FPGA Central

Hello,

I have worked in Analog Tx/Rx. Past 4 years I am working in Data acq
systems. Hence I have very good knowledge of A2D and sampling rate. But
still don't understand how a low-pass or high pass filtering is achieve
using digital technique.

I know that in case of analog filter, if I design a low-pass filter, it'
cut-off freq. is given by 1/2piRC,which means about below this freq. al
signals will pass through the filter & above this freq. will b
attenuated.

But how can I come to this same conclusion using digital filters lik
FIR,IIR etc.

In case of digital filter, I know that we will be sampling at Fs rate an
digital data of analog signal is present in memory but how can thes
samples behave like a low-pass or high-pass filter.

Hope my question is clear, if not, please let me know I will send you mor
details.

yatinkamal wrote:
> Hello,
>
> I have worked in Analog Tx/Rx. Past 4 years I am working in Data acq.
> systems. Hence I have very good knowledge of A2D and sampling rate. But I
> still don't understand how a low-pass or high pass filtering is achieved
> using digital technique.
>
> I know that in case of analog filter, if I design a low-pass filter, it's
> cut-off freq. is given by 1/2piRC,which means about below this freq. all
> signals will pass through the filter & above this freq. will be
> attenuated.
>
> But how can I come to this same conclusion using digital filters like
> FIR,IIR etc.
>
> In case of digital filter, I know that we will be sampling at Fs rate and
> digital data of analog signal is present in memory but how can these
> samples behave like a low-pass or high-pass filter.
>
> Hope my question is clear, if not, please let me know I will send you more
> details.

Yatinkamal,

The question is quite clear, but the answer is long. If I try to write a
treatise in the time I can devote to it, I will not be clear enough.
Many people have written books about this, and we can recommend a few.

There is a bibliography at http://dspguru.com/, and one of the books in
that list is available on line at http://dspguide.com/. Try reading a
bit of it. The question you asked is dealt with in Chapter 14, but I
imagine you will want to read some of what comes before.

If specific difficulties puzzle you along the way, ask here about them.
Many people will be glad to help.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Jerry Avins wrote:
> yatinkamal wrote:

snip

>> I know that in case of analog filter, if I design a low-pass filter, it's
>> cut-off freq. is given by 1/2piRC,which means about below this freq. all
>> signals will pass through the filter & above this freq. will be
>> attenuated.
>>
>> But how can I come to this same conclusion using digital filters like
>> FIR,IIR etc.
>>
>> In case of digital filter, I know that we will be sampling at Fs rate and
>> digital data of analog signal is present in memory but how can these
>> samples behave like a low-pass or high-pass filter.

snip

> There is a bibliography at http://dspguru.com/, and one of the books in
> that list is available on line at http://dspguide.com/. Try reading a
> bit of it. The question you asked is dealt with in Chapter 14, but I
> imagine you will want to read some of what comes before.
>
> If specific difficulties puzzle you along the way, ask here about them.
> Many people will be glad to help.
>
> Jerry

The books on the list that Jerry linked to are all excellent (I own a
copy of many of them). Another book which, in my opinion, is great for
somebody with a background in analog is "Digital Filters" by Hamming.
It's very cheap, and along with the Smith or Lyons books, should give
you what you are looking for.

Cheers

Marc

This is my second reply to this message, but Google hickuped on my
first try. I quickly copied my entry before the Google page
disappeared. My message hasn't been posted yet, so here it goes again
(apologies if it appears twice):

Jerry Avins wrote:
> yatinkamal wrote:
> > Hello,
>
> > I have worked in Analog Tx/Rx. Past 4 years I am working in Data acq.
> > systems. Hence I have very good knowledge of A2D and sampling rate. But I
> > still don't understand how a low-pass or high pass filtering is achieved
> > using digital technique.
>
> > I know that in case of analog filter, if I design a low-pass filter, it's
> > cut-off freq. is given by 1/2piRC,which means about below this freq. all
> > signals will pass through the filter & above this freq. will be
> > attenuated.
>
> > But how can I come to this same conclusion using digital filters like
> > FIR,IIR etc.
>
> > In case of digital filter, I know that we will be sampling at Fs rate and
> > digital data of analog signal is present in memory but how can these
> > samples behave like a low-pass or high-pass filter.
>
> > Hope my question is clear, if not, please let me know I will send you more
> > details.Yatinkamal,
>
> The question is quite clear, but the answer is long.

I think not. Essentially, all that digital filters do is add weighted
and time-delayed copies of the input signal to produce the output
signal.

A simple example: Add the input signal and the input signal delayed by
T, where T is the sampling period. We can normalize T = 1, and
therefore the maximum allowable frequency for an input signal is w_N =
pi. Look at the effect that this process has on an input signal cos(w
t), with w < w_N:

y(t) = cos(w t) + cos(w (t-1) ).

By using a trigonometric identity
(http://en.wikipedia.org/wiki/Trigono...uct_identities),
this becomes

y(t) = 2 cos( w (t + t - 1) / 2) cos(w (t - t + 1) / 2 )
= 2 cos(w t - w / 2 ) cos(w / 2).

Notice how y(t) is still a cosine, but with new ampltidue 2 cos(w/2),
which depends on w, and an additional phase shift of (- w / 2), which
also depends (linearly) on w. When you look at the function f(w) = |2
cos(w /2)|, 0 < w < pi, you get the magnitude response of the filter,
as a function of the frequency w. It's a lowpass filter.

In general, a digital filter is characterized by its impulse response
which specifies how one has to add up the delayed versions of the input
signal, and you can do the same analysis like I have just done and will
always get the same result: for a sinusoid input, you get a sinusoid
output at the same frequency, weighted and phase-shifted depending on
the frequency. For a general input signal, imagine the Fourier
transform decomposition, and that each frequency component of the input
signal is scaled by the magnitude function and shifted by the phase
function of the filter to produce the output signal.

Hope this gets you started,
Regards,
Andor

Jerry Avins wrote:
> yatinkamal wrote:
> > Hello,
>
> > I have worked in Analog Tx/Rx. Past 4 years I am working in Data acq.
> > systems. Hence I have very good knowledge of A2D and sampling rate. But I
> > still don't understand how a low-pass or high pass filtering is achieved
> > using digital technique.
>
> > I know that in case of analog filter, if I design a low-pass filter, it's
> > cut-off freq. is given by 1/2piRC,which means about below this freq. all
> > signals will pass through the filter & above this freq. will be
> > attenuated.
>
> > But how can I come to this same conclusion using digital filters like
> > FIR,IIR etc.
>
> > In case of digital filter, I know that we will be sampling at Fs rate and
> > digital data of analog signal is present in memory but how can these
> > samples behave like a low-pass or high-pass filter.
>
> > Hope my question is clear, if not, please let me know I will send you more
> > details.Yatinkamal,
>
> The question is quite clear, but the answer is long.

I think not. Essentially, all that digital filters do is add weighted
and time-delayed copies of the input signal to produce the output
signal.

A simple example: Add the input signal and the input signal delayed by
T, where T is the sampling period. We can normalize T = 1, and
therefore the maximum allowable frequency for an input signal is w_N =
pi. Look at the effect that this process has on an input signal cos(w
t), with w < w_N:

y(t) = cos(w t) + cos(w (t-1) ).

By using a trigonometric identity
(http://en.wikipedia.org/wiki/Trigono...uct_identities),
this becomes

y(t) = 2 cos( w (t + t - 1) / 2) cos(w (t - t + 1) / 2 )
= 2 cos(w t - w / 2 ) cos(w / 2).

Notice how y(t) is still a cosine, but with new ampltidue 2 cos(w/2),
which depends on w, and an additional phase shift of (- w / 2), which
also depends (linearly) on w. When you look at the function f(w) = |2
cos(w /2)|, 0 < w < pi, you get the magnitude response of the filter,
as a function of the frequency w. It's a lowpass filter.

In general, a digital filter is characterized by its impulse response
which specifies how one has to add up the delayed versions of the input
signal, and you can do the same analysis like I have just done and will
always get the same result: for a sinusoid input, you get a sinusoid
output at the same frequency, weighted and phase-shifted depending on
the frequency. For a general input signal, imagine the Fourier
transform decomposition, and that each frequency component of the input
signal is scaled by the magnitude function and shifted by the phase
function of the filter to produce the output signal.

Hope this gets you started,
Regards,
Andor

"Andor" <[email protected]> wrote in news:1164900353.759077.223830@
16g2000cwy.googlegroups.com:

> I think not. Essentially, all that digital filters do is add weighted
> and time-delayed copies of the input signal to produce the output
> signal.
>

Well, in many cases, weighted and time delayed copies of the input AND
OUTPUT signals.

--
Scott
Reverse name to reply

Andor wrote:
> Jerry Avins wrote:
>> yatinkamal wrote:
>>> Hello,
>>> I have worked in Analog Tx/Rx. Past 4 years I am working in Data acq.
>>> systems. Hence I have very good knowledge of A2D and sampling rate. But I
>>> still don't understand how a low-pass or high pass filtering is achieved
>>> using digital technique.
>>> I know that in case of analog filter, if I design a low-pass filter, it's
>>> cut-off freq. is given by 1/2piRC,which means about below this freq. all
>>> signals will pass through the filter & above this freq. will be
>>> attenuated.
>>> But how can I come to this same conclusion using digital filters like
>>> FIR,IIR etc.
>>> In case of digital filter, I know that we will be sampling at Fs rate and
>>> digital data of analog signal is present in memory but how can these
>>> samples behave like a low-pass or high-pass filter.
>>> Hope my question is clear, if not, please let me know I will send you more
>>> details.Yatinkamal,
>> The question is quite clear, but the answer is long.
>
> I think not. Essentially, all that digital filters do is add weighted
> and time-delayed copies of the input signal to produce the output
> signal.

Great! All an alternator does is rotate wires in a magnetic field. From
this we deduce that is has an inherent current limit and will not be
damaged by driving into a short circuit. The logic of that follows a
simple path, but one needs a few pointers along the way.

> A simple example: Add the input signal and the input signal delayed by
> T, where T is the sampling period. We can normalize T = 1, and
> therefore the maximum allowable frequency for an input signal is w_N =
> pi. Look at the effect that this process has on an input signal cos(w
> t), with w < w_N:
>
> y(t) = cos(w t) + cos(w (t-1) ).
>
> By using a trigonometric identity
> (http://en.wikipedia.org/wiki/Trigono...uct_identities),
> this becomes
>
> y(t) = 2 cos( w (t + t - 1) / 2) cos(w (t - t + 1) / 2 )
> = 2 cos(w t - w / 2 ) cos(w / 2).
>
> Notice how y(t) is still a cosine, but with new ampltidue 2 cos(w/2),
> which depends on w, and an additional phase shift of (- w / 2), which
> also depends (linearly) on w. When you look at the function f(w) = |2
> cos(w /2)|, 0 < w < pi, you get the magnitude response of the filter,
> as a function of the frequency w. It's a lowpass filter.

"Still a cosine, but with shifted phase?" I'm confused. Suppose the
phase shift is pi/2: is that a cosine, or a sine?

> In general, a digital filter is characterized by its impulse response
> which specifies how one has to add up the delayed versions of the input
> signal, and you can do the same analysis like I have just done and will
> always get the same result: for a sinusoid input, you get a sinusoid
> output at the same frequency, weighted and phase-shifted depending on
> the frequency. For a general input signal, imagine the Fourier
> transform decomposition, and that each frequency component of the input
> signal is scaled by the magnitude function and shifted by the phase
> function of the filter to produce the output signal.

Impulse response? What's an impulse response? For that matter, what's an
impulse? I always thought an impulse was infinite in height and of zero
width. Since no digital signal can be infinite, and the duration of all
samples is one sample period (doesn't everybody know that?) how can
there be a digital impulse?

All tongue in cheek, of course. My aim was pointing out that you were
assuming a facility with basic concepts that might best be acquired with
some quiet reading.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

yatinkamal wrote:
> I have worked in Analog Tx/Rx. Past 4 years I am working in Data acq.
> systems. Hence I have very good knowledge of A2D and sampling rate. But I
> still don't understand how a low-pass or high pass filtering is achieved
> using digital technique.
>
> I know that in case of analog filter, if I design a low-pass filter, it's
> cut-off freq. is given by 1/2piRC,which means about below this freq. all
> signals will pass through the filter & above this freq. will be
> attenuated.
>
> But how can I come to this same conclusion using digital filters like
> FIR,IIR etc.
>
> In case of digital filter, I know that we will be sampling at Fs rate and
> digital data of analog signal is present in memory but how can these
> samples behave like a low-pass or high-pass filter.
>
> Hope my question is clear, if not, please let me know I will send you more
> details.

An analog filter works because energy (charge or field) is
stored in it's reactive elements. This stored history built
up by previous input then effects the present output. The
resulting output can be modeled by linear differential
equations.

A digital filter just stores the history of the previous input
in a different form. The effects can be modeled by linear
difference equations. If the input signal is "slow enough"
relative to the sample rate, and you have enough bits relative
to your desired S/N ratio, then the difference equations
produce about the same results as the differential equations.


IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

Ron N. wrote:

> An analog filter works because energy (charge or field) is
> stored in it's reactive elements. This stored history built
> up by previous input then effects the present output. The
> resulting output can be modeled by linear differential
> equations.
>
> A digital filter just stores the history of the previous input
> in a different form. The effects can be modeled by linear
> difference equations. If the input signal is "slow enough"
> relative to the sample rate, and you have enough bits relative
> to your desired S/N ratio, then the difference equations
> produce about the same results as the differential equations.

Nicely put --- perhaps worth adding that the mathematical tools
that we use to analyze/understand/model/design the behaviour of
these systems that we call digital filters are different from
those that we use for analog filters. That is, there is no
notion of RC time constant being the inverse of the cutoff
frequency; the signal analysis principles are the same, but
the way they're applied are different.

For the OP: still find it confusing? Then it would be the
time to look up some Digital Filters literature to try and
understand things more in detail, and not simply from an
"intuitive" point of view.

Carlos
--

Jerry Avins wrote:
> Andor wrote:
>> Jerry Avins wrote:
>>> yatinkamal wrote:
>>>> Hello,
>>>> I have worked in Analog Tx/Rx. Past 4 years I am working in Data acq.
>>>> systems. Hence I have very good knowledge of A2D and sampling rate.
>>>> But I
>>>> still don't understand how a low-pass or high pass filtering is
>>>> achieved
>>>> using digital technique.
>>>> I know that in case of analog filter, if I design a low-pass filter,
>>>> it's
>>>> cut-off freq. is given by 1/2piRC,which means about below this freq.
>>>> all
>>>> signals will pass through the filter & above this freq. will be
>>>> attenuated.
>>>> But how can I come to this same conclusion using digital filters like
>>>> FIR,IIR etc.
>>>> In case of digital filter, I know that we will be sampling at Fs
>>>> rate and
>>>> digital data of analog signal is present in memory but how can these
>>>> samples behave like a low-pass or high-pass filter.
>>>> Hope my question is clear, if not, please let me know I will send
>>>> you more
>>>> details.Yatinkamal,
>>> The question is quite clear, but the answer is long.
>>
>> I think not. Essentially, all that digital filters do is add weighted
>> and time-delayed copies of the input signal to produce the output
>> signal.
>
> Great! All an alternator does is rotate wires in a magnetic field. From
> this we deduce that is has an inherent current limit and will not be
> damaged by driving into a short circuit. The logic of that follows a
> simple path, but one needs a few pointers along the way.
>
>> A simple example: Add the input signal and the input signal delayed by
>> T, where T is the sampling period. We can normalize T = 1, and
>> therefore the maximum allowable frequency for an input signal is w_N =
>> pi. Look at the effect that this process has on an input signal cos(w
>> t), with w < w_N:
>>
>> y(t) = cos(w t) + cos(w (t-1) ).
>>
>> By using a trigonometric identity
>> (http://en.wikipedia.org/wiki/Trigono...uct_identities),
>>
>> this becomes
>>
>> y(t) = 2 cos( w (t + t - 1) / 2) cos(w (t - t + 1) / 2 )
>> = 2 cos(w t - w / 2 ) cos(w / 2).
> >
>> Notice how y(t) is still a cosine, but with new ampltidue 2 cos(w/2),
>> which depends on w, and an additional phase shift of (- w / 2), which
>> also depends (linearly) on w. When you look at the function f(w) = |2
>> cos(w /2)|, 0 < w < pi, you get the magnitude response of the filter,
>> as a function of the frequency w. It's a lowpass filter.
>
> "Still a cosine, but with shifted phase?" I'm confused. Suppose the
> phase shift is pi/2: is that a cosine, or a sine?
>
>> In general, a digital filter is characterized by its impulse response
>> which specifies how one has to add up the delayed versions of the input
>> signal, and you can do the same analysis like I have just done and will
>> always get the same result: for a sinusoid input, you get a sinusoid
>> output at the same frequency, weighted and phase-shifted depending on
>> the frequency. For a general input signal, imagine the Fourier
>> transform decomposition, and that each frequency component of the input
>> signal is scaled by the magnitude function and shifted by the phase
>> function of the filter to produce the output signal.
>
> Impulse response? What's an impulse response? For that matter, what's an
> impulse? I always thought an impulse was infinite in height and of zero
> width. Since no digital signal can be infinite, and the duration of all
> samples is one sample period (doesn't everybody know that?) how can
> there be a digital impulse?
>
> All tongue in cheek, of course. My aim was pointing out that you were
> assuming a facility with basic concepts that might best be acquired with
> some quiet reading.
>
> Jerry

Bravo I say!

As someone who has taught and is in the process of attempting to master
the digital domain of filtering, I admit to having already read many of
the books you noted. I am now doing implementations in a FPGA (at home)
to get a more intuitive feel. (I've been doing electronics for almost 4
decades and never had to get that close to the details in the digital
filtering domain - now I find I need to, so I'll learn it).

A student, by _definition_, does not know the subject, and can not be
expected to know the background. As a longtime student of whatever comes
my way, and a teacher on occasion (well, for some years), it is
heartening to know others see it that way.

Cheers

PeteS

PeteS wrote:

...

> As someone who has taught and is in the process of attempting to master
> the digital domain of filtering, I admit to having already read many of
> the books you noted. I am now doing implementations in a FPGA (at home)
> to get a more intuitive feel. (I've been doing electronics for almost 4
> decades and never had to get that close to the details in the digital
> filtering domain - now I find I need to, so I'll learn it).
>
> A student, by _definition_, does not know the subject, and can not be
> expected to know the background. As a longtime student of whatever comes
> my way, and a teacher on occasion (well, for some years), it is
> heartening to know others see it that way.

Pete,

I bought a DSK as a toy when I retired too long ago to count the years,
and I asked in this group to learn how to use it. In many years as an
instrument maker, analog and digital designer, I had implemented some
projects using sampled data, but mostly with ad hoc methods of my own
devising*. Always in the back of my mind was the notion that I really
ought to learn how to do these things right, and retirement was my
chance. I didn't actually lurk, but I read a lot more than I asked.

I soon found questions from people who lacked basic EE theory that I
could do my share by answering. The answers to my early questions led me
to books that helped to round out the little I know so that the many
gaps don't keep me awake. Somehow, people I respect got the impression
that I actually know something useful, so I suppose I do.

You seem to be about where I was back in 1998 or so. Keep plugging at
the questions. The things you care about will probably come clear pretty
soon.

Jerry
_____________________________________
* One project used a signal buried in noise to control a feedback loop.
There was over a 40 dB Noise-to-signal ratio, but my home-made filters
(and novel loop configuration) made it work. I still don't know how I
could have done that one better using a Z-80 coupled by I/O to an HP85
for operator interface and data logging.
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

"Ron N." <[email protected]> wrote in message
news:[email protected] ups.com...
>
> yatinkamal wrote:
>> I have worked in Analog Tx/Rx. Past 4 years I am working in Data acq.
>> systems. Hence I have very good knowledge of A2D and sampling rate. But I
>> still don't understand how a low-pass or high pass filtering is achieved
>> using digital technique.
>>
>> I know that in case of analog filter, if I design a low-pass filter, it's
>> cut-off freq. is given by 1/2piRC,which means about below this freq. all
>> signals will pass through the filter & above this freq. will be
>> attenuated.
>>
>> But how can I come to this same conclusion using digital filters like
>> FIR,IIR etc.
>>
>> In case of digital filter, I know that we will be sampling at Fs rate and
>> digital data of analog signal is present in memory but how can these
>> samples behave like a low-pass or high-pass filter.
>>
>> Hope my question is clear, if not, please let me know I will send you
>> more
>> details.
>
> An analog filter works because energy (charge or field) is
> stored in it's reactive elements. This stored history built
> up by previous input then effects the present output. The
> resulting output can be modeled by linear differential
> equations.
>
> A digital filter just stores the history of the previous input
> in a different form. The effects can be modeled by linear
> difference equations. If the input signal is "slow enough"
> relative to the sample rate, and you have enough bits relative
> to your desired S/N ratio, then the difference equations
> produce about the same results as the differential equations.

This is the sort of response I had in mind also Ron.

If you like notions of system "state" then the state of an analog system is
made up of capacitor voltages and inductor currents and the contents of
delay lines at one instant (the devices that tend to "hold" their state over
short periods of time) and the state of a digital system is made up memory
contents (which may also be modeled as delays or delay lines).

The "system" further connects and combines the state values and the inputs
to form the output at any instant or sample time (and, thus by implication,
the output values too). Analog filters use resistors, amplifiers, etc. to
combine the state values. Digital filters use summers and multipliers to
accomplish much the same thing. Thus, analog and digital filters are rather
similar in how they work.

Fred

Ron N. wrote:
> yatinkamal wrote:
> > ... I
> > still don't understand how a low-pass or high pass filtering is achieved
> > using digital technique.
> >
> > I know that in case of analog filter, if I design a low-pass filter, it's
> > cut-off freq. is given by 1/2piRC,which means about below this freq. all
> > signals will pass through the filter & above this freq. will be
> > attenuated.
> >
> > But how can I come to this same conclusion using digital filters like
> > FIR,IIR etc.
....
>
> An analog filter works because energy (charge or field) is
> stored in it's reactive elements. This stored history built
> up by previous input then effects the present output.
....
> A digital filter just stores the history of the previous input
> in a different form.

a third vote for Ron's answer and another little conceptual or
philosophical addition. because it's how signals vary in *time* is
what constitutes the difference between low-frequency and
high-frequency (or mid-frequency or "frequency-about-100Hz") signals.
operating on it at a single instant, x(t0) of the signal does not give
you any idea how such a signal varies, so you have no idea what kind
(in terms of frequency) of signal it is. also operating on multiple
instants of it at many different times gives you little inference of
the frequency content if you do it WITHOUT relative reference of one
value x(t0) to the signal at other times x(t1), x(t2), etc. so, in
both cases you need *memory* to be able to differentiate (not
necessarily the calculus meaning of the term) signal content of one
frequency to another.

the means of memory, of storing the values of past instants of the
signal is just different in analog vs. digital filters. in digital
filters the past values of the signal are stored simply as that: the
past values are stored as numbers in the mind of the computer. in
analog filters it is certain weighted sums of past values of a signal
that is stored in the reactive components (i.e. capacitors and
inductors). in digital filters the math done to combine these
different values to do this frequency discrimination is "simply"
programmed into the computer relatively straight-forward. in analog
filters, the math operations are done with summing junctions (Kirchoff
Current Law) and summing ladders of components (Kirchoff Voltage Law)
and scaling of values is done with amplifiers (usually op-amps,
now-a-daze) or sometimes resistive attenuators.

that's my philosophical spin of maybe where Ron was going with this.

r b-j

Ron N. wrote:

(snip)

> A digital filter just stores the history of the previous input
> in a different form. The effects can be modeled by linear
> difference equations. If the input signal is "slow enough"
> relative to the sample rate, and you have enough bits relative
> to your desired S/N ratio, then the difference equations
> produce about the same results as the differential equations.

and it is just that. That you don't have enough bits, "slow
enough" signals, and always want a better S/N ratio that
makes digital filters interesting. If it were just the same
old differential equations, it would be much less interesting.

-- glen

glen herrmannsfeldt wrote:
> Ron N. wrote:
>
> (snip)
>
>> A digital filter just stores the history of the previous input
>> in a different form. The effects can be modeled by linear
>> difference equations. If the input signal is "slow enough"
>> relative to the sample rate, and you have enough bits relative
>> to your desired S/N ratio, then the difference equations
>> produce about the same results as the differential equations.
>
> and it is just that. That you don't have enough bits, "slow
> enough" signals, and always want a better S/N ratio that
> makes digital filters interesting. If it were just the same
> old differential equations, it would be much less interesting.
>
> -- glen
>

Actually, I see it as an extension to the various approximation
techniques we used to use, such as Simpson's rule.

The difference here is that due to the processing power available, the
approximation technique is better than the analogue (or direct integral
/ differential) technique because of the limitations of circuit elements
to deal with it.

Just my $0.02

Cheers

PeteS

Jerry Avins wrote:
> PeteS wrote:
>
> ...
>
>> As someone who has taught and is in the process of attempting to
>> master the digital domain of filtering, I admit to having already read
>> many of the books you noted. I am now doing implementations in a FPGA
>> (at home) to get a more intuitive feel. (I've been doing electronics
>> for almost 4 decades and never had to get that close to the details in
>> the digital filtering domain - now I find I need to, so I'll learn it).
>>
>> A student, by _definition_, does not know the subject, and can not be
>> expected to know the background. As a longtime student of whatever
>> comes my way, and a teacher on occasion (well, for some years), it is
>> heartening to know others see it that way.
>
> Pete,
>
> I bought a DSK as a toy when I retired too long ago to count the years,
> and I asked in this group to learn how to use it. In many years as an
> instrument maker, analog and digital designer, I had implemented some
> projects using sampled data, but mostly with ad hoc methods of my own
> devising*. Always in the back of my mind was the notion that I really
> ought to learn how to do these things right, and retirement was my
> chance. I didn't actually lurk, but I read a lot more than I asked.
>
> I soon found questions from people who lacked basic EE theory that I
> could do my share by answering. The answers to my early questions led me
> to books that helped to round out the little I know so that the many
> gaps don't keep me awake. Somehow, people I respect got the impression
> that I actually know something useful, so I suppose I do.
>
> You seem to be about where I was back in 1998 or so. Keep plugging at
> the questions. The things you care about will probably come clear pretty
> soon.
>
> Jerry
> _____________________________________
> * One project used a signal buried in noise to control a feedback loop.
> There was over a 40 dB Noise-to-signal ratio, but my home-made filters
> (and novel loop configuration) made it work. I still don't know how I
> could have done that one better using a Z-80 coupled by I/O to an HP85
> for operator interface and data logging.

I've done a lot of stuff with sampled data, and indeed I have designed
many a system using digitised data. I see a full understanding (and
anything less would be laziness) of digital filtering as the logical
extension.

I lurk here (mostly) listening and learning; on a linkedin page I state:

"but there is -always- something new to learn. That's what makes it fun,
after all"

That has been my guide over the last *cough* years, and I have always
said that when there was nothing new to learn I would choose a different
career.

Thank you very much for your comments, because they give me hope I shall
master the subject (as much as anyone can master it).

Cheers

PeteS

yatinkamal skrev:

> In case of digital filter, I know that we will be sampling at Fs rate and
> digital data of analog signal is present in memory but how can these
> samples behave like a low-pass or high-pass filter.

"Digital data" consist of a sequence of numbers, x(n). Digital filters
are mathematical operations that act on that sequence of numbers
to produce a new sequence of numbers, y(n).

One naive way of producing a new sequence of numbers is to
start at the beginning of x(n) and compute the average of pairs
of consecutive samples:

y(0) = (x(0))/2
y(1) = (x(0)+ x(1))/2
y(2) = (x(1)+x(2))/2
:
y(n) = (x(n-1)+x(n))/2

It does not take a lot of thought to understand that this type of
operation tends to average out any sample-to-sample variation;
this is a low-pass filter.

Similarly, computing the differences between consecutive
samples,

y(0) = (x(0))/2
y(1) = (x(1)- x(0))/2
y(2) = (x(2)-x(1))/2
:
y(n) = (x(n)-x(n-1))/2

tends to remove steady-state trends, i.e. this is a high-pass
filter.

And that's it. By varying the coefficients and introducing
feedback from the output, one can design various forms of
digital filters.

Instead of reactive elements that contain signal energy for
a certain amount of time, we here have delay elements
that contain intermediate variables, no big deal.

Just be very careful about projecting "analog-world"
concepts to the discretes data case. Difference equations
are *similar* to differential equations, not "the same as".
There are some very subtle differences(!) that you need
to keep track of.

Rune