# ref: pg 174, Oppenheim, Schafer and Buck, Discrete-Time Signal Processing, second ed.

Started by May 14, 2014
Subject: δ(ω/ T) = T.δ(ω)
I'm sorry that this email turned out long. My apologies that I couldn't figure out my exact question. So only managed to disperse all the information I could that hopefully revolves around it.

To start with, I wish to understand the difference between the unit impulse function (or dirac-delta function δ(t) ) and the unit sample sequence (or discrete-time impulse, not a function though, δ(n) ). For the same, a brief prelude follows:
When x(t) becomes x(nT), I thought it was sampling. But no. I realized that it was just an impulse-train modulation, whereby x(nT) is still an analog SIGNAL (not a sequence); just that x(t) was sifted by an impulse train comprising the dirac-delta impulse function.
So x(nT) is x(t) found at intervals of time T, and not found (or zero) at all other values of t -- still continuous and analog. So when we plot x(nT), it is still against t on the abscissa, telling me that I'm still in the analog domain: the signal x(nT) is a continuous signal in 't', though it once appeared to me that it looks like a sample sequence x[n], which is not continuous. So the difference.
Transpiring the frequency domain, t -> Ω (i.e. post-impulse train modulation, t found at nT), I realize --
is not the same as t -> f (pre- impulse train modulation), f = 1/ t.
When we talk of Ω, we also talk of Fs = 1/T. And though its called sampling frequency, we're not yet in the discrete-time domain.
Moving from the impulse train modulation -> discrete-time domain:-
- x(nT) becomes x[n], i.e. a time normalization by a factor of T, and correspondingly
- In the frequency domain, Ω gets scaled/ normalized by an inverse factor 1/T or Fs = 1/T.
So Ω/Fs = ΩT = ω. Then may I conclude, that:
{ Ω : t :: ω : n }
I don't know if I should call Ω an analog frequency and ω a discrete-time frequency. If yes, then what I should I call f, where f = 1/t before we knew time as t = nT !
Actually, it is Fs which we call the sampling frequency expressed as Hertz, while Ω is expressed as radians/ sec.
For e.g. Ωs = 2π/ T. And in general Ω = ω/ T. So maybe I shouldn't even term it 'frequency'.
In the transition from the impulse-train modulated signal -> discrete-time sequence, may I comment on what we know as the impulse:
if { δ(Ω) or δ(ω/ T) : Area of the dirac-delta impulse function }, then
{ δ(ω) : the sample value or the height of the impulse sequence, and not the area }
Then w.r.t the equation in the subject (as the title) of this email, how can we equate δ(ω/ T) with δ(ω)! I mean, the author states that scaling the independent variable, also scales its AREA. Here I presume that both Ω and ω are independent variables.
First of all δ(ω) is not an area. So scaling it by T and then assigning to δ(ω/ T) which is actually an area doesn't make any sense to me. Notably, the author says δ(ω/ T) = T.δ(ω)
Only a dirac-delta impulse FUNCTION is ascribed an area. A unit sample impulse sequence is not a function, but just a floating value in height, because:
when we scale Ω by T, i.e. ω/ T * T, we're loosing information about time in toto, i.e. when we reach the abscissa scaled in terms of
ω, we only know the value of the sample in frequency domain, not its position in time, i.e. we've lost even its spacing because space-time being a continuum.
I dunno. Maybe using the sample value δ(ω) and the value of ω, we can calculate x[n] in the time domain,
(where ω = ωo +/- nωs; ωo = ΩoT = 2πFoT, Fo being the fundamental frequency of the signal x(t); ωs = ΩsT = 2πFsT = 2π).
And by the way, I'm thinking that { δ(ω) : frequency domain :: δ(n) : time domain }. Moreover, as we poked Ω, can ω be even termed 'frequency'! I mean, if we call ω discrete-time frequency, then what should I call -- some variable = 1/n in dealing with ω, just as we know Fs = 1/T in dealing with Ω. Just that we've dropped information about T before we're left purely with n. I mean, does it imply that sampling implies dropping time info in moving from nT -> n or ω/ T -> ω, having nothing to do with impulse-time modulation.
Drawing attention to pg. 174, the author seems to state that δ(ω/ T) = π and δ(ω) = π/ T (from the equation δ(ω/ T) = T.δ(ω)), which doesn't make sense to me at all, because the the graphs on the page clearly show that δ(ω) = π and δ(ω/ T) = δ(Ω) = π/ T. I mean:
ref. pg. 173 says x(t) = cos(4000πT), so in the frequency domain we have
X(jΩ) = πδ(Ω - 4000π) + πδ(Ω + 4000π). So that's what I was referring to when I wrote δ(ω) = π above.
And by the way, I forgot: "The area of an impulse is either '0' or '1'; I mean the dirac-delta impulse function δ(t). Just alike, the same applies even to the unit sample impulse δ(n), i.e. it is either '0' or '1'. Then what about { δ(Ω) : δ(t) :: δ(ω) : δ(n) }. Do the same properties also apply to δ(Ω) and δ(ω)??"
If yes, then why not keep the area of the impulse seperate from 'π', in X(jΩ) = 'π' δ(Ω - 4000π) + 'π' δ(Ω + 4000π). To me, the π is like A in the function x(t) = A cos(Ωot). It has nothing to do with the area of the impulse. I can understand if π gets multiplied with δ(ω), because in this case, it is not an area that is getting scaled, but just the value of the sample, which is '1', being multiplied by π which modifies the height of the sample. Well, you can go ahead and multiply π with δ(Ω) too, but don't say that the area is now scaled, because here its a dirac-delta function involved, whose area is ALWAYS 1. Multiplying it by π doesn't change its area.
And coming back to the same question, "How can you equate an area with a sample". I mean, is the author trying to say that multiplying a unit sample impulse with T mutates it into a dirac-delta impulse?? That is ridiculous. Once you've lost information about T in moving from nT -> n, you can never go back. You just know the value of the samples, but you don't know the rate at which they arrive. Which is like saying, "I know that the second sample proceeds the first, but I don't know at what interval."
Or should I think that though I've lost T, the bastion of intervals is compensated by ω! Because as said earlier,
ω = ωo +/- nωs. Or for what matter, Ω = Ωo +/- kΩs.
So with a quantity δ(Ω - (Ω)), i.e. the points at which t = nT (in the frequency domain ofcourse), we've
δ(Ω - Ωo +/- kΩs) = δ(ω/T - 2πFo +/- k.2π/T), because ΩT = ω; and when Ω = Ωs, ω = 2π. So Ωs = 2π/ T.
So when in moving from Ω -> ω, the above quantity becomes δ(ω - 2πFoT +/- k.2π). Which to me, as I see it, implies that we've really not lost any information about time, because it is retained in the quantity 2πFoT, where it does not get cancelled. Because the fundamental frequency is independent of Ω or ω. It was in existence even before they came into picture.
I mean, if the information about the interval is lost, there is no way that we can recover the original signal x(t).
Sorry. I know that I'm only able to envisage. But I'm unable to vivify the exact problem that lurks behind this email. If somebody could please help me identify the same.

_____________________________________