Question regarding attenuation

Thread Starter

kiwi00

Joined Jul 30, 2015
7
Hi! I was wondering what causes more attenuation, the guided media (ex. coaxial cables, twisted pairs, fiber optics) or the unguided media (ex. microwave, radiowave, infrared). They both have differences as they have different types, but as whole, what do you think gives more attenuation than the other?

It might get discussed tomorrow on class but I wanna get a head start on this. Thanks! :)
 

wayneh

Joined Sep 9, 2010
17,498
Broadcast energy suffers from the inverse square law. The total energy isn't really attenuated, but the amount of energy seen at the receiver sure is. A cable avoids that, but introduces other mechanisms for energy loss.

Your question made me picture the probe passing Pluto. Would a cable to Pluto improve the signal, or destroy it?
 

Papabravo

Joined Feb 24, 2006
21,225
Lemme see....7.5e12 meters * 0.0842 Ohms/meter = 631500 Megohms

If you started with 120 VAC you have about 100 picovolts at the load
 

Thread Starter

kiwi00

Joined Jul 30, 2015
7
@wayneh
The other mechanisms for energy loss, is it the electromagnetic field and crosstalk? So does this mean unguided media produces more attenuation (since distance also is one of the main problems of attenuation and it seems to have more range)?

@Papabravo
Umm sorry we haven't studied formulas yet so I'm confused lol

I'm a newbie student so I don't know much yet umm could you explain a bit more simpler? But don't worry I'm eager to learn :D
 

alfacliff

Joined Dec 13, 2013
2,458
the use of transmission lines is different from radiating signals. the transmission lines are more for controling the signal, keeping it inside the line, than when radiating. also, most transmssion lines prevent other signals from getting into the signal, causing crosstalk, and distortion. free space transmission would probably have less loss than even the best transmission line, if formed into a "beam", but would be more susceptable to interfearance.
 

Thread Starter

kiwi00

Joined Jul 30, 2015
7
The free space transmissions radiate in a lot of directions, varying in performance on their patterns. Does this affect the attenuation? Will it still have less loss if it wasn't formed into a beam?
 

nsaspook

Joined Aug 27, 2009
13,270
The free space transmissions radiate in a lot of directions, varying in performance on their patterns. Does this affect the attenuation? Will it still have less loss if it wasn't formed into a beam?
You have path loss (or path attenuation) for a transmitted signal as the reduction in power density at the receiver. There is the Free-space_path_loss and other factors like antenna gain in the Friis_transmission_equation.
 

BR-549

Joined Sep 22, 2013
4,928
It also depends on what you are transferring.

With information, we only need a speck(integral) of the source to be successful.

So even though emission always has non recoverable loss, it's quite successful for information exchange.

Power transfer is different. We have to transfer mass(current). We need it all, not just a speck or sample.

For this we must use conduction. We have loss in conduction, but it can be quantified and managed, plus some of the loss is recovered.

Every emission since the beginning has reduced the amount of cosmos mass. This can never be recovered.

The cosmos cools and gravity decays. Continuously.
 

BR-549

Joined Sep 22, 2013
4,928
The premise of acceleration is based on 13 billion old film. I don't agree with that time scale, but to me it means that in the beginning there was acceleration. And there are other things that effect redshift. We can only see the past, not the present.

I couldn't help myself, I'll be quiet.
 

Papabravo

Joined Feb 24, 2006
21,225
@wayneh
The other mechanisms for energy loss, is it the electromagnetic field and crosstalk? So does this mean unguided media produces more attenuation (since distance also is one of the main problems of attenuation and it seems to have more range)?

@Papabravo
Umm sorry we haven't studied formulas yet so I'm confused lol

I'm a newbie student so I don't know much yet umm could you explain a bit more simpler? But don't worry I'm eager to learn :D
No formula involved -- it is just distance times resistance per unit length.
You must admit the 631,500 Megohms is fearsomely large resistance.
How big a voltage do you think you might need to apply to get something to the other end of a cable that long?
 
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