The Cyber-Spy.Com Usenet Archive Feeds Directly
From The Open And Publicly Available Newsgroup
This Group And Thousands Of Others Are Available
On Most IS NNTP News Servers On Port 119.
Cyber-Spy.Com Is NOT Responsible For Any Topic,
Opinions Or Content Posted To This Or Any Other
Newsgroup. This Web Archive Of The Newsgroup And
Posts Are For Informational Purposes Only.
From: Fred Bloggs
X-Mailer: Mozilla 4.7 [en]C-CCK-MCD EBM-Compaq1 (Win95; U)
Subject: Re: How to write the transfer function of the circuit?
Date: Thu, 12 Sep 2002 13:41:39 GMT
NNTP-Posting-Date: Thu, 12 Sep 2002 06:41:39 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
> what we want to know is as follow:
> The outpt function of above is as follow:
> Vo=(1/2)*Vcc*(1+R2/R1) - (R2/R1)*V1 ;
> If we use block diagram in a control system as follow, now the
> question is what the transfer function is ?
> Vi _____| H(s) |____Vo
> | |
> H(s) = Vo/Vi = ?
In this case you can redefine Vi'=Vi-0.5*Vcc and Vo'=Vo-0.5*Vcc. Then
since Vo=0.5*Vcc*(1+R2/R1)-Vi*(R2/R1) this is rewritten as
Vo'=-R2/R1*Vi'. Then the transfer function H(s)=Vo'/Vi'=-R2/R1. In
words, define new variables Vo' and Vi' as the offset of the respective
variables from 0.5*Vcc, and the transfer function of the new variables
becomes -R2/R1 as you would expect. This works for ANY bias voltage
applied to IN(+) and not just 0.5*Vcc and you can see that it reduces to
the usual -R2/R1 for IN(+) at 0V.
Go Back To The Cyber-Spy.Com
Usenet Web Archive Index Of
The sci.electronics.design Newsgroup