Hi Mike,
I’m sorry, if you found my language disrespectful. The funny thing is,
I spent a fair amount of time on that last passage, trying to find
language, which was civil and, yet, still made the point that a
condescending dismissal does not a successful rebuttal to a
mathematical argument make. My apologies for any offense taken, and
I’d be happy to buy you a drink at the next conference we both attend,
in order to try and make amends.
Now, if you’re still game for a few more rounds, I think there’s some
really juicy stuff, below; stuff that’s interesting to a lot of folks
in this community. I think it’d be very productive for us to trade a
few more challenges/rebuttals on this topic; I feel like we’re really
close to blowing the lid off of this one and gaining some clarity.
I’ve made some very pointed challenges to your repost, below. Will you
parry, sir?
Thanks, in advance, for reconsidering,
-db
*From:*Mike Steinberger [mailto:msteinb@xxxxxxxxxx]
*Sent:* Tuesday, June 25, 2013 12:59 PM
*To:* David Banas
*Cc:* ibis-macro@xxxxxxxxxxxxx
*Subject:* Re: [ibis-macro] Re: On impulse and step responses.
Dave-
I believe that this type of disrespectful language has no place in any
professional environment, especially a public reflector. If you're
unwilling to live up to this expectation, then this conversion is
concluded.
Mike Steinberger
On 06/25/2013 02:53 PM, David Banas wrote:
Hi Mike,
I’ve responded in-line, below, in order to preserve context.
Thanks,
-db
*From:*Mike Steinberger [mailto:msteinb@xxxxxxxxxx]
*Sent:* Tuesday, June 25, 2013 12:23 PM
*To:* David Banas
*Cc:* ibis-macro@xxxxxxxxxxxxx <mailto:ibis-macro@xxxxxxxxxxxxx>
*Subject:* Re: [ibis-macro] Re: On impulse and step responses.
Dave-
Even though "Kronecker Delta" has the word "delta" in it, it's not
relevant to this conversation.
The problem to be solved here is that we need a sampled time domain
function that expresses the transfer function of a linear, time
invariant process in a way that
1. Allows us to calculate the output of the process in response to a
particular sampled time domain input waveform
*/[David Banas] Isn’t this parroting my post, below? For instance,
isn’t this just a rewording of this excerpt from my text, below?:/*
That is, what input sequence to a discrete time, linear, shift
invariant (LSI) system will elicit a response from that system, which
will then allow us to predict the output of that system, in response
to any arbitrary input, via convolution?
2. Allows us to perform the same calculation and get a re-sampled
version of the same result if we change the sample interval and
re-sample the input waveform.
I submit that you can't solve this problem by ignoring the sample
interval.
*/[David Banas] And you are correct. Each time Init() is called, we
must re-calculate the tap weights of our digital filter, according to
the value we find in the “sample_interval” parameter. I’m not sure how
anything I said, previously, would suggest that I was intending to
ignore the sample interval. What I claim is that there is no need for
the EDA tool to scale the values in the “impulse_matrix” parameter by
the sample rate. At the time those values are about to be processed by
the filter, Init() has already adjusted the filter tap weights to the
new sample rate./*
Your example assumes that the sample interval is equal to the symbol
duration.
*/[David Banas] No, it doesn’t. Please show me, precisely, where it
makes this assumption./*
*//*
That is, the delay between x(i-1) and x(i) is one symbol time.
*/[David Banas] I make no such assumption. The {x_i } are intended as
samples, not bits. In fact, the notion of “bit time”, or “unit
interval”, is nowhere to be found in my previous post./*
*//*
In some contexts, that's a perfectly fine problem to solve, but that
doesn't happen to be the problem we need to solve. We need waveforms
which have been uniformly sampled with a sample interval that is a
small fraction of a symbol duration.
*/[David Banas] I agree with you, and claim that nothing I have said,
previously, is inconsistent with this assertion./*
You will find it instructive to re-derive the equations in your
example in a way that makes the sample interval an explicit variable.
*/[David Banas] Those equations aren’t so much “derived” as they are
“quoted” from a very well respected graduate level text on DSP. So, if
you’re right and they’re missing the sample rate as a necessary
parameter, then DSP must be being taught wrong in a fair number of
American post-graduate courses. I suspect that this is not the case./*
*//*
I'm not going to derive those equations for you, but if you want to
re-derive them, I'm willing to check your math.
*/[David Banas] Thanks for the offer, but at this point I lack the
confidence in your understanding of the fundamental principles at
hand, which would be required, in order for me to refer to you as an
authoritative reference on this topic./*
Respectfully yours,
Mike S.
On 06/25/2013 01:17 PM, David Banas wrote:
Hi Mike,
Thanks for the reply. Respectfully, you’re wrong.
This notion of enforcing an area of 1, when we move into the discrete
time domain, is very misguided.
As is noted in most DSP texts, the discrete time equivalent of the
continuous time Dirac delta function is the Kronecker delta sequence:
{1, 0, 0, …}. And the magnitude of the first element of that sequence
is /not/ scaled by the system sampling rate.
https://en.wikipedia.org/wiki/Kronecker_delta#Digital_signal_processing
(Before you criticize me for relying upon Wikipedia for my
understanding of DSP fundamentals, let me assure you that I am posting
the above only because it is extremely convenient for all to view
quickly, and that I have verified what it says against my own graduate
school DSP text: “Digital Signal Processing” by Richard A. Roberts and
Clifford T. Mullis.)
I believe we can understand why, by going back to the original
motivation for deriving the Dirac delta function and then following
the same motivation for deriving the Kronecker delta sequence. So,
then, why was the Dirac delta function useful? Well, because it
produces output from a continuous time system, which allows us to
predict the output of that same system to /any arbitrary input/, via
continuous time convolution:
/yt=τ=0∞h( 64;)∙x(t-τ)dτ/
Where y(t) is the output of the system, in response to any arbitrary
input, x(t), and h(t) is the response of the system to the Dirac delta
function as input; the so called “impulse response” of the system. (I
have assumed a causal system.)
Now, what is the discrete time equivalent of the Dirac delta function?
That is, what input sequence to a discrete time, linear, shift
invariant (LSI) system will elicit a response from that system, which
will then allow us to predict the output of that system, in response
to any arbitrary input, via convolution? We have, for convolution in
discrete time:
/yi=k=0N-1hk∙xi-k/
Where {y_i } is the output sequence of the order-N system, in response
to input sequence, {x_i }, and {h_k } is the discrete time equivalent
of the system impulse response. That is, it is that sequence, which
uniquely describes our LSI system, such that we can use the equation,
above, in order to find the output sequence of the system, in response
to any arbitrary input sequence.
Now, how do we find {h_k } for some particular LSI system? Let’s take
a simple, 2-tap FIR filter for example:
/yi=b0∙xi+b1∙xi-1=k=01bk∙xi-k/
And we find that, for a 2-tap FIR filter, {h_k } = {b_k }.
Now, we have to assume that we do not know {b_k }, a priori. And the
final question that remains then is, “What sequence should we use as
input to the 2-tap FIR filter, in order that it generate {b_k } as its
output?” When we find the answer to this question, we will have found
the discrete time equivalent of the Dirac delta function.
Let’s try, first, the Kronecker delta sequence: {x_k } = {1, 0, 0, …}:
/yi=k=01bk∙xi-k=bi,i//∈0,1;0 otherwise=b0,b1,0,0,…=bi(Ignoring
trailing zeros.)/
//
/(since, in this case, x_k = 0 for k /= 0)./
//
/And, so, we find that the discrete time equivalent to the Dirac delta
function is not the sequence, {<sample_rate>, 0, 0, …}, but simply,
{1, 0, 0, …}./
//
/If you’ve read this far, thanks! ;-)/
/-db/
//
//
*/From:/*/ibis-macro-bounce@xxxxxxxxxxxxx
<mailto:ibis-macro-bounce@xxxxxxxxxxxxx>
[mailto:ibis-macro-bounce@xxxxxxxxxxxxx] *On Behalf Of *Mike Steinberger
*Sent:* Tuesday, June 25, 2013 9:29 AM
*To:* ibis-macro@xxxxxxxxxxxxx <mailto:ibis-macro@xxxxxxxxxxxxx>
*Subject:* [ibis-macro] Re: On impulse and step responses./
//
/Dave-
I clarified that point for Greg Edlund on this e-mail reflector.
Here's a repeat of my response to him (minus the snippy comments):
If you want to use a narrow pulse of whatever shape, that's fine;
however it is essential that the pulse always has unit area (volts *
seconds). Therefore, as your pulse gets narrower and narrower, its
amplitude has to get greater and greater. In fact, the Dirac delta
function has, by definition, unit area, in that it's defined as the
limit of your narrow pulse (with unit area) as the width of the pulse
goes to zero.
In the sampled data World, we don't actually take the width of the
pulse to zero. Rather, we leave it one sample wide, as being the
narrowest pulse we can generate in that domain. The sampled data
equivalent of the (continuous time domain) Dirac delta function
therefore has a width of one sample and an amplitude of one over the
sample interval.
Hope this helps.
Mike S.
On 06/25/2013 11:04 AM, David Banas wrote: /
/Hi Fangyi,/
//
/Thanks for the reply./
/Please, see below./
//
/Thanks,/
/-db/
//
//
*/From:/*/fangyi_rao@xxxxxxxxxxx <mailto:fangyi_rao@xxxxxxxxxxx>
[mailto:fangyi_rao@xxxxxxxxxxx]
*Sent:* Thursday, June 20, 2013 8:18 AM
*To:* David Banas; ibis-macro@xxxxxxxxxxxxx
<mailto:ibis-macro@xxxxxxxxxxxxx>
*Subject:* RE: On impulse and step responses./
//
/David;/
//
/Step response has an unit of volt. Impulse response, which is the
derivative of step response by definition, has an unit of volt/sec./
*/[David Banas] If this discussion pertained to the continuous time
domain, I would agree with you, but it doesn’t. This discussion
pertains to the discrete time domain. (It has to, since we’re sending
in a discrete set of samples, taken at a uniform sampling interval, to
Init().) And, in that domain, both quantities must have the same unit,
since we require:/*
*/uk=i=0khi/*
*//*
*/where {u_k } is the “unit step response sequence” and {h_i } the
“unit pulse response sequence” of the LSI discrete time system being
discussed, and I have taken the liberty of assuming we’re only
interested in describing causal systems. In our particular
application, the most reasonable unit for these two sequences seems to
be “Volt”, which is why I’m very perplexed as to why several of us
seem to feel that “Volts/sec.” is the proper unit to be sending into
Init(). Does our current spec. call out the exact units to be used?/*
*//*
/Also, please keep in mind that the Dirac delta function has an unit
of 1/sec. /
//
/Regards,/
/Fangyi/
//
*/From:/*/ibis-macro-bounce@xxxxxxxxxxxxx
<mailto:ibis-macro-bounce@xxxxxxxxxxxxx>
[mailto:ibis-macro-bounce@xxxxxxxxxxxxx] *On Behalf Of *David Banas
*Sent:* Thursday, June 20, 2013 7:50 AM
*To:* ibis-macro@xxxxxxxxxxxxx <mailto:ibis-macro@xxxxxxxxxxxxx>
*Subject:* [ibis-macro] On impulse and step responses./
//
/Hi all,/
//
/In our work, we often take as a priori that the impulse response is
the time derivative of the step response. As I puzzle over this
further, I realize that I’m stumped by something very fundamental,
which is this:/
//
/A quantity, which is the time derivative of some other quantity,
cannot have the same units as that other quantity. And, yet, when we
discuss/measure/simulate either a step response or an impulse
response, we expect to be talking about / measuring / viewing a
voltage as a function of time, in both cases! How can this be?/
//
/Thanks,/
/-db/
//
//
/
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