Discrete-Time Feedback System — Transfer Function, Stability, Impulse Response: 17-Phys-B4 Signals and Communications, December 2015, Q1

Question 1: Discrete-Time Feedback System — Transfer Function, Stability, Impulse Response (1/6 of paper)

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. Let $w(n)$ be the first adder's output. Reading the diagram: the once-delayed node $w(n-1)$ feeds back into the first adder with gain $-3/4$ and forward into the second adder with gain $1/2$; the twice-delayed node $w(n-2)$ feeds back into the first adder with gain $\alpha$.

Find. (a) $H(z)=Y(z)/X(z)$; (b) the range of $\alpha$ for BIBO stability; (c) $h(n)$ for $\alpha=1/2$.

Reconstructed block diagram: w(n) is the first adder’s output; the once-delayed node w(n-1) feeds the second adder with gain 1/2 (forward path) and the first adder with gain -3/4 (feedback); the twice-delayed node w(n-2) feeds the first adder with gain α.

Approach. Write the two difference equations directly off the diagram, z-transform both, eliminate $W(z)$ to get $H(z)$; apply the Jury stability test to the resulting quadratic denominator; for the specific $\alpha$, factor the denominator and invert by partial fractions.

Part (a) — difference equations and transfer function. From the diagram, $w(n)=x(n)-\tfrac34 w(n-1)+\alpha w(n-2)$ and $y(n)=w(n)+\tfrac12 w(n-1)$. Z-transforming (zero initial conditions), $$W(z)\Big(1+\tfrac34z^{-1}-\alpha z^{-2}\Big)=X(z),\qquad Y(z)=W(z)\Big(1+\tfrac12z^{-1}\Big)$$ Dividing out $W(z)$, $$H(z)=\boxed{\dfrac{1+\tfrac12z^{-1}}{1+\tfrac34z^{-1}-\alpha z^{-2}}}$$
Part (b) — stability. The poles satisfy $z^2+\tfrac34z-\alpha=0$ (multiplying the denominator by $z^2$). For a causal system, BIBO stability needs both roots strictly inside the unit circle. For a quadratic $z^2+a_1z+a_0$ the Jury conditions are $|a_0|<1$, $D(1)=1+a_1+a_0>0$, $D(-1)=1-a_1+a_0>0$. With $a_1=3/4$, $a_0=-\alpha$: $$|-\alpha|<1\ \Rightarrow\ -1<\alpha<1,\qquad 1+\tfrac34-\alpha>0\ \Rightarrow\ \alpha<1.75,\qquad 1-\tfrac34-\alpha>0\ \Rightarrow\ \alpha<0.25$$ The tightest pair of bounds governs: $$\boxed{-1<\alpha<\tfrac14}$$ confirmed by sampling pole locations numerically on both sides of each boundary.
Part (c) — impulse response at $\alpha=1/2$. Note $\alpha=1/2$ lies outside the part (b) stability range, so the closed form below is a valid but growing (unstable) sequence — the question asks for $h(n)$ regardless of stability. The poles are $$z_{1,2}=\dfrac{-\tfrac34\pm\sqrt{\tfrac{9}{16}+2}}{2}=\dfrac{-3\pm\sqrt{41}}{8}\ \Rightarrow\ z_1=0.42539,\ \ z_2=-1.17539$$ Partial-fraction expansion of $H(z)=\dfrac{1+0.5z^{-1}}{(1-z_1z^{-1})(1-z_2z^{-1})}=\dfrac{A}{1-z_1z^{-1}}+\dfrac{B}{1-z_2z^{-1}}$ with $A=\dfrac{z_1+0.5}{z_1-z_2}$, $B=\dfrac{z_2+0.5}{z_2-z_1}$ gives $A=0.5781$, $B=0.4219$, so $$h(n)=\boxed{\big[0.5781(0.4254)^n+0.4219(-1.1754)^n\big]u(n)}$$ Checked two independent ways: the closed form matches a direct recursion of $w(n)=\delta(n)-0.75w(n-1)+0.5w(n-2)$, $y(n)=w(n)+0.5w(n-1)$ term-by-term ($h(0){=}1$, $h(1){=}-0.25$, $h(2){=}0.6875$, $h(3){=}-0.6406$, ...), and the growing $|z_2|^n=1.1754^n$ term is exactly what part (b)'s instability verdict predicts.

Question 1 — final results
Quantity	Result
$H(z)$	$\dfrac{1+0.5z^{-1}}{1+0.75z^{-1}-\alpha z^{-2}}$
Stability range	$-1<\alpha<0.25$
Poles at $\alpha=1/2$	$z_1=0.4254,\ z_2=-1.1754$
$h(n)$, $\alpha=1/2$	$[0.5781(0.4254)^n+0.4219(-1.1754)^n]u(n)$

Concept & Theory

Topic. Discrete-time LTI feedback systems: transfer function from a block diagram, BIBO stability via pole location, impulse response by partial fractions.

Key relations. $z$-transform of a delay: $x(n-k)\leftrightarrow z^{-k}X(z)$; BIBO stability $\Leftrightarrow$ all poles of a causal system's $H(z)$ satisfy $|z_i|<1$; Jury test for a quadratic $z^2+a_1z+a_0$; partial-fraction inversion $\dfrac{A}{1-pz^{-1}}\leftrightarrow Ap^nu(n)$.

Why this works. Each delay block contributes one $z^{-1}$ factor, so reading gains straight off the diagram and z-transforming turns the block diagram directly into an algebraic transfer function; the poles of that transfer function are the natural modes $z_i^n$ whose growth or decay determines BIBO stability.

Common pitfall. Forgetting that the diagram's "$-3/4$" label is the gain the branch multiplies by (so the difference equation carries $-\tfrac34w(n-1)$, not $+\tfrac34w(n-1)$), and computing an impulse response without first checking whether the requested $\alpha$ is even in the stable range from part (b) — here it is not, which explains the diverging $(-1.1754)^n$ term rather than signalling an arithmetic error.

Source. A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed., §10.3 (z-transform properties), §10.7 (stability from pole locations); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed., §3.3 (partial-fraction inverse z-transform), §3.6 (BIBO stability).

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

Question 1: Discrete-Time Feedback System — Transfer Function, Stability, Impulse Response (1/6 of paper)