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*To*: <linrad@xxxxxxxxxxxxxxxxxxxxxx>*Subject*: RE: [linrad] DSP Question*From*: "Robert McGwier" <rwmcgwier@xxxxxxxxxxx>*Date*: Fri, 23 Apr 2004 08:18:43 -0000*Importance*: Normal*In-reply-to*: <057f01c428ef$3ac6fa30$2e2c42d4@CAT>*Reply-to*: linrad@xxxxxxxxxxxxxxxxxxxxxx*Sender*: owner-linrad@xxxxxxxxxxxxxxxxxxxxxx

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 UR3IQOLINRADDARNIL

**References**:**Re: [linrad] DSP Question***From:*Oleg Skydan

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