# 1b. Products Involving Unit Step Functions

When combined with other functions defined for `t > 0`, the unit step function "turns off" a portion of their graph.

### Later, on this page...

The concept is related to having a switch in an electronic circuit **open** for a period of time (so there is no current flow), then the switch is **closed** (so the current begins to flow).

### Example 1 - Products with Unit Functions

(a) If `f(t) = sin\ t`, then the graph of `g(t) = sin\ t · u(t − 2π)` is

The `sin\ t` portion starts at `t = 2π`, because we have multiplied `sin\ t` by `u(t − 2π)`.

We use the dot (`·`) for multiplication so that it is easier to read.

(b) If `f(t) = 10e^(-2t)`, then the graph of `g(t) = 10e^(-2t)· u(t − 5)` is

The portion `10e^(-2t)` starts at `t = 5`.

## Product of *u*(*t*) vs. Shifting the Function Along the *t*-axis

Note the differences between the following:

`f(t) · u(t)`, where the `f(t)` part begins at `t = 0`.

`f(t) · u(t − a)`, where the `f(t)` part begins at `t = a`.

`f(t − a) · u(t)`, where the `f(t)` part has been **shifted** to the right by `a` units and begins at `t = 0`.

`f(t − a) · u(t − a)`, where the `f(t)` part has been **shifted** to the right by `a` units and begins at `t = a`.

Let's see some examples.

### Example 2

Let `f(t) = 4t + 2` and `a = 1`. We see different combinations of shifting with different starting points.

(a) `g_1(t) = f(t) · u(t) = (4t + 2) · u(t)`

In this example, the `4t + 2` part starts at `t = 0`.

(b) `g_2(t) = f(t) · u(t − a) = (4t + 2) · u(t − 1)`

In this example, the `4t + 2` part starts at `t = 1`.

(c) `g_3(t) = f(t − a) · u(t) = (4(t − 1) + 2) · u(t) = (4t − 2) · u(t)`

In this example, the `4t + 2` part has been shifted 1 unit to the right and starts at `t = 0`.

(d) `g_4(t) = f(t − a) · u(t − a) = (4t − 2) · u(t − 1)`

In this example, the `4t + 2` part has been shifted 1 unit to the right (like example (c)) and starts at `t = 1`.

### Example 3

Let `f(t) = sin\ t` and `a = 0.7` and we combine them to shift our graph and start at different times, similar to what we did in Example 1.

(a) `g_1(t) = sin\ t · u(t)`

In this example, the `sin\ t`* *part starts at* *`t= 0`.

(b) `g_2(t) = sin\ t · u(t − 0.7)`

In this example, the `sin\ t`* *part starts at `t = 0.7`.

(c) `g_3(t) = sin (t − 0.7) · u(t)`

In this example, the `sin\ t`* *part has been shifted `0.7` units to the right, and it starts at `t=0`.

(d) `g_4(t) = sin (t − 0.7) · u(t − 0.7)`

In this example, the `sin\ t`* *part has been shifted `0.7` units to the right, and it starts at `t = 0.7`.

## Exercises

### Need Graph Paper?

Rewrite the following functions in a suitable way and then sketch the functions:

1. `f(t) = u(t) + (1 − t) · u(t − 1)` `+ (t − 2) · u(t − 2)`

2. `f(t) = t^2 · u(t) − (t^2− 4) · u(t − 2)`

3. `f(t) = u(t) + (sin\ t − 1) · u(t − π/2) − (sin\ t + 1) · u(t − (3π)/2) + u(t − 2π)`

4. `f(t) = 3t^2· u(t)` `+ (12 − 3t^2) · u(t − 2)` ` + (4t − 40) · u(t − 4)` ` − 4(t − 7) · u(t − 7)`

### Scientific Notebook Aside...

NOTE: To graph unit step functions using Scientific Notebook, we must realise that SNB recognises "Heaviside(*t*)", but not *u*(*t*).

So we need to **define** `u(t)` as `"Heaviside"(t)`, so SNB will graph it properly. Simply type `u(t) = "Heaviside"(t)`, and click on the "New definition" button. Nothing seems to happen, but if you click on the "Show Definitions" button you will see that it is defined. Now you can graph unit step functions in terms of `u(t)`.

Graph of `y = 3·u(t − 4)`

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