--- title: "LTI properties; BIBO stability condition" subject: "Signals and Systems" chapter: 6 key: "semester_4/signals_and_systems:lti-bibo-stability" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-bibo-stability site: BCT Engineering --- # LTI properties; BIBO stability condition Signals and Systems · Chapter 6: Discrete-Time Systems - Asked in: 2070 Magh, 8 marks - Video: [Condition for stability of LTI Systems (DSP Lecture-43)](https://www.youtube.com/watch?v=yQIQJRbtJbw) — Engineers choice tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-bibo-stability - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6 ## Question What are the properties of LTI system? Show that the output of an LTI system is stable if the impulse response is absolutely summable. ## Short answer For an LTI system, BIBO stability holds iff the impulse response is absolutely summable. If ∑|h[n]|=S<∞ and |x[n]|≤B, then |y[n]|≤B·S, so bounded output. The converse holds using a bounded input sign(h[-n]) to show necessity. ## Answer #### Properties of LTI systems For an LTI system with impulse response $h[n]$ (output $y=x*h$): 1. **Commutative:** $x*h=h*x$. 2. **Associative:** $(x*h_1)*h_2=x*(h_1*h_2)$ (cascade connection). 3. **Distributive:** $x*(h_1+h_2)=x*h_1+x*h_2$ (parallel connection). 4. **Memoryless** if and only if $h[n]=K\delta[n]$. 5. **Causal** if and only if $h[n]=0$ for $n<0$. 6. **Stable (BIBO)** if and only if $\sum_n|h[n]|<\infty$. 7. **Invertible** if an $h_i[n]$ exists with $h*h_i=\delta[n]$. 8. **Step response:** $s[n]=\sum_{k=-\infty}^{n}h[k]$. #### Proof: absolutely summable $h[n]$ gives a stable system A system is **BIBO stable** if every bounded input produces a bounded output. Let the input be bounded: $|x[n]|\le B_x<\infty$ for all $n$. The output is $$ y[n]=\sum_{k=-\infty}^{\infty}h[k]\,x[n-k] $$ Take the magnitude and use the triangle inequality ($|\sum a_k|\le\sum|a_k|$): $$ \begin{aligned} |y[n]| &= \left|\sum_{k}h[k]\,x[n-k]\right| \le \sum_{k}|h[k]|\,|x[n-k]| \\ &\le B_x\sum_{k=-\infty}^{\infty}|h[k]| \end{aligned} $$ If $h[n]$ is absolutely summable, $\sum_k|h[k]|=S<\infty$, then $$ |y[n]|\le B_xS<\infty \quad\text{for all } n $$ so the output is bounded and the system is stable. **Converse (necessity):** if $S=\infty$, choose the bounded input $x[n]=\mathrm{sgn}(h[-n])$ (values $\pm1$). Then $y[0]=\sum_k h[k]\,\mathrm{sgn}(h[k])=\sum_k|h[k]|=\infty$, so a bounded input gives an unbounded output. Hence **absolute summability is necessary and sufficient** for BIBO stability of an LTI system. #### Examples - $h[n]=(0.5)^n u[n]$: $S=\frac{1}{1-0.5}=2<\infty$, **stable**. - $h[n]=u[n]$ (accumulator): $S=\sum_{n\ge0}1=\infty$, **unstable** (input $u[n]$ gives $y[n]=(n+1)u[n]$, which grows without bound). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does BIBO stable mean? A. Every bounded input gives a bounded output B. Every input gives zero output eventually C. The impulse response is zero for n<0 D. The output is bounded only for the unit step input Answer: A. Bounded-input bounded-output stability. (Key point: BIBO definition) 2. Which step bounds |y[n]| in the proof? A. Triangle inequality and |x[n-k]|<=Bx give |y[n]|<=Bx sum|h[k]| B. Parseval's theorem gives |y[n]|=Bx sum h[k] C. Cauchy product gives |y[n]|<=Bx / sum|h[k]| D. Differentiation of y[n] gives |y[n]|<=Bx Answer: A. Taking magnitudes inside the sum gives the product of the input bound and the sum of |h|. (Key point: Proof inequality) 3. Which bounded input shows that a non-summable h gives an unbounded output? A. x[n]=delta[n] B. x[n]=sgn(h[-n]) C. x[n]=h[n] D. x[n]=u[n] always Answer: B. With this input y[0]=sum|h[k]|=infinity. (Key point: Necessity) 4. Why is h[n]=u[n] unstable? A. sum|h|=2; so it is stable B. h[n] is non-zero for n<0 C. sum|h|=infinity; input u[n] gives y=(n+1)u[n] which grows without bound D. Its output is bounded by 1 for the unit step input Answer: C. The infinite sum means some bounded input blows up the output. 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