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...

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).

EXAMPLES of PRODUCTS with UNIT FUNCTIONS

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

1_lap_unitstepfns_18pt__53.gif

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

1_lap_unitstepfns_18pt__56.gif

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)

The f(t) part begins at t = 0.


f(t) • u(t − a)

The f(t) part begins at t = a.


f(t − a) • u(t)

The f(t) part has been shifted to the right by a units and begins at t = 0.


f(t − a) • u(t − a)

The f(t) part has been shifted to the right by a units and begins at t = a.

Let's see some examples.

Example 1

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

(a) g1(t) = f(t) • u(t) = (4t + 2) • u(t)

1_lap_unitstepfns_18pt__64.gif

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


(b) g2(t) = f(t) • u(t − a) = (4t + 2) • u(t − 1)
1_lap_unitstepfns_18pt__66.png

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


(c) g3(t) = f(t − a) • u(t) = (4(t − 1) + 2) • u(t) = (4t − 2) • u(t)
1_lap_unitstepfns_18pt__68.png

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


(d) g4(t) = f(t − a) • u(t − a) = (4t − 2) • u(t − 1)
1_lap_unitstepfns_18pt__70.png

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


Example 2

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) g1(t) = sin t • u(t)

1_lap_unitstepfns_18pt__74.gif

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


(b) g2(t) = sin t • u(t − 0.7)

1_lap_unitstepfns_18pt__76.gif

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


(c) g3(t) = sin (t − 0.7) • u(t)

1_lap_unitstepfns_18pt__78.gif

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


(d) g4(t) = sin (t − 0.7) • u(t − 0.7)

1_lap_unitstepfns_18pt__80.gif

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


Exercises

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

(a) f(t) = u(t) + (1 − t) • u(t − 1) + (t − 2) • u(t − 2)


Answer


(b) f(t) = t2 • u(t) − (t2 − 4) • u(t − 2)


Answer


(c) f(t) = u(t) + (sin t − 1) • u(t − π/2) − (sin t + 1) • u(t − 3π/2) + u(t − 2π)


Answer


(d) f(t) = 3t2 • u(t) + (12 − 3t2) • u(t − 2) + (4t − 40) • u(t − 4) − 4(t − 7) • u(t − 7)


Answer


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) = 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 = 3u(t − 4)

lastdiag__2.png



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