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RE: [linrad] DSP Question
- Subject: RE: [linrad] DSP Question
- From: "Robert McGwier" <rwmcgwier@xxxxxxxxxxxxxxx
- Date: Fri, 23 Apr 2004 08:18:43 -0000
Here h[k] represents a low pass filter as Oleg states.
It's design parameters besides length are where the cutoff
is. If you want a complex bandpass filter, centered at w0
as in Oleg's formula to have bandwidth B, then the cutoff
frequency for your lowpass filter with taps h[k] is B/2.
This is because a real lowpass filter is nothing more than a bandpass
filter with center frequency at 0. It has as much bandwidth below
zero frequency as above.
Bob
-----Original Message-----
From: owner-linrad@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-linrad@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Oleg Skydan
Sent: Friday, April 23, 2004 4:56 AM
To: Linrad List
Subject: Re: [linrad] DSP Question
Hi, Mark !
> I assume that t represents the sample interval
No, t is just the time.
You need just to do the following:
0<=k<=N-1
hi[k]=h[k]*cos(w0*(k-N/2+1/2)*T)
hq[k]=h[k]*sin(w0*(k-N/2+1/2)*T)
N - FIR length
T - sample period
h - LPF FIR taps
w0 - the center of the resulting BPFs
All the best !
Oleg UR3IQO
LINRADDARNIL