# 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) = (4(t − 1) + 2) · u(t-1) = (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)`

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