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I'm currently in the process of going over the $\mathcal Z$-transform and more specifically its derivation. I understand and I am able to follow it up until the final step whereby involving the Laplace transform.

The step that is confusing me is indicated as the step within the red box, derivation taken from Wikipedia page on the $\mathcal Z$-transform. I do not understand how this integral solves to $e^{-nsT}$. I have searched for possible identities relating to this but have not been able to find any help.

\begin{align} X_q(s)&=\int_{0^-}^\infty x_q(t)e^{-st} dt\\ &=\int_{0^-}^\infty \sum_{n=0}^\infty x[n]\delta (t-nT)e^{-st} dt\\ &=\sum_{n=0}^\infty x[n]\color{red}{\boxed{\color{black}{\int_{0^-}^\infty \delta (t-nT)e^{-st} dt}}}\\ &= \sum_{n=0}^\infty x[n]\color{red}{\boxed{\color{black}{e^{-nsT}}}} \end{align}

Thanks for taking the time for looking at my question. Any help would be greatly appreciated!

Regards

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Remember the definition of the delta function: it integrates $1$ all over the t-axis but it is zero for any $t\neq 0$. This means that if we multiply any function by $\delta (t)$, then the function is going to be zero everywhere except for $t=0$. In this case, we have

$$ \delta (t-nT)e^{-st}= \left\{ \begin{array}{ll} \delta (t-nT)e^{-snT} & \mbox{if } t = nT \\ 0 & \mbox{if } t\neq nT \end{array} \right.$$

So if we integrate this function, we have that the integral is zero for every point except for $t=nT$. So

$$\int_0^\infty \delta (t-nT)e^{-st}dt=\int_0^\infty \delta (t-nT)e^{-snT}dt=e^{-snT}\int_0^\infty \delta (t-nT)dt$$

But by definition, $$\int_0^\infty \delta (t-nT)dt=\int_{-\infty}^\infty \delta (t-nT)dt=1$$

So $$\int_0^\infty \delta (t-nT)e^{-st}dt=e^{-snT}$$

Sidenote: For this to be true, $n$ must be positive. If it is negative, then the integral will be $0$ (as the function inside the integral will be zero for all the integration interval).

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  • $\begingroup$ Cleared that up for me - thanks very much! $\endgroup$ – Tricks May 14 '16 at 22:26
  • $\begingroup$ @Tricks it's great to provide feedback to how helpful an answer is to you, but comments are not really the right way to go about this. Click the green tick to the left of this answer to mark it as the accepted answer. This way people from the future will know that this was what solved the problem for you. Comments are second class citizens on any stackexchange site. Think of them as if they could be deleted at any time. $\endgroup$ – null May 14 '16 at 22:35
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$\delta(t)$ is an impulse that's infinitely thin and infinitely high. The area under it is 1 though. $\delta(t - nT)$ places the impulse at time $nT$.

Now this is being multiplied by some function, in your case $e^{-st}$ and that product is integrated.

Integrating is just a glorified sum. Mathematicians try to disguise it as something fancy by using a special symbol, but in the end it's really just plain old addition. What is summed up in an integral? The values of the function to be integrated within the given range.

The result of an integral is the area under the plot of a function. $\delta(t - nT)$ has area 1 and is so extremely thin, it can be thought of to exist only at one position ($nT$) and to be 0 elsewhere. Kind of like rectangle in a histogram representing one value, but so thin, it could almost work at the fashion week.

Multiplying a function by a scalar value basically means scaling the function (stretching it along the y axis). You have a rectangle of area 1 and you scale one dimension by a factor. This factor is the value of the function you are multiplying it with: $e^{-st}$ and the value is evaluated at the time the impulse occurs: $nT$, so the area of the rectangle is $1 \cdot e^{-snT}$, which is the result of the integral.

If you have been doing programming with embedded systems you are probably familiar with how masking individual bits works with bitwise operators:

  0101100101110
& 0000000100000
---------------
  0000000100000

You can think about the impulse as the single bit that masks out all values of a function except the single value at its position.

$$\int^\infty _{0^-}\delta(t-nT)e^{-st}dt = e^{-nsT}$$

Is just a special form of

$$\int^{+\infty}_{-\infty}\delta(t-T)f(t)~dt = f(T)$$

which is sometimes called

the sifting property or the sampling property. The delta function is said to "sift out" the value at $t = T$.

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  • $\begingroup$ @chromatically aberrated "Integrating is just a glorified sum" This notation was introduced by Leibniz adapting a long s (ſ), standing for summa (ſumma, latin for "sum" or "total"), you can find it in ancient texts $\endgroup$ – Laurent Duval May 15 '16 at 12:08

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