Question 6 of 7: Discrete-time system: difference equation, transfer function, impulse response and stability (20 marks)

Question 6 — Discrete-time system: difference equation, transfer function, impulse response and stability (20 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. Two cascaded single-pole recursive stages, redrawn below from the examination figure. The input $x(n)$ enters the first adder; the adder output $w(n)$ is fed back through a unit delay $D$ and a gain of $-3/4$; $w(n)$ then enters the second adder, whose output $y(n)$ is fed back through a unit delay and a gain of $\alpha$. Both junctions are additive, and the system is causal.

[Figure not reproduced: The examination block diagram redrawn: two cascaded first-order recursive stages with feedback gains −3/4 and α. See the official exam paper.]

Find. (a) a difference equation relating $y(n)$ to $x(n)$; (b) $H(z)$; (c) $h(n)$; (d) the range of $\alpha$ for stability.

Approach. Write one node equation per adder in terms of the intermediate signal $w(n)$, transform both, eliminate $W(z)$ to obtain $H(z)$, then invert by partial fractions. Reading the two stages separately keeps the algebra trivial; only at the end are they combined.

1. Write the node equation at each adder. At the first summing junction the input is added to the delayed, scaled feedback of the junction’s own output: $$w(n) = x(n) - \tfrac{3}{4}\,w(n-1).$$ At the second junction $w(n)$ is added to the delayed, scaled output: $$y(n) = w(n) + \alpha\,y(n-1).$$
2. Eliminate the intermediate signal. From the second equation $w(n) = y(n) - \alpha y(n-1)$; substituting into the first and shifting where needed, $$y(n) - \alpha y(n-1) = x(n) - \tfrac{3}{4}\bigl[y(n-1) - \alpha y(n-2)\bigr].$$ Collecting terms in $y$, $$\boxed{\;y(n) + \left(\tfrac{3}{4} - \alpha\right)y(n-1) - \tfrac{3}{4}\alpha\,y(n-2) = x(n)\;}$$ a second-order recursion with no non-trivial input terms. As a check, at $n = 0$ it gives $h(0) = 1$ and at $n = 1$ it gives $h(1) = \alpha - 3/4$, both of which the block diagram confirms by inspection.
3. Transform to obtain the transfer function. Taking $z$-transforms of the two node equations with zero initial conditions, $$W(z)\left(1 + \tfrac{3}{4}z^{-1}\right) = X(z), \qquad Y(z)\left(1 - \alpha z^{-1}\right) = W(z),$$ so the cascade multiplies: $$\boxed{\;H(z) = \frac{Y(z)}{X(z)} = \frac{1}{\left(1 + \tfrac{3}{4}z^{-1}\right)\left(1 - \alpha z^{-1}\right)} = \frac{z^{2}}{\left(z + \tfrac{3}{4}\right)\left(z - \alpha\right)},\quad |z| \gt \max\left(\tfrac{3}{4},|\alpha|\right)\;}$$ Expanding the denominator reproduces $1 + (3/4 - \alpha)z^{-1} - (3/4)\alpha z^{-2}$, in agreement with part (a). The system has a double zero at the origin and poles at $z = -3/4$ and $z = \alpha$.

4. Invert by partial fractions. For $\alpha \ne -3/4$ write $$H(z) = \frac{z^{2}}{(z + \tfrac{3}{4})(z-\alpha)} \quad\Longrightarrow\quad \frac{H(z)}{z} = \frac{A_1}{z + \tfrac34} + \frac{A_2}{z - \alpha},$$ with residues $A_1 = -\tfrac{3}{4}/(-\tfrac34 - \alpha)$ and $A_2 = \alpha/(\alpha + \tfrac34)$. Combining and inverting term by term with the causal pair $z/(z-p) \leftrightarrow p^{n}u(n)$ gives $$\boxed{\;h(n) = \frac{\alpha^{\,n+1} - \left(-\tfrac{3}{4}\right)^{n+1}}{\alpha + \tfrac{3}{4}}\;u(n)\;}$$ Equivalently $h(n)$ is the convolution of the two single-pole responses, $h(n) = \bigl(-\tfrac34\bigr)^{n}u(n) * \alpha^{n}u(n)$, which is the geometric-sum form written above. Special case: if $\alpha = -3/4$ the poles coincide and the limit gives the repeated-pole response $h(n) = (n+1)\bigl(-\tfrac34\bigr)^{n}u(n)$.
5. Verify against the recursion. Driving the block diagram with a unit sample gives $h(0) = 1$, $h(1) = \alpha - 3/4$, $h(2) = \alpha^{2} - \tfrac34\alpha + \tfrac{9}{16}$, and the closed form reproduces these and every later term exactly (checked to $n = 39$ for $\alpha = 0.5$, $-0.4$ and $0.9$). Note that $h(0)$ and $h(1)$ are fixed by the structure, so any candidate expression failing them is wrong.
6. Apply the stability condition. A causal LTI system is BIBO stable if and only if every pole lies strictly inside the unit circle. One pole is fixed at $z = -3/4$, with $|-3/4| = 0.75 \lt 1$, so it never threatens stability. The other pole is at $z = \alpha$, so the requirement is $|\alpha| \lt 1$: $$\boxed{\;-1 \lt \alpha \lt 1\;}$$ Equivalently $\sum_n |h(n)| \lt \infty$ exactly on this range. At $|\alpha| = 1$ the response is marginally stable (it neither grows nor decays — a constant or an alternating sequence persists forever), and for $|\alpha| \gt 1$ the impulse response grows without bound.

The plot uses $\alpha = 0.5$ as a representative stable value: the response starts at $h(0) = 1$, dips to $h(1) = -0.25$ because the two feedback paths oppose one another at the first step, and then decays with the alternating signature of the $z = -3/4$ pole superimposed on the smooth decay of the $z = \alpha$ pole.