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Re: [TenTec] INRAD Filter Question

To: geraldj@isunet.net, tentec@contesting.com
Subject: Re: [TenTec] INRAD Filter Question
From: Bill Fuqua <wlfuqu00@uky.edu>
Reply-to: tentec@contesting.com
Date: Fri, 23 May 2003 16:05:07 -0400
List-post: <mailto:tentec@contesting.com>
Ringing is in the nature of narrow band pass filters. Ringing "length of decay" is more a function of bandwidth than that of
skirts.

If you have a 1 Hz bandwidth filter you can't detect pulses spaced less than 1 second apart. In other words the product of bandwidth and timing resolution is always greater than 1 or equal to one. This is related to Heisenberg's Uncertainty Principle.

So with a 1 Hz bandwidth filter you can have 2 seconds of timing resolution ( .5 Hz) or even 1 second (1 Hz) but not .5 seconds (2 Hz) . So if dits are occurring more frequently than 1 per second you cannot distinguish one from another. The output from the filter will be a continuous sinusoidal signal. If the dit rate is less than one Hz or at a period of greater than a second then they are distinguishable.

73
Bill wa4lav



At 11:27 AM 5/23/2003 -0500, you wrote:
If anything its the other way around. Steeper skirts and squarer
passband corners mean more ringing, and so the Inrad should ring more
than the TT.

73, Jerry, K0CQ

--
Entire content copyright Dr. Gerald N. Johnson, electrical engineer.
Reproduction by permission only.
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