Finding Voltages and Power?

[Newbie 2 Electronics]

Hi All

I have been given the following question from University and I am abit stuck
on it, if anyone could point me in the right direction, please shout....

Q) The input power of a line is Pi = 18mW and the loss Ndb = -16dB. If the
resistance of the line is 50?, calculate:

1) The input voltage Vi=?
2) The output voltage Vo=?
3) The output power Po=?

I have found the equation of how to calculate the loss: -

Loss Ndb= = 10log Po/Pi = dB.

I have tried looking through a number of books, but I cant seem to get what
is going on. The lecturer has said that he will go through it next week, but
I would like to be able to understand it before this.

Any ideas or hints would be good.

Thanks. Geoff

 

[Joe McElvenney]

Hi,

I have been given the following question from University and I am abit stuck
on it, if anyone could point me in the right direction, please shout....

Q) The input power of a line is Pi = 18mW and the loss Ndb = -16dB. If the
resistance of the line is 50?, calculate:

1) The input voltage Vi=?
2) The output voltage Vo=?
3) The output power Po=?

I have found the equation of how to calculate the loss: -

Loss Ndb= = 10log Po/Pi = dB.

I have tried looking through a number of books, but I cant seem to get what
is going on. The lecturer has said that he will go through it next week, but
I would like to be able to understand it before this.

You haven't stated the line's load impedance which in the real world
affects all of these things, so I will assume that it is also 50 ohms.

1) This is just Ohm's Law as the line is acting like a 50 ohm load resistor
as far as the input power is concerned. So, if you know the relationship
between power, voltage and resistance, it is straightforward.

2) Transform the formula for voltage loss/gain in a matched system, which
is 20log(Vo/Vi) to put Vo on the left. It will come out as negative for a loss
but that need not concern you here.

3) Knowing Vo, it is Ohm's Law once again to find Po.


If you have any problems with this - just email.


Cheers - Joe

 

[Newbie 2 Electronics]

Thank you for your help, I will sit down with this question tomorrow and try
and see if I can make any more sense of it!

Thanks again

Geoff