While I was learning signal theory, I have come across different definitions for the unit step signal. For example.
$u(t)=\begin{cases} 1 & t\geq0\\ 0 & t<0\\ \end{cases}$
$u(t)=\begin{cases} 1 & t>0\\ 0 & t<0\\ \frac{1}{2} & t=0\\ \end{cases}$
$u(t)=\begin{cases} 1 & t>0\\ 0 & t\leq0\\ \end{cases}$
However, in the context of LTI systems the outputs are not different. i.e LTI systems with these inputs produce same outputs. How can one justify this?

  • 3
    $\begingroup$ Compute $\int_{-1}^1 u(t) \mathrm dt$ for the three definitions of $u(t)$ and report back on the difference. The integral is the output (at $t=0$) of a (non causal) short-term integrator, which is an example of an LTI system. $\endgroup$ Oct 12 '17 at 2:21
  • $\begingroup$ @Dilip-I agree what u say but that does not justify for all LTI systems. $\endgroup$
    – JaalaP
    Oct 12 '17 at 8:18
  • $\begingroup$ And that is why I did not write an answer to your question but merely posted a comment. Note that what you wish to prove is false for all LTI systems that are pure delays. $\endgroup$ Oct 12 '17 at 10:40
  • $\begingroup$ @Dilip- the integral result is 1 by the way, I forgot to add. Please don't mind. $\endgroup$
    – JaalaP
    Oct 12 '17 at 10:40
  • $\begingroup$ @Dilip-Thanks. So, we can say for all (LCCDE) continuous LTI systems this is true. I mean systems whose i/p-o/p relation can be represented as Linear Constant-Coefficient Differential Equation. $\endgroup$
    – JaalaP
    Oct 12 '17 at 10:44

$$\begin{align} u(t) &= \begin{cases} 1 & t\geq0\\ 0 & t<0\\ \end{cases} \\ \\ u(t) &= \begin{cases} 1 & t>0\\ 0 & t<0\\ \tfrac{1}{2} & t=0\\ \end{cases} \\ \\ u(t) &= \begin{cases} 1 & t>0\\ 0 & t\leq0\\ \end{cases} \end{align}$$

the reason these all produce the same output when $u(t)$ is used as an input or when $u(t)$ multiplies something else that doesn't have a dirac impulse $\delta(t)$ at $t=0$, is because this expression will find itself in an integral, specifically the convolution integral.

there is zero difference between those three expressions in the integral. there is zero area underneath the point $u(t)\Big|_{t=0}$ no matter what the value is of $u(0)$.

  • $\begingroup$ (+1) recently in a question I had to sample the unit step function $u(t)$ and I did it as $u[n]$ which didn't of course work. Then I redefined $u[0]=0.5$ then it worked ok. Of course $u(t)$ is not a bandlimited signal. But in some problems where one has to discretize an analog equation you find yourself in need of replacing $u(t)$ with something discrete? $\endgroup$
    – Fat32
    Oct 13 '17 at 11:26
  • $\begingroup$ usually the discrete-time unit step is $$ u[n] = \begin{cases} 1 & n \ge 0\\ 0 & n < 0\\ \end{cases} \qquad n \in \mathbb{Z} \\ $$ which works with sampling the top definition of $u(t)$ if you wanted a bandlimited continuous-time unit step, it would have to be the integral of a sinc function and would be equal to $\tfrac12$ at $t=0$. or, i s'pose, you could reconstruct a $u(t)$ directly with the sum of sinc functions and the definition of the samples $u[n]$ above. $\endgroup$ Oct 13 '17 at 20:57

The Wikipedia article on Heaviside has a section that discusses the 3 different conventions on $H(0)$:


Essentially, the step function is used in probability theory as well but is called something different.

I believe that the "symmetric" $H(0)=\frac{1}{2}$ version is probably most suitable in Linear Systems Theory. In discussions on the Gibbs phenomena, the truncated Fourier series interpolates discontinuities at the mid-value which in this case is $\frac{1}{2}$.

Also, if you look at the article, there are a number of limits of continuous functions like Erf, that approach the step as some parameter approaches some value. If you were to set up a physical experiment, you would never be able to realize an ideal step function, you would need infinite bandwidth, so in reality, the best you could achieve would be something like a one of those continuous functions. $H(0)=1/2$ is a vestige of physical realizability.


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