PE Exam Question Help

[gray.one]

I disagree with the sample problem solution in the book.

The losses will be directly proportional to length of the line. The terms inside the log in the definition of db should be multiplied by 10. The db should not be multiplied by 10.

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[gray.one]

Never mind. The book is right, and I can't read.


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[Designer69]

holy crap the PE is going to be easy. Thanks for posting this, I was thinking it would be impossible questions.

you can pick D without even thinking about it.

 

[gray.one]

I'm back to thinking the book's answer is wrong. Unless losses increase exponentially with length.


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[Smart $]

I'm back to thinking the book's answer is wrong.

Don't do that.

Each 100m has 0.95dB loss.
1km is 10 times 100m.
10 times 0.95dB loss totals 9.5dB loss over 1km of transmission line..

 

[Ingenieur]

Doesn't make sense
-0.95 dB
loss = 10 log Pout/Pin
or P ratio 0.8035
about 19.65% loss in 100 m

-9.5 dB
P ratio 0.1122 or 88.78% loss in 1000 m

100 is 2% / m
1000 is 0.89% / m
10 times the length but only 88.78/19.65 = 4.5 the losses

the losses are not linear
if you lost 20% every 100 m it would be 0 at 500
so all you can do is mupltiply it out
but they represent it as linear since units are dB/m
bad question

 

[gray.one]

Just multiplying means that losses increase exponentially with length. That does make sense.


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[gray.one]

-0.95 dB

Question didn't have a negative.




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[gray.one]

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[Ingenieur]

Question didn't have a negative.




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it's assumed since it is a loss
but works out either way
P in / P out gives positive, ie, > 1 typically associated with a gain not loss
Or
P out / P in < 1 so loss and gives a neg value , this is the std with reference input on bottom

 

[Ingenieur]

Just multiplying means that losses increase exponentially with length. That does make sense.


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As in not linear
decreaes exponential or log
20% in the first 100 m
70% in the next 900 m

first 100 20%
at 200 35% total
so second 100 15%
third 100 13% more

 

[Smart $]

Just multiplying means that losses increase exponentially with length.

In a sense. It's more that the 'losses' are compounded...

A 0.95dB loss is a drop to 89.6396%. If you take that % to the tenth power (for ten segments) you get 33.4865%... which is equivalent to -9.5dB.

 

[GoldDigger]

In a sense. It's more that the 'losses' are compounded...

A 0.95dB loss is a drop to 89.6396%. If you take that % to the tenth power (for ten segments) you get 33.4865%... which is equivalent to -9.5dB.

When you have two lengths of cable in series you multiply the two loss ratios. Since dB is alog og loss ratio you add the logs to get the log of the product.
Slightly confusing is that to add .95 ten times you multiply it by ten.

 

[steve66]

The loss is .95 dB per 100 M. The total loss is 10 * .95 = 9.5 dB. Its that easy.

Harder is determining what you have left at the end of the line. That would take an actual calculation, but every -3 dB is half power.

So -6 dB is 1/4 power, and -9 dB is 1/8 power.

So you have slightly less than 1/8th the power at the output than you do at the input.

 

[Smart $]

...every -3 dB is half power.

So -6 dB is 1/4 power, and -9 dB is 1/8 power.

So you have slightly less than 1/8th the power at the output than you do at the input.

While you are probably correct that the dB unit used in the book is in reference to power, one has to be aware of whether the unit references power or 'signal' (aka amplitude). The former is half the latter: -3dB_POWER = -6dB_SIGNAL = 50%. See: https://en.wikipedia.org/wiki/Decibel

I was using signal percentage in my earlier post. In terms of power, the percentages in that post would be:

-0.95dB = 80.3526 %
-9.5dB = 11.2202 %​

 

[steve66]

While you are probably correct that the dB unit used in the book is in reference to power, one has to be aware of whether the unit references power or 'signal' (aka amplitude). The former is half the latter: -3dB_POWER = -6dB_SIGNAL = 50%. See: https://en.wikipedia.org/wiki/Decibel

I was using signal percentage in my earlier post. In terms of power, the percentages in that post would be:

-0.95dB = 80.3526 %
-9.5dB = 11.2202 %​

I almost mentioned that, but since the equations in the notes show power, I assumed the question was about power.

If its about voltage or current, you are correct. If I remember correctly, the equations would be 20 log (voltage ratio) or 20 log (current ratio), and 6 dB would be half the original signal, so -9.5 dB would be somewhere between 1/2 the signal and 1/4 the signal at the far end. (

 

[gray.one]

Thanks for the responses. I was having trouble seeing what caused the losses to not be linear while thinking of the simplified series and shunt impedance model.

I've come to the conclusion that shunt losses diminish as the line gets longer because the voltage decreases with the line length.


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[gray.one]

Doesn't make sense
-0.95 dB
loss = 10 log Pout/Pin
or P ratio 0.8035
about 19.65% loss in 100 m

-9.5 dB
P ratio 0.1122 or 88.78% loss in 1000 m

100 is 2% / m
1000 is 0.89% / m
10 times the length but only 88.78/19.65 = 4.5 the losses

the losses are not linear
if you lost 20% every 100 m it would be 0 at 500
so all you can do is mupltiply it out
but they represent it as linear since units are dB/m
bad question

Thanks, the negative makes sense and I completely missed it.


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