Numerical implementation issues

In terms of numerical implementation of the algorithm, the time domain pressure travelling waves are stored in vectors with elements referred to by an index (which we shall call $i$) which runs from 1 to $N$. Care must be taken over what time is represented by the index. On surfaces 1 and 2 we know the pressure at $t=0$, $T$, $2T$, ..., $(N-1)T$ so the $i$th element of the vector in the numerical implementation therefore corresponds to the pressure at time $t=(i-1)T$.

Equations (5.9) and (5.15) are then simply

$$r_{1,2} = \frac{p_{-}^{(1)}(i=1)}{p_{+}^{(1)}(i=1)},$$ (5.21)

$$\left( \begin{array}{c} p_{+}^{(2)}(i) \\ p_{-}^{(2)}(i) \en... ...rray}{c} p_{+}^{(1)}(i) \\ p_{-}^{(1)}(i) \end{array} \right).$$ (5.22)

The forward going pressure wave on surface 3 is sampled at $t=T/2$, $3T/2$, $5T/2$, ..., $(N-1/2)T$ and the $i$th element refers to the time $t=(i-1/2)T$. Comparison with equation (5.17) then yields

$$p_{+}^{(3)}(i) = p_{+}^{(2)}(i).$$ (5.23)

For the pressure on surface 3, we know the backward going pressure at $t=-T/2$, $T/2$, $3T/2$, ..., $(N-3/2)T$. At $t=-T/2$ the backward going pressure is zero, so we only need to carry on using the non-zero pressures at $t=T/2$ ,$3T/2$, ..., $(N-3/2)T$. The $i$th element refers to the time at $t=(i-1/2)T$. Comparison with equation (5.18) gives

$$p_{-}^{(3)}(i) = p_{-}^{(2)}(i+1).$$ (5.24)

The pressure waves $p_{+}^{(3)}(i)$ and $p_{-}^{(3)}(i)$ are then the forward going and backward going sequences for the left hand side of the junction at surface 3. We have therefore peeled off the first layer in the object. Equations (5.21) to (5.24) can then be used to peel off the next layer. First the primary reflection coefficient for the junction from surface 3 to surface 4 is calculated, then the forward and backward travelling waves on surface 4 are obtained and then the delay for travel along the cylinder to next discontinuity is added. The algorithm repeats as necessary, peeling off successive layers in the object, calculating the new cross-section using equation (5.10) each time.