Waves and Fourier sums

FOURIER TRANSFORM

We first examine the two ways to visualize polynomial multiplication. The two ways lead us to the most basic principle of Fourier analysis that

A product in the Fourier domain is a convolution in the physical domain

Look what happens to the coefficients when we multiply polynomials.

$$X(Z)\, B(Z) \quad =\quad Y(Z)$$ (1)

$$(x_0 + x_1 Z + x_2 Z^2 + \cdots )\, (b_0 + b_1 Z + b_2 Z^2) \quad =\quad y_0 + y_1 Z + y_2 Z^2 + \cdots$$ (2)

Identifying coefficients of successive powers of $Z$, we get

$$y_0 \quad =\quad x_0 b_0$$

$$y_1 \quad =\quad x_1 b_0 + x_0 b_1$$

$$y_2 \quad =\quad x_2 b_0 + x_1 b_1 + x_0 b_2$$ (3)

$$y_3 \quad =\quad x_3 b_0 + x_2 b_1 + x_1 b_2$$

$$y_4 \quad =\quad x_4 b_0 + x_3 b_1 + x_2 b_2$$

$$\quad =\quad \cdots\cdots\cdots\cdots\cdots\cdots$$

In matrix form this looks like

$$\left[ \begin{array}{c} y_0 \\ y_1 \\ y_2 \\ y_3 \\ ... ...[ \begin{array}{c} b_0 \\ b_1 \\ b_2 \end{array} \right]$$ (4)

The following equation, called the ``convolution equation,'' carries the spirit of the group shown in (3)

$$y_k \quad =\quad \sum_{i = 0} x_{k - i} b_i$$ (5)

The second way to visualize polynomial multiplication is simpler. Above we did not think of $Z$ as a numerical value. Instead we thought of it as ``a unit delay operator''. Now we think of the product $X(Z) B(Z) = Y(Z)$ numerically. For all possible numerical values of $Z$, each value $Y$ is determined from the product of the two numbers $X$ and $B$. Instead of considering all possible numerical values we limit ourselves to all values of unit magnitude $Z=e^{i\omega}$ for all real values of $\omega$. This is Fourier analysis, a topic we consider next.

Waves and Fourier sums