Question 3 of 6: Direct form II and transposed flow graphs

Question 3: Direct form II and transposed flow graphs (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A causal LTI system with $H(z) = (1 - 2z^{-1})(1 - 4z^{-1}) \big/ \big[z(1 - \tfrac12 z^{-1})\big]$. Causality fixes the region of convergence at $|z| \gt \tfrac12$.

Find. A direct form II signal flow graph, and its transpose.

Approach. Put $H(z)$ into a single ratio of polynomials in $z^{-1}$, read the difference equation off it, realise the all-pole and all-zero halves in the shared-delay order that defines direct form II, then apply the transposition theorem.

1. Write $H(z)$ as a ratio of polynomials in $z^{-1}$. Expanding the numerator, $$(1 - 2z^{-1})(1 - 4z^{-1}) = 1 - 6z^{-1} + 8z^{-2},$$ and noting that the factor $z$ in the denominator is simply $z^{-1}$ moved upstairs, $$H(z) = \frac{z^{-1}\left(1 - 6z^{-1} + 8z^{-2}\right)}{1 - \tfrac12 z^{-1}} = \boxed{\;\frac{z^{-1} - 6z^{-2} + 8z^{-3}}{1 - \tfrac12 z^{-1}}\;}$$ so the numerator coefficients are $b_0 = 0$, $b_1 = 1$, $b_2 = -6$, $b_3 = 8$ and the single denominator coefficient is $a_1 = \tfrac12$. The system has one pole at $z = \tfrac12$ and three zeros, at $z = 2$, $z = 4$ and $z = 0$ (the last contributed by the explicit $1/z$).
2. Read off the difference equation. Cross-multiplying $Y(z)\left(1 - \tfrac12 z^{-1}\right) = X(z)\left(z^{-1} - 6z^{-2} + 8z^{-3}\right)$ and inverting term by term, $$y[n] = \tfrac{1}{2}\,y[n-1] + x[n-1] - 6\,x[n-2] + 8\,x[n-3].$$ This is the input–output relation the flow graph must realise.
3. Split into the two cascaded halves and count delays. Direct form II writes $H(z) = \dfrac{1}{1 - \tfrac12 z^{-1}} \cdot \left(z^{-1} - 6z^{-2} + 8z^{-3}\right)$, i.e. the all-pole section first. Introducing the intermediate signal $w[n]$, $$w[n] = x[n] + \tfrac{1}{2}\,w[n-1], \qquad y[n] = w[n-1] - 6\,w[n-2] + 8\,w[n-3].$$ Because both halves now read from the same delay chain, the structure needs only $$\boxed{\;\max(N, M) = \max(1, 3) = 3 \ \text{delay elements}\;}$$ rather than the $1 + 3 = 4$ that direct form I would use. It is canonic in delays.
4. Draw the direct form II graph. One adder receives $x[n]$ and the feedback branch $\tfrac12 w[n-1]$; its output $w[n]$ heads a chain of three unit delays; the four tap points $w[n], \dots, w[n-3]$ feed gains $0, 1, -6, 8$ into the output adder. The $b_0 = 0$ tap is simply absent, which is why $h[0] = 0$: the system is strictly causal, responding one sample after the input arrives.
5. Apply the transposition theorem for part (b). Transposition reverses the direction of every branch, exchanges adders with branch (fan-out) nodes, and interchanges the input and output. The branch gains are unchanged, and because the theorem preserves the transfer function of a single-input single-output graph, the transposed structure realises the same $H(z)$ with the same three delays. In the transposed graph the input fans out to all three gain branches at once and the delays sit inside the accumulator chain rather than ahead of it.
6. Verify both structures against the closed form. Driving either graph with $\delta[n]$ gives $$h[n] = \left(\tfrac12\right)^{n-1} - 6\left(\tfrac12\right)^{n-2} + 8\left(\tfrac12\right)^{n-3}$$ for the respective shifted terms, i.e. $$h[0] = 0, \quad h[1] = 1, \quad h[2] = -5.5, \quad h[3] = 5.25, \quad h[4] = 2.625,$$ after which the response is a pure geometric decay, $h[n] = 5.25\left(\tfrac12\right)^{n-3}$ for $n \ge 3$. Both the direct form II and the transposed recursion reproduce this sequence exactly, confirming the two graphs are equivalent.