22-Elec-A3 Signals and Communications · December 2015

Question 1 of 6: Discrete-Time System — Transfer Function, Stability and Impulse Response

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Discrete-Time System — Transfer Function, Stability and Impulse Response (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. The block diagram is a Direct-Form II structure: the input enters a first summing junction, whose output $w(n)$ passes through two cascaded unit delays $D$. The output of the first delay is fed back to the input summer through a gain $-3/4$ and forward to the output summer through a gain $1/2$; the output of the second delay is fed back to the input summer through the gain $\alpha$. Both junctions are plain adders.

Find. (a) $H(z) = Y(z)/X(z)$; (b) the range of $\alpha$ for which the system is BIBO stable; (c) the impulse response $h(n)$ when $\alpha = 1/2$.

[Figure not reproduced: Figure 1.1 — The system redrawn from the examination paper. The internal node $w(n)$ is the Direct-Form II state variable; the two delay taps supply one feedback path each and one feed-forward path. See the official exam paper.]

Approach. Write one difference equation at each summing junction in terms of the internal node $w(n)$, transform both, eliminate $W(z)$ to obtain $H(z)$, apply the second-order Jury stability test to the denominator, and finally invert $H(z)$ by partial fractions at the requested value of $\alpha$.

1. Write the difference equation at the input summer. The adder receives $x(n)$, the first-delay output scaled by $-3/4$, and the second-delay output scaled by $\alpha$: $$w(n) = x(n) - \tfrac{3}{4}\,w(n-1) + \alpha\,w(n-2)$$
2. Write the difference equation at the output summer. It sums the forward path (unity gain from $w(n)$) with the first-delay tap scaled by $1/2$: $$y(n) = w(n) + \tfrac{1}{2}\,w(n-1)$$
3. Transform and eliminate the internal node. Taking $z$-transforms of both relations gives $W(z)\left[1 + \tfrac{3}{4}z^{-1} - \alpha z^{-2}\right] = X(z)$ and $Y(z) = W(z)\left[1 + \tfrac{1}{2}z^{-1}\right]$. Dividing the second by the first removes $W(z)$ entirely: $$\boxed{\,H(z) = \frac{1 + \tfrac{1}{2}z^{-1}}{1 + \tfrac{3}{4}z^{-1} - \alpha z^{-2}} = \frac{z\left(z + \tfrac{1}{2}\right)}{z^{2} + \tfrac{3}{4}z - \alpha}\,}$$ The system has one finite zero at $z = -1/2$, a zero at the origin cancelled by the numerator factor $z$, and two poles given by the roots of the denominator.
4. Apply the stability test to the denominator. For a causal system the poles must lie strictly inside the unit circle. Writing the denominator as $z^{2} + a_{1}z + a_{2}$ with $a_{1} = 3/4$ and $a_{2} = -\alpha$, the Jury conditions for a second-order polynomial are $$|a_{2}| \lt 1, \qquad 1 + a_{1} + a_{2} \gt 0, \qquad 1 - a_{1} + a_{2} \gt 0$$ Substituting: $|\alpha| \lt 1$; $1 + \tfrac{3}{4} - \alpha \gt 0 \Rightarrow \alpha \lt 7/4$; and $1 - \tfrac{3}{4} - \alpha \gt 0 \Rightarrow \alpha \lt 1/4$.
5. Intersect the three conditions. The second condition is weaker than the third, so the binding pair is $|\alpha| \lt 1$ together with $\alpha \lt 1/4$: $$\boxed{\,-1 \lt \alpha \lt \tfrac{1}{4}\,}$$ At $\alpha = 1/4$ a pole sits exactly at $z = -1$ and at $\alpha = -1$ the pole pair lands on the unit circle; both boundaries are marginally stable, not stable.

Part (c) now asks for the impulse response at $\alpha = 1/2$. This value lies outside the range just derived, so the system is unstable at the requested setting — the impulse response still exists and is perfectly well defined for a causal system, but it grows without bound. This is worth stating explicitly in the answer book, because it is the connection the examiner is testing between parts (b) and (c).

6. Locate the poles at $\alpha = 1/2$. The denominator becomes $z^{2} + \tfrac{3}{4}z - \tfrac{1}{2}$, whose roots are $$p_{1,2} = \frac{-\tfrac{3}{4} \pm \sqrt{\tfrac{9}{16} + 2}}{2} = \frac{-0.75 \pm 1.60078}{2}$$ giving $p_{1} = 0.42539$ and $p_{2} = -1.17539$. Since $|p_{2}| = 1.1754 \gt 1$, the second pole is outside the unit circle, confirming instability.
7. Expand in partial fractions. Because $H(z)$ carries a factor $z$ in the numerator, expand $H(z)/z$ so that each term inverts to a pure geometric sequence: $$\frac{H(z)}{z} = \frac{z + \tfrac{1}{2}}{(z - p_{1})(z - p_{2})} = \frac{A}{z - p_{1}} + \frac{B}{z - p_{2}}$$ with $A = (p_{1} + 0.5)/(p_{1} - p_{2}) = 0.92539/1.60078 = 0.57808$ and $B = (p_{2} + 0.5)/(p_{2} - p_{1}) = (-0.67539)/(-1.60078) = 0.42192$. As a check, $A + B = 1$ exactly, which must equal $h(0)$.
8. Invert term by term. Multiplying back by $z$ and using the causal pair $\dfrac{z}{z-p} \leftrightarrow p^{n}u(n)$: $$\boxed{\,h(n) = \left[0.57808\,(0.42539)^{n} + 0.42192\,(-1.17539)^{n}\right]u(n)\,}$$
9. Verify against the recursion. Driving the block diagram directly with $x(n) = \delta(n)$ gives $w(0) = 1$, $w(1) = -0.75$, hence $h(0) = 1$ and $h(1) = w(1) + 0.5w(0) = -0.25$. The closed form returns $h(0) = 0.57808 + 0.42192 = 1$ and $h(1) = 0.57808(0.42539) + 0.42192(-1.17539) = 0.24591 - 0.49591 = -0.25$. The two agree, and every subsequent sample matches as well.