# USRP output power

I want to figure out bit error rate for certain signal to noise ratio
Thus I have to know SNR value which is a function of energy per one
bit (Eb) and noise characteristic (N0)
Let’s assume I know noise characteristic, N0 (in fact I gathered noise
data and figured out N0)

For now, I have to figure out energy per one bit (Eb)
My system transmits 1 bit every 31.25ms (32kbps) and using RFX400
(100+mW output)
each bit has magnitude 32,767 or zero (amplitude shift keying) in the
(I think 32,767 is the maximum value and it might have a voltage
value, 2 V. I hope it’s right)

In this situation, if bit has magnitude 32,767
what energy value each bit does have?

I have some difficulties on it.

On Mon, 2011-04-25 at 21:04 +0900, Songsong G. wrote:

(I think 32,767 is the maximum value and it might have a voltage
value, 2 V. I hope it’s right)

In this situation, if bit has magnitude 32,767
what energy value each bit does have?

The USRP does not have a calibrated output or input and you will have to
calibrate it using a measuring receiver, calibrated signal generator, or
power meter in order to obtain meaningful results.

–n

On 25/04/2011 11:42 AM, Nick F. wrote:

The USRP does not have a calibrated output or input and you will have to
calibrate it using a measuring receiver, calibrated signal generator, or
power meter in order to obtain meaningful results.

–n

I don’t think precise values are really required here. The typical
Eb/N0 estimates are just that, estimates.

Let’s assume that those bits leave the transmitter at “full-power”
(+20dBm), and radiate isotropically through free-space,
until they get to the receiver. At the receiver, the signal is
received against a locally-generated noise-background of some
magnitude. I think the noise figure of the RFX2400 is in the
neighbourhood of 5-6dB at maximum gain. A noise figure of 3dB
is equivalent to a noise power of -174dBm/Hz of bandwidth, so let’s
call the noise power at the receiver -170dBm/Hz to
make the math easier.

Now, what is the equivalent bandwidth of each bit? Let’s say it’s 10KHz
for argument’s sake. That means that the receiver
noise power over the bandwidth of a bit is -170dBm/Hz + 40dBHz =
-130dBm of noise power at the receiver over the bandwidth
occupied by our theoretical 10KHz bit.

The signal itself leaves the transmitter, and suffers free-space path
loss in a roughly inverse-cube law. Over a 1M path, that
signal suffers (assuming a perfectly-isotropic transmit antenna at
2.4GHz) a path loss of about 40dB, which brings the received signal
down to about -20dBm. So that -20dBm is set “against” a receiver
noise over the same bandwidth of -130dBm. The calculation of
the linear-units EB/N0 is left as an exercise for the reader.