# Impulse Train vs Sampling Function

1. Mar 29, 2009

### HasuChObe

How come these two functions are used interchangebly? A sampling function is supposed to be a square wave with a very small duty cycle, but at least the height is still finite. A delta function has infinite height and zero width. It seems like both square wave and dirac train are used as sampling functions interchangebly.

In my signals book, an 'impulse train' or dirac train is used to sample a function. In lab, a rectangle wave with a very small duty cycle (so that it resembles an impulse train) is used to sample a function. The rectangle wave makes sense to me because the height of the rectangle waves is finite. The infinite height dirac train is what confuses me. Explain?

What is delta(t)*h(t)?

2. Mar 29, 2009

### JamesGoh

My understanding is that delta(t) is formed from a rect() function that has its height continually increased and its width decreased (by reducing this period). This could explain how the infinite height comes about, as T approaches infinity itself.

delta(t)*h(t) = h(tau) where tau is the variable used in the convolution integral

3. Mar 29, 2009

### HasuChObe

I think they squished a gaussian function to make the delta function, but you are right. It has zero width and infinite height.

Btw, I meant delta(t) x h(t) (multiplication, not convolution). Sry for the confusion.

4. Mar 29, 2009

### JamesGoh

delta(t)xh(t) = h(0) if you consider the mathematical properties of a dirac function

5. Mar 29, 2009

### HasuChObe

I thought it was the integral of delta(t)xh(t) = h(0).

6. Mar 30, 2009

### Eidos

The heaviside step function ($u(t)$) is 1 for t>0 and 0 elsewhere.

The dirac delta function, by definition is the derivative of the heaviside unit step function.
$\delta(t) = \frac{d}{dt}u(t)$.

Hence,$\delta(t)$ is undefined at t=0 (since $u(t)$ changes instantaneously at t=0), and zero everywhere else (since $u(t)$ does not have any other rate of change).

Why its useful is that:
$\int_{-\infty}^{\infty}\delta(t) dt = 1$
which is easy to prove from the definition above and the fundamental theorem of calculus.

It also has the property that any function multiplied by it, is only non-zero where the dirac delta is non-zero. Hence it samples.

$f(t)\delta(t-\gamma)=f(\gamma)d(t-\gamma)$

We say that this dirac delta is weighted by $f(\gamma)$ at $t=\gamma$.

So if we wish to sample a function, f(t) in intervals of T.
$f(t) \overbrace{\sum_{k=0}^{\infty}\delta(t-kT)}^{Impulse Train} = \sum_{k=0}^{\infty}f(kT)\delta(t-kT)$

In answer to your question, the sampling function you are referring to is an approximation to this ideal sampling function. In the limit as the height of the square wave tends to infinity and the width of each square goes to zero we recover this ideal.

7. Mar 30, 2009

### JamesGoh

Sorry my bad. Eidos definition is correct.

8. Apr 19, 2009

### dkumar

So, the square wave is just an approximation of impulse train , right?
What other waveforms can be used as a good approximation? and why square wave is so holy that it has been used for such a long time?

and if we use another waveform type , say sine wave for samling, then what problems/ differences we might see?

thanks

9. Apr 19, 2009

### JamesGoh

Don't you mean an impulse is an approximation of a square wave ?

A "square" in a square wave has uniform area and height (typically an area of 1), which makes it easy to apply to create more complex waveforms. To create say a triangular waveform you just simply multiply a ramp function by a "square" to create the triangle

Impulse trains are ideal for sampling, because you have a stronger chance of reproducing the original wave, when one considers at the sifting property of delta functions.

Does this make sense ?

10. Apr 19, 2009

Delta functions for sampling make the math look nice -- easier to look at, easier to remember. When you sample in real life, you model the hardware that performs the sampling with (at least) a rectangle function.

Check out <http://en.wikipedia.org/wiki/Sampling_(signal_processing)>, [Broken] in particular:

"The conventional, practical digital-to-analog converter (DAC) does not output a sequence of dirac impulses (such that, if ideally low-pass filtered, result in the original signal before sampling) but instead output a sequence of piecewise constant values or rectangular pulses. This means that there is an inherent effect of the zero-order hold on the effective frequency response of the DAC resulting in a mild roll-off of gain at the higher frequencies (a 3.9224 dB loss at the Nyquist frequency). This zero-order hold effect is a consequence of the hold action of the DAC and is not due to the sample and hold that might precede a conventional ADC as is often misunderstood. The DAC can also suffer errors from jitter, noise, slewing, and non-linear mapping of input value to output voltage.

Jitter, noise, and quantization are often analyzed by modeling them as random errors added to the sample values. Integration and zero-order hold effects can be analyzed as a form of low-pass filtering. The non-linearities of either ADC or DAC are analyzed by replacing the ideal linear function mapping with a proposed nonlinear function"

Last edited by a moderator: May 4, 2017
11. Apr 25, 2009

### Ragnarok22788

The reason that the dirac delta is a sampling function is a matter of the definition. There are two ways to define it:

infinite height, zero width

or, you create the dirac sampling function by defining it as the limit

width epsilon, height 1/epsilon as epsilon --> 0

then, when integrated it simply gives 1 so when you multiply the dirac defined this way by something and integrate you simply get the value of that function only where the delta exists (a single point).