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## Re: [Discuss-gnuradio] USRP output power

 From: Marcus D. Leech Subject: Re: [Discuss-gnuradio] USRP output power Date: Mon, 25 Apr 2011 12:13:34 -0400 User-agent: Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US; rv:1.9.2.15) Gecko/20110303 Thunderbird/3.1.9

```On 25/04/2011 11:42 AM, Nick Foster 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.

```