[Severe cuts applied to get to the essence here.]
Active8 said:
I didn't get that from the app note. It's the cutoff freq for a
single pole lowpass filter. I dug up the appnote for the OP.
Ok, I assumed you got it from the app note you quoted
because it seemed too long a coincidence that the same
misconception would appear independently.
It's also the freq where the reactance equals the resistance, which
is the cutoff freq of a one pole RC filter. Notice the equation you
gave is similar to the eq for a simple passive RC integrator/LPF ?
Similar, perhaps, in the appearance of a similar
term, (2 pi R C), but not similar enough to turn a
low-pass filter configuration into an integrator.
Yeah, they called it an integrator in the Army - when that was it's
intended purpose. Good enough for gov't work?
![Smile :) :)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
You don't need to
point out that the charging current will not be constant as the cap
charges up.
That is is an important aspect of the difference between
a low pass filter and an integrator. In the passband of
an RC LPF, the cap does more or less charge up so
that there is little loss across the R and, more importantly,
the output magnitude asymptotically approaches the
input magnitude. That never happens in an integrator
except, in an op-amp realization, at the absurdly low
frequency where the amplifier operates nearly open
loop.
They used RC passive "differentiators" too. Square in,
sharp pulses out.
Calling an RC high-pass filter a differentiator is much like
calling an RC low-pass filter an integrator. So I have to
hand it to you for consistency. said:
---/\/\/\/---+--- Vo/Vi = 1/(1 + RCs)
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Strictly speaking, (and going along with considering
only the magnitude), it is 1/sqrt(2) at Fc for that circuit.
But there is no passband! The pole is at 0.
There is simply no pole at 1/(2.pi.R.C) as you
have claimed, which is what got me into this.
[big snip]
There was nothing to be gained by quoting text irrelevant to my
reply. As I said, I checked the mfg and AN # for the OP. First you
ream me for not cutting enough chaff from my code, now you're
complaining about me snipping posts
Well, if I was complaining it was only about this
short and almost simple statement that vanished:
[T]he integrator gain is -1/(s*(R*C)) for each input R.
That really says it all. There is no cutoff frequency
because there is no passband with a single pole at 0.
And I'm sorry if you felt reamed. I often try to
persuade programmers that they can often solve
their own problems by trying to reduce their code
to a minimal problematic version.
Likewise,