> Hello,
>
> the green signal is the Input, the violet signal is the output
> Which of two work well?
>
What works well depends on your objectives. Only you can say.
That said, one can make all sorts of subjective observations looking at
the two images. But it seems silly to go there.
Maybe I missed something....
Fred
Reply by gpezzella●May 25, 20122012-05-25
Hello,
the green signal is the Input, the violet signal is the output
Which of two work well?
Single 3 order or 2 third order in cascade?
Here the screenshot:
https://www.dropbox.com/sh/5r28c1tbh61p0mg/luNus0Po-y
Thanks
Reply by gpezzella●May 24, 20122012-05-24
>Hello dear friends,
>
>thanks for your help.
>
>I have resolved the problem... in a sense that the noise source not was
my
>device but the oscilloscope himself (picoscope toy!!)
>
>Now with Professional Agilent MSO-X 2024A no noise is present at output
of
>my high gain preamplifier.
>
>Just for curiosity, I have follow one of yours tricks putting two filter
in
>cascade.
>
>The filter is a 3° LowPass Filter with Fcut= 200Hz
>
>Here the screenshot:
https://www.dropbox.com/sh/5r28c1tbh61p0mg/luNus0Po-y
I have forgot the question: they work good?
Reply by gpezzella●May 24, 20122012-05-24
Hello dear friends,
thanks for your help.
I have resolved the problem... in a sense that the noise source not was my
device but the oscilloscope himself (picoscope toy!!)
Now with Professional Agilent MSO-X 2024A no noise is present at output of
my high gain preamplifier.
Just for curiosity, I have follow one of yours tricks putting two filter in
cascade.
The filter is a 3° LowPass Filter with Fcut= 200Hz
Here the screenshot: https://www.dropbox.com/sh/5r28c1tbh61p0mg/luNus0Po-y
P.s.
Very good Tim, I live in Naples, Italy.
Moreover:
Vettore means Array
Filtrato means Filtered
Vettore_Filtrato = Filtered_Array
Reply by HardySpicer●May 22, 20122012-05-22
On May 22, 1:57�am, "gpezzella" <gpezzella@n_o_s_p_a_m.yahoo.com>
wrote:
xv[0] = xv[1]; xv[1] = xv[2]; xv[2] = xv[3]; xv[3] = xv[4]; xv[4] =
xv[5]; xv[5] = xv[6]
Should be the other way round ie
xv[6]=xv[5]
etc
Reply by David Drumm●May 21, 20122012-05-21
I don't see the transfer function error, but I do see:
Vettore_y(7) = Vettore_y(7) + (Vettore_x(1) + Vettore_x(7))
should be:
Vettore_y(7) = (Vettore_x(1) + Vettore_x(7))
unless Vettore_y(7) is set to zero by a mechanism I don't understand.
Reply by Les Cargill●May 21, 20122012-05-21
Tim Wescott wrote:
> On Mon, 21 May 2012 08:57:30 -0500, gpezzella wrote:
>>
>> It return me overflow on last row --> Sample_Filtrato =
>> Vettore_y(7)[HERE]
>>
>> The range of samples come from 1 to 1024
>
> Just out of curiosity -- from what language do the variable names come?
> Your English is impeccable, but "Sample_Filtrato" clearly isn't English.
> I'm thinking Italian ("Vettore" somehow says "Italian" to me), but I'm
> lame enough in the Romance languages that it could just as well be some
> Spanish language.
>
> And -- more pointless curiosity -- what does "Vettore" mean? ("Filtrato"
> I can guess at!).
>
I will guess "vector", Victor. Over.
--
Les Cargill
Reply by Tim Wescott●May 21, 20122012-05-21
On Mon, 21 May 2012 08:57:30 -0500, gpezzella wrote:
Aha -- you have made a mistake in translating your transfer function into
a difference equation. As a consequence you've made a 7-th order
unstable system instead of a 6th-order stable one. The really big hint
here is that you have seven states where you should only have six -- I
suspect that your mistake came in misunderstanding how to handle the
leading z^6 term in the denominator polynomial: that coefficient = 1 is
implied if you do the difference equation correctly, it DOES NOT have to
be explicit.
To give an example with a second-order system, if your transfer function
is
foo * z^2 + bar * z
H(z) = ----------------------
z^2 + blah * z + tee
then the difference equation is
y(t+2) = foo * x(t+2) + bar * x(t+1) - blah * y(t+1) - tee * y(t)
Shift this back in time by two, so you're getting today's outputs from
today's values:
y(t) = foo * x(t) + bar * x(t-1) - blah * y(t-1) - tee * y(t-2)
So you have a _2nd_ order transfer function, which gives you a difference
equation with _2_ state variables (y(t-1) and y(t-2)). Make sense?
Go do some web searching until you find some code for a 2nd-order direct-
form filter (either direct form 1 or 2 -- each has its advantages, either
will work better than what you're doing). Decompose your transfer
function _correctly_, implement the difference equations _correctly_, and
I think you'll find joy.
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com
Reply by David Drumm●May 21, 20122012-05-21
You might want to change your variables to double precision to see if the
overflow goes away. If it does, follow the advice given above.
If you're trying to remove 50 Hz AC interference, you might want to
consider an LMS adaptive filter approach instead of the IIR.
Reply by Tim Wescott●May 21, 20122012-05-21
On Mon, 21 May 2012 08:57:30 -0500, gpezzella wrote:
>
> It return me overflow on last row --> Sample_Filtrato =
> Vettore_y(7)[HERE]
>
> The range of samples come from 1 to 1024
Just out of curiosity -- from what language do the variable names come?
Your English is impeccable, but "Sample_Filtrato" clearly isn't English.
I'm thinking Italian ("Vettore" somehow says "Italian" to me), but I'm
lame enough in the Romance languages that it could just as well be some
Spanish language.
And -- more pointless curiosity -- what does "Vettore" mean? ("Filtrato"
I can guess at!).
--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?
Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com