Re: Why isn't Cancel more or less automatic?
- To: mathgroup at smc.vnet.net
- Subject: [mg35441] Re: Why isn't Cancel more or less automatic?
- From: "John Doty" <jpd at space.mit.edu>
- Date: Fri, 12 Jul 2002 04:29:10 -0400 (EDT)
- Organization: MIT Center for Space Research
- References: <ag6eh2$bo5$1@smc.vnet.net> <agbejo$jcb$1@smc.vnet.net> <ageghu$p7b$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <ageghu$p7b$1 at smc.vnet.net>, "AES" <siegman at stanford.edu> wrote: > If you really wanted to be sure you got all the system degrees of > freedom, however, wouldn't you just look at the roots of the determinant > of the system, without going on to the transfer function for one > particular input/output pair, which might indeed suffer from this > problem. Well, sure, you could do that. The advantage of the transfer function is you can ask it other questions, and it's a little less abstract (which makes it less likely you'll make a mistake). It's easy to type in a bunch of equations, KCL and Ohm's law generalizations, and grind them through Solve[] to get the transfer function. Your determinant is lurking in there, of course, but it's a bit of extra work to reveal it. In general, it comes down to what 0/0 means to you. In numerical calculation, singularities must be avoided or removed, but in this sort of symbolic work they are the keys to what's going on. Is a pole with a zero residue still a pole? I think so (is an empty glass still a glass?). If you see such a thing in an ideal calculation, you ought to be prepared to see the signature of a pole with small but nonzero residue in the physical process you're modeling. -- |John Doty "You can't confuse me, that's my job." |Home: jpd at w-d.org |Work: jpd at space.mit.edu