02 Chapter Map - LTI System Properties & Time-Domain Convolution

Chapter 2 Overview & Map of Content (MOC)

Superposition linearity tests, graphical convolution, recursive differential difference equations, and feedback blocks.

📚 Study Notes Index

Read in order — each note assumes the previous one.

#	Note	What it covers
2.00	2.00 LTI System Properties and Time-Domain Convolution Compa	LTI Systems Compact Review, Convolution Formula Sheet
2.01	2.01 Systems Classification LTI Properties and Stability	System Properties, Linearity & Stability, BIBO Stability Proof
2.02	2.02 The Continuous Convolution Integral	Convolution Integral, Continuous Convolution, LTI Convolution
2.03	2.03 Properties of the Convolution Integral	Convolution Properties, Shifting Property of Convolution, Convolution Algebraic Theorems
2.04	2.04 Systems Described by Differential and Difference Equati	Differential Equations, Difference Equations, LCCDE, Homogeneous Solution, Particular Solution
2.05	2.05 Block Diagram Representations and System Interconnectio	Block-Diagrams, System Realization, Direct Form I, Direct Form II, Canonic Realization, Cascade, Parallel

🎯 Exam Weight

ECE 2107 Exam Relevance

Master the core derivations, mathematical definitions, and problem-solving techniques. Refer to ECE 2107 - Signals and Systems for syllabus boundaries and past year questions.

Course Teaching Plan: ECE 2107 - Signals and Systems
Previous chapter: 01 Chapter Map - Signal Fundamentals & Operations
Next chapter: 03 Chapter Map - State-Space Representation of CT Systems

← Chapter 1 Compact Review | Chapter 2 Map | Chapter 3 Compact Review →

Chapter 2: LTI System Properties & Time-Domain Convolution - Compact Review

This compact review card synthesizes all core concepts, mathematical definitions, final-line derivations, and architectural realization structures for continuous-time and discrete-time systems under Chapter 2: LTI System Properties & Time-Domain Convolution.

1. Systems Classification & Mathematical Operators

1.1 Linearity: Superposition & Homogeneity

*(Target: Theory Descriptive / Linearity Verification)*

Concept: A system is linear if and only if it simultaneously satisfies the principles of additivity (superposition) and homogeneity (scaling). A linear system has the property that a zero input strictly produces a zero output.
Governing Equations:
- Additivity: $T [x_{1} (t) + x_{2} (t)] = y_{1} (t) + y_{2} (t)$
- Homogeneity: $T [a \cdot x_{1} (t)] = a \cdot y_{1} (t)$
- Unified Linearity Condition: $T [a \cdot x_{1} (t) + b \cdot x_{2} (t)] = a \cdot y_{1} (t) + b \cdot y_{2} (t) e x t (d er i v a t i o n)$

1.2 Time-Invariance (Shift-Invariance)

*(Target: Theory Descriptive / Time-Invariance Verification)*

Concept: A system is time-invariant if its internal parameters do not change over time. A time shift applied to the input signal must produce an identical time shift in the output response.
Governing Formula: $y (t - t_{0}) = T [x (t - t_{0})] e x t (d er i v a t i o n)$
The Two-Step Verification Checklist:
1. Step 1: Delay the system output $y (t)$ by $t_{0}$ directly: $y (t, t_{0}) = y (t - t_{0})$ .
2. Step 2: Pass a delayed input $x_{1} (t) = x (t - t_{0})$ through the operator: $y_{1} (t) = T [x (t - t_{0})]$ .
3. Check: If $y_{1} (t) \equiv y (t - t_{0})$ , the system is Time-Invariant. If they are not equal, the system is Time-Variant.

1.3 Causality

*(Target: Theory Descriptive / Causality Boundary Check)*

Concept: A system is causal if its output at any instant depends strictly on current and past input values, but is completely independent of future input values. A continuous-time LTI system described by a differential equation is causal when it is initially relaxed {if input $x (t) = 0$ for $t \leq t_{0}$ , then output $y (t) = 0$ for $t \leq t_{0}$ }.
s-Domain/z-Domain Causal Boundary: The impulse response must satisfy: $h (t) = 0 e x t f or t < 0$ $h [n] = 0 e x t f or n < 0$

1.4 Memory (Dynamic) vs. Memoryless (Static)

*(Target: Theory Descriptive / Memoryless Bound)*

Concept: A system is memoryless (static) if its output at any instant $t_{0}$ depends strictly on the input applied at that exact instant $t_{0}$ . If the output depends on inputs at other times (past or future), the system has memory (dynamic).
Memoryless LTI Bound: An LTI system is memoryless if and only if its impulse response is a scaled Dirac delta function: $h (t) = c \cdot δ (t) e x t an d h [n] = c \cdot δ [n]$

1.5 Bounded-Input Bounded-Output (BIBO) Stability

*(Target: Theory Descriptive / Stability Test)*

Concept: A system is BIBO stable if every bounded input sequence produces a bounded output response.
Input-Output Bounded Limits: $e x t I f ∣ x (t) ∣ \leq M_{x} < \infty ⟹ ∣ y (t) ∣ \leq M_{y} < \infty e x t f or a ll t$

2. Continuous-Time Convolution Integral

2.1 The Representation of Signals in Terms of Impulses

*(Target: Mathematical Proof / 5-Mark derivation)*

Concept: Any arbitrary continuous-time signal $x (t)$ can be represented as a weighted superposition of infinite, shifted continuous impulse functions.
Governing Formula: $x (t) = \int_{- \infty}^{\infty} x (a u) δ (t - a u) d a u e x t (d er i v a t i o n)$

2.2 The Continuous Convolution Integral

*(Target: Numerical Solving / System Response)*

Concept: The continuous-time convolution operates on an input $x (t)$ and system impulse response $h (t)$ to calculate the exact time-domain system response $y (t)$ .
Governing Formula: $y (t) = x (t) * h (t) = \int_{- \infty}^{\infty} x (a u) h (t - a u) d a u$

2.3 Mathematical Properties of Continuous Convolution

*(Target: Theory Descriptive & Proofs)*

Commutative Property: $x (t) * h (t) = h (t) * x (t) e x t (d er i v a t i o n)$
Distributive Property: $x (t) * [h_{1} (t) + h_{2} (t)] = x (t) * h_{1} (t) + x (t) * h_{2} (t) e x t (d er i v a t i o n)$
Associative Property: $[x (t) * h_{1} (t)] * h_{2} (t) = x (t) * [h_{1} (t) * h_{2} (t)] e x t (d er i v a t i o n)$
Convolution with a Unit Impulse: $x (t) * δ (t) = x (t) e x t (d er i v a t i o n)$
Convolution with Shifted / Delayed Inputs: $e x t I f y (t) = x (t) * h (t) ⟹ x (t - t_{1}) * h (t - t_{2}) = y (t - t_{1} - t_{2}) e x t (d er i v a t i o n)$
Convolution Time-Scaling Identity:

ight] \quad ext{for } a eq 0 $* * Sp ec ia lC a se (Do u b l e S c a l in g) : *$ y(2t) = 2 \left[ x(2t) * h(2t) ight] \quad ext{(derivation)}$$

Differentiation Property of Convolution: $rac{dy(t)}{dt} = rac{dx(t)}{dt} * h(t) = x(t) * rac{dh(t)}{dt} \quad ext{(derivation)}$

3. Discrete-Time Convolution Sum

3.1 The Discrete Convolution Sum

*(Target: Numerical Solving / Sequence Inversion)*

Concept: The discrete counterpart of the convolution integral, evaluating the output sequence $y [n]$ by summing the weighted and shifted impulse response sequences triggered by each input sample.
Governing Formula: $y [n] = x [n] * h [n] = \sum_{k = - \infty}^{\infty} x [k] h [n - k]$

3.2 Sequence Length Theorem

*(Target: Numerical Solving)*

Concept: Convolving two finite-duration sequences of lengths $N_{1}$ and $N_{2}$ yields a resulting convolved sequence of a larger finite duration.
Governing Formula: $L = N_{1} + N_{2} - 1$

3.3 Periodic (Circular) vs. Linear Convolution

*(Target: Theory Descriptive / 5-Mark comparison)*

Concept: Linear convolution operates on infinite or finite aperiodic sequences. If signals $x_{1} (t)$ and $x_{2} (t)$ are periodic with common period $T$ , linear convolution does not converge. Periodic convolution integrates/sums only over a single period $T$ (or $N$ samples), forcing the output response to be strictly periodic.
Continuous Periodic Convolution: $y_{p} (t) = \int_{0}^{T} x_{1, p} (a u) x_{2, p} (t - a u) d a u e x t (d er i v a t i o n)$
Discrete Periodic Convolution:

angle_N]$$

4. Continuous & Discrete BIBO Stability Proofs

4.1 Continuous-Time Absolute Integrability Proof

*(Target: Mathematical Proof / 8-Mark Derivation)*

Premise: A continuous-time LTI system is BIBO stable if and only if its impulse response is absolutely integrable.
Governing Formula: $\int_{- \infty}^{\infty} ∣ h (a u) ∣ d a u < \infty e x t (d er i v a t i o n)$

4.2 Discrete-Time Absolute Summability Proof

*(Target: Mathematical Proof / 5-Mark Derivation)*

Premise: A discrete-time LTI system is BIBO stable if and only if its impulse response sequence is absolutely summable.
Governing Formula: $\sum_{n = - \infty}^{\infty} ∣ h [n] ∣ < \infty e x t (d er i v a t i o n)$

5. Systems Described by Differential & Difference Equations

5.1 Linear Constant-Coefficient Differential Equations (LCCDE)

*(Target: Numerical Solving / Continuous-Time System)*

Concept: Represents a continuous-time system where the relationship between the output $y (t)$ and input $x (t)$ is governed by a linear combination of derivatives.
Governing Equation: $\sum_{n=0}^{N} a_n rac{d^n y(t)}{dt^n} = \sum_{m=0}^{M} b_m rac{d^m x(t)}{dt^m}$

5.2 Linear Constant-Coefficient Difference Equations

*(Target: Numerical Solving / Discrete-Time System)*

Concept: Governs discrete-time LTI recursive (IIR) or non-recursive (FIR) systems.
Governing Equation: $\sum_{k = 0}^{N} a_{k} y [n - k] = \sum_{m = 0}^{M} b_{m} x [n - m] (e x t w i t h a_{0} = 1)$

5.3 Complete Response Decompositions

*(Target: Theory Descriptive)*

The Output Splitting Law: $y (t) = y_{c} (t) + y_{p} (t) e x t an d y [n] = y_{c} [n] + y_{p} [n]$
The Complementary (Homogeneous) Solution: Evaluated by setting the input terms to zero ( $x (t) = 0$ or $x [n] = 0$ ).
The Particular Solution: Represents the system’s steady-state output corresponding to the specific profile of the input signal.

5.4 Characteristic Roots & Homogeneous Solution Forms

*(Target: Numerical Solving / Algebraic Rules)*

Distinct Roots ( $λ_{i}$ ): $y_c[n] = lpha_1 \lambda_1^n + lpha_2 \lambda_2^n + \dots + lpha_N \lambda_N^n$
Repeated Roots (Root $λ_{1}$ of multiplicity $m$ ):

ight) \lambda_1^n + eta_2 \lambda_2^n + \dots$$

Complex Conjugate Roots ( $λ_{1, 2} = a \pm jb$ ):

ight) \quad ext{where } r = \sqrt{a^2 + b^2}, ; heta = an^{-1}\left(rac{b}{a} ight)$$

6. Block Diagram Realizations & Interconnections

6.1 Direct Form Realizations

*(Target: Block Diagram Realization / 5-to-8 Mark Draw)*

Direct Form I Realization: Separately realizes input zeros (feedforward delays) and output poles (feedback delays). It is non-canonic because it uses a separate set of delay/integrator elements for both parts.
Direct Form II Realization: Merges the delay paths for both input and output sections into a single, unified delay/integrator chain. It is canonic because it minimizes the total number of delay elements required ( $N$ delays for an $N$ -th order system).

6.2 System Interconnections

*(Target: Theory Descriptive / Cascade vs. Parallel)*

  Cascade (Series Connection):      Parallel Connection:
                                        ┌───────┐
       ┌────────┐    ┌────────┐      ┌─>│ h₁(t) ├──┐
  x(t)─│  h₁(t) ├───>│  h₂(t) ─>y(t) │  └───────┘  v  ┌───┐
       └────────┘    └────────┘  x(t)┤             ├──>│ + ├──> y(t)
                                     │  ┌───────┐  ^  └───┘
                                     └─>│ h₂(t) ├──┘
                                        └───────┘

Cascade (Series) Connection: The overall impulse response is the convolution of individual subsystem responses: $h_{e x t o v er a ll} (t) = h_{1} (t) * h_{2} (t) e x t an d h_{e x t o v er a ll} [n] = h_{1} [n] * h_{2} [n]$
Parallel Connection: The overall impulse response is the direct algebraic sum of subsystem responses: $h_{e x t o v er a ll} (t) = h_{1} (t) + h_{2} (t) e x t an d h_{e x t o v er a ll} [n] = h_{1} [n] + h_{2} [n]$

6.3 Inverse Systems & Deconvolution

*(Target: Theory Descriptive)*

Concept: A system $h (t)$ is invertible if its input $x (t)$ can be completely recovered from its output response $y (t)$ using an inverse system $g (t)$ .
Governing Formula: $h (t) * g (t) = δ (t) e x t an d h [n] * g [n] = δ [n]$

7. Master Comparison Tables

Table 2.1: System Properties Test Matrix

*(Pre-Exam Quick Lookup)*

Property	Continuous Condition	Discrete Condition	Practical Checklist
Linearity	$T [a x_{1} + b x_{2}] = a y_{1} + b y_{2}$	$T [a x_{1} + b x_{2}] = a y_{1} + b y_{2}$	Zero input must produce zero output. No nonlinear terms (squares, logs, sines of signals).
Time-Invariance	$T [x (t - t_{0})] = y (t - t_{0})$	$T [x [n - n_{0}]] = y [n - n_{0}]$	Check independent variable coefficient. If scaled (e.g., $x (2 t)$ , $x [- n]$ ), it is Time-Variant.
Causality	$h (t) = 0 e x t f or t < 0$	$h [n] = 0 e x t f or n < 0$	Output at $t_{0}$ must not depend on inputs at $t > t_{0}$ (e.g., $x (t + 3)$ , $x [n + 1]$ are noncausal).
Memory	$h(t)
eq c\delta(t)$	$h[n]
eq c\delta[n]$	If output depends only on the present input, it is memoryless; else, it is dynamic.
BIBO Stability	$\int_{-\infty}^{\infty} \lvert h(t)
vert dt < \infty$	$\sum_{-\infty}^{\infty} \lvert h[n]
vert < \infty$	Every bounded input must produce a bounded output (e.g., integrations, step responses are unstable).

Table 2.2: Continuous vs. Discrete Convolution Properties

*(Symmetry and Sizing Mapping)*

Parameter	Continuous-Time Convolution	Discrete-Time Convolution
Governing Operator	Integral: $\int_{- \infty}^{\infty} x (a u) h (t - a u) d a u$	Infinite Summation: $\sum_{k = - \infty}^{\infty} x [k] h [n - k]$
Aperiodic Sequence Bounds	Evaluated via piecewise continuous limits.	Sequence Length Theorem: $L = N_{1} + N_{2} - 1$ .
Unit Impulse Response	$x (t) * δ (t) = x (t)$	$x [n] * δ [n] = x [n]$
Integrability/Summability	Area: $A_{y} = A_{x} \cdot A_{h}$	Sum: $\sum y[n] = \left( \sum x[n]
ight) \cdot \left( \sum h[n]
ight)$

Table 2.3: Recursive (IIR) vs. Non-Recursive (FIR) Difference Equations

*(Structural Properties)*

Attribute	Non-Recursive (FIR) Systems	Recursive (IIR) Systems
Feedback Paths	None. No output feedback loop is present.	Present. Output terms $y [n - k]$ are fed back.
Impulse Response Duration	Finite Duration (Finite Impulse Response).	Infinite Duration (Infinite Impulse Response).
Stability	Always stable (since $h [n]$ is finite and bounded).	Conditionally stable (depends on pole locations inside the unit circle).
Memory Elements	Fixed memory storage proportional to order $N$ .	Requires infinite decay tracking of feedback states.

8. Verbatim Chapter 2 PYQ Bank

2025/2022 Exam Section B Q. 1

If $y (t) = x (t) * h (t)$ , then show that: $y (2 t) = 2 x (2 t) * h (2 t) e x t (03 M a r k s)$ Answer Plan: Use variable substitutions inside the convolution integral of $y (a t)$ with $a = 2$ , scaling both dummy variable $a u$ and actual time $t$ .

2023/2017/2015 Exam Section B

Show that the output response of an LTI system is the convolution sum of the input signal and the impulse response of the system. $e x t (03 t o 04 M a r k s)$ Answer Plan: Represent the arbitrary input signal $x [n]$ as a weighted summation of discrete impulses: $x [n] = \sum x [k] δ [n - k]$ . Apply the system operator $T [\cdot]$ , utilizing Linearity (superposition) to move the operator inside the sum, and Time-Invariance ( $T [δ [n - k]] = h [n - k]$ ) to yield the convolution sum.

2024 Exam Section B Q. 2

Draw the block diagram of the system described by: $rac{d^2 y(t)}{dt^2} + 3y(t) = rac{dx(t)}{dt} + 2 rac{d^2 x(t)}{dt^2} \quad ext{(05 Marks)}$ Answer Plan: Integrate the second-order equation twice to avoid differentiators, isolate $y (t)$ , and construct the Direct Form II canonic block diagram utilizing two integrators, multipliers, and adders.

2021 Exam Section B Q. 3

Derive the expression of convolution with delayed input and delayed impulse response. $e x t (09 M a r k s)$ Answer Plan: Starting from $y (t) = x (t) * h (t) = \int x (a u) h (t - a u) d a u$ , substitute delayed variables $x (t - t_{1})$ and $h (t - t_{2})$ into the integral, use variable substitution $v = a u - t_{1}$ , and match the resulting expression directly to $y (t - t_{1} - t_{2})$ .

9. Common Mistakes That Cost Marks

Critical Exam Pitfalls

Direct Form II Feedback Coefficient Sign Swap: When sketching Direct Form II diagrams, failing to negate the feedback coefficients. Remember: feedback terms in difference/differential equations must be subtracted on summing junctions to represent negative feedback multipliers.

Evaluating Causal Overlap Boundaries incorrectly: Performing graphical convolution of piecewise signals without checking overlap boundaries. Always verify the active integration interval for each segment of time $t$ before computing areas.

Discrete Origin ( $n = 0$ ) Arrow Omission: When evaluating discrete tabular convolution, failing to locate and mark the $n = 0$ index in convolved sequences.

Applying Value Theorems to Unstable LTI Systems: Evaluating the Final Value Theorem on unstable systems with poles outside or on the imaginary axis.

10. Pre-Exam Self-Check Checklist

Can you verify linearity and time-invariance for any continuous-time system equations?
Can you prove that a continuous LTI system is BIBO stable if and only if $\int_{- \infty}^{\infty} ∣ h (t) ∣ d t < \infty$ ?
Can you calculate the length $L = N_{1} + N_{2} - 1$ of convolved finite discrete arrays?
Can you draw canonic Direct Form II representations of second-order continuous differential and discrete difference equations?

2.01 Systems Classification, LTI Properties & Stability

1. Conceptual Framework of Systems

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

$Continuous-Time: y (t) = T [x (t)] Discrete-Time: y [n] = T [x [n]]$

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).

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: $h (t) = cδ (t) ⟹ y (t) = c x (t)$
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: $h (t) = 0 for all t < 0$
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 $T^{- 1} [\cdot]$ can be cascaded with the original system to perfectly recover the input signal $x (t)$ : $T^{- 1} [y (t)] = T^{- 1} [T [x (t)]] = x (t)$
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.

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:

Define two distinct inputs $x_{1} (t)$ and $x_{2} (t)$ which produce outputs $y_{1} (t) = T [x_{1} (t)]$ and $y_{2} (t) = T [x_{2} (t)]$ .
Construct a weighted linear combination of these inputs: $x_{3} (t) = a x_{1} (t) + b x_{2} (t)$
Evaluate the system’s output due to this combined input: $y_{3} (t) = T [x_{3} (t)] = T [a x_{1} (t) + b x_{2} (t)]$
Compare $y_{3} (t)$ with the weighted combination of the individual outputs: $y_{target} (t) = a y_{1} (t) + b y_{2} (t)$
- If $y_{3} (t) = y_{target} (t)$ , the system is Linear.
- If $y_{3} (t) \neq = y_{target} (t)$ , 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 $x (t) = 0$ ).

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)

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 $x (t - t_{0})$ : $y_{1} (t) = T [x (t - t_{0})]$
Pathway 2 (Shifted system output): Take the original output equation $y (t) = T [x (t)]$ and delay the entire signal by replacing every instance of the independent variable $t$ with $(t - t_{0})$ : $y_{2} (t) = y (t - t_{0})$
Compare:
- If $y_{1} (t) = y_{2} (t)$ , the system is Time-Invariant.
- If $y_{1} (t) \neq = y_{2} (t)$ , 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., $y (t) = t x (t)$ ), Pathway 2 will shift that coefficient while Pathway 1 will not. This makes the system Time-Variant.

5. Standard Core Solved Numericals

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

Determine if this system is Linear and Time-Invariant.

1. Linearity Proof:

Let $y_{1} (t) = 3 x_{1} (- t) + 2 x_{1} (t)$ and $y_{2} (t) = 3 x_{2} (- t) + 2 x_{2} (t)$ .
Apply combined input $x_{3} (t) = a x_{1} (t) + b x_{2} (t)$ : $y_{3} (t) = 3 [a x_{1} (- t) + b x_{2} (- t)] + 2 [a x_{1} (t) + b x_{2} (t)]$ $y_{3} (t) = a [3 x_{1} (- t) + 2 x_{1} (t)] + b [3 x_{2} (- t) + 2 x_{2} (t)]$ $y_{3} (t) = a y_{1} (t) + b y_{2} (t)$
Result: The system is Linear.

2. Time-Invariance Check:

Pathway 1: Shift the input signal $x (t) \to x (t - t_{0})$ . $y_{1} (t) = 3 x (- t - t_{0}) + 2 x (t - t_{0})$
Pathway 2: Shift the output signal $t \to t - t_{0}$ . $y_{2} (t) = y (t - t_{0}) = 3 x (- (t - t_{0})) + 2 x (t - t_{0}) = 3 x (- t + t_{0}) + 2 x (t - t_{0})$
Compare: $y_{1} (t) \neq = y_{2} (t)$ because $- t - t_{0} \neq = - t + t_{0}$ .
Result: The system is Time-Variant.

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

Determine if this system is Linear.

Let input $x_{1} (t)$ yield output $y_{1} (t)$ satisfying: $\overset{y}{˙}_{1} (t) + 3 y_{1} (t) = x_{1} (t) \overset{x}{˙}_{1} (t)$ .
Let input $x_{2} (t)$ yield output $y_{2} (t)$ satisfying: $\overset{y}{˙}_{2} (t) + 3 y_{2} (t) = x_{2} (t) \overset{x}{˙}_{2} (t)$ .
Let input be $x_{3} (t) = a x_{1} (t) + b x_{2} (t)$ . The system output $y_{3} (t)$ must satisfy: $\overset{y}{˙}_{3} (t) + 3 y_{3} (t) = [a x_{1} (t) + b x_{2} (t)] \frac{d}{d t} [a x_{1} (t) + b x_{2} (t)]$ $\overset{y}{˙}_{3} (t) + 3 y_{3} (t) = a^{2} x_{1} (t) \overset{x}{˙}_{1} (t) + b^{2} x_{2} (t) \overset{x}{˙}_{2} (t) + ab [x_{1} (t) \overset{x}{˙}_{2} (t) + x_{2} (t) \overset{x}{˙}_{1} (t)]$
Evaluate the target response $y_{target} (t) = a y_{1} (t) + b y_{2} (t)$ : $\frac{d}{d t} [a y_{1} (t) + b y_{2} (t)] + 3 [a y_{1} (t) + b y_{2} (t)] = a [\overset{y}{˙}_{1} (t) + 3 y_{1} (t)] + b [\overset{y}{˙}_{2} (t) + 3 y_{2} (t)]$ $a [\overset{y}{˙}_{1} (t) + 3 y_{1} (t)] + b [\overset{y}{˙}_{2} (t) + 3 y_{2} (t)] = a x_{1} (t) \overset{x}{˙}_{1} (t) + b x_{2} (t) \overset{x}{˙}_{2} (t)$
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, $y_{3} (t) \neq = a y_{1} (t) + b y_{2} (t)$ .
Result: The system is Non-linear.

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 $[y_{1} [n]]^{2} = n [x_{1} [n]]^{2} ⟹ y_{1} [n] = n x_{1} [n]$ .
Let $[y_{2} [n]]^{2} = n [x_{2} [n]]^{2} ⟹ y_{2} [n] = n x_{2} [n]$ .
Apply combined input $x_{3} [n] = a x_{1} [n] + b x_{2} [n]$ : $y_{3} [n] = n (a x_{1} [n] + b x_{2} [n]) = a (n x_{1} [n]) + b (n x_{2} [n]) = a y_{1} [n] + b y_{2} [n]$
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: $y_{1} [n] \neq = y (n - n_{0})$ due to the un-shifted time multiplier coefficient $n$ on Pathway 1.
Result: The system is Time-Variant.

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.

$Condition: If ∣ x (t) ∣ \leq M_{x} < \infty ⟹ ∣ y (t) ∣ \leq M_{y} < \infty$

🚨 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 $∣ x (t) ∣ \leq M_{x} < \infty$ for all $t$ . The output of a continuous LTI system is governed by the continuous convolution integral: $y (t) = \int_{- \infty}^{\infty} h (τ) x (t - τ) d τ$

Step 1: Upper Bound Expansion (Sufficiency Proof)

Take the absolute value of the output $y (t)$ : $∣ y (t) ∣ = \int_{- \infty}^{\infty} h (τ) x (t - τ) d τ$

Apply the integral triangle inequality property {the absolute value of an integral is less than or equal to the integral of the absolute value}: $∣ y (t) ∣ \leq \int_{- \infty}^{\infty} ∣ h (τ) x (t - τ) ∣ d τ$

$⟹ ∣ y (t) ∣ \leq \int_{- \infty}^{\infty} ∣ h (τ) ∣ \cdot ∣ x (t - τ) ∣ d τ$

Step 2: Substitute the Input Bound

Since $∣ x (t) ∣ \leq M_{x}$ for all $t$ , we can substitute this bound into the inequality: $∣ y (t) ∣ \leq M_{x} \int_{- \infty}^{\infty} ∣ h (τ) ∣ d τ$

For $∣ y (t) ∣ \leq M_{y} < \infty$ to hold (ensuring the output is bounded), the integral must evaluate to a finite value: $\int_{- \infty}^{\infty} ∣ h (τ) ∣ d τ < \infty (Sufficient Condition Proved) [299]$

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: $x (t - τ) = sgn (h (τ)) = {+ 1, - 1, h (τ) \geq 0 h (τ) < 0$ Note that this input is strictly bounded: $∣ x (t) ∣ \leq 1 < \infty$ . Evaluate the system output $y (t)$ at the origin ( $t = 0$ ): $y (0) = \int_{- \infty}^{\infty} h (τ) x (- τ) d τ = \int_{- \infty}^{\infty} h (τ) sgn (h (τ)) d τ = \int_{- \infty}^{\infty} ∣ h (τ) ∣ d τ$ If $\int_{- \infty}^{\infty} ∣ h (τ) ∣ d τ = \infty$ , then $y (0) = \infty$ , meaning a bounded input has produced an infinite output. Thus, absolute integrability is a necessary condition.

$\int_{- \infty}^{\infty} ∣ h (t) ∣ d t < \infty (Proved)$

7. Core Solved Stability Traps

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

Let the input sequence be bounded: $∣ x (n) ∣ \leq M_{x} < \infty$ for all $n$ .
Evaluate the absolute boundary of the output sequence: $∣ y (n) ∣ = ∣ a x (n + 1) + b x (n - 1) ∣$
Apply the triangle inequality: $∣ y (n) ∣ \leq ∣ a ∣ \cdot ∣ x (n + 1) ∣ + ∣ b ∣ \cdot ∣ x (n - 1) ∣$
Substitute the bound $M_{x}$ : $∣ y (n) ∣ \leq ∣ a ∣ M_{x} + ∣ b ∣ M_{x} = (∣ a ∣ + ∣ b ∣) M_{x}$
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: $I = \int_{- \infty}^{\infty} ∣ h (t) ∣ d t = \int_{1}^{\infty} e^{- t} d t$ $I = [- e^{- t}]_{1}^{\infty} = 0 - (- e^{- 1}) = e^{- 1} \approx 0.368$
Since the integral yields a finite value ( $0.368 < \infty$ ), the system is stable.
Result: The system is BIBO Stable.

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 $y (t) = x (2 t)$ , students often assume it is causal because $t > 0 ⟹ 2 t > t$ . 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.

PYQ Bank — Verbatim Questions & Answer Plans

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

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 $∣ x (t) ∣ \leq M_{x} < \infty$ and express output $y (t)$ using the continuous convolution integral.
2. Apply the integral triangle inequality to prove the sufficiency condition ( $∣ y (t) ∣ \leq M_{x} \int ∣ h (τ) ∣ d τ < \infty$ ).
3. Set up the necessity proof by defining the bounded test input $x (t - τ) = sgn (h (τ))$ to show the system output blows up if the absolute integrability integral is infinite.

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

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 $- 2 x [n + 1]$ depends on a future input sample at index $n + 1 ⟹$ Non-Causal.

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).

2.02 The Continuous Convolution Integral

Core Definition & Objective

For any continuous-time Linear Time-Invariant {LTI} system, the relationship between the input signal $x (t)$ and the output response $y (t)$ is completely governed by the system’s impulse response $h (t)$ . This relationship is mathematically expressed by the Convolution Integral: $y (t) = x (t) * h (t) = \int_{- \infty}^{\infty} x (τ) h (t - τ) d τ$ This note focuses on the physical derivation of this integral, its mathematical properties, the step-by-step graphical sliding workflow, and analytical RC transient solutions.

1. Mathematical Derivation of the Convolution Integral

To understand why convolution works, we must look at how we can represent any arbitrary continuous-time signal $x (t)$ as an infinite, dense sum of weighted and shifted unit impulses $δ (t)$ .

Step 0: Setup & Impulse Representation of $x (t)$

Let us approximate an arbitrary continuous-time signal $x (t)$ by dividing it into narrow rectangular slices, each of width $Δ$ .

[GRAPH: A continuous curved signal x(t) approximated by a staircase-like grid of narrow rectangular pulses of width \Delta - source: Rabiul Sir Notes, Page 5]

Mathematically, a single narrow pulse at time $t = k Δ$ with width $Δ$ and height $x (k Δ)$ can be represented as: $\overset{x}{^} (t) = \sum_{k = - \infty}^{\infty} x (k Δ) δ_{Δ} (t - k Δ) Δ$ where $δ_{Δ} (t)$ is a rectangular pulse of height $1/Δ$ and width $Δ$ centered at the origin.

As we let the slice width approach zero ( $Δ \to 0$ ):

The discrete summation index $k Δ$ transitions into a continuous variable of integration $τ$ .
The width step $Δ$ becomes an infinitesimal differential $d τ$ .
The approximation pulse $δ_{Δ} (t)$ converges to the perfect Dirac delta function $δ (t)$ .
The staircase approximation $\overset{x}{^} (t)$ becomes the exact original signal $x (t)$ .

This yields the sifting property representation {representing a signal as a continuous sweep of impulses}: $x (t) = lim_{Δ \to 0} \sum_{k = - \infty}^{\infty} x (k Δ) δ_{Δ} (t - k Δ) Δ = \int_{- \infty}^{\infty} x (τ) δ (t - τ) d τ$

Step 1: Applying the LTI System Operator

Let $T [\cdot]$ represent the transformation operator of a system. Feeding the representation of $x (t)$ into the system yields: $y (t) = T [x (t)] = T [\int_{- \infty}^{\infty} x (τ) δ (t - τ) d τ]$

Because $T [\cdot]$ operates strictly on the independent time variable $t$ (and not on the integration dummy variable $τ$ ), we can exploit the system’s Linearity properties (superposition and homogeneity) to slide the operator inside the integral: $y (t) = \int_{- \infty}^{\infty} x (τ) T [δ (t - τ)] d τ$

Step 2: Incorporating Time-Invariance

By definition, the response of a system to a perfect unit impulse at the origin $δ (t)$ is its impulse response $h (t)$ : $h (t) = T [δ (t)]$

If the system is Time-Invariant {system characteristics do not change over time}, a shift in the input impulse by $τ$ units must yield an identical shift in the output impulse response: $T [δ (t - τ)] = h (t - τ)$

Substituting this back into our linearity equation yields the Convolution Integral: $y (t) = \int_{- \infty}^{\infty} x (τ) h (t - τ) d τ (Q.E.D.)$

2. Rigorous Proofs of Core Properties

2.1 Time Scaling Property

Theorem (Time Scaling in Convolution)

If $y (t) = x (t) * h (t)$ , then: $y (a t) = ∣ a ∣ \cdot [x (a t) * h (a t)]$ For $a = 2$ , this proves the classic textbook relation: $y (2 t) = 2 \cdot [x (2 t) * h (2 t)]$ .

Step-by-Step Derivations:

Start with the standard definition of the scaled output $y (a t)$ : $y (a t) = \int_{- \infty}^{\infty} x (a t - τ) h (τ) d τ$

We want to transform the integration variable so that the arguments of both $x$ and $h$ are scaled by $a$ . Let us substitute $τ = aλ ⟹ d τ = a d λ$ .

Case 1: $a > 0$ (Integration limits remain $- \infty$ to $\infty$ ): $y (a t) = \int_{- \infty}^{\infty} x (a t - aλ) h (aλ) \cdot a d λ = a \int_{- \infty}^{\infty} x (a (t - λ)) h (aλ) d λ$
Case 2: $a < 0$ (Limits flip, and restoring them to normal order introduces a negative sign): $y (a t) = \int_{\infty}^{- \infty} x (a t - aλ) h (aλ) \cdot a d λ = - a \int_{- \infty}^{\infty} x (a (t - λ)) h (aλ) d λ$

Combining both cases using the absolute value operator yields: $y (a t) = ∣ a ∣ \int_{- \infty}^{\infty} x (a (t - λ)) h (aλ) d λ = ∣ a ∣ \cdot [x (a t) * h (a t)] (Proved)$

2.2 Even Symmetry of Convolved Odd Signals

Theorem (Convolution of Two Odd Signals)

If $x (t)$ and $h (t)$ are both purely odd signals, their convolved output $y (t) = x (t) * h (t)$ is a purely even signal.

Step-by-Step Derivations:

By definition, odd signals satisfy $x (- t) = - x (t)$ and $h (- t) = - h (t)$ . Let us evaluate $y (- t)$ : $y (- t) = \int_{- \infty}^{\infty} x (τ) h (- t - τ) d τ$

Substitute $λ = - τ ⟹ d τ = - d λ$ . The limits of integration flip from $[- \infty, \infty]$ to $[\infty, - \infty]$ : $y (- t) = \int_{\infty}^{- \infty} x (- λ) h (- t + λ) (- d λ) = \int_{- \infty}^{\infty} x (- λ) h (- (t - λ)) d λ$

Now, substitute the odd properties $x (- λ) = - x (λ)$ and $h (- (t - λ)) = - h (t - λ)$ into the integral: $y (- t) = \int_{- \infty}^{\infty} [- x (λ)] [- h (t - λ)] d λ = \int_{- \infty}^{\infty} x (λ) h (t - λ) d λ$

Since this is the exact definition of $y (t)$ : $y (- t) = y (t) ⟹ y (t) is Even. (Proved)$

3. Graphical Convolution: The Sliding Window Workflow

To convolve two piecewise signals graphically under exam time pressure, follow this rigid, error-proof sequence:

graph LR
    A[1. Re-map variables to tau] --> B[2. Fold h to h(-tau)]
    B --> C[3. Shift to h(t-tau)] --> D[4. Find boundary overlap intervals]
    D --> E[5. Integrate product for each interval]
    E --> F[6. Plot final output y(t)]

3.1 Graphical Piecewise Example

Question: Convolve the rectangular pulse $x (t) = u (t - 1) - u (t - 3)$ with the symmetrical double-square impulse response $h (t) = [u (t + 1) - u (t - 1)] - [u (t - 1) - u (t - 3)]$ .

[GRAPH: Waveform of input x(t) starting at t=1, ending at t=3, height 1. Waveform of h(t) showing positive pulse from t=-1 to t=1 (height 1) and negative pulse from t=1 to t=3 (height -1) - source: Rabiul Sir Notes, Page 14]

Step 1: Variable Mapping & Folding

Map both functions to the dummy integration variable $τ$ :

$x (τ)$ is active for $1 \leq τ \leq 3$ .
$h (- τ)$ is folded around the y-axis, spanning from $τ = - 3$ to $τ = 1$ .

[GRAPH: Folded impulse response h(-\tau) showing positive pulse from \tau=-1 to \tau=1, and negative pulse from \tau=-3 to \tau=-1 - source: Rabiul Sir Notes, Page 14]

Step 2: Shifting & Interval Sweeping

Slide $h (t - τ)$ from the far left (representing $t \to - \infty$ ) towards the right:

Interval 1 ( $t < 0$ ): No Overlap The leading edge of $h (t - τ)$ (which is at $τ = t + 1$ ) has not reached the starting edge of $x (τ)$ (at $τ = 1$ ). $t + 1 < 1 ⟹ t < 0 ⟹ y (t) = 0$
Interval 2 ( $0 \leq t < 2$ ): Partial Positive Overlap The positive lobe of $h (t - τ)$ slides into $x (τ)$ . The overlap region spans from $τ = 1$ to $τ = t + 1$ : $y (t) = \int_{1}^{t + 1} (1) (1) d τ = [τ]_{1}^{t + 1} = t$
Interval 3 ( $2 \leq t < 4$ ): Full Positive & Partial Negative Overlap The positive lobe has fully entered, and the negative lobe (spanning from $t - 1$ to $t + 1$ ) begins overlapping $x (τ)$ . $y (t) = \int_{t - 1}^{3} (1) d τ + \int_{3}^{t + 1} (- 1) d τ = (4 - t) - (t - 2) = 4 - 2 t$
Interval 4 ( $4 \leq t < 6$ ): Pure Negative Overlap The positive lobe has exited the boundaries of $x (τ)$ completely. Only the negative lobe overlaps, spanning from $t - 1$ to $3$ : $y (t) = \int_{t - 1}^{3} (- 1) d τ = - [3 - (t - 1)] = t - 4$
Interval 5 ( $t \geq 6$ ): No Overlap The trailing edge of $h (t - τ)$ (at $τ = t - 3$ ) has passed the trailing edge of $x (τ)$ (at $τ = 3$ ). $t - 3 \geq 3 ⟹ t \geq 6 ⟹ y (t) = 0$

[GRAPH: The final convolved output y(t) showing a triangular positive peak of height 2 at t=2, dipping down to a negative peak of height -2 at t=4, and returning to zero at t=6 - source: Rabiul Sir Notes, Page 14]

4. High-Yield Analytical Solutions (RC Circuits)

4.1 Classical Low-Pass RC Transient Response

Standard Exam Problem (RC Step Transient)

An analog low-pass RC filter circuit has an impulse response $h (t) = \frac{1}{RC} e^{- t / RC} u (t)$ . Find the output response $y (t)$ when the input signal is $x (t) = e^{- t / RC} u (t)$ .

[CIRCUIT: A classic analog low-pass RC network showing input voltage V_in(t), series resistor R, parallel capacitor C, and output voltage V_out(t) across the capacitor - source: Senior Notes]

Analytical Convolution Integration:

By definition: $y (t) = \int_{- \infty}^{\infty} x (τ) h (t - τ) d τ$

Substitute the active step functions $u (τ)$ and $u (t - τ)$ into the limits of integration. The product $u (τ) u (t - τ)$ is non-zero if and only if $0 \leq τ \leq t$ : $y (t) = \int_{0}^{t} (e^{- τ / RC}) (\frac{1}{RC} e^{- (t - τ) / RC}) d τ, t \geq 0$

Factor out the terms that do not depend on the dummy integration variable $τ$ : $y (t) = \frac{1}{RC} e^{- t / RC} \int_{0}^{t} e^{- τ / RC} e^{τ / RC} d τ$ $y (t) = \frac{1}{RC} e^{- t / RC} \int_{0}^{t} (1) d τ$ $y (t) = \frac{t}{RC} e^{- t / RC} u (t) (Q.E.D.)$

[GRAPH: Transient response y(t) starting at zero, rising to a smooth peak of height 0.368 at t = RC, and slowly decaying towards zero for t > RC - source: Rabiul Sir Notes]

5. Common Mistakes That Cost Marks

The Critical Integration Limits Trap

The Mistake: Students often write down the default integration limits $- \infty$ to $\infty$ on their exam scripts and integrate blindly, ignoring the step functions $u (τ)$ and $u (t - τ)$ .

The Correction: Always rewrite the step functions as limits of integration before doing calculus: $u (τ) ⟹ Lower limit becomes 0$ $u (t - τ) ⟹ Upper limit becomes t$

The Volts-unit Penalty: Failing to normalize exponential power coefficients (e.g. omitting the $1/ RC$ amplitude factor in RC networks) leads to a deduction of 2–3 marks.

6. Verbatim PYQ Bank

[PYQ 2025, 2022]

If $y (t) = x (t) * h (t)$ , then show that $y (2 t) = 2 x (2 t) * h (2 t)$ . (12 Marks)

Answer Plan: Refer to Section 2.1 of this note. Use the coordinate transformation substitution $τ = 2 λ$ to scale the dummy variable and factor out the amplitude multiplier of 2.

[PYQ 2021]

Determine the output response of a low-pass RC network for an input signal $x (t) = e^{- t / RC} u (t)$ . (10 Marks)

Answer Plan: Refer to Section 4.1. Set up the KVL impulse response $h (t) = \frac{1}{RC} e^{- t / RC} u (t)$ , convolve it with $x (t)$ , and show how the exponential terms cancel out inside the integral to leave a linear ramp term $t$ .

[PYQ 2016]

Determine the output response $y (t)$ of the RC low-pass network due to an input $x (t) = t e^{- t / RC} u (t)$ by convolution. (12 Marks)

Answer Plan: Set up the convolution integral with limits from $0$ to $t$ : $y (t) = \int_{0}^{t} τ e^{- τ / RC} \cdot \frac{1}{RC} e^{- (t - τ) / RC} d τ = \frac{e ^{- t / RC}}{RC} \int_{0}^{t} τ d τ = \frac{t ^{2}}{2 RC} e^{- t / RC} u (t)$

7. Self-Check Before Moving On

Can you derive the continuous-time convolution integral from first principles (impulse representation)?
Can you prove why convolving two odd functions yields an even function?
Do you know how to graphically convolve piecewise rectangular and triangular waveforms without losing track of active interval limits?

Sources:

K. Deergha Rao, “Signals and Systems”, Section 2.3 (The Convolution Integral).
Rabiul Sir Continuous-Time Lecture Handnotes (Pages 14–15).
University Past Year Papers (2016, 2021, 2022, 2025).

2.03 Properties of the Convolution Integral

Core Theorem Overview

Convolution {the mathematical folding, shifting, and integrating of two signals} is not just a blind computational tool; it behaves like a standard algebraic product. By utilizing the algebraic properties of convolution—namely the Commutative, Distributive, Associative, and Impulse-Shifting properties—we can bypass complex integrations entirely and solve cascaded or parallel LTI networks instantly.

1. The Commutative Property

Alright — let’s start with a highly comforting fact: the order in which you convolve two signals does not matter. The system’s output remains identical whether you fold-and-shift the input signal $x (t)$ or the system’s impulse response $h (t)$ .

Mathematically, this is stated as: $x (t) * h (t) = h (t) * x (t)$

This is incredibly high-signal for exams! If you are convolving a simple rectangular pulse with a complex decaying exponential, the commutative property lets you choose the simpler function to fold and shift, saving you valuable time and reducing algebraic errors.

Step-by-Step Derivation

Let’s prove this formally using a classic change of variables.

Step 0: Setup and Definition

Start with the standard definition of the continuous convolution integral: $x (t) * h (t) = \int_{- \infty}^{\infty} x (τ) h (t - τ) d τ$

Step 1: Change of Variables

We want to swap the arguments of our functions. Let us define a new dummy integration variable, $λ$ : $λ = t - τ ⟹ τ = t - λ$

Step 2: Evaluate Limits and Differentials

Now, find the differential $d τ$ by differentiating both sides with respect to $λ$ : $d τ = - d λ$

Next, transform the limits of integration from $τ$ to $λ$ :

When $τ \to - \infty$ , then $λ = t - (- \infty) \to \infty$ .
When $τ \to \infty$ , then $λ = t - (\infty) \to - \infty$ .

Step 3: Substitute and Simplify

Substitute these variables back into our original integral setup: $\int_{- \infty}^{\infty} x (τ) h (t - τ) d τ = \int_{\infty}^{- \infty} x (t - λ) h (λ) (- d λ)$

We can use the negative sign of the differential to flip the integration limits back to their standard order: $\int_{\infty}^{- \infty} x (t - λ) h (λ) (- d λ) = \int_{- \infty}^{\infty} h (λ) x (t - λ) d λ$

Since $λ$ is just a dummy variable {a temporary variable name used strictly for the duration of the integration}, we can rename it back to $τ$ : $\int_{- \infty}^{\infty} h (τ) x (t - τ) d τ = h (t) * x (t) (Proved)$

2. The Distributive and Associative Properties

These two properties allow us to analyze interconnected LTI networks by simplifying parallel and cascaded systems into a single equivalent impulse response block.

2.1 The Distributive Property (Parallel Systems)

Convolution is distributive over addition: $x (t) * [h_{1} (t) + h_{2} (t)] = x (t) * h_{1} (t) + x (t) * h_{2} (t)$

Physical Significance

This property describes Parallel System Interconnections. If an input signal $x (t)$ is applied simultaneously to two parallel LTI subsystems with impulse responses $h_{1} (t)$ and $h_{2} (t)$ , the overall system’s impulse response is simply the sum of the individual impulse responses: $h_{total} (t) = h_{1} (t) + h_{2} (t)$

          +--------+
     +--->| h1(t)  |---+
     |    +--------+   |
x(t)-+                 v  (+)---> y(t)
     |    +--------+   ^
     +--->| h2(t)  |---+
          +--------+

2.2 The Associative Property (Cascaded Systems)

Convolution is associative: $[x (t) * h_{1} (t)] * h_{2} (t) = x (t) * [h_{1} (t) * h_{2} (t)]$

Physical Significance

This property describes Cascaded System Interconnections. If a signal passes through subsystem $h_{1} (t)$ and its output immediately feeds into subsystem $h_{2} (t)$ , the overall cascaded system behaves as a single block with an equivalent impulse response: $h_{total} (t) = h_{1} (t) * h_{2} (t)$

x(t) ---> [ h1(t) ] ---> y1(t) ---> [ h2(t) ] ---> y(t)
                   Equivalently:
x(t) --------------> [ h1(t) * h2(t) ] -------------> y(t)

By the commutative property, we can swap the order of blocks in a cascade without changing the output: $h_{1} (t) * h_{2} (t) = h_{2} (t) * h_{1} (t)$

3. Convolution with an Impulse ( $δ (t)$ )

Convolution with a unit impulse behaves exactly like multiplication by 1 in standard algebra. The impulse acts as an identity operator under convolution.

Mathematically, this is expressed as: $x (t) * δ (t) = x (t)$

Step-by-Step Derivation

Let’s prove this using the sifting property {the mathematical ability of the Dirac delta function to select a single, discrete value of a signal when integrated}.

Step 0: Setup and Definition

Apply the standard convolution integral definition: $x (t) * δ (t) = \int_{- \infty}^{\infty} x (τ) δ (t - τ) d τ$

Step 1: Utilize the Sifting Property

The impulse function $δ (t - τ)$ is active {non-zero} only at the single point where its argument becomes zero: $t - τ = 0 ⟹ τ = t$

Step 2: Evaluate the Integral

Because the impulse is zero everywhere except at $τ = t$ , the integral evaluates directly to the value of the function $x (τ)$ at that specific point: $\int_{- \infty}^{\infty} x (τ) δ (t - τ) d τ = x (τ)_{τ = t} = x (t) (Proved)$

4. The Time-Delay Shifting Property

This property is a massive, high-yield shortcut for your continuous-time signal examinations! It states that convolving two shifted signals simply shifts the final convolved output by the sum of their individual delays.

The Time-Delay Shifting Theorem

If we know that: $y (t) = x (t) * h (t)$ Then convolving their delayed versions yields: $x (t - t_{1}) * h (t - t_{2}) = y (t - t_{1} - t_{2})$

Step-by-Step Derivation

Let’s prove this mathematically.

Step 1: Setup the Convolution of Shifted Inputs

Using the convolution integral definition, let’s write out the left-hand side of our equation: $x (t - t_{1}) * h (t - t_{2}) = \int_{- \infty}^{\infty} x (τ - t_{1}) h (t - τ - t_{2}) d τ$

Step 2: Define a Variable Substitution

To align this with our standard convolution form, let’s substitute the argument of our first function: $υ = τ - t_{1} ⟹ τ = υ + t_{1} and d τ = d υ$

The limits of integration remain unchanged at $- \infty$ and $\infty$ .

Step 3: Substitute and Simplify the Integral

Substitute these variables back into our shifted integral expression: $\int_{- \infty}^{\infty} x (υ) h (t - (υ + t_{1}) - t_{2}) d υ = \int_{- \infty}^{\infty} x (υ) h ((t - t_{1} - t_{2}) - υ) d υ$

Step 4: Map to the Base Convolution

Notice that the resulting integral is identical to the definition of $y (t) = \int x (υ) h (t - υ) d υ$ , except that the time variable $t$ has been replaced by $(t - t_{1} - t_{2})$ .

Therefore: $x (t - t_{1}) * h (t - t_{2}) = y (t - t_{1} - t_{2}) (Proved)$

5. High-Yield Worked Numerical

Let’s see how we can use these algebraic properties to solve complex, piecewise exam problems in seconds without doing any calculus.

Worked Exam Problem

Find the output $y (t)$ of an LTI system with impulse response $h (t) = u (t - 1) - u (t - 3)$ when the input signal is $x (t) = 4 δ (t - 2) + 2 δ (t - 5)$ .

Step-by-Step Solution

Step 1: Write down the Algebraic Expression

Using the distributive and commutative properties, we can write the overall convolution as: $y (t) = x (t) * h (t) = [4 δ (t - 2) + 2 δ (t - 5)] * [u (t - 1) - u (t - 3)]$

Distribute the terms: $y (t) = 4 δ (t - 2) * [u (t - 1) - u (t - 3)] + 2 δ (t - 5) * [u (t - 1) - u (t - 3)]$

Step 2: Apply the Shifting Property of the Impulse

We know that convolving any signal $f (t)$ with a shifted impulse $δ (t - t_{0})$ simply shifts the signal by $t_{0}$ : $f (t) * δ (t - t_{0}) = f (t - t_{0})$

Let’s apply this to our two distributed terms:

First term: $4 δ (t - 2) * u (t - 1) = 4 u (t - 1 - 2) = 4 u (t - 3)$ $4 δ (t - 2) * (- u (t - 3)) = - 4 u (t - 3 - 2) = - 4 u (t - 5)$
Second term: $2 δ (t - 5) * u (t - 1) = 2 u (t - 1 - 5) = 2 u (t - 6)$ $2 δ (t - 5) * (- u (t - 3)) = - 2 u (t - 3 - 5) = - 2 u (t - 8)$

Step 3: Combine and Write the Final Expression

Combine the terms to get our final output: $y (t) = 4 u (t - 3) - 4 u (t - 5) + 2 u (t - 6) - 2 u (t - 8)$

By using convolution properties, we solved a complex, multi-interval piecewise problem with zero calculus integrations!

[GRAPH: x-y axis plotting y(t). It is 0 for t < 3. It jumps to amplitude 4 at t = 3, stays at 4 until t = 5, where it drops to 0. It stays at 0 until t = 6, where it jumps to amplitude 2, stays at 2 until t = 8, and drops back to 0 for all t > 8.]

Common Mistakes That Cost Marks

Critical Exam Pitfalls

The Time-Scaling Phase Trap: Remember that the time-scaling property introduces a scaling amplitude multiplier: $y (a t) = ∣ a ∣ \cdot [x (a t) * h (a t)]$ . Forgetting the $∣ a ∣$ factor when compressing or expanding a convolved signal is a very common source of lost marks.

Applying Shifting to the Wrong Variable during Integration: In the proof of commutative property, when substituting $V = t - τ$ , remember that the differential is with respect to $τ$ (meaning $d t = 0$ , so $d τ = - d V$ ). Treating $t$ as a variable instead of a constant parameter during integration will ruin the entire proof.

PYQ Bank — Verbatim Questions & Answer Plans

[PYQ 2025] [12 Marks]

Question: If $y (t) = x (t) * h (t)$ , then show that $y (2 t) = 2 x (2 t) * h (2 t)$ .
Answer Plan:
1. Write out the basic convolution integral for $y (2 t)$ .
2. Set up the variable substitution $τ = 2 λ$ to scale the dummy variable.
3. Compute $d τ = 2 d λ$ and substitute it into the integral.
4. Extract the scaling factor of 2 outside the integral to prove the theorem.

[PYQ 2021] [10 Marks]

Question: Prove the time-shifting property of convolution: $x (t - t_{1}) * h (t - t_{2}) = y (t - t_{1} - t_{2})$ .
Answer Plan:
1. Follow the 4-step proof outlined in Section 4 of this note verbatim.
2. Include a brief physical explanation of how delay in system blocks accumulates.

Self-Check Before Moving On

Can you derive the commutative property of convolution from scratch?
Do you understand how parallel and cascaded LTI systems map to the distributive and associative properties?
Can you apply the time-delay shifting property to solve delta-convolution numericals rapidly?

Source: (k.Deergha Rao) signals and systems.pdf Section 2.3.1

2.04 Systems Described by Differential & Difference Equations

1. The Classical Modeling Paradigm

Alright — today we are diving into the mathematical engine room of LTI systems. Up until now, we’ve treated systems as “black boxes” represented by their impulse response $h (t)$ or $h [n]$ . But how do we model the physical reality of circuits containing inductors, capacitors, or discrete digital feedback loops?

We use Linear Constant-Coefficient Differential Equations (LCCDEs) for continuous-time systems and Linear Constant-Coefficient Difference Equations for discrete-time systems. These models are the mathematical bedrock of physical engineering.

2. Time-Domain Solutions of Continuous LCCDEs

A general $N$ -th order continuous-time system is modeled by the differential equation: $\sum_{k = 0}^{N} a_{k} \frac{d ^{k} y ( t )}{d t ^{k}} = \sum_{m = 0}^{M} b_{k} \frac{d ^{m} x ( t )}{d t ^{m}}$

To find the output response $y (t)$ for a given input $x (t)$ , classical time-domain analysis splits the solution into two distinct parts: $y (t) = y_{h} (t) + y_{p} (t)$

A. The Homogeneous (Complementary) Solution, $y_{h} (t)$

The homogeneous solution $y_{h} (t)$ {the natural response of the system representing its behavior without any external excitation} is found by setting the input $x (t) = 0$ : $\sum_{k = 0}^{N} a_{k} \frac{d ^{k} y _{h} ( t )}{d t ^{k}} = 0$

To solve this, we assume a trial solution of the form $y_{h} (t) = C e^{λ t}$ . Substituting this trial solution yields the characteristic equation: $\sum_{k = 0}^{N} a_{k} λ^{k} = 0 ⟹ a_{N} λ^{N} + a_{N - 1} λ^{N - 1} + \dots + a_{1} λ + a_{0} = 0$

Solving this polynomial gives the characteristic roots (or system eigenvalues) $λ_{1}, λ_{2}, \dots, λ_{N}$ . The form of $y_{h} (t)$ depends entirely on the nature of these roots:

Distinct Real Roots ( $λ_{1} \neq = λ_{2} \neq = \dots$ ): $y_{h} (t) = C_{1} e^{λ_{1} t} + C_{2} e^{λ_{2} t} + \dots + C_{N} e^{λ_{N} t}$
Repeated Real Roots ( $λ_{1} = λ_{2} = λ$ with multiplicity $r$ ): $y_{h} (t) = (C_{1} + C_{2} t + C_{3} t^{2} + \dots + C_{r} t^{r - 1}) e^{λ t}$
Complex Conjugate Pairs ( $λ = α \pm j β$ ): $y_{h} (t) = e^{α t} [A cos (βt) + B sin (βt)]$

Physical Meaning of Roots

The real part $α$ of the characteristic roots dictates the damping of the natural response. If $α < 0$ , the natural response decays exponentially, representing a stable physical system. The imaginary part $β$ dictates the frequency of natural oscillations.

B. The Particular Solution, $y_{p} (t)$

The particular solution $y_{p} (t)$ {the forced response of the system representing its behavior driven entirely by the input signal} satisfies the differential equation for the given input $x (t)$ . We find $y_{p} (t)$ using the Method of Undetermined Coefficients by matching the mathematical form of the input $x (t)$ :

Input Shape $x (t)$	Assumed Particular Solution $y_{p} (t)$
Constant: $A$	$K$ (Constant)
Exponential: $A e^{s t}$	$K e^{s t}$ (provided $s$ is not a characteristic root)
Polynomial: $A t^{n}$	$K_{n} t^{n} + K_{n - 1} t^{n - 1} + \dots + K_{0}$
Sinusoidal: $A cos (ω t + θ)$	$K_{1} cos (ω t) + K_{2} sin (ω t)$

3. Step-by-Step Continuous Numerical Solution

Let’s apply this classical pipeline to solve a standard exam-scoring continuous-time system.

Continuous-Time Step and Impulse Responses

Question: Consider the continuous LTI system described by the first-order differential equation: $\frac{d y ( t )}{d t} + y (t) = x (t)$ Determine the Step Response and the Impulse Response of this system, assuming the system is initially relaxed ( $y (0) = 0$ at $t = 0$ ).

Part I: Solving for the Step Response, $s (t)$

For the step response, the input is $x (t) = u (t)$ . We analyze the system for $t > 0$ , where $x (t) = 1$ .

Step 1: Find the Homogeneous Solution, $y_{h} (t)$

Set the right-hand side of the differential equation to zero: $\frac{d y _{h} ( t )}{d t} + y_{h} (t) = 0$ Assume $y_{h} (t) = C e^{λ t}$ . The characteristic equation is: $λ + 1 = 0 ⟹ λ = - 1$ Thus, the homogeneous solution is: $y_{h} (t) = C e^{- t}$

Step 2: Find the Particular Solution, $y_{p} (t)$

Since the input for $t > 0$ is a constant ( $x (t) = 1$ ), we assume a constant particular solution: $y_{p} (t) = K$ Substitute $y_{p} (t) = K$ into our governing differential equation: $\frac{d ( K )}{d t} + K = 1 ⟹ 0 + K = 1 ⟹ K = 1$ Thus, our particular solution is: $y_{p} (t) = 1$

Step 3: Combine and Apply Initial Conditions

The total response is the sum of the homogeneous and particular responses: $y (t) = y_{h} (t) + y_{p} (t) = C e^{- t} + 1 for t > 0$

Now, we apply the initial condition $y (0) = 0$ to find the undetermined constant $C$ : $y (0) = C e^{0} + 1 = 0 ⟹ C + 1 = 0 ⟹ C = - 1$

Substituting $C = - 1$ back into our total response yields our step response: $s (t) = (1 - e^{- t}) u (t)$

[GRAPH: Step response curve y(t) rising exponentially from 0 at t=0 and asymptotically flattening toward 1 as t approaches infinity. — governing equation: y(t) = (1 - e^-t)u(t) — source: Rabiul Sir Notes, Lec 6]

Part II: Solving for the Impulse Response, $h (t)$

The impulse response $h (t)$ is the system output when the input is a unit impulse $x (t) = δ (t)$ . Since a unit impulse $δ (t)$ is the derivative of a unit step $u (t)$ , and the system is linear and time-invariant, the impulse response is the time derivative of the step response: $h (t) = \frac{d}{d t} [s (t)] = \frac{d}{d t} [(1 - e^{- t}) u (t)]$

Applying the product rule of calculus: $h (t) = \frac{d}{d t} (1 - e^{- t}) \cdot u (t) + (1 - e^{- t}) \cdot \frac{d}{d t} [u (t)]$ $h (t) = e^{- t} u (t) + (1 - e^{- t}) δ (t)$

Since the term $(1 - e^{- t})$ evaluated at the impulse location $t = 0$ is exactly zero, the second term vanishes completely: $h (t) = e^{- t} u (t)$

[GRAPH: Impulse response curve h(t) starting abruptly at a peak of 1 at t=0 and decaying exponentially toward 0 as t approaches infinity. — governing equation: h(t) = e^-t * u(t) — source: Rabiul Sir Notes, Lec 6]

4. Time-Domain Solutions of Discrete Difference Equations

A discrete-time LTI system of order $N$ is represented by the difference equation: $\sum_{k = 0}^{N} a_{k} y [n - k] = \sum_{m = 0}^{M} b_{k} x [n - m]$

Just like its continuous counterpart, we solve this classical recurrence relation by decomposing it: $y [n] = y_{h} [n] + y_{p} [n]$

A. The Discrete Homogeneous Solution, $y_{h} [n]$

We set $x [n] = 0$ and assume a trial solution of the form $y_{h} [n] = C λ^{n}$ . Substituting this trial form gives the discrete characteristic equation: $\sum_{k = 0}^{N} a_{k} λ^{N - k} = 0 ⟹ a_{0} λ^{N} + a_{1} λ^{N - 1} + \dots + a_{N - 1} λ + a_{N} = 0$

Solving for the roots $λ_{1}, λ_{2}, \dots, λ_{N}$ determines the shape of $y_{h} [n]$ :

Distinct Roots: $y_{h} [n] = C_{1} λ_{1}^{n} + C_{2} λ_{2}^{n} + \dots$
Repeated Roots: $y_{h} [n] = (C_{1} + C_{2} n + C_{3} n^{2}) λ^{n}$

Discrete Stability Condition

For a discrete system to be stable, the natural response must decay over time as $n \to \infty$ . This requires the magnitude of all characteristic roots to be strictly less than unity ( $∣ λ_{i} ∣ < 1$ ). On the complex z-plane, this means all poles must lie inside the unit circle.

B. The Discrete Particular Solution, $y_{p} [n]$

The trial particular solution $y_{p} [n]$ matches the discrete shape of the input $x [n]$ :

Input Shape $x [n]$	Assumed Particular Solution $y_{p} [n]$
Constant: $A$	$K$ (Constant)
Exponential: $A β^{n}$	$K β^{n}$ (provided $β$ is not a characteristic root)
Polynomial: $A n^{k}$	$K_{k} n^{k} + K_{k - 1} n^{k - 1} + \dots + K_{0}$

5. Step-by-Step Discrete Numerical Solution

Let’s tackle a highly-tested discrete recurrence problem featuring initial conditions.

Discrete Second-Order Difference Equation

Question: Solve the second-order recursive discrete system described by: $y [n] - \frac{1}{9} y [n - 2] = x [n - 1]$ with step input $x [n] = u [n]$ , and initial auxiliary conditions {conditions specifying the starting state of the system’s memory delay buffers} given as $y [- 1] = 1$ and $y [- 2] = 0$ .

Step 1: Establish the Homogeneous Solution, $y_{h} [n]$

Set the right-hand side of our difference equation to zero: $y_{h} [n] - \frac{1}{9} y_{h} [n - 2] = 0$ Assume $y_{h} [n] = C λ^{n}$ . The characteristic polynomial equation is: $λ^{2} - \frac{1}{9} = 0 ⟹ λ = \pm \frac{1}{3}$ Thus, our homogeneous solution has the form: $y_{h} [n] = α_{1} (\frac{1}{3})^{n} + α_{2} (- \frac{1}{3})^{n}$

Step 2: Establish the Particular Solution, $y_{p} [n]$

Our input is the step function $x [n] = u [n]$ , meaning $x [n - 1] = u [n - 1]$ . For $n \geq 1$ , the input is a constant ( $x [n - 1] = 1$ ), so we assume a constant trial particular solution: $y_{p} [n] = K$ Substitute $y_{p} [n] = K$ into our difference equation for $n \geq 1$ : $K - \frac{1}{9} K = 1 ⟹ \frac{8}{9} K = 1 ⟹ K = \frac{9}{8}$ Thus, our particular solution is: $y_{p} [n] = \frac{9}{8} for n \geq 1$

Step 3: Compute Discrete Boundary States via Recursion

To apply our initial values to find $α_{1}$ and $α_{2}$ , we need the system output states at $n = 0$ and $n = 1$ . We calculate these directly using the system difference equation and the given initial conditions $y [- 1] = 1$ , $y [- 2] = 0$ .

For $n = 0$ : $y [0] - \frac{1}{9} y [- 2] = x [- 1] ⟹ y [0] - \frac{1}{9} (0) = u [- 1]$ Since step $u [- 1] = 0$ in negative time: $y [0] = 0$
For $n = 1$ : $y [1] - \frac{1}{9} y [- 1] = x [0] ⟹ y [1] - \frac{1}{9} (1) = u [0]$ Since step $u [0] = 1$ : $y [1] - \frac{1}{9} = 1 ⟹ y [1] = 1 + \frac{1}{9} = \frac{10}{9}$

Step 4: Combine and Solve for Coefficients

Our total response is modeled as: $y [n] = α_{1} (\frac{1}{3})^{n} + α_{2} (- \frac{1}{3})^{n} + \frac{9}{8} u [n - 1]$

Now we apply our calculated boundary states $y [0]$ and $y [1]$ to set up a system of linear equations:

At $n = 0$ : $y [0] = α_{1} (\frac{1}{3})^{0} + α_{2} (- \frac{1}{3})^{0} + \frac{9}{8} u [- 1] = 0$ $α_{1} + α_{2} + 0 = 0 ⟹ α_{2} = - α_{1}$
At $n = 1$ : $y [1] = α_{1} (\frac{1}{3})^{1} + α_{2} (- \frac{1}{3})^{1} + \frac{9}{8} u [0] = \frac{10}{9}$ $\frac{1}{3} α_{1} - \frac{1}{3} α_{2} + \frac{9}{8} = \frac{10}{9}$

Substitute $α_{2} = - α_{1}$ into the equation: $\frac{1}{3} α_{1} - \frac{1}{3} (- α_{1}) = \frac{10}{9} - \frac{9}{8}$ $\frac{2}{3} α_{1} = \frac{80 - 81}{72} = - \frac{1}{72}$ $α_{1} = - \frac{1}{72} \times \frac{3}{2} = - \frac{1}{48}$

Since $α_{2} = - α_{1}$ , we find: $α_{2} = \frac{1}{48}$

Step 5: Final Total Solution

Substituting our solved constants back into our equation yields our final complete response: $y [n] = - \frac{1}{48} (\frac{1}{3})^{n} + \frac{1}{48} (- \frac{1}{3})^{n} + \frac{9}{8} u [n - 1]$

6. Natural vs. Forced Response (An Critical Distinction)

In exams, students often confuse the classical homogeneous/particular split with the physical Natural/Forced split. They are not the same!

graph TD
    TR[Total Solution Response] --> HS_PS[Method 1: Homogeneous + Particular]
    TR --> NR_FR[Method 2: Natural + Forced]
    
    HS_PS --> HS[Homogeneous Response]
    HS_PS --> PS[Particular Response]
    
    NR_FR --> NR[Natural/Zero-Input Response]
    NR_FR --> FR[Forced/Zero-State Response]
    
    style TR fill:#f9f,stroke:#333,stroke-width:2px
    style HS_PS fill:#bbf,stroke:#333
    style NR_FR fill:#bfb,stroke:#333

Natural Response (Zero-Input Response, $y_{zi} (t)$ ): The system response due only to initial conditions, setting the input $x (t) = 0$ .
Forced Response (Zero-State Response, $y_{zs} (t)$ ): The system response due only to the input $x (t)$ , setting all initial conditions to zero.

7. ECE 2108 Laboratory Connection

In Experiment 03 of your lab, you will model these continuous differential equations and discrete systems using MATLAB:

% Continuous system simulation
num = [1]; den = [1, 1];          % Represents dy/dt + y = x
sys = tf(num, den);
step(sys);                       % Plots the step response
 
% Discrete recurrence simulation using filter
b = [0, 1]; a = [1, 0, -1/9];     % Represents y[n] - 1/9 y[n-2] = x[n-1]
x = ones(1, 20);                  % Step input array
y = filter(b, a, x);              % Computes recursive response
stem(y);                         % Plots discrete stems

Common Mistakes That Cost Marks

Exam Pitfall Checklist

Applying Initial Conditions Too Early: This is the #1 source of lost marks in difference/differential equation questions! Students frequently apply initial conditions to the homogeneous solution $y_{h}$ before adding the particular solution $y_{p}$ . Always apply initial conditions to the total combined response $y (t) = y_{h} (t) + y_{p} (t)$ .

Ignoring the Duplicate Root Trap: If the input signal matches one of your characteristic roots, your standard particular trial solution will fail. For example, if your characteristic root is $λ = - 1$ and your input is $x (t) = e^{- t}$ , your particular trial solution cannot be $K e^{- t}$ (as it is already absorbed by the homogeneous solution). You must multiply the trial solution by the independent variable: $K t e^{- t}$ .

Using the Wrong Auxiliary Limits: For continuous-time systems, remember that physical variables (like current through an inductor or voltage across a capacitor) cannot change instantaneously. Thus, boundary states at $t = 0^{-}$ are equated to states at $t = 0^{+}$ . For discrete systems, compute $y [0], y [1]$ recursively from the delay terms.

PYQ Bank — Verbatim Questions & Answer Plans

Q1: Continuous System Representation [10 Marks] [PYQ 2024]

Represent the following differential equation in a state model: $\frac{d ^{3} y}{d t ^{3}} + \frac{d ^{2} y}{d t ^{2}} + 6 \frac{d y}{d t} + 7 y = 2 u (t)$

Answer Plan:
1. Define state variables based on the phase-variable method: $x_{1} = y$ , $x_{2} = \overset{y}{˙}$ , $x_{3} = \overset{y}{¨}$ .
2. Write the derivatives of the state variables: $\overset{x}{˙}_{1} = x_{2}$ , $\overset{x}{˙}_{2} = x_{3}$ , and $\overset{x}{˙}_{3} = - 7 x_{1} - 6 x_{2} - x_{3} + 2 u (t)$ .
3. Express the equations in matrix form $\dot{X} = AX + BU$ and $Y = CX + DU$ .

Q2: System Realization and Real-World Traps [15 Marks] [PYQ 2018]

Determine the output of the system described by: $y [n] - \frac{1}{9} y [n - 2] = x [n - 1]$ where $x [n] = u [n - 1]$ , and initial conditions are given as $y [- 1] = 1, y [- 2] = 0$ .

Answer Plan:
1. Find the homogeneous solution $y_{h} [n] = α_{1} (1/3)^{n} + α_{2} (- 1/3)^{n}$ .
2. Note that the input is $x [n] = u [n - 1]$ , meaning $x [n - 1] = u [n - 2]$ . For $n \geq 2$ , the input is 1, so particular solution $y_{p} [n] = 9/8$ .
3. Compute boundary outputs $y [0]$ and $y [1]$ using recursive equations from the difference relation.
4. Combine homogeneous and particular solutions, and solve for constants $α_{1}$ and $α_{2}$ using the boundary states.

Self-Check Before Moving On

Can you explain the physical difference between the homogeneous response and the particular response?
Do you know how to identify and avoid the “duplicate root particular solution trap”?
Can you solve a first-order continuous differential equation step-by-step for a unit step input?
Can you recursively calculate discrete output states $y [0]$ and $y [1]$ given arbitrary past initial conditions $y [- 1]$ and $y [- 2]$ ?

Source: (K. Deergha Rao) Signals and Systems, Chapter 2 & 6; Rabiul Sir class notes, Lecture 7 & 8.

2.05 Block-Diagram Representations & System Interconnections

[!abstract] Block-Diagram Realization A graphical representation mapping the algebraic relations of linear constant-coefficient differential equations (LCCDEs) or difference equations into elementary physical hardware blocks. It serves as the physical bridge between abstract time-domain equations and practical analog circuit computer simulations {using operational amplifiers} or digital signal processing (DSP) hardware chips {using registers and adders}.

1. Basic Building Blocks of LTI Systems

Before drawing system architectures, we must master the standard graphical symbols. These blocks are the elementary operations we perform on signals in analog and digital domains.

1.1 Continuous-Time (CT) Realization Blocks

Continuous-time systems are physically built using operational amplifiers (op-amps), resistors, and capacitors. The three fundamental building blocks are:

Element	Algebraic Relation	Graphical representation (ASCII)	Physical Implementation
CT Integrator	$y (t) = \int_{- \infty}^{t} x (τ) d τ$	`x(t) ----> [ 1/s ] ----> y(t)`	Op-amp integrator circuit with capacitor feedback
CT Adder	$y (t) = x_{1} (t) + x_{2} (t)$	`x1(t) ----\` `( + ) ----> y(t)` `x2(t) ----/`	Op-amp summing amplifier circuit
CT Multiplier	$y (t) = c \cdot x (t)$	`x(t) ----> ( c ) ----> y(t)`	Op-amp inverting/non-inverting amplifier with gain $c$

1.2 Discrete-Time (DT) Realization Blocks

Discrete-time systems do not use capacitors or integration. Instead, they operate on digital clock cycles using registers, logic adders, and multipliers:

Element	Algebraic Relation	Graphical representation (ASCII)	Physical Implementation
DT Unit Delay	$y [n] = x [n - 1]$	`x[n] ----> [ z^-1 ] ----> y[n]`	Digital register / flip-flop clocked delay element
DT Adder	$y [n] = x_{1} [n] + x_{2} [n]$	`x1[n] ----\` `( + ) ----> y[n]` `x2[n] ----/`	Logic adder circuit
DT Multiplier	$y [n] = c \cdot x [n]$	`x[n] ----> ( c ) ----> y[n]`	Digital hardware multiplier circuit

2. Realization of Continuous-Time Systems (LCCDEs)

For a general $N$ -th order continuous-time linear constant-coefficient differential equation: $\sum_{n = 0}^{N} a_{n} \frac{d ^{n} y ( t )}{d t ^{n}} = \sum_{m = 0}^{M} b_{m} \frac{d ^{m} x ( t )}{d t ^{m}}$

To draw the block diagram, we must solve for the highest-order derivative of the output, $\frac{d ^{N} y ( t )}{d t ^{N}}$ , by isolating it on the left-hand side: $\frac{d ^{N} y ( t )}{d t ^{N}} = \frac{1}{a _{0}} [\sum_{m = 0}^{M} b_{m} \frac{d ^{m} x ( t )}{d t ^{m}} - \sum_{n = 1}^{N} a_{n} \frac{d ^{N - n} y ( t )}{d t ^{N - n}}]$

The $a_{0}$ Division Trap (Rabiul Sir's Lab Warning)

Always check if the coefficient of the highest-order derivative of the output, $a_{0}$ , is equal to 1. If $a_{0} \neq = 1$ , you must divide every single coefficient in the equation by $a_{0}$ before drawing! Forgetting this division is the #1 source of lost marks in both the exam and the lab [Rabiul Sir Class Note].

2.1 Direct Form I Realization (Non-Canonic)

Direct Form I represents a literal translation of the differential equation. It realizes the input operations (feedforward zeros) first, followed by the output operations (feedback poles) in series:

Input x(t) ---> [ Feedforward zeros (System 1) ] ---> [ Feedback poles (System 2) ] ---> Output y(t)

Characteristics: It keeps the input and output scaling operations completely independent.
Drawback: For an $N$ -th order system, it requires $M + N$ integrators, which is hardware-inefficient.

2.2 Direct Form II Realization (Canonic)

Direct Form II is obtained by reversing the order of the cascaded subsystems: realizing the feedback poles first, followed by the feedforward zeros:

Input x(t) ---> [ Feedback poles (System 2) ] ---> [ Feedforward zeros (System 1) ] ---> Output y(t)

By swapping the cascade order, the intermediate state nodes are identical. We can merge the parallel branches, sharing the exact same integrators down the center column of the diagram!

Canonic Property: A system realization is called canonic if it uses the absolute minimum number of integrators/delays possible, which is exactly equal to the system order, $N$ [Textbook Sec 2.6].
Hardware Savings: Direct Form II cuts the required number of integrators exactly in half!

3. Realization of Discrete-Time Systems (Difference Equations)

In the discrete-time domain, the difference equation relates input and output sequences: $\sum_{n = 0}^{N} a_{n} y [n - n] = \sum_{m = 0}^{M} b_{m} x [n - m]$

Assuming $a_{0} = 1$ , we solve for the present output $y [n]$ : $y [n] = \sum_{m = 0}^{M} b_{m} x [n - m] - \sum_{n = 1}^{N} a_{n} y [n - n]$

3.1 Discrete Direct Form I

Realizes the delay lines for the input sequence $x [n]$ and output sequence $y [n]$ on separate delay paths.

3.2 Discrete Direct Form II (Canonic)

Creates a single center delay line using an intermediate digital state sequence $w [n]$ : $w [n] = x [n] - \sum_{n = 1}^{N} a_{n} w [n - n]$ $y [n] = \sum_{m = 0}^{M} b_{m} w [n - m]$

Structure:
- Left-hand side (Feedback): Feeds back delayed states of $w [n]$ multiplied by negated coefficients $- a_{n}$ into the main summing junction.
- Right-hand side (Feedforward): Feeds forward delayed states of $w [n]$ multiplied by $b_{m}$ coefficients into the output summing junction.

4. Step-by-Step Solved Numericals

📝 Numerical 1: Continuous-Time Realization (KUET 2024, 2022)

Question: Obtain the block-diagram representation of a continuous-time system described by: $\frac{d ^{2} y ( t )}{d t ^{2}} + 2 \frac{d y ( t )}{d t} + y (t) = \frac{d ^{2} x ( t )}{d t ^{2}} - \frac{d x ( t )}{d t} - 6 x (t)$

Step-by-Step Solution:

Step 1: Check $a_{0}$ and identify coefficients The highest derivative coefficient is $a_{0} = 1$ . The system order is $N = 2$ .

Output coefficients: $a_{1} = 2$ , $a_{2} = 1$ .
Input coefficients: $b_{0} = 1$ , $b_{1} = - 1$ , $b_{2} = - 6$ .

Step 2: Set up the intermediate feedback equation Define the intermediate state variable $w (t)$ in the s-domain or time-domain: $\overset{w}{¨} (t) = x (t) - 2 \overset{w}{˙} (t) - w (t)$

Step 3: Set up the output feedforward equation $y (t) = \overset{w}{¨} (t) - \overset{w}{˙} (t) - 6 w (t)$

Step 4: Draw the Direct Form II Block Diagram We cascade two CT integrators down the center to generate $\overset{w}{˙} (t)$ from $\overset{w}{¨} (t)$ , and $w (t)$ from $\overset{w}{˙} (t)$ .

               Input x(t)
                   |
                   v
               +-------+
       +-----> |   +   | <----------------+
       |       +-------+                  |
       |           | \                    |
       |           |  \                   |
       |           v   \ ( b0 = 1 )       |
       |      ddot w(t) \                 |
       |           |     v                |
       |        [ 1/s ] --->( + )         |
       |           |         ^            |
       |           v         |            | ( -a2 = -1 )
       | ( -a1 = -2 )        |            |
       +<----( -2 )+--dot w(t)            |
       |           |         |            |
       |           |   ( b1 = -1 )        |
       |           v         |            |
       |        [ 1/s ] ---->+            |
       |           |         |            |
       |           v         v            |
       +<----( -1 )+--- w(t)              |
                   |                      |
                   +------>( -6 )-------->+-----> Output y(t)
                         ( b2 = -6 )

📝 Numerical 2: Discrete-Time Realization with “Trap” Coefficients (KUET 2025, 2019)

Question: Draw the canonic Direct Form II realization for the system described by the difference equation: $y [n] + \frac{1}{3} y [n - 1] + \frac{1}{5} y [n - 3] = x [n - 1]$

Step-by-Step Solution:

Step 1: Identify coefficients and system order The highest delay is 3, so system order is $N = 3$ . This requires exactly 3 delay elements ( $z^{- 1}$ ) for canonic realization.

Output coefficients: $a_{0} = 1$ , $a_{1} = 1/3$ , $a_{2} = 0$ , $a_{3} = 1/5$ . (Note: $a_{2} = 0$ is a trap!)
Input coefficients: $b_{0} = 0$ , $b_{1} = 1$ , $b_{2} = 0$ , $b_{3} = 0$ . (Note: Only $b_{1}$ is non-zero, another trap!)

Step 2: Formulate the canonic node equations $w [n] = x [n] - \frac{1}{3} w [n - 1] - \frac{1}{5} w [n - 3]$ $y [n] = w [n - 1]$

Step 3: Draw the Direct Form II realization We lay out three delays in series down the center. Because $a_{2} = 0$ , we omit the feedback multiplier tap at the second delay stage. Because $b_{0}, b_{2}, b_{3} = 0$ , we do not tap those nodes for the output sum; the output $y [n]$ is taken directly from the output of the first delay node $w [n - 1]$ !

               Input x[n]
                   |
                   v
               +-------+
               |   +   | <-------------------------+
               +-------+                           |
                   |                               |
                   v w[n]                          |
               +-------+                           |
               |  z^-1 |                           |
               +-------+                           |
                   |---- w[n-1] ------------------>+-----> Output y[n]
                   |             \ (b1 = 1)        |
                   v              v                | ( -a3 = -1/5 )
             ( -a1 = -1/3 )      ( + )             |
       +<----( -1/3 )              ^               |
       |                           |               |
       |           |               | (b2 = 0)      |
       |           v               |               |
       |       +-------+           |               |
       |       |  z^-1 |           |               |
       |       +-------+           |               |
       |           |               |               |
       |           v w[n-2] -------+               |
       |           |                               |
       |           v                               |
       |       +-------+                           |
       |       |  z^-1 |                           |
       |       +-------+                           |
       |           |---- w[n-3]                    |
       |           |                               |
       +<----------+------( -1/5 )---------------->+

5. System Interconnections

LTI systems are connected in multi-block structures. We analyze their combined behavior using mathematical operators:

graph LR
    subgraph Cascade Connection
    c_in[x] --> c_h1[h1] --> c_mid[w] --> c_h2[h2] --> c_out[y]
    end
    
    subgraph Parallel Connection
    p_in[x] --> p_h1[h1] --> p_add((+))
    p_in --> p_h2[h2] --> p_add --> p_out[y]
    end

5.1 Cascade (Series) Connection

Time-Domain Impulse Response: The overall impulse response is the convolution of individual impulse responses: $h_{total} (t) = h_{1} (t) * h_{2} (t)$
Laplace/Z-Domain Transfer Function: Corresponds to algebraic multiplication of their individual transfer functions: $H_{total} (s) = H_{1} (s) \cdot H_{2} (s) ⟺ H_{total} (z) = H_{1} (z) \cdot H_{2} (z)$
Commutative Benefit: Because convolution is commutative, the order of blocks in a cascade does not affect the overall LTI system response.

5.2 Parallel Connection

Time-Domain Impulse Response: The overall impulse response is the algebraic sum of individual impulse responses: $h_{total} (t) = h_{1} (t) + h_{2} (t)$
Laplace/Z-Domain Transfer Function: Corresponds to the sum of their transfer functions: $H_{total} (s) = H_{1} (s) + H_{2} (s) ⟺ H_{total} (z) = H_{1} (z) + H_{2} (z)$

6. Obsidian Desmos Visualization: Integrator & Delay Magnitude Response

To understand how cascading blocks impacts system gain (like low-pass filtering), view the frequency magnitude responses:

height=300; width=100%;
---
y = 1 / sqrt(1 + x^2) | GREEN | label: Single Stage RC
y = 1 / (1 + x^2) | BLUE | label: Cascaded 2-Stage RC

7. End-of-Note Master Review

⚠️ Common Mistakes That Cost Marks

The Feedback Sign Flip Trap: When moving from a difference equation to a block diagram, feedback coefficients are negated. For example, if the equation has $+ 3 y [n - 1]$ , the multiplier block gain must be $- 3$ . If the equation has $- 2 y [n - 1]$ , the multiplier block gain is $+ 2$ .
The Integrator-Delay Swap: Drawing continuous-time blocks (integrators $[1/ s]$ ) when asked to realize discrete difference equations, or drawing delay registers $[z^{- 1}]$ for continuous differential equations.
Non-Canonic Bloat: Drawing a Direct Form I structure when the exam question explicitly asks for a canonic representation. Direct Form II is canonic; Direct Form I is not.

⚗️ ECE 2108 Lab Connection

In the ECE 2108 Laboratory, block realizations are modeled in MATLAB using state-space variables and the tf (transfer function) object commands:

% MATLAB script to define and test a transfer function block cascade
num1 = [1 -1 -6]; den1 = [1 2 1];
sys1 = tf(num1, den1); % Create system TF
is_stable = isstable(sys1); % Check system stability
bode(sys1); % Plot frequency magnitude and phase spectra

🔍 Verbatim PYQ Bank

KUET ECE 2107:

Draw the block diagram of the system described by: $\frac{d ^{2} y ( t )}{d t ^{2}} + 3 y (t) = \frac{d x ( t )}{d t} + 2 \frac{d ^{2} x ( t )}{d t ^{2}}$ Answer Plan: Divide by highest derivative coefficient ( $a_{0} = 1$ ). Note that $a_{1} = 0$ , $a_{2} = 3$ , $b_{0} = 2$ , $b_{1} = 1$ , $b_{2} = 0$ . Draw Direct Form II center delay with shared integrators.

KUET ECE 2107:

Draw the block diagram for: $\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} + y (t) = \frac{d ^{2} x}{d t ^{2}} - \frac{d x}{d t} - 6 x (t)$ Answer Plan: Refer directly to the fully solved Numerical 1 in Section 4.

🏆 Self-Check Before Moving On

Can you explain the exact difference between canonic (Direct Form II) and non-canonic (Direct Form I) realizations?
Do you remember to divide all coefficients by $a_{0}$ if $a_{0} \neq = 1$ ?
Can you trace the feedback paths and apply the correct sign negations?

References: (K. Deergha Rao) Signals and Systems, Ch 2, Pg 93–95; Rabiul Sir Lecture Notes, Lec 7, Pg 27–28; KUET ECE 2107 Past Papers (2025, 2024, 2022).

007_index

Chapter 2 - LTI System Properties & Time-Domain Convolution

02 Chapter Map - LTI System Properties & Time-Domain Convolution

📚 Study Notes Index

🎯 Exam Weight

🔗 Related Resources

Chapter 2: LTI System Properties & Time-Domain Convolution - Compact Review

1. Systems Classification & Mathematical Operators

1.1 Linearity: Superposition & Homogeneity

1.2 Time-Invariance (Shift-Invariance)

1.3 Causality

1.4 Memory (Dynamic) vs. Memoryless (Static)

1.5 Bounded-Input Bounded-Output (BIBO) Stability

2. Continuous-Time Convolution Integral

2.1 The Representation of Signals in Terms of Impulses

2.2 The Continuous Convolution Integral

2.3 Mathematical Properties of Continuous Convolution

3. Discrete-Time Convolution Sum

3.1 The Discrete Convolution Sum

3.2 Sequence Length Theorem

3.3 Periodic (Circular) vs. Linear Convolution

4. Continuous & Discrete BIBO Stability Proofs

4.1 Continuous-Time Absolute Integrability Proof

4.2 Discrete-Time Absolute Summability Proof

5. Systems Described by Differential & Difference Equations

5.1 Linear Constant-Coefficient Differential Equations (LCCDE)

5.2 Linear Constant-Coefficient Difference Equations

5.3 Complete Response Decompositions

5.4 Characteristic Roots & Homogeneous Solution Forms

6. Block Diagram Realizations & Interconnections

6.1 Direct Form Realizations

6.2 System Interconnections

6.3 Inverse Systems & Deconvolution

7. Master Comparison Tables

Table 2.1: System Properties Test Matrix

Table 2.2: Continuous vs. Discrete Convolution Properties

Table 2.3: Recursive (IIR) vs. Non-Recursive (FIR) Difference Equations

8. Verbatim Chapter 2 PYQ Bank

9. Common Mistakes That Cost Marks

10. Pre-Exam Self-Check Checklist

2.01 Systems Classification, LTI Properties & Stability

1. Conceptual Framework of Systems

2. Basic System Classifications & Properties

2.1 Memoryless vs. Dynamic (Memory) Systems

2.2 Causal, Non-Causal, and Anticausal Systems

2.3 Invertible vs. Non-Invertible Systems

3. The Linearity Test (Superposition & Homogeneity)

Step-by-Step Algebraic Verification Template:

4. The Time-Invariance Test

Step-by-Step Algebraic Verification Template:

5. Standard Core Solved Numericals

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

5.2 Case Study: dtdy(t)​+3y(t)=x(t)dtdx(t)​ [PYQ: 2022]

5.3 Case Study: [y[n]]2=n[x[n]]2 [HW / Lab]

6. BIBO Stability & Absolute Integrability Derivation

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

Step 0: Setup and Definitions

Step 1: Upper Bound Expansion (Sufficiency Proof)

Step 2: Substitute the Input Bound

Step 3: Necessity Proof

7. Core Solved Stability Traps

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

7.2 Stability check of h(t)=e−tu(t−1)

Common Mistakes That Cost Marks

PYQ Bank — Verbatim Questions & Answer Plans

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

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

Q3: System parameter evaluation of y[n]=3y[n−1]−n2x[n]+4x[n−1]−2x[n+1]

Self-Check Before Moving On

2.02 The Continuous Convolution Integral

1. Mathematical Derivation of the Convolution Integral

Step 0: Setup & Impulse Representation of x(t)

Step 1: Applying the LTI System Operator

Step 2: Incorporating Time-Invariance

2. Rigorous Proofs of Core Properties

2.1 Time Scaling Property

Step-by-Step Derivations:

2.2 Even Symmetry of Convolved Odd Signals

Step-by-Step Derivations:

3. Graphical Convolution: The Sliding Window Workflow

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

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

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

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

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

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

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

Step 0: Setup & Impulse Representation of $x (t)$

3. Convolution with an Impulse ( $δ (t)$ )

A. The Homogeneous (Complementary) Solution, $y_{h} (t)$

B. The Particular Solution, $y_{p} (t)$

Part I: Solving for the Step Response, $s (t)$

Step 1: Find the Homogeneous Solution, $y_{h} (t)$

Step 2: Find the Particular Solution, $y_{p} (t)$

Part II: Solving for the Impulse Response, $h (t)$

A. The Discrete Homogeneous Solution, $y_{h} [n]$

B. The Discrete Particular Solution, $y_{p} [n]$

Step 1: Establish the Homogeneous Solution, $y_{h} [n]$

Step 2: Establish the Particular Solution, $y_{p} [n]$