[Top][All Lists]
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: [Discuss-gnuradio] Pulse shaping filter advice needed
From: |
Linus Gasser |
Subject: |
Re: [Discuss-gnuradio] Pulse shaping filter advice needed |
Date: |
Mon, 4 Aug 2003 11:29:37 +0200 |
User-agent: |
KMail/1.5.2 |
On Lundi, 4 Août 2003 09.25, Ian Wraith wrote:
> You will see the corner is set to 9 KHz. As the web page recommends this be
> set to half the baud rate. As my symbol rate is 18 KHz I set this to 9
> KHz. The trouble is when I look at the raw data and the filtered data in
> graphical form it looks to me as though the data is being pulse shaped to
> to low a frequency i.e the pulses are to wide. You can see what I mean by
> looking at ..
Hi,
I think I'm missing a point here... You take the Q/I inputs on Left/Right of
your soundcard, right? And then you want to filter your signals so that you
can undersample w/o having frequency-overlaps. So, in general:
- chose the sampling-frequency to be a multiple (in N) of your symbol-rate
- baudrate is VERY misleading and has been used in 20 different contexts.
Usually one uses 'bandwith' as a measure for the information. So the corner
frequency should be half the bandwith. But this is only the case if you have
a complex input-signal. Let me show you:
|A|
^
_I___
/ I I
/ I I
------+------> f
-B/2 B/2
This graph tries to show a signal A in it's frequency-domain. On the y-axis is
the amplitude, while on the x-axis is the frequency. This is a COMPLEX
signal, as it is non-symmetric. It's bandwith is B. What you want to do is to
filter this signal with a RRC, so that you can be sure that upon downsampling
you won't get any extra-signal (look up some signal-theory book for this). On
the web-page you indicated, the corner-frequency is B/2.
Now, if you have Q and I seperation, you get two pictures:
|Q(A)|
^
_I_
/ I \
/ I \
------+-----> f
-B B
|I(A)|
^
___I___
I I I
I I I
------+-----> f
-B B
So, why did I denote the limits now as B instead of B/2? Well, in a REAL
signal, the complex part is always 0, and the frequency-diagram (the FFT) is
always symmetric. So, in this case, the corner frequency would also be 9000.
OK, after this little theory, I insist on the fact that you should chose a
downsampling-factor that is in N, so for 18kSymbols/s, chose 18kHz (if you
can take I and Q seperatly) or 36kHz as a sampling-rate. If not it will be
quite difficult to get your signals...
Linus
--
----------------------------------------------------------------
Linus Gasser Phone: +41 21 693 5635
EPFL FAX : +41 21 693 4312
Mobile Communications Lab. email: address@hidden
CH-1015 Lausanne WEB : http://lcmwww.epfl.ch
SWITZERLAND Map: http://plan.epfl.ch/?room=INR037
----------------------------------------------------------------
Message not available