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

GNU Radio scope plot

(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:

GNU Radio scope plot

(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.