The subject line describes what I want to do.
I want to be able to call methods created within the GRC-created code
from my own imported functions.
For example, if the GRC created flowgraph created a method called
set_foonly(), I'd like to be able to
call set_foonly() from within my own importe
Sorry for the bump, but I'm confused about the output power of the BasicTX,
especially in regards to 1 channel or both.
Any thoughts?
William
On Mon, Jun 21, 2010 at 10:46 AM, William Cox wrote:
> I'm getting conflicting info on how much power the BasicTX can transmit.
> Ettus FAQ says, "Yes, t
what you can do is pass a string representing the name of your variable;
the block wrapper template can expand that to self.set_$(var), but you
will never get grc to reference the methods of blocks and classes at
edit time.
-Josh
On 06/25/2010 10:35 AM, Marcus D. Leech wrote:
The subject lin
Hi ,
I am beginner in gnuradio, I have installed gnuradio 3.2.2 in my ubuntu
lucid system.I am trying to understand the various files and packages and
how they coordinate among them, and there implementation details like
threading implementation in python level,scheduler etc ,shortly i am trying
Hi guys,
I have two questions about the USRP's architecture that I have been unable to
answer from looking around.
1) I understand that when a daughter board is tuned for Rx, the exact frequency
is not always possible, so the receiver tunes to the nearest possible frequency
and the 'requested' f
On 06/25/2010 01:31 AM, Matt Robert wrote:
> Hi guys,
>
> I have two questions about the USRP's architecture that I have been
> unable to answer from looking around.
>
> 1) I understand that when a daughter board is tuned for Rx, the exact
> frequency is not always possible, so the receiver tunes t
On 06/25/2010 02:07 PM, Josh Blum wrote:
> what you can do is pass a string representing the name of your
> variable; the block wrapper template can expand that to
> self.set_$(var), but you will never get grc to reference the methods
> of blocks and classes at edit time.
>
> -Josh
>
So, I have to
It is worth noting that your calculation of power is missing a divide by
2, since you specify a peak-to-peak voltage, and not a peak voltage.
The equation for peak-to-peak voltage to power (given a 50 ohm
impedance) is P (watts) = (.707*V/2)^2/50, which when I calculated
yields something like
On Fri, Jun 25, 2010 at 11:19 PM, Jeffrey Lambert wrote:
> It is worth noting that your calculation of power is missing a divide by 2,
> since you specify a peak-to-peak voltage, and not a peak voltage. The
> equation for peak-to-peak voltage to power (given a 50 ohm impedance) is P
> (watts) = (
