Convolution and Spectra

When Fourier transforms are applicable, it means the ``earth response'' now is the same as the earth response later. Switching our point of view from time to space, the applicability of Fourier transformation means that the ``impulse response'' here is the same as the impulse response there. An impulse is a column vector full of zeros with somewhere a one, say $(0,0,1,0,0,\cdots)'$(where the prime ()' means transpose the row into a column.) An impulse response is a column from the matrix

 

$$\bold q \eq \left[ \begin{array} {c} q_0 \\ q_1 \\ q... ..._3 \\ p_4 \\ p_5 \end{array} \right] \eq \bold B \bold p$$ (1)

The impulse response is the $\bold q$ that comes out when the input $\bold p$is an impulse. In a typical application, the matrix would be about $1000\times 1000$and not the simple $8\times 6$ example that I show you above. Notice that each column in the matrix contains the same waveform (b₀,b₁,b₂). This waveform is called the ``impulse response''. The collection of impulse responses in equation (1) defines the the convolution operation.

Not only do the columns of the matrix contain the same impulse response, but each row likewise contains the same thing, and that thing is the backwards impulse response (b₂,b₁,b₀). Suppose (b₂,b₁,b₀) were numerically equal to $(1,-2,1)/\Delta t^2$.Then equation (1) would be like the differential equation ${d^2\over dt^2}p=q$.Equation (1) would be a finite-difference representation of a differential equation. Two important ideas are equivalent; either they are both true or they are both false:

1.

The columns of the matrix all hold the same impulse response.

2.

The differential equation has constant coefficients.

Let us take a quick peek ahead. The relationship of equation (1) with Fourier transforms is that the k-th row in (1) is the k-th power of Z in a polynomial multiplication Q(Z)=B(Z)P(Z). The relationship of any polynomial such as Q(Z) to Fourier Transforms results from the relation $Z=e^{i\omega\Delta t}$, as we will see.