Related Concepts: 2.01 Systems Classification, LTI Properties & Stability | 2.02 The Continuous Convolution Integral | 2.04 Systems Described by Differential & Difference Equations

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.

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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: $x(t)\ast h(t)=h(t)\ast x(t)$

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: $x(t)\ast h(t)={\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau$

Step 1: Change of Variables

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

Step 2: Evaluate Limits and Differentials

Now, find the differential $d\tau$ by differentiating both sides with respect to $\lambda$: $d\tau =-d\lambda$

Next, transform the limits of integration from $\tau$ to $\lambda$:

• When $\tau \to -\infty$, then $\lambda =t-(-\infty )\to \infty$.
• When $\tau \to \infty$, then $\lambda =t-(\infty )\to -\infty$.

Step 3: Substitute and Simplify

Substitute these variables back into our original integral setup: ${\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau ={\int }_{\infty }^{-\infty }x(t-\lambda )h(\lambda )(-d\lambda )$

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

Since $\lambda$ is just a dummy variable {a temporary variable name used strictly for the duration of the integration}, we can rename it back to $\tau$: ${\int }_{-\infty }^{\infty }h(\tau )x(t-\tau )d\tau =h(t)\ast x(t)\text{(Proved)}$

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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: $x(t)\ast [h_1(t)+h_2(t)]=x(t)\ast h_1(t)+x(t)\ast h_2(t)$

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: $h_{\text{total}}(t)=h_1(t)+h_2(t)$

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

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2.2 The Associative Property (Cascaded Systems)

Convolution is associative: $[x(t)\ast h_1(t)]\ast h_2(t)=x(t)\ast [h_1(t)\ast h_2(t)]$

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: $h_{\text{total}}(t)=h_1(t)\ast h_2(t)$

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: $h_1(t)\ast h_2(t)=h_2(t)\ast h_1(t)$

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3. Convolution with an Impulse ($\delta (t)$)

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: $x(t)\ast \delta (t)=x(t)$

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: $x(t)\ast \delta (t)={\int }_{-\infty }^{\infty }x(\tau )\delta (t-\tau )d\tau$

Step 1: Utilize the Sifting Property

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

Step 2: Evaluate the Integral

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

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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: $y(t)=x(t)\ast h(t)$ Then convolving their delayed versions yields: $x(t-t_1)\ast h(t-t_2)=y(t-t_1-t_2)$

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: $x(t-t_1)\ast h(t-t_2)={\int }_{-\infty }^{\infty }x(\tau -t_1)h(t-\tau -t_2)d\tau$

Step 2: Define a Variable Substitution

To align this with our standard convolution form, let’s substitute the argument of our first function: $\upsilon =\tau -t_1⟹\tau =\upsilon +t_1\text{and}d\tau =d\upsilon$

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: ${\int }_{-\infty }^{\infty }x(\upsilon )h(t-(\upsilon +t_1)-t_2)d\upsilon ={\int }_{-\infty }^{\infty }x(\upsilon )h((t-t_1-t_2)-\upsilon )d\upsilon$

Step 4: Map to the Base Convolution

Notice that the resulting integral is identical to the definition of $y(t)=\int x(\upsilon )h(t-\upsilon )d\upsilon$, except that the time variable $t$ has been replaced by $(t-t_1-t_2)$.

Therefore: $x(t-t_1)\ast h(t-t_2)=y(t-t_1-t_2)\text{(Proved)}$

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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 $x(t)=4\delta (t-2)+2\delta (t-5)$.

Step-by-Step Solution

Step 1: Write down the Algebraic Expression

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

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

Step 2: Apply the Shifting Property of the Impulse

We know that convolving any signal $f(t)$ with a shifted impulse $\delta (t-t_0)$ simply shifts the signal by $t_0$: $f(t)\ast \delta (t-t_0)=f(t-t_0)$

Let’s apply this to our two distributed terms:

1. First term: $4\delta (t-2)\ast u(t-1)=4u(t-1-2)=4u(t-3)$ $4\delta (t-2)\ast (-u(t-3))=-4u(t-3-2)=-4u(t-5)$
2. Second term: $2\delta (t-5)\ast u(t-1)=2u(t-1-5)=2u(t-6)$ $2\delta (t-5)\ast (-u(t-3))=-2u(t-3-5)=-2u(t-8)$

Step 3: Combine and Write the Final Expression

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

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.]

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Common Mistakes That Cost Marks

Critical Exam Pitfalls

• The Time-Scaling Phase Trap: Remember that the time-scaling property introduces a scaling amplitude multiplier: $y(at)=|a|\cdot [x(at)\ast h(at)]$. 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 $V=t-\tau$, remember that the differential is with respect to $\tau$ (meaning $dt=0$, so $d\tau =-dV$). Treating $t$ as a variable instead of a constant parameter during integration will ruin the entire proof.

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PYQ Bank — Verbatim Questions & Answer Plans

[PYQ 2025] [12 Marks]

• Question: If $y(t)=x(t)\ast h(t)$, then show that $y(2t)=2x(2t)\ast h(2t)$.
• Answer Plan:
  1. Write out the basic convolution integral for $y(2t)$.
  2. Set up the variable substitution $\tau =2\lambda$ to scale the dummy variable.
  3. Compute $d\tau =2d\lambda$ 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: $x(t-t_1)\ast h(t-t_2)=y(t-t_1-t_2)$.
• 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.

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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