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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
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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

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