2.01 Systems Classification LTI Properties and Stability

2.01 Systems Classification, LTI Properties & Stability

Related Concepts: 1.01 Signals, Systems & Singularity Functions | 2.02 The Continuous Convolution Integral | 2.05 Block-Diagram Representations & System Interconnections

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1. Conceptual Framework of Systems

A system is mathematically defined as an operator $\mathbb{T}[\cdot ]$ {a transformation rule} that maps an input excitation signal $x(t)$ or $x[n]$ to a unique output response signal $y(t)$ or $y[n]$.

$\text{Continuous-Time: }y(t)=\mathbb{T}[x(t)]|\text{Discrete-Time: }y[n]=\mathbb{T}[x[n]]$

graph LR
    Input["x(t) / x[n] <br> (Excitation / Independent Variable)"] --> System["System Operator <br> <b>T[·]</b>"]
    System --> Output["y(t) / y[n] <br> (Response / Dependent Variable)"]
    style System fill:#2a9d8f,stroke:#1e6059,stroke-width:2px,color:#fff

To prevent cognitive overload, let’s break down the basic classifications of system behaviors before jumping into algebraic proofs.

Terminology & Concept Breakdown

• Excitation: The external input signal applied to a system to drive its state.
• Response: The resulting output signal generated by the system as a direct consequence of the excitation.
• Memory (State): The system’s capacity to store energy or data, allowing its current output to be influenced by past or future excitations.
• Canonic System: A system designed with the absolute minimum number of memory storage components (e.g., integrators or delays).

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2. Basic System Classifications & Properties

2.1 Memoryless vs. Dynamic (Memory) Systems

• Memoryless (Static) System: The current output $y(t)$ at any instant $t_0$ depends strictly on the input $x(t)$ at that same instant $t_0$. It contains no energy-storage elements (like capacitors or inductors).
  • LTI Characterization: A continuous-time LTI system is memoryless if and only if its impulse response $h(t)$ is a scaled impulse: $h(t)=c\delta (t)⟹y(t)=cx(t)$
• Dynamic (Memory) System: The current output depends on present inputs as well as past or future inputs. These systems contain energy-storage elements.
  • Examples: Integrators, differentiators, capacitors, inductors, and unit delays.

2.2 Causal, Non-Causal, and Anticausal Systems

This property dictates whether a system is physically realizable {buildable in the real world using physical components}.

graph TD
    SystemC["System Causality Classifications"] --> Causal["Causal <br> (Real-world buildable)"]
    SystemC --> NonCausal["Non-Causal <br> (Theoretical / Offline)"]
    SystemC --> Anticausal["Anticausal <br> (Strictly future-dependent)"]

    Causal --> CDef["Depends on: <br> Present & Past inputs <br> <i>y(t) = x(t) + x(t-1)</i>"]
    NonCausal --> NDef["Depends on: <br> Future inputs as well <br> <i>y(t) = x(t+1) + x(t-1)</i>"]
    Anticausal --> ADef["Depends on: <br> Strictly future inputs <br> <i>y(t) = x(t+2)</i>"]

• Causal System: The output at any time depends strictly on the present and/or past values of the input, never on future values.
  • Mathematical Condition: $y(t_0)$ is independent of $x(t)$ for all $t>t_0$.
  • LTI Impulse Response Boundary: A continuous LTI system is causal if and only if: $h(t)=0\text{for all }t<0$
• Non-Causal System: The output depends on future values of the input. These can only be implemented in offline processing {processing recorded data stored in a database}.
• Anticausal System: The output depends strictly on the future values of the input.
  • LTI Impulse Response Boundary: $h(t)=0$ for all $t>0$.

2.3 Invertible vs. Non-Invertible Systems

• Invertible System: A system where distinct inputs produce distinct outputs. An inverse system ${\mathbb{T}}^{-1}[\cdot ]$ can be cascaded with the original system to perfectly recover the input signal $x(t)$: ${\mathbb{T}}^{-1}[y(t)]={\mathbb{T}}^{-1}[\mathbb{T}[x(t)]]=x(t)$
• Non-Invertible System: Multiple distinct inputs map to the same output (e.g., $y(t)=x^2(t)$, where $x(t)=\pm 2$ both yield $y(t)=4$). The original input cannot be uniquely recovered.

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3. The Linearity Test (Superposition & Homogeneity)

To prove a system is linear, it must simultaneously satisfy the Additive (superposition) property and the Homogeneous (scaling) property.

graph TD
    LTest["Linearity Check (Superposition)"] --> Add["1. Additivity Test <br> T[x_1(t) + x_2(t)] = T[x_1(t)] + T[x_2(t)]"]
    LTest --> Hom["2. Homogeneity Test <br> T[a·x_1(t)] = a·T[x_1(t)]"]
    Add --> Combined["Unified Superposition Condition: <br> T[a·x_1(t) + b·x_2(t)] = a·y_1(t) + b·y_2(t)"]
    Hom --> Combined

Step-by-Step Algebraic Verification Template:

1. Define two distinct inputs $x_1(t)$ and $x_2(t)$ which produce outputs $y_1(t)=\mathbb{T}[x_1(t)]$ and $y_2(t)=\mathbb{T}[x_2(t)]$.
2. Construct a weighted linear combination of these inputs: $x_3(t)=a{x}_1(t)+b{x}_2(t)$
3. Evaluate the system’s output due to this combined input: $y_3(t)=\mathbb{T}[x_3(t)]=\mathbb{T}[a{x}_1(t)+b{x}_2(t)]$
4. Compare $y_3(t)$ with the weighted combination of the individual outputs: $y_{\text{target}}(t)=a{y}_1(t)+b{y}_2(t)$
   • If $y_3(t)=y_{\text{target}}(t)$, the system is Linear.
   • If $y_3(t)\ne y_{\text{target}}(t)$, the system is Non-linear.

The Zero-Input Linearity Hack

A necessary (but not sufficient) condition for a system to be linear is that a zero input must yield a zero output. If evaluating the system with $x(t)=0$ results in a non-zero output, the system is immediately Non-linear (e.g., $y(t)=x(t)+3⟹y(t)=3$ when $x(t)=0$).

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4. The Time-Invariance Test

A system is time-invariant {shift-invariant} if the behavioral characteristics of the system are constant over time. A delay or advance in the input signal must produce an identical shift in the output signal.

Step-by-Step Algebraic Verification Template:

To prevent shift-scale confusion, always use this rigid two-pathway comparison:

                 PATHWAY 1: Delay the Input, then pass through System
Input x(t) ------------> Delay by t_0 ------> x(t - t_0) ------> [ System ] ------> y_1(t) = T[x(t - t_0)]
 
                 PATHWAY 2: Pass through System, then delay the Output
Input x(t) ------------> [ System ] --------> y(t) ------------> Delay by t_0 ------> y_2(t) = y(t - t_0)

1. Pathway 1 (System output due to shifted input): Evaluate the system response by replacing every instance of $x(t)$ in the system equation with a delayed version $x(t-t_0)$: $y_1(t)=\mathbb{T}[x(t-t_0)]$
2. Pathway 2 (Shifted system output): Take the original output equation $y(t)=\mathbb{T}[x(t)]$ and delay the entire signal by replacing every instance of the independent variable $t$ with $(t-t_0)$: $y_2(t)=y(t-t_0)$
3. Compare:
   • If $y_1(t)=y_2(t)$, the system is Time-Invariant.
   • If $y_1(t)\ne y_2(t)$, the system is Time-Variant.

The Variable Coefficient Trap

If the independent variable $t$ appears as an explicit multiplier or coefficient outside the input bracket (e.g., $y(t)=tx(t)$), Pathway 2 will shift that coefficient while Pathway 1 will not. This makes the system Time-Variant.

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5. Standard Core Solved Numericals

5.1 Case Study: $y(t)=3x(-t)+2x(t)$ [PYQ: 2025, 2022]

Determine if this system is Linear and Time-Invariant.

1. Linearity Proof:

• Let $y_1(t)=3x_1(-t)+2x_1(t)$ and $y_2(t)=3x_2(-t)+2x_2(t)$.
• Apply combined input $x_3(t)=a{x}_1(t)+b{x}_2(t)$: $y_3(t)=3[a{x}_1(-t)+b{x}_2(-t)]+2[a{x}_1(t)+b{x}_2(t)]$ $y_3(t)=a[3x_1(-t)+2x_1(t)]+b[3x_2(-t)+2x_2(t)]$ $y_3(t)=a{y}_1(t)+b{y}_2(t)$
• Result: The system is Linear.

2. Time-Invariance Check:

• Pathway 1: Shift the input signal $x(t)\to x(t-t_0)$. $y_1(t)=3x(-t-t_0)+2x(t-t_0)$
• Pathway 2: Shift the output signal $t\to t-t_0$. $y_2(t)=y(t-t_0)=3x(-(t-t_0))+2x(t-t_0)=3x(-t+t_0)+2x(t-t_0)$
• Compare: $y_1(t)\ne y_2(t)$ because $-t-t_0\ne -t+t_0$.
• Result: The system is Time-Variant.

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5.2 Case Study: $\frac{dy(t)}{dt}+3y(t)=x(t)\frac{dx(t)}{dt}$ [PYQ: 2022]

Determine if this system is Linear.

• Let input $x_1(t)$ yield output $y_1(t)$ satisfying: ${\dot{y}}_1(t)+3y_1(t)=x_1(t){\dot{x}}_1(t)$.
• Let input $x_2(t)$ yield output $y_2(t)$ satisfying: ${\dot{y}}_2(t)+3y_2(t)=x_2(t){\dot{x}}_2(t)$.
• Let input be $x_3(t)=a{x}_1(t)+b{x}_2(t)$. The system output $y_3(t)$ must satisfy: ${\dot{y}}_3(t)+3y_3(t)=[a{x}_1(t)+b{x}_2(t)]\frac{d}{dt}[a{x}_1(t)+b{x}_2(t)]$ ${\dot{y}}_3(t)+3y_3(t)=a^2{x}_1(t){\dot{x}}_1(t)+b^2{x}_2(t){\dot{x}}_2(t)+ab[x_1(t){\dot{x}}_2(t)+x_2(t){\dot{x}}_1(t)]$
• Evaluate the target response $y_{\text{target}}(t)=a{y}_1(t)+b{y}_2(t)$: $\frac{d}{dt}[a{y}_1(t)+b{y}_2(t)]+3[a{y}_1(t)+b{y}_2(t)]=a[{\dot{y}}_1(t)+3y_1(t)]+b[{\dot{y}}_2(t)+3y_2(t)]$ $a[{\dot{y}}_1(t)+3y_1(t)]+b[{\dot{y}}_2(t)+3y_2(t)]=a{x}_1(t){\dot{x}}_1(t)+b{x}_2(t){\dot{x}}_2(t)$
• Since the RHS of the actual combined response equation contains squared parameters ($a^2,b^2$) and cross-terms ($ab$) which do not equal the target response RHS, $y_3(t)\ne a{y}_1(t)+b{y}_2(t)$.
• Result: The system is Non-linear.

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5.3 Case Study: $[y[n]{]}^2=n[x[n]{]}^2$ [HW / Lab]

Determine if this discrete-time system is Linear and Time-Invariant.

1. Linearity Proof:

• Let $[y_1[n]{]}^2=n[x_1[n]{]}^2⟹y_1[n]=\sqrt{n}{x}_1[n]$.
• Let $[y_2[n]{]}^2=n[x_2[n]{]}^2⟹y_2[n]=\sqrt{n}{x}_2[n]$.
• Apply combined input $x_3[n]=a{x}_1[n]+b{x}_2[n]$: $y_3[n]=\sqrt{n}(a{x}_1[n]+b{x}_2[n])=a(\sqrt{n}{x}_1[n])+b(\sqrt{n}{x}_2[n])=a{y}_1[n]+b{y}_2[n]$
• Squaring both sides shows: $[y_3[n]{]}^2=n[a{x}_1[n]+b{x}_2[n]{]}^2$.
• Since the scaled superposition holds perfectly in amplitude, the system is Linear. (Note: This verifies the senior’s lab note checklist stating $[y[n]{]}^2=n[x[n]{]}^2$ behaves as a linear system under amplitude transformation).

2. Time-Invariance Check:

• Pathway 1 (Shifted input): $[y_1[n]{]}^2=n[x[n-n_0]{]}^2$.
• Pathway 2 (Shifted output): $[y[n-n_0]{]}^2=(n-n_0)[x[n-n_0]{]}^2$.
• Compare: $y_1[n]\ne y(n-n_0)$ due to the un-shifted time multiplier coefficient $n$ on Pathway 1.
• Result: The system is Time-Variant.

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6. BIBO Stability & Absolute Integrability Derivation

A system is defined as Bounded-Input Bounded-Output (BIBO) stable if every bounded input signal applied to it produces a bounded output response.

$\text{Condition: If }|x(t)|\le M_x<\infty ⟹|y(t)|\le M_y<\infty$

🚨 MANDATORY DERIVATION: Sufficient and Necessary Stability Bound [PYQ: 2017, 2015]

Prove that a continuous-time LTI system is BIBO stable if and only if its impulse response $h(t)$ is absolutely integrable.

Step 0: Setup and Definitions

An input signal $x(t)$ is bounded such that $|x(t)|\le M_x<\infty$ for all $t$. The output of a continuous LTI system is governed by the continuous convolution integral: $y(t)={\int }_{-\infty }^{\infty }h(\tau )x(t-\tau )d\tau$

Step 1: Upper Bound Expansion (Sufficiency Proof)

Take the absolute value of the output $y(t)$: $|y(t)|=\left|{\int }_{-\infty }^{\infty }h(\tau )x(t-\tau )d\tau \right|$

Apply the integral triangle inequality property {the absolute value of an integral is less than or equal to the integral of the absolute value}: $|y(t)|\le {\int }_{-\infty }^{\infty }\left|h(\tau )x(t-\tau )\right|d\tau$

$⟹|y(t)|\le {\int }_{-\infty }^{\infty }|h(\tau )|\cdot |x(t-\tau )|d\tau$

Step 2: Substitute the Input Bound

Since $|x(t)|\le M_x$ for all $t$, we can substitute this bound into the inequality: $|y(t)|\le M_x{\int }_{-\infty }^{\infty }|h(\tau )|d\tau$

For $|y(t)|\le M_y<\infty$ to hold (ensuring the output is bounded), the integral must evaluate to a finite value: ${\int }_{-\infty }^{\infty }|h(\tau )|d\tau <\infty \text{(Sufficient Condition Proved)}[299]$

Step 3: Necessity Proof

To prove this condition is also necessary, we must show that if the integral is infinite, we can construct a bounded input that forces the output to blow up to infinity. Let the bounded input signal be defined as the sign-reversal of the shifted impulse response: $x(t-\tau )=\text{sgn}(h(\tau ))=\{\begin{array}{ll}+1,& h(\tau )\ge 0\\ -1,& h(\tau )<0\end{array}$ Note that this input is strictly bounded: $|x(t)|\le 1<\infty$. Evaluate the system output $y(t)$ at the origin ($t=0$): $y(0)={\int }_{-\infty }^{\infty }h(\tau )x(-\tau )d\tau ={\int }_{-\infty }^{\infty }h(\tau )\text{sgn}(h(\tau ))d\tau ={\int }_{-\infty }^{\infty }|h(\tau )|d\tau$ If ${\int }_{-\infty }^{\infty }|h(\tau )|d\tau =\infty$, then $y(0)=\infty$, meaning a bounded input has produced an infinite output. Thus, absolute integrability is a necessary condition.

${\int }_{-\infty }^{\infty }|h(t)|dt<\infty (\mathbf{Proved})$

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7. Core Solved Stability Traps

7.1 Stability check of $y(n)=ax(n+1)+bx(n-1)$ [PYQ: 2020, 2019]

• Let the input sequence be bounded: $|x(n)|\le M_x<\infty$ for all $n$.
• Evaluate the absolute boundary of the output sequence: $|y(n)|=|ax(n+1)+bx(n-1)|$
• Apply the triangle inequality: $|y(n)|\le |a|\cdot |x(n+1)|+|b|\cdot |x(n-1)|$
• Substitute the bound $M_x$: $|y(n)|\le |a|M_x+|b|M_x=(|a|+|b|)M_x$
• Since $a$ and $b$ are finite system constants, $(|a|+|b|)M_x$ is a finite value.
• Result: The system is BIBO Stable.

7.2 Stability check of $h(t)=e^{-t}u(t-1)$

• Evaluate the absolute integrability integral: $I={\int }_{-\infty }^{\infty }|h(t)|dt={\int }_1^{\infty }{e}^{-t}dt$ $I={\left[-e^{-t}\right]}_1^{\infty }=0-(-e^{-1})=e^{-1}\approx 0.368$
• Since the integral yields a finite value ($0.368<\infty$), the system is stable.
• Result: The system is BIBO Stable.

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Common Mistakes That Cost Marks

• Applying Shift to Coefficients: When checking time-invariance for systems like $y(t)=t^{2}x(t)$, students often write Pathway 1 as $y_1(t)=(t-t_0{)}^{2}x(t-t_0)$. This is incorrect. In Pathway 1, only the input signal $x(t)$ is delayed, yielding $y_1(t)=t^{2}x(t-t_0)$.
• Flipping Signs in Reversal Shifting: For the system $y(t)=x(-t)$, Pathway 1 (shifted input) is $y_1(t)=x(-t-t_0)$. Pathway 2 (shifted output) is $y_2(t)=y(t-t_0)=x(-(t-t_0))=x(-t+t_0)$. Mixing up these signs is the #1 reason students lose marks in time-invariance proofs.
• Causality Boundary Confusion: For the system $y(t)=x(2t)$, students often assume it is causal because $t>0⟹2t>t$. However, we must test for all real time. At $t=-1$, $y(-1)=x(-2)$, which depends on past data. But at $t=1$, $y(1)=x(2)$, which depends on future data. Thus, the system is Non-Causal.

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PYQ Bank — Verbatim Questions & Answer Plans

Q1: Linearity and Time-Invariance check of $y(t)=3x(-t)+2x(t)$

• Metadata: Type: Analytical Proof | Years: 2025, 2022 | Marks: 10/08 | Frequency: ⭐⭐⭐⭐⭐
• Answer Plan:
  1. Perform the linearity proof using the weighted combined input $x_3(t)=a{x}_1(t)+b{x}_2(t)$. Show that $y_3(t)=a{y}_1(t)+b{y}_2(t)$.
  2. Perform the time-invariance check using Pathway 1 and Pathway 2. Explicitly show the sign difference ($x(-t-t_0)$ vs. $x(-t+t_0)$) to prove the system is time-variant.

Q2: Derive the Necessary and Sufficient Condition for BIBO Stability of an LTI system

• Metadata: Type: Mathematical Derivation | Years: 2017, 2015 | Marks: 12/10 | Frequency: ⭐⭐⭐⭐
• Answer Plan:
  1. Define bounded input $|x(t)|\le M_x<\infty$ and express output $y(t)$ using the continuous convolution integral.
  2. Apply the integral triangle inequality to prove the sufficiency condition ($|y(t)|\le M_x\int |h(\tau )|d\tau <\infty$).
  3. Set up the necessity proof by defining the bounded test input $x(t-\tau )=\text{sgn}(h(\tau ))$ to show the system output blows up if the absolute integrability integral is infinite.

Q3: System parameter evaluation of $y[n]=3y[n-1]-n^{2}x[n]+4x[n-1]-2x[n+1]$

• Metadata: Type: Discrete System Proof | Years: 2024, 2019 | Marks: 15/12 | Frequency: ⭐⭐⭐⭐⭐
• Answer Plan:
  1. Test Linearity: The difference equation contains only first-power scaling terms of $y$ and $x$ with no added offsets. Superposition holds $⟹$ Linear.
  2. Test Time-Invariance: The coefficient multiplier term $n^2$ depends explicitly on time $⟹$ Time-Variant.
  3. Test Causality: The term $-2x[n+1]$ depends on a future input sample at index $n+1⟹$ Non-Causal.

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Self-Check Before Moving On

• Can you prove why $y(t)=x(t)+5$ is non-linear using the zero-input test?
• Do you know the exact algebraic steps to prove why $y(t)=x(-t)$ is time-variant?
• Can you write out the complete sufficiency and necessity steps for the LTI BIBO stability derivation on a blank sheet of paper?

Source: (k.Deergha Rao) signals and systems.pdf Section 2.2; 1 intro to Signal & systems.pdf Pages 15–18; ECE 2107 Past Term Examination Papers (2015-2025).