Question 4 of 6: Parallel and cascade realisations of an IIR system (12 marks)

Question 4: Parallel and cascade realisations of an IIR system (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 system already written as the sum of a first-order and a second-order section.

Find. (a) a stability verdict with reasons; (b) a parallel signal flow graph using direct form II for the second-order section; (c) a cascade signal flow graph of a first-order section and a transposed direct form II second-order section; (d) a comparison of multiplier and delay counts and a recommendation.

Approach. Factor the second-order denominator to expose all three poles, which settles part (a); draw the parallel structure directly from the given decomposition; combine the two fractions over a common denominator to obtain the single numerator needed for the cascade; then count multipliers and delays in each drawing.

1. Locate all three poles (part a). The first section contributes a pole at $z=\tfrac12$. For the second, multiply the denominator by $z^{2}$: $$z^{2}+\tfrac34 z-\tfrac14=0 \quad\Longrightarrow\quad z=\frac{-\tfrac34\pm\sqrt{\tfrac{9}{16}+1}}{2}=\frac{-0.75\pm 1.25}{2},$$ giving $z=\tfrac14$ and $z=-1$. Equivalently, in $z^{-1}$, $$1+\tfrac34 z^{-1}-\tfrac14 z^{-2}=\left(1+z^{-1}\right)\left(1-\tfrac14 z^{-1}\right).$$
2. Apply the stability test (part a). The complete pole set is $\left\{\tfrac12,\ \tfrac14,\ -1\right\}$. For a causal system the ROC is $|z| \gt \max|p| = 1$, which does not contain the unit circle because the pole at $z=-1$ lies exactly on it. Hence $$\boxed{\,\text{poles at } z=\tfrac12,\ \tfrac14,\ -1 \ \Longrightarrow\ \text{the system is NOT stable}\,}$$ The pole at $z=-1$ is a marginal, undamped mode: it contributes a term $(-1)^{n}$ to $h[n]$, which neither decays nor grows, so $h[n]$ is not absolutely summable and a bounded input at $\omega=\pi$ produces an unbounded output.

Part (b): parallel form. The system function is already a sum of two sections, so the parallel structure is drawn directly from the given expression. The first-order branch realises $$y_1[n]=x[n]+\tfrac12\,y_1[n-1],$$ and the second-order branch, in direct form II, splits into the recursive and non-recursive halves $$w[n]=x[n]-\tfrac34\,w[n-1]+\tfrac14\,w[n-2],\qquad y_2[n]=w[n]+2\,w[n-1],$$ with $y[n]=y_1[n]+y_2[n]$. Direct form II shares one delay chain between the two halves, which is why only two delays appear in that branch.

Part (c): cascade form. A cascade needs the two sections multiplied rather than added, so the two fractions must first be placed over a common denominator.

3. Combine the two sections into one rational function. Using the factorisation from step 1, the common denominator is $$D(z)=\left(1-\tfrac12 z^{-1}\right)\left(1+z^{-1}\right)\left(1-\tfrac14 z^{-1}\right),$$ and the numerator is $$\begin{aligned} N(z)&=\left(1+\tfrac34 z^{-1}-\tfrac14 z^{-2}\right)+\left(1+2z^{-1}\right)\left(1-\tfrac12 z^{-1}\right)\\ &=\left(1+\tfrac34 z^{-1}-\tfrac14 z^{-2}\right)+\left(1+\tfrac32 z^{-1}-z^{-2}\right)\\ &=2+\tfrac94 z^{-1}-\tfrac54 z^{-2}. \end{aligned}$$
4. Split the result into a first-order and a second-order factor. Keeping the numerator whole with the second-order denominator gives the cleanest pairing: $$\boxed{\,H(z)=\underbrace{\frac{1}{1-\tfrac12 z^{-1}}}_{\text{section A, 1st order}}\times\underbrace{\frac{2+\tfrac94 z^{-1}-\tfrac54 z^{-2}}{1+\tfrac34 z^{-1}-\tfrac14 z^{-2}}}_{\text{section B, 2nd order}}\,}$$ For reference the numerator's own roots are $z=\left(-9\pm\sqrt{241}\right)/16$, that is $z\approx 0.4078$ and $z\approx -1.5328$; they are irrational, so pairing the whole numerator with the second-order denominator avoids introducing ugly section coefficients for no benefit.
5. Write the transposed direct form II recursion for section B. With $b_0=2$, $b_1=\tfrac94$, $b_2=-\tfrac54$ and $a_1=-\tfrac34$, $a_2=\tfrac14$ (using $H=\left(b_0+b_1z^{-1}+b_2z^{-2}\right)/\left(1-a_1z^{-1}-a_2z^{-2}\right)$), the transposed structure computes $$\begin{aligned} y[n] &= b_0\,v[n]+s_1[n-1],\\ s_1[n] &= b_1\,v[n]+a_1\,y[n]+s_2[n-1],\\ s_2[n] &= b_2\,v[n]+a_2\,y[n], \end{aligned}$$ where $v[n]$ is the output of section A. Both realisations were checked: running a unit impulse through each structure reproduces the impulse response of the original $H(z)$ sample for sample.

Part (d): comparison. Counting a multiplier only for coefficients that are not $\pm 1$ (a unity coefficient is a plain wire), and one storage element per $z^{-1}$ block:

Both structures are canonic in storage — three delays for a third-order system, which is the theoretical minimum — so storage does not separate them. The multiplier counts do: $$\boxed{\,\text{parallel: 4 multiplies, 3 delays}\quad\text{vs}\quad\text{cascade: 6 multiplies, 3 delays}\ \Rightarrow\ \text{the PARALLEL form is preferable}\,}$$ The saving arises because the parallel decomposition was handed to us with a unity leading numerator coefficient in each branch, whereas forming the common numerator for the cascade produced three non-trivial feed-forward coefficients. Two secondary arguments reinforce the same choice: in a parallel structure the round-off noise generated in one section is not filtered by the other sections, so noise does not accumulate along a chain, and the two branches can be evaluated concurrently in a pipelined processor. The one advantage the cascade retains is that its zeros are controlled explicitly section by section, which matters when the zero locations must be held accurately under coefficient quantisation.