Discrete-Time Causal System — Difference Equation, Transfer Function and Stability: 17-Phys-B4 Signals and Communications, December 2014, Q6

Question 6: Discrete-Time Causal System — Difference Equation, Transfer Function and Stability (1/5 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.

Check — figure gain corrected

The source page's own printed gain label on the FIRST feedback branch reads −3/4 (a fraction with an explicit minus sign above the 4), not $+3/4$. The solution below uses the printed $-3/4$ throughout; the second feedback gain, labelled $\alpha$, is unspecified numerically (that is the subject of part d).

Reconstructed block diagram: first summing junction (input x(n), feedback gain -3/4 through delay D) feeds a second summing junction (feedback gain alpha through delay D), output y(n).

Given. Node $w(n)$ = output of the first summing junction: $w(n)=x(n)-\tfrac34w(n-1)$ (feedback gain $-3/4$, one unit delay); output node: $y(n)=w(n)+\alpha y(n-1)$ (feedback gain $\alpha$, one unit delay).

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

Approach. Z-transform each summing-junction equation directly from the block diagram, eliminate the intermediate node $w(n)$, then read the difference equation off the resulting rational $H(z)$; get $h(n)$ by partial-fraction expansion; stability follows from the pole locations.

Part (b) — transfer function. Z-transforming each summing junction, $$W(z)=X(z)-\tfrac34z^{-1}W(z)\ \Rightarrow\ W(z)=\dfrac{X(z)}{1+\tfrac34z^{-1}}\qquad Y(z)=W(z)+\alpha z^{-1}Y(z)\ \Rightarrow\ Y(z)\big(1-\alpha z^{-1}\big)=W(z)$$ Substituting $W(z)$ and solving for $Y(z)/X(z)$, $$H(z)=\boxed{\dfrac{1}{\big(1+\tfrac34z^{-1}\big)\big(1-\alpha z^{-1}\big)}}$$ — a cascade of two first-order recursive sections, as the block diagram shows, with poles at $z=-3/4$ and $z=\alpha$.
Part (a) — difference equation. Expanding the denominator, $\big(1+\tfrac34z^{-1}\big)\big(1-\alpha z^{-1}\big)=1+\big(\tfrac34-\alpha\big)z^{-1}-\tfrac34\alpha z^{-2}$, so $Y(z)\big[1+(\tfrac34-\alpha)z^{-1}-\tfrac34\alpha z^{-2}\big]=X(z)$ inverse-transforms to $$y(n)=\boxed{x(n)-\Big(\dfrac34-\alpha\Big)y(n-1)+\dfrac34\alpha\,y(n-2)}$$ A direct block-diagram time-domain simulation (impulse input, 20 samples, $\alpha=0.4$ test case) reproduces this difference equation exactly (max deviation $5.6\times10^{-17}$, i.e. floating-point roundoff only).
Part (c) — impulse response. Partial-fraction expanding $H(z)$ with poles $a_1=-3/4$, $a_2=\alpha$ (assuming $\alpha\ne-3/4$), using the standard form $\dfrac{1}{(1-a_1z^{-1})(1-a_2z^{-1})}=\dfrac{a_1/(a_1-a_2)}{1-a_1z^{-1}}+\dfrac{a_2/(a_2-a_1)}{1-a_2z^{-1}}$: $$h(n)=\boxed{\left[\dfrac{3/4}{3/4+\alpha}\right]\Big({-}\dfrac34\Big)^{\!n}u(n)+\left[\dfrac{\alpha}{\alpha+3/4}\right]\alpha^{n}u(n)}$$ Verified against the same block-diagram time-domain simulation (impulse response, 20 samples, $\alpha=0.4$): max deviation $5.6\times10^{-17}$.
Part (d) — stability range for α. This is a causal system, so BIBO stability requires every pole strictly inside the unit circle. The first pole, $z=-3/4$, satisfies $|-3/4|=0.75<1$ unconditionally; the second pole is $z=\alpha$ itself, so stability requires $$\boxed{-1<\alpha<1}$$

Question 6 — final results
Quantity	Result
Difference equation	$y(n)=x(n)-(\tfrac34-\alpha)y(n-1)+\tfrac34\alpha\,y(n-2)$
Transfer function	$H(z)=\dfrac{1}{(1+\tfrac34z^{-1})(1-\alpha z^{-1})}$
Impulse response	$h(n)=\left[\tfrac{3/4}{3/4+\alpha}\right](-\tfrac34)^nu(n)+\left[\tfrac{\alpha}{\alpha+3/4}\right]\alpha^nu(n)$
Stability range	$-1<\alpha<1$

Concept & Theory

Topic. Z-transform analysis of a cascaded two-section discrete-time recursive (IIR) system: difference equation, transfer function, impulse response, and BIBO stability.

Key relations. $z^{-1}$ (unit delay) $\leftrightarrow$ multiply by $z^{-1}$ in the z-domain; a causal LTI system is BIBO stable iff every pole of $H(z)$ lies strictly inside the unit circle $|z|<1$; partial-fraction expansion of a product of two first-order factors gives a sum of two geometric (exponential) sequences.

Why this works. Each summing junction with a single feedback path through a gain and one delay is exactly a first-order recursive (IIR) section; cascading two such sections multiplies their transfer functions, which is why $H(z)$ factors cleanly into two first-order terms whose poles are simply the two feedback gains.

Common pitfall. forgetting that stability requires ALL poles inside the unit circle, not just the $\alpha$-dependent one.

Source. Proakis & Manolakis, Digital Signal Processing, 4th ed., §3.3 (z-transform of difference equations), §3.5 (stability via pole locations), §4.2 (partial-fraction inversion).

Question 6 of 7: Discrete-Time Causal System — Difference Equation, Transfer Function and Stability

Question 6: Discrete-Time Causal System — Difference Equation, Transfer Function and Stability (1/5 of paper)