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Consider in the discrete-time domain, the unit-step function : $$ u[n]=\begin{cases} 1&\text{if $n\geq 0$}\\ 0&\text{if $n<0$} \end{cases} $$ We know that the first-order difference equation describes a relation between the unit-impulse function and the unit-step function as follow : $$ \delta[n]=u[n]-u[n-1] $$ I was wondering that in general, how can we express a difference equation of this kind : $$ u[n-a]-u[n-b] $$ In terms of unit impulses $\delta$? With $a,b\in\mathbb{Z}^{+}$

Joumana L. Kincaid
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  • Is $ a-b$ an integer? If so, you can chain together $\delta[n] + \delta[n-1] + \ldots$. – Calvin Lin Mar 12 '21 at 11:26
  • Hello, yes I have fixed the question, can you clarify more? – Joumana L. Kincaid Mar 12 '21 at 11:29
  • Can you write out what $ \delta[n] + \delta[n-1]$ is? How about $ \delta[n] + \delta[n-1] + \delta[n-2]$? Do you see a pattern? Now, express $u[n-a] - u[n-b]$ in terms of these $\delta[n-i]$. – Calvin Lin Mar 12 '21 at 11:30
  • So it appears that $\displaystyle u[n]-u[n-b]=\sum_{k=0}^{b}\delta[n-k]$ so this means $\displaystyle u[n-a]-u[n-b]=\sum_{k=0}^{a}\delta[n-k]-\sum_{k=0}^{b}\delta[n-k]$ Can we improve this sum? @CalvinLin – Joumana L. Kincaid Mar 12 '21 at 11:38
  • So I assume this should give us $\displaystyle\sum_{k=a}^{b}\delta[n-k]$ assuming $a<b$ right? – Joumana L. Kincaid Mar 12 '21 at 11:51

1 Answers1

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$u(x)=\int\limits\limits_{-\infty}^x\delta(x)\,dx$ where $u(x)$ is the Heaviside step function and $\delta(x)$ is the Dirac delta function.

So $u(x-a)-u(x-b)=\int\limits\limits_{-\infty}^x(\delta(x-a)-\delta(x-b))\,dx$

Steven Clark
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