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From: sgcpal@watdcsu.UUCP (P.A. Layman [EE-Silcon Devices Research])
Newsgroups: net.analog
Subject: Re: filter circuit Q
Message-ID: <1981@watdcsu.UUCP>
Date: Wed, 18-Dec-85 08:41:38 EST
Article-I.D.: watdcsu.1981
Posted: Wed Dec 18 08:41:38 1985
Date-Received: Thu, 19-Dec-85 03:06:09 EST
References: <35500001@ICO.UUCP>
Reply-To: sgcpal@watdcsu.UUCP (P.A. Layman [EE-Silcon Devices Research])
Organization: U of Waterloo, Ontario
Lines: 44
Summary: 

In article <35500001@ICO.UUCP> chris@ICO.UUCP writes:
>
>There is a question that has been puzzling me
>about filter circuits that i was hoping someone
>could explain to me. In physics they talked about the
>"Q" of a mechanical circuit, which was a measure of its
>ability to reject off resonant frequency signals. I see
>how an LC circuit forms a filter, but i don't see how to
>change the Q of the circuit. Anybody have an explanation
>for me?
>
Essentially and L-C  resonant circuit will be driven by a circuit
with some finite output resistance. Thus,

	Z = R + j( wL - 1/wC ) ,

or
			  2
	Z = R + j( wL - w  L/w ),
	     o           o
and Q is defined as,

	Q = w L/R
	     o
w  is the resonant frequency, and R  is the effective resistance at 
 o				   o
resonance. Note this definition also applies for an inductance which
includes a parasitic resistance R.  Note also that for parallel resonance
the w term appears in the denominator.

	Z = R  ( R/R   + jQ ( w/w  - w /w ) )
	     o	    o            o    o

Clearly if the half power points occur at w1 and w2,

	Q = w  /(w1-w2)
	     o

Paul A. Layman

University of Waterloo, Electrical Engineering,
Silicon Devices and Integrated Circuits Research Group (SiDIC)

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