# Power output

Discussion in 'Homework Help' started by GirishC, Jan 29, 2010.

1. ### GirishC Thread Starter Active Member

Jan 23, 2009
58
0
This has equivalent circuit as a voltage source connectoed to load via 200ft cable.

First we need to calculate the input voltage based on the given load output.
output voltage.

Step 1:

Po = 56.5dBuV = 20*log(V0/1exp-6)

solving, we get

Vo = 668.34uV

therefore Po in dB is

Po = 10*log(668.34exp-6/1exp-6) = 28.25dBuW

Step 2:
The cable loss can be calculated as

Cable Loss = 4.5 * 200/100 = 9dB

Step 3:
This is power in the load of 50 Ohms.

The power at the input of the load is give by

Po = cable loss + Pin

Pin = 28.25 + 9 = 37.25dBuW = 10*log(Pin/1exp-6)

Pin = 5.31mW

Vin = sqrt (Pin*50) = 515.21mV

Step4:
No with this input, reaplce 50 Ohm load and cable by 100 Ohms load
and calculate Vo

Vo = (100 * Vin)/150 = 343.47mV
Vo = 110.72dBuV

I do not understand what is wrong with the procedure I am following.

The answer is given in the bracket in problem statement.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Remember dBuV is a voltage expression - not a power expression.

I would proceed as follows

(1) Work out the voltage at the cable input [Zo=50Ω].

Voltage at cable input = 56.5dBuV + 9dB= 65.5 dBuV

(2) Convert 65.5 dBuV to absolute voltage. 65.5dBuV <=> U uV.

(3) Remember this absolute voltage is half the source voltage since the source has 50Ω output impedance.

(4) Source voltage = 2*U volts

(5) Remove the cable and place the 100Ω load.

(6) V_100Ω_load = 2*U*(100/150) - it's a voltage divider.

(7) Convert Vload to dBuV - I get 67.999 dBuV rather than 67.98 dBuV

You've made several mistakes along the way.

For instance .....

You calculated Po with the cable in place as 28.25dBuW

You (correctly) have Vo = 668.34uV = 668.34e-6 Volts

So Po=Vo^2/50=(668.34e-6)^2/50 Watt = 8.9337 nW = 8.9337e-3 uW

Po=8.9337e-3 uW <=> -20.49 dBuW

In reality you don't need to go to the trouble of calculating powers - you can do it all in volts. But however you do it, take note of step (4) above in my method.

3. ### GirishC Thread Starter Active Member

Jan 23, 2009
58
0
Thanks a lot...I really did not know where I was going wrong.

The reason I converted to power is to make sure I do not mess with voltage ration in dB...though it is essentially a power.

4. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,316
1,109
0 dB/uV is 1 uV.
0 dBm is 1 mW. Converting that to a voltage it is (0.001 * 50)^.5 or 223.6 mV. Converting 0 dBm to dB/uV is 20 log(223.6 mV / 1 uV) is approximately 106.9897 dB/uV or rounded off to 107 dB/uV.

The question asked "Determine the reading of the meter on the source dBm."

65.5 dB/uV - 107 dB/uV = -41.5 dBm

The cable length, attenuation, and resultant dB/uV were distractors.

Last edited: Feb 4, 2010