# 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

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)

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

Expanding where possible:

= u(t) + u(t − 1) − t · u(t − 1) + t · u(t − 2)  − 2 · u(t − 2)

= [u(t) − u(t − 1)]  + 2 · u(t − 1)  − t · [u(t − 1) − u(t − 2)]   − 2 · u(t − 2)

[We wrote u(t − 1) as  − u(t − 1) + 2 · u(t − 1), to get the expression in the form we need.]

= [u(t) − u(t − 1)]  + 2 · [ u(t − 1) − u(t − 2)]   − t · [u(t − 1) − u(t − 2)]

[We simply moved the last term.]

= 1 · [u(t) − u(t − 1)]  + (2 − t) · [ u(t − 1) − u(t − 2)]

[Collecting like terms.]

From this expression, we can graph the function. It has value:

1 between 0 < t < 1

2 − t between 1 < t < 2

0 thereafter

NOTE: We could have obtained the graph of

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

by adding ordinates (y-values) of the 3 parts as follows:

u(t) → red

(1 − t) · u(t − 1) → blue

(t − 2) · u(t − 2) → magenta

The green part is the answer.

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

Expanding:

f(t)= t^2 · u(t) -t^2 · u(t-2)  + 4 · u(t-2)

f(t)= t^2 · {u(t)  - u(t-2)}  + 4 · u(t-2)

So our graph is as follows. It is a parabola before t=2, and then has constant value 4 thereafter.

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

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

Expanding:

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

f(t) = [u(t) − u(t − π/2)]  + sin t · [u(t − π/2) − u(t − (3π)/2)]   − [u(t − (3π)/2) - u(t − 2π)]

From this expression, we can see that the function has value:

0 for t < 0

1 between 0 < t < π/2

sin t (it is a curve) between π/2 < t < (3π)/2

-1 between (3π)/2 < t < 2π

0 for t > 2π

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

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

Expanding:

= 3t^2· u(t) + 12 · u(t − 2)  − 3t^2 · u(t − 2)  + 4t · u(t − 4)   − 40 · u(t − 4)   − 4t · u(t − 7)  + 28 · u(t − 7)

Collecting like terms:

= 3t^2· [u(t) − u(t − 2)]  + 12 · [u(t − 2) − u(t − 4)]  − 28 · u(t − 4) + 28 · u(t − 7) + 4t · u(t − 4) − 4t · u(t − 7)

Collecting terms and re-expressing into a usable form gives:

= 3t^2· [u(t) − u(t − 2)]  + 12 · [u(t − 2) − u(t − 4)]  − 28 · [u(t − 4) − u(t − 7)] + 4t · [u(t − 4) − u(t − 7)]

= 3t^2· [u(t) − u(t − 2)] + 12 · [u(t − 2) − u(t − 4)]  + 4(t − 7)· [u(t − 4) − u(t − 7)]