Sampling with impulse train

Question

There are many times when the textbooks used by all university students like me make an introduction to the notion of sampling of a continuous-time signal with an impulse train as shown below.

Why do we use this approach? Impulse train is something theoretical and of course it cannot be achieved in real life due to its infinite magnitude. Why don't we approach the problem by just saying that $$x_d[k] = x_c(kT)$$ and extract the relationship between z-Transform and Laplace-Transform from it?

Thnaks in advance for your answers!

But doesn't your formula do an exact same thing? Taking infinitesimally narrow samples of continuous signal at each sample interval? So how is it different from multiplying with impulse train first and then storing height of each impulse into discrete samples? They both are impossible with practical components, so a practical sample/hold circuitry can be modeled better as a train of wider pulses and thus aperture effect can be analyzed. — Justme, Aug 07 '22 at 01:11
In figure you already put magnitude as one. Then how it will be infinite? Did you mean the infinite frequency components on unit impulse function? — abhilash, Aug 07 '22 at 04:26
why only impulse signal for convolution? , Why unit impulse function is used to find impulse response of an LTI system? go through these answers and you will get an idea. — abhilash, Aug 07 '22 at 04:38

score 1 · Accepted Answer · answered Aug 06 '22 at 19:03

I would like to know the historical reason why this is done. But I suspect it's this:

Introducing sampling as multiplication by an impulse train unifies the Laplace and $z$ transforms, and it unifies all four variations of the Fourier transform (Fourier integral, Fourier series, Discrete-time Fourier transform, and Discrete Fourier Transform).

With it, the $z$ transform is just a special case of the Laplace transform; if you know the Laplace transform inside an out, you can derive characteristics of the $z$ transform from the Laplace transform characteristics and from sampling. Without it, they're two independent mathematical entities which happen to share a rather large number of properties.

With it, the Fourier transforms are a family of four. You can start with the Fourier integral and sampling, and you can derive the other three. As with the Laplace and $z$ transforms, you can derive properties of the other three transforms from properties of the Fourier integral, and sampling. Without it, they're four independent mathematical entities which happen to share a large number of properties, and those properties all have to be derived separately.

So even though $\delta (t)$ is a hard thing to get your head wrapped around (and the laxity with which engineers tend to treat it drives mathematicians up the wall), the unity it brings to the various flavors of transform is worth it when you get into the deep math of signal processing.

If someone out there has actual references that give reasons, please don't let my answer stop you from supplying you -- better -- one. — TimWescott, Aug 06 '22 at 19:05
In my mind it would be better to define a function $a(t)$ which takes value $1$ when $t = 0$ and value $0$ everywhere else. Thsi function is the same with the dirac delta except from the fact that we don;t have infinte amplitude now. If instead, I take an impulse train with period $T$ of the newly defined function $a(t)$, isn't it possible to make conclusions about the relationship beteween continuous signals and their sampled counterparts? — Anastassis Kapetanakis, Aug 07 '22 at 01:29
@AnastassisKapetanakis If you closely look at your diagram, the p(t) is not the Dirac delta but a pulse train with amplitude 1, which is basically what you are after with your signal a(t). — Justme, Aug 07 '22 at 01:46

Sampling with impulse train

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