Question 3 of 6: Impulse Response, Output and Stability of a First-Order System

Question 3: Impulse Response, Output and Stability of a First-Order 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 first-order system driven by a two-sided input made of a decaying causal exponential and an anti-causal unit step.

Find. The impulse response, the complete output sequence, and a justified verdict on bounded-input bounded-output stability.

Approach. Split $H(z)$ into a constant plus a single pole term to invert it, transform the two halves of the input on their own strips, multiply, and recognise that the zero of $H$ at $z=1$ cancels the input pole at $z=1$.

1. Split the improper first-order system function. Numerator and denominator have the same degree in $z^{-1}$, so write $H(z)=\alpha+\dfrac{\beta}{1+\frac{3}{4}z^{-1}}$ and match: $$\tfrac{3}{4}\alpha=-1\ \Rightarrow\ \alpha=-\tfrac{4}{3},\qquad \alpha+\beta=1\ \Rightarrow\ \beta=\tfrac{7}{3}.$$
2. Invert term by term. The system is causal, so the pole at $z=-3/4$ yields a right-sided exponential: $$\boxed{h[n]=-\tfrac{4}{3}\,\delta[n]+\tfrac{7}{3}\left(-\tfrac{3}{4}\right)^{n}u[n]}$$ The first two samples check against long division of $H(z)$: $h[0]=-\tfrac43+\tfrac73=1$ and $h[1]=\tfrac73\left(-\tfrac34\right)=-\tfrac74$, matching the expansion $(1-z^{-1})(1-\tfrac34z^{-1}+\dots)=1-\tfrac74z^{-1}+\dots$.
3. Transform the input on its own strip. From the standard pairs, $(1/3)^{n}u[n]\leftrightarrow 1/(1-\tfrac13z^{-1})$ for $|z| \gt 1/3$ and $-u[-n-1]\leftrightarrow 1/(1-z^{-1})$ for $|z| \lt 1$, so $u[-n-1]$ contributes the same pole with a minus sign: $$X(z)=\frac{1}{1-\frac{1}{3}z^{-1}}-\frac{1}{1-z^{-1}},\qquad \tfrac13 \lt |z| \lt 1 .$$ The two strips overlap, so the two-sided input does have a transform.
4. Form the output transform and watch the cancellation. Multiplying, the second term collapses because $H$ has a zero exactly at the input pole $z=1$: $$\frac{1-z^{-1}}{1+\frac34 z^{-1}}\cdot\left(-\frac{1}{1-z^{-1}}\right)=-\frac{1}{1+\frac34 z^{-1}} .$$ The first term expands in partial fractions as $\dfrac{21/13}{1+\frac34z^{-1}}-\dfrac{8/13}{1-\frac13z^{-1}}$ (the coefficients follow from $A+B=1$ and $-A/3+3B/4=-1$).
5. Collect the terms. Adding the two contributions, the residue at the system pole becomes $21/13-1=8/13$: $$Y(z)=\frac{8/13}{1+\frac34 z^{-1}}-\frac{8/13}{1-\frac13 z^{-1}} .$$ The pole at $z=1$ has vanished, so the region of convergence is no longer an annulus: it extends to $|z| \gt 3/4$ and both terms are right-sided. Hence $$\boxed{y[n]=\frac{8}{13}\left[\left(-\frac{3}{4}\right)^{n}-\left(\frac{1}{3}\right)^{n}\right]u[n]}$$
6. Sanity-check the first two samples. The formula gives $y[0]=0$ and $y[1]=\tfrac{8}{13}\left(-\tfrac34-\tfrac13\right)=-\tfrac23$. Direct convolution agrees: because the input equals 1 for all $n\le-1$, $$y[0]=\sum_{k\ge0}h[k]x[-k]=h[0]\cdot1+\sum_{k\ge1}h[k]\cdot1=H(z)\big|_{z=1}=\frac{1-1}{1+3/4}=0 ,$$ so the vanishing first sample is a direct consequence of the zero at $z=1$: the system is a differencer at DC and the anti-causal step is a DC signal.
7. Part (c): test stability. The system is causal with its only pole at $z=-3/4$, so its region of convergence $|z| \gt 3/4$ contains the unit circle, which is the frequency-domain criterion for BIBO stability. Equivalently, summing the impulse response absolutely, $$\sum_{n=-\infty}^{\infty}|h[n]|=|h[0]|+\frac{7}{3}\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^{n}=1+\frac{7}{3}\cdot3=\boxed{8 \lt \infty}$$ so the system is stable. Note that the $n=0$ term must be evaluated as $|h[0]|=|-\tfrac43+\tfrac73|=1$, not as the sum of the magnitudes of the two pieces — they partially cancel there.