# What happens to the frequency response as a single sample Kronecker delta impulse widens into a square pulse, and then into two distinct steps?

## 1) Kronecker

I was instructed here that a single sample Kronecker delta unit impulse function (goes from 0 to $$A$$ to 0 again in single sample) has a white noise type frequency response - all frequencies are equally excited. Is that true?

## 2) Pulse

If you widen that into 2, 3, or $$n$$ number of samples width, it becomes a pulse (which I suppose could be thought of as 1/2 of a square wave or a square wave with a shifted baseline) for which there will be a high frequency rolloff as you widen it.

But in what manner? A square wave has a frequency response of $$f_0/f$$:

At what $$n$$ of sample width does it go from being a full spectrum "white" Dirac, to a square wave type amplitude response, which presumably also has a harmonic fundamental frequency? Is it an immediate change that happens as soon as you are 2 samples wide? Or what width does this happen at and it becomes "harmonic"?

## 3) Steps

If you widen the square pulse even further, eventually, at some point, I presume it is no longer a pulse, but rather becomes a pair of two steps: one step up to $$A$$ then eventually another step back to $$0$$.

A step function I believe excites at $$1/f$$ with no harmonic content, right? What is the cutoff between it being a harmonic pulse and two non-harmonic steps?

## Review

So is it true that if we start with a single sample Dirac, we get full spectrum "white" excitation, then as we widen it we get the result of a harmonic pulse/square wave, and then eventually we get a full spectrum $$1/f$$ explosion of two distinct steps?

Are there specific thresholds for these transformations? Are they discrete or does the transition happen slowly at any stage?

This is all in the context of audio signals at a given sample rate (say 44.1 kHz). Thanks for any clarification.

• You're really throwing around terminology in unhealthy ways. You cannot say "this is a Dirac", then speak of samples, claiming you widen the Dirac. These are incompatible concepts! Jan 24 at 1:56
• Okay, but you obviously understand what I mean. Is a Dirac unit impulse in the sample domain not a single sample amplitude spike? What happens when you widen that from a single sample width? It becomes a pulse right? And if you widen that it becomes two steps right? Hence my question. If you think the terminology should be different tell me what to change it to. I don't know any more than this.
– mike
Jan 24 at 2:13
• no, that's not a Dirac; that's a Kronecker Delta, and it corresponds to a sin(x)/x in the continuous-time domain, not to a Dirac Delta. If you sample a Dirac Delta, then multiple things are true: 1. stochastically, it's certain that you miss it, but let's ignore that, 2. there's really no concept of widening a Dirac Delta, by the very definition of what a Dirac Delta is. We can talk about things like a rectangular or a Gaussian-bell-shaped signal with fixed area that we make wider, but you'll see that any of these inherently leak into infinitely many samples – that's very much in line with Jan 24 at 11:27
• the basic theory of sampling! Jan 24 at 11:30
• "Is a Dirac unit impulse in the sample domain not a single sample amplitude spike?" No. A Dirac is unlimited in bandwidth and hence cannot be sampled without aliasing at any sample rate. You need to be very clear in your question whether your talking about the time continuous domain to time discrete domain. This can make a big difference. Jan 24 at 13:59

Kronecker

In the sampled-time domain a Kronecker delta, $$\delta(n)$$, has a perfectly flat spectrum. If you look at a table of discrete-time Fourier transform pairs, this is at the top of the list: $$\mathcal F^{-1}\left \{ \delta(n) \right \} = A$$, where $$A$$ is a positive real number whose value depends on how the author likes to treat their Fourier transform pairs.

Pulse, Step

As soon as there's more than one non-zero sample, it's a pulse. If it goes up and then down again, it's always a pulse. If it goes up at $$k = 0$$ and down after you stop caring, then you can probably model it like a step -- unless you were wrong, in which case you really did care about when it comes down.

The only specific threshold is that a Kronecker delta is, specifically, one at $$n = 0$$, and zero otherwise, and if you have something that's not zero for $$n \ne 0$$, then it's not a Kronecker delta.

Context

Whether you're sampling audio at $$44.1 \mathrm{kHz}$$, sea level at $$44.1 \mathrm{\mu Hz}$$, or radio at $$44.1 \mathrm{GHz}$$, once you get it into the sampled-time domain the math is all the same, and the only time scale is in "samples".