# 1a. The Unit Step Function (Heaviside Function)

In engineering applications, we frequently encounter functions whose values change abruptly at specified values of time *t*. One common example is when a voltage is switched on or off in an electrical circuit at a specified value of time *t*.

The value of *t* = 0 is usually taken as a convenient time to switch on or off the given voltage.

The switching process can be described mathematically by the function called the **Unit Step Function** (otherwise known as the **Heaviside function** after Oliver Heaviside).

## The Unit Step Function

Definition: The unit step function, `u(t)`, is defined as

`u(t)={: {(0, t < 0), (1, t > 0) :}`

That is, *u* is a function of time *t*, and *u* has value **zero** when time is negative (before we flip the switch); and value **one **when time is positive (from when we flip the switch).

Graph of `f(t)=u(t)`, the unit step function.

### Value at *t* = 0?

In some text books you will see the unit step function defined as having value 1 at *t* = 0, as follows:

`u(t)={: {(0, t < 0), (1, t >= 0) :}`

We would indicate the discontinuity on our graph like this:

Graph of `f(t)=u(t)`, the unit step function, with `f(0) = 1`.

Also, sometimes you'll see the value given as `f(0) = 0.5`.

In this work, it doesn't make a great deal of difference to our calculations, so we'll continue to use the first interpretation, and draw our graphs accordingly.

## Shifted Unit Step Function

In many circuits, waveforms are applied at specified intervals other than `t=0`. Such a function may be described using the **shifted** (aka **delayed**) unit step function.

### Definition of Shifted Unit Step Function

A function which has value `0` up to the time `t = a` and thereafter has value `1`, is written:

`u(t-a)={{: (0, if, t < a), (1, if, t > a) :}`

### Example 1 - Shifted Unit Step Function

`f(t) = u(t − 3)`

The equation means *f*(*t*) has value of `0` when `t < 3` and `1` when `t > 3`.

The sketch of the waveform is as follows:

Graph of `f(t)=u(t-3)`, a shifted unit step function.

## Rectangular Pulse

A common situation in a circuit is for a voltage to be applied at a particular time (say *t = a*) and removed later, at *t = b* (say). We write such a situation using unit step functions as:

`V(t) = u(t − a) − u(t − b)`

This voltage has strength `1`, duration `(b − a)`.

### Example 2 - Rectangular Pulse

The graph of `V(t) = u(t − 1.2) − u(t − 3.8)` is as follows. Here, the duration is `3.8 − 1.2 = 2.6`.

Graph of `V(t)=u(tâˆ’1.2)âˆ’u(tâˆ’3.8)`, an example of a rectangular pulse.

## Exercises

### Need Graph Paper?

Write the following functions in terms of **unit step** function(s). Sketch each waveform.

(a) A 12-V source is switched on at *t* = 4 s.

Answer

Since the voltage is turned on at *t* = 4, we need to use *u*(*t* − 4). We multiply by 12 since that is the voltage.

We write the function as follows:

`V(t)=12·u(t-4)`.

Here's the graph:

Graph of `V(t)=12·u(t-4)`, a shifted step function.

(b) `V(t)={{: (1, 0 < t < a),(0, t > a) :}`

(Assume *a* > 0.)

Answer

In words, the voltage has value `1` up until time `t = a`. Then it is turned off.

We have a "rectangular pulse" situation and need to use this formula:

`V(t) = u(t − a) − u(t − b)`

In our example, the pulse starts at `t = 0`, so we use `u(t)`, and finishes at `t = a`, so we use `u(t − a)`.

So the required function is:

`V(t)=1·[u(t)-u(t-a)]` `=u(t)-u(t-a)`

Graph of `V(t)=u(t)-u(t-a)`, a shifted unit step function.

(c) One cycle of a square wave, `f(0) = 4`, amplitude = `4`, period = `2` seconds.

Answer

`f(0) = 4` means we start at value `4`.

If the whole wave has period `2`, and it is a square wave, then it means for half of the time, the value is (positive) `4` and the other half it is `-4`.

So for the first second, it has value `4`, for the second second, the function value is `-4`.

We write this, using the "rectangular pulse" formula from before:

`f(t) = 4·{u(t) − u(t − 1)}` `-4·{u(t-1) − u(t − 2)}`

`=4·u(t)-8·u(t-1)+4·u(t-2)`

The graph of this first cycle is:

Graph of `f(t)=4·u(t)-8·u(t-1)+4·u(t-2)`, a square wave.

(d) The unit Ramp function (i.e.`f(t) = t` for `t > 0`)

Answer

The unit ramp function has slope `1` [so the function is simply `V(t) = t`], starting from `t = 0` [so we need to multiply by `u(t)`], and passes through `(0, 0)`.

So the voltage function is given by:

`V(t) = t · u(t)`

The graph of the function is:

Graph of `V(t)=t · u(t)`, the unit ramp function.

(e) One cycle of a sawtooth waveform (i.e. `f(t)=a/b t` for `0 < t < b`. Assume `a > 0`.)

Answer

Our graph starts at `t = 0` and has slope `a / b`. It finishes at `t = b`.

So our function will be:

`f(t)=a/b t · {u(t)-u(t-b)}`

The graph of our function:

Graph of `f(t)=a/b t · {u(t)-u(t-b)}`, a sawtooth waveform.

(f) `V(t)={ {: (0,t < 3), (2t+8,3 < t < 5), (0,t > 5) :}`

Answer

In this example, our function is `V(t) = 2t + 8` which has slope `2` and `V`-intercept `8`.

The signal is only turned on between `t = 3` and `t = 5`. The rest of the time it is off.

So our voltage function will be:

`V(t) ` `= (2t+8) ·{u(t − 3)− u(t − 5)}`

The graph is as follows:

Graph of `V(t) = (2t+8) ·{u(t − 3)− u(t − 5)}`. The dashed line is `V(t) = 2t + 8`.

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