nonetheless it is a wave. And what do all waves do? They reflect and refract in different media. The last piece of the puzzle is the impedance, which in the context of a this E&M wave can be thought of as an index of refraction. Thus if the wave hits a barrier with index of refraction (impedance) much greater from the index of the material in which it was traveling, then it will mostly reflect. Thus the impedance needs to be as close as possible to the material's characteristic impedance.

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