Skip to the content of the web site.

Lecture 17

Lecture 16 | Lecture 18


We have already used the unit step function tex:$$u(t) = \cases{ $0$ & $t < 0$ \cr $1$ & $t \ge 0$ }$$ which can be used to act as a switch for a single or as a switch, double-switch, or delay.

Using it as a switch, we turn a function on at some time tex:$$t = a$$ by using multiplication; for example, Figure 1 shows a function tex:$$f(t)$$ and tex:$$f(t)u(t - a)$$.


Figure 1. The function tex:$$f(t)u(t - a)$$.

If tex:$$b > a$$, we can create a double switch which turns the function on at time tex:$$t = a$$ and off again at time tex:$$t = b$$; for example, Figure 2 shows a function tex:$$f(t)$$ and tex:$$f(t)\left( u(t - a) - u(t - b) \right)$$.


Figure 2. The function tex:$$f(t)\left( u(t - a) - u(t - b) \right)$$.

To delay a signal, both the function and the unit step function must be delayed; for example, Figure 3 shows a function tex:$$f(t)$$ and tex:$$f(t - a)u(t - a)$$.


Figure 3. The function tex:$$f(t - a)u(t - a)$$.

Because these functions arise so often in engineering, we require their Laplace transforms:

tex:$$u(t) \leftrightarrow \frac{1}{s}$$
tex:$$u(t - a) \leftrightarrow \frac{e^{-sa}}{s}$$
tex:$$u(t - a) - u(t - b) \leftrightarrow \frac{1}{s}\left( e^{-as} - e^{-bs}\right)$$

Finally, the second shift theorem states that if tex:$$f(t) \leftrightarrow F(s)$$, it follows that

tex:$$f(t - a)u(t - a) \leftrightarrow e^{-as}F(s)$$.

The proofs are straight forward and are given in AMEM.

Periodic Functions

It is possible to use the unit step functions to define periodic functions other than the usual trigonometric functions. For example, Figure 4 shows the function

tex:$$f(t) = Au\left(t\right) - Au\left(t - \frac{1}{2}T\right) + Au\left(t - T\right) - Au\left(t - \frac{3}{2}T\right) + Au\left(t - 2T\right) - Au\left(t - \frac{5}{2}T\right) + \cdots $$.

We may now take the Laplace transform of this infinite series simply by applying it term-by-term and each term is simply a shifted unit step function. Therefore, we have

tex:$$F(s) = \frac{A}{s} - A\frac{e^{-0.5Ts}{s} + A\frac{e^{-\frac{T}s}{s} - A\frac{e^{-1.5Ts}{s} + A\frac{e^{-2Ts}{s} - A\frac{e^{-2.5Ts}{s} + \cdots$$.

If we factor out tex:$$\frac{A}{s}$$, we can write this sum as

tex:$$F(s) = \frac{A}{s}\left ( 1 - e^{-0.5Ts} + e^{-\frac{T}s} - e^{-1.5Ts} + e^{-2Ts} - e^{-2.5Ts} + \cdots \right .$$

or

tex:$$F(s) = \frac{A}{s}\left ( 1 - e^{-0.5Ts} + e^{-\frac{0.5T}s}^2 - e^{-0.5Ts}^3 + e^{-0.5Ts}^4 - e^{-0.5Ts}^5 + \cdots \right .$$.

This is a geometric series tex:$$1 + r + r^2 + r^3 + r^4 + r^5 + \cdots = \frac{1}{1 - r}$$ where in this case tex:$$r = -e^{-\frac{T}{2}s}$$ and therefore, we have that

tex:$$F(s) = \frac{A}{s} \frac{1}{1 + e^{-\frac{T}{2}s}}$$.

This is a special case of the following more general theorem:

Theorem

If tex:$$f(t)$$ is defined for all tex:$$t \ge 0$$ and it is a periodic function with period tex:$$T$$, then

tex:$$f(t) \leftrightarrow \frac{1}{1 + e^{-Ts}} \int_0^T e^{-st}f(t)dt$$.

Example

Give the example we previously looked at, tex:$$f(t) = \cases{ $A$ & $0 \le t < \frac{T}{2}$ \cr $A$ & $\frac{T}{2} \le t < T$ }$$ with tex:$$f(t + nT) = f(t)$$. We must therefore calculate

tex:$$\int_0^T e^{-st}f(t) dt = A\int_0^\frac{T}{2} e^{-st}dt = \left. -\frac{A}{s}e^{-st}\right |_{t = 0}^{t = \frac{T}{2}} = 1 - e^{-\frac{T}{2}s} $$

and therefore the Laplace transform is

tex:$$f(t) \leftrightarrow \frac{A}{s} \frac{1 - e^{-\frac{T}{2}s}}{1 - e^{-Ts}} = \frac{A}{s}\frac{1}{1 + e^{-\frac{T}{2}s}}$$

as before.

Further examples are in the assignments.