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I'm trying to make sense of different notations used in DSP. This is sometimes quite hard since notation brackets like in x[n] are used both to denote a single value or a sequence.

In mathematics, the generalized Iverson bracket notation is:

$$[P]=\begin{cases} 1 & \textrm{if P is true}\\ 0 & \textrm{otherwise} \end{cases}$$

and Kronecker delta is noted either: $$\delta_{n,m}=\begin{cases} 1 & \textrm{if } n=m\\ 0 & \textrm{otherwise} \end{cases}$$

or:

$$[n=m]=\begin{cases} 1 & \textrm{if } n=m\\ 0 & \textrm{otherwise} \end{cases}$$

In DSP (example), $\delta_{n,m}$ is noted $\delta[n-m]$, a mix of the two mathematics forms I find confusing because as $\delta[n-m]$ is legitimate, then is $\delta[n-1-m]$ legitimate too?

My questions to experienced DSP community:

  • Why this notation and not the mathematical one which seems more clear?
  • Is $\delta[n-1-m]$ legitimate? If so, what is its signification?
mins
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1 Answers1

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The fact that one symbol (in this case $x[n]$) is used to denote a sequence or a single number is nothing that is specific to DSP. If you have a continuous function $f(t)$, then this symbol is used to denote the function but also to refer to its value at $t$, if $t$ is some fixed number in the domain of $f$.

Since we use the notation $x[n]$ to denote sequences (or some specific value of that sequence), it seems natural to define a sequence $\delta[n]$, which is zero everywhere except for $n=0$ where it equals $1$. That's why this sequence is also called the unit impulse (sequence).

Referring to your last question, of course $\delta[n-1-m]$ is legitimate. It is zero everywhere except when $n-1-m=0$.

As a final comment, note that $\delta[n]$ is mathematically totally unrelated to the distribution $\delta(t)$, also called Dirac impulse, even though $\delta[n]$ plays a similar role in discrete time as does $\delta(t)$ in continuous time.

Matt L.
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  • Re. to "f(t), then this symbol is used to denote the function but also to refer to its value at t, if t is some fixed number", Is it legitimate to substitute $n$ in an equation as we do for any mathematical function? E.g for a convolution, $y[n]=x_{1}[n]\ast x_{2}[n]\Longrightarrow y[4]=x_{1}[4]\ast x_{2}[4]$? – mins Mar 17 '23 at 11:20
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    @mins: The problem here is that the notation $x_1[n]x_2[n]$ doesn't make much sense in the first place. It should be $(x_1x_2)[n]$. Then $(x_1x_2)[4]$ does make sense. Same in continuous time: $f(t_0)g(t_0)$ is meaningless. – Matt L. Mar 17 '23 at 11:27
  • "The problem here is that the notation x1[n]∗x2[n] doesn't make much sense in the first place", this notation is commonly seen, e.g. here, so the comment about confusing use of brackets in my question. You do use the correct notation. – mins Mar 17 '23 at 11:31
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    @mins: Yes, it's quite common, and if just used to denote convolution - without the need to refer to a specific element of the resulting sequence - then it's kind of OK. However, it has obvious shortcomings which become obvious when you refer to a specific element. The result of convolution is a sequence and in general you need to denote as such, i.e., with a single independent variable: $(x_1*x_2)[n]$. But again, this is a general problem, regardless whether you refer to discrete time or continuous time. – Matt L. Mar 17 '23 at 11:37