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Lecture 14

Lecture 13 | Lecture 15

Introduction to Transforms

Simple Example

Suppose we are trying to solve a problem which in itself may require effort. For example, suppose you are asked to calculate the product 914523098123515 × 52354239523952323553. You could use standard multiplication techniques to get the answer 47879161329345458034788898437648795. This is a direct method of solving this problem, but due to the limited processing power of the human brain, this is both tedious and error prone. An alternate solution is to program a computer to solve these problems; however, the computer calculates arithmetic in binary precision. Therefore, it is necessary to transform both numbers into base 2, carry out the multiplication, and then transform back to base 10. In this example, we would convert the multiplier and multiplicand to 110011111111000001000000000110011100001100111110112 and 1011010110100011111001111101110110010111101110010011101011111000012, allow the computer to take the product, 100100111000100111111111101000001100111011101110011011100111001100000100011100111101101011010000100011010001100110112, and then transform back to 47879161329345458034788898437648795. This process is shown in Figure 1.

Figure 1. Calculating the product of two integers.

An average or poor high school student will always pick the most straight-forward route: just multiply the two numbers together. A good high school student, however, will realize that while programming the conversion may be more difficult, and the two numbers could not be stored as a single integer, a little effort will produce significant rewards in the end. A good engineering student may even look up the GNU GIMP package for large-integer multiplication.

Matrix Example

To give a second transformation, suppose you have an tex:$$n \times n$$ matrix tex:$${\bf M}$$ and you wish to calculate tex:$${\bf M}^3 - 4{\bf M}^2 + 5{\bf M} + 2{\bf Id}_n$$. From your other courses, you will understand that this is a very expensive operation both in terms of memory and time. If, however, the matrix is diagonalizable, that is, we may write tex:$${\bf M} = {\bf PDP}^{-1}$$ where tex:$${\bf D}$$ is a diagonal matrix, we may substitute this into the equation above to get

tex:$$\left( {\bf PDP}^{-1} \right )^3 - 4\left( {\bf PDP}^{-1} \right )^2 + 5\left( {\bf PDP}^{-1} \right ) + 2\left( {\bf PDP}^{-1} \right )$$

and using the properties of matrices, we can simplify this to

tex:$${\bf P}\left( {\bf D}^3 - 4{\bf D}^2 + 5{\bf D} + 2{\bf Id}_n \right) {\bf P}^{-1}$$.

This new calculation is now relatively easy because tex:$${\bf D}$$ is a diagonal matrix where the diagonal entries are the roots of the characteristic equation of the matrix.

The Laplace Transform

We have up to this point seen how we can solve a relatively simple differential equation to get a solution to an initial-value problem. This straight-forward computation is shown in Figure 2.

Figure 2. The direct methods of solving initial-value problems using solutions to the homogeneous equation and particular solutions.

Unfortunately, as problems become larger, it becomes more and more difficult to set up the correct equations to find the solution. Therefore, we will follow an alternate route. By now, it should be obvious that the characteristics of a solution to an initial-value problem depend on the placement of the roots. Therefore, it would be reasonable to assume that we could perform calculations by focusing on the the characteristic equation. As it turns out, this is exactly what the Laplace transform allows us to do. If you take a look at Figure 3, rather than solving the IVP directly, we convert the problem using the Laplace transform into an algebraic problem; we perform some algebraic manipulation; and we transform back to the solution.

Figure 3. The indirect methods of solving initial-value problems using the Laplace transform.

The other benefit of the Laplace transform is that it abstracts out the essential information about the differential equation and its solution. You, as a student, will need to learn a complete new approach to solving problems: by this point, most students focus on solving problems directly. Initially, the Laplace transform may seem unnecessary, but it is worth the effort and later, you will see the Fourier transform, as well.

The ability of the Laplace transform to simplify the finding of solutions is used in numerous fields, including:

  • Finding transient and steady-state responses of electrical circuits
  • Applications of dynamical problems (vibrations, acoustics)
  • Applications to structural problems

We will look at these issues but we will start with some definitions.

The Definition of the Laplace Transform

The Laplace transform of a function tex:$$f(t)$$ is the improper integral

tex:$$L\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) dt$$

where tex:$$s$$ is assumed to be a complex variable.

Recall that tex:$$\int_0^\infty e^{-st} f(t) dt = \lim_{T \rightarrow \infty} \int_0^T e^{-st} f(t) dt$$.

In general, if the function tex:$$f(t)$$ has a Laplace transform tex:$$F(s)$$, we will write tex:$$f(t) \leftrightarrow F(s)$$.

The calculation of the Laplace transform is reasonably straight-forward for certain simple functions which we will do here.

The Unit Step Function

The unit step function is a function defined as

The Laplace transform of the unit step function is

tex:$$L\{u(t)\} = \int_0^\infty e^{-st} dt = \left.-\frac{1}{s} e^{-st}\right|_0^\infty = \frac{1}{s}$$.

Therefore, tex:$$u(t) \leftrightarrow \frac{1}{s}$$.

Ramp Function

The ramp function is the unit step function multiplied by the function tex:$$t$$:

= \int_0^\infty e^{-st} t dt
= -\left.  \frac{1}{s} t e^{-st}|_0^\infty + \frac{1}{s} \int_0^\infty e^{-st} dt
= \left. -\frac{1}{s^2} e^{-st}\right|_0^\infty
= \frac{1}{s^2}$$.

Therefore, tex:$$t u(t) \leftrightarrow \frac{1}{s^2}$$.

Exponential Function

The exponential function has a straight-forward transformation:

= \int_0^\infty e^{-st} e^{kt} dt
= \int_0^\infty e^{-(s - k)t} e^{kt} dt
= \left. -\frac{1}{s - k} e^{-(s - k)t}\right|_0^\infty
= \frac{1}{s - k}$$.

Therefore, tex:$$e^{kt}u(t) \leftrightarrow \frac{1}{s - k}$$.

Ramped Exponential Function

= \int_0^\infty e^{-st} te^{kt} dt
= -\left.  \frac{1}{s - k} t e^{-(s - k)t}|_0^\infty + \frac{1}{s - k} \int_0^\infty e^{-(k - s)t} dt
= \left. -\frac{1}{(s - k)^2} e^{-(s - k)t}\right|_0^\infty
= \frac{1}{(s - k)^2}$$.

Therefore, tex:$$te^{kt}u(t) \leftrightarrow \frac{1}{(s - k)^2}$$.


At this point, you should have already recognized some patterns. Notice that if tex:$$k = 0$$, then tex:$$e^{kt}u(t) = u(t)$$ and the Laplace transform tex:$$\frac{1}{s - k}$$ equals tex:$$\frac{1}{s}$$. At this point, many students will try to plot the Laplace transforms and attempt to understand the relationships. Avoid this temptation at least for the balance of this course until you have a clearer understanding of integral transforms: at this point, it is better to focus on understanding the consequences of the transform.

Tables of Laplace Transforms

In general, it is easiest to build a table of various Laplace transforms. Once we do this, we can can then use the table as a look-up and not have to repetitively calculate various integrals.

Table 1. A table of simple Laplace transforms.

tex:$$e^{kt}u(t)$$tex:$$\frac{1}{s - k}$$
tex:$$te^{kt}\,u(t)$$tex:$$\frac{1}{(s - k)^2}$$

Properties of the Laplace Transform

There are five properties of the Laplace transform which we will look at here:


If tex:$$f(t) \leftrightarrow F(s)$$, tex:$$g(t) \leftrightarrow G(s)$$, and tex:$$\alpha$$ and tex:$$\beta$$ are real or complex constants, it follows that

tex:$$\alpha f(t) + \beta g(t) \leftrightarrow \alpha F(s) + \beta G(s)$$.

First Shift Theorem

If tex:$$f(t) \leftrightarrow F(s)$$ and tex:$$a$$ is a real or complex constant, it follows that

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

Derivative of the Transform

If tex:$$f(t) \leftrightarrow F(s)$$ and tex:$$n$$ is a positive integer, it follows that

tex:$$t^n f(t) \leftrightarrow (-1)^n \frac{d^n}{ds^n} F(s)$$.

We may prove this with induction:

For the case tex:$$n = 1$$, it is straight forward to see that

tex:$$t f(t) \leftrightarrow \int_0^\infty e^{-st} t f(t) dt = \int_0^\infty \left( -\frac{d}{ds} e^{-st} \right ) f(t) dt = -\frac{d}{ds} \int_0^\infty e^{-st}f(t) dt = -\frac{d}{ds}F(s)$$.

Assume it is true for the case tex:$$n = k$$. It follows that the case tex:$$n = k + 1$$ may be found using integration by parts to get tex:$$t^{k + 1} f(t) \leftrightarrow -\frac{d}{ds} \left( (-1)^k \frac{d^k}{ds^k} F(s) \right ) = (-1)^{k + 1} \frac{d^{k + 1}}{ds^{k + 1}} F(s)$$.

It is always useful to test that our properties hold for what we have already seen: We know tex:$$u(t) \leftrightarrow \frac{1}{s}$$, and therefore tex:$$t\,u(t) \leftrightarrow -\frac{d}{ds} \frac{1}{s} = \frac{1}{s^2}$$. Similarly, we know tex:$$e^{kt}u(t) \leftrightarrow \frac{1}{s - k}$$, and therefore tex:$$te^{kt}\,u(t) \leftrightarrow -\frac{d}{ds} \frac{1}{s - k} = \frac{1}{(s - k)^2}$$.

Transform of the Derivative

If tex:$$f(t) \leftrightarrow F(s)$$ and tex:$$n$$ is a positive integer, it follows that

tex:$$\frac{d^n}{dt^n} f(t) \leftrightarrow s^n F(s) - s^{n - 1}f(0) - s^{n - 2}f^{(1)}(0) - s^{n - 3}f^{(2)}(0) - \cdots - f^{(n - 1)}(0)$$.

Again, we may prove this with induction:

For the case tex:$$n = 1$$, it is straight forward to see that

tex:$$\frac{d}{dt} f(t) \leftrightarrow \int_0^\infty e^{-st} \left(\frac{d}{dt} f(t)\right) dt = \left. e^{-st} f(t) \right |_0^\infty - \int_0^\infty f(t)\left( -s e^{st} \right )dt = -f(0) + s F(s)$$.

Assume it is true for the case tex:$$n = k$$. It follows that the case tex:$$n = k + 1$$ may be found using integration by parts to get tex:$$\frac{d^{k + 1}}{dt^{k + 1}} f(t) = \frac{d}{dt} \frac{d^k}{dt^k} f(t) \leftrightarrow s\left ( s^k F(s) - s^{k - 1}f(0) - s^{k - 2}f^{(1)}(0) - \cdots f^{(k - 1)}(0) \right ) - \frac{d^n}{dt^n}f(0)$$ which expands into our desired result.

Transform of the Indefinite Integral

If tex:$$f(t) \leftrightarrow F(s)$$, it follows that

tex:$$\int_0^t f(\tau)d\tau \leftrightarrow \frac{1}{s} F(s)$$.

This may be shown by substituting tex:$$g(t) = \int_0^t f(\tau)d\tau$$ and therefore tex:$$\frac{d}{dt} g(t) = f(t)$$. Using the previous theorem, we get that

Example: Transform of the Trigonometric Functions

Recall that tex:$$\left( \cos(at) + j\sin(at) \right)u(t) = e^{jat}u(t) \leftrightarrow \frac{1}{s - ja} = \frac{1}{s - ja} \frac{s + ja}{s + ja} = \frac{s + ja}{s^2 + a^2} = \frac{s}{s^2 + a^2} + j\frac{a}{s^2 + a^2}$$.

Because the real and imaginary components must match, it follows that

tex:$$\cos(at)u(t) \leftrightarrow = \frac{s}{s^2 + a^2}$$


tex:$$\sin(at)u(t) \leftrightarrow = \frac{a}{s^2 + a^2}$$.