# 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.

Time shifting

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

Graph of g(t) = sin t · u(t − 2π), the product of a shifted unit function.

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

Graph of g(t) = 10e^(-2t)· u(t − 5), the product of a shifted unit function..

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)

Graph of g_1(t) = (4t + 2) · u(t), the product of a unit function..

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)

Graph of g_2(t) = (4t + 2) · u(t-1), the product of a shifted unit function..

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)

Graph of g_3(t) = (4t - 2) · u(t), the product of a unit function..

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)

Graph of g_4(t) = (4t - 2) · u(t-1), the product of a shifted unit function..

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)

Graph of g_1(t) = sin t · u(t), the product of a unit function..

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

(b) g_2(t) = sin t · u(t − 0.7)

Graph of g_2(t) = sin t · u(t-0.7), the product of a shifted unit function..

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

(c) g_3(t) = sin (t − 0.7) · u(t)

Graph of g_3(t) = sin (t-0.7) · u(t), the product of a shifted unit function..

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)

Graph of g_4(t) = sin (t-0.7) · u(t-0.7), the product of a shifted unit function..

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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