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To the extent you can factor the transfer function into individual integrator sections of the general form $\frac{1}{s}$ you can make this substitution, which is an approximation of the Matched-$z$ Transform where you substitute every $s$ for $s=\frac{\ln(z)}{T}$. (map from $s$ to $z$ using $z =e^{sT}$). This results in first order forms given by $H_\mathrm{... 1 You can't implement the transfer function H(z) directly, you need to convert it to a difference equation. However, the process is trivial, so once you understand it you'll see the connection between the diagram and transfer function better. First, we need to unroll the summation. For example, we get this with m=2, for a second-order equation:$H(z) = \...
It would help to apply some specificity to the general equation: $H(z) = \frac{b_0 + b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$ Rearrange into a difference equation, skipping the steps for brevity: $y[n] = b_0x[n] +b_1x[n-1]+b_2x[n-1]-a_1y[n-1]-a_2y[n-2]$ Does the difference equation form show now that the Direct Form structure is derived purely by ...