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When analyzing simple circuits to understand the theory of their operation, we often make use of the Heaviside step function, \$u(t)\$.

When we calculate a convolution integral, we must understand the time-reversed and delayed function \$u(t-\tau)\$.

How do we plot \$u(t−τ) \$ ?

What is the range of \$\tau\$ for which the value of \$u(t−τ)\$ is 1.

\$u(t−τ)\$=\$u(−(τ-t))\$.

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    \$\begingroup\$ Not sure why this was closed as "off topic". It is directly related to "the theory and simulation of electromagnetic forces" which is on-topic per the FAQ. \$\endgroup\$
    – The Photon
    Feb 23, 2013 at 15:54
  • \$\begingroup\$ @ThePhoton I flagged this as off topic since the first revision looked like math to me and I didn't see it actually was the Heaviside step function. Sorry about that! \$\endgroup\$
    – user17592
    Feb 23, 2013 at 17:25

2 Answers 2

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The step function is one when it's argument is greater than or equal to zero, as the wiki page you linked to says.

Therefore, to determine when the function outputs a 1, you need to determine when it's argument meets the above definition. So you have the equation

$$t-\tau \ge 0$$

When is this true? When

$$t \ge \tau$$

So really \$u(t-r)\$ is just a delay with respect to the variable t. There are some very good interactive tools for understanding this kind of stuff here. As you can see in the "Joy of Convolution" applet on the linked page, the step function is frequently used with respect to tau, in which case it is time-reversed as you mentioned.

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  • \$\begingroup\$ One thing I'd maybe reword, though. You say "u(t-r) is just a shift". But usually when we use this we're actually integrating with respect to r. And if you look at this as a function of r, with t as a parameter, then it's both time-reversed and shifted. \$\endgroup\$
    – The Photon
    Feb 23, 2013 at 17:32
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How do we plot \$u(t-\tau)\$ ?

Like Nick points out, it's a very simple function. Whenever \$t-\tau\$ is greater than zero, the value is 1, otherwise it's 0.

One more thing to point out though, is that when we use this function, we're usually calculating a convolution integral like

\$\int_a^bh(\tau)u(t-\tau)\mathrm{d}\tau\$

To visualize what's going on in this integral we really want to plot \$u(t-\tau)\$ as a function of \$\tau\$, not t. That means you want to think of the result as being 1 when \$\tau < t\$ and 0 when \$\tau \ge t\$, instead of the other way around.

enter image description here

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