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

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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 $\delta (t)$.

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 $\Delta$.

[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 $t=k\Delta$ with width $\Delta$ and height $x(k\Delta )$ can be represented as: $\hat{x}(t)={\sum }_{k=-\infty }^{\infty }x(k\Delta ){\delta }_{\Delta }(t-k\Delta )\Delta$ where ${\delta }_{\Delta }(t)$ is a rectangular pulse of height $1/\Delta$ and width $\Delta$ centered at the origin.

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

1. The discrete summation index $k\Delta$ transitions into a continuous variable of integration $\tau$.
2. The width step $\Delta$ becomes an infinitesimal differential $d\tau$.
3. The approximation pulse ${\delta }_{\Delta }(t)$ converges to the perfect Dirac delta function $\delta (t)$.
4. The staircase approximation $\hat{x}(t)$ becomes the exact original signal $x(t)$.

This yields the sifting property representation {representing a signal as a continuous sweep of impulses}: $x(t)={\lim }_{\Delta \to 0}{\sum }_{k=-\infty }^{\infty }x(k\Delta ){\delta }_{\Delta }(t-k\Delta )\Delta ={\int }_{-\infty }^{\infty }x(\tau )\delta (t-\tau )d\tau$

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Step 1: Applying the LTI System Operator

Let $\mathbb{T}[\cdot ]$ represent the transformation operator of a system. Feeding the representation of $x(t)$ into the system yields: $y(t)=\mathbb{T}[x(t)]=\mathbb{T}\left[{\int }_{-\infty }^{\infty }x(\tau )\delta (t-\tau )d\tau \right]$

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

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Step 2: Incorporating Time-Invariance

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

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

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

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2. Rigorous Proofs of Core Properties

2.1 Time Scaling Property

Theorem (Time Scaling in Convolution)

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

Step-by-Step Derivations:

Start with the standard definition of the scaled output $y(at)$: $y(at)={\int }_{-\infty }^{\infty }x(at-\tau )h(\tau )d\tau$

We want to transform the integration variable so that the arguments of both $x$ and $h$ are scaled by $a$. Let us substitute $\tau =a\lambda ⟹d\tau =ad\lambda$.

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

Combining both cases using the absolute value operator yields: $y(at)=|a|{\int }_{-\infty }^{\infty }x(a(t-\lambda ))h(a\lambda )d\lambda =|a|\cdot [x(at)\ast h(at)]\text{(Proved)}$

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

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

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

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

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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 $\tau$:

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

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

• Interval 1 ($t<0$): No Overlap The leading edge of $h(t-\tau )$ (which is at $\tau =t+1$) has not reached the starting edge of $x(\tau )$ (at $\tau =1$). $t+1<1⟹t<0⟹y(t)=0$

• Interval 2 ($0\le t<2$): Partial Positive Overlap The positive lobe of $h(t-\tau )$ slides into $x(\tau )$. The overlap region spans from $\tau =1$ to $\tau =t+1$: $y(t)={\int }_1^{t+1}(1)(1)d\tau =[\tau {]}_1^{t+1}=t$

• Interval 3 ($2\le t<4$): Full Positive & Partial Negative Overlap The positive lobe has fully entered, and the negative lobe (spanning from $t-1$ to $t+1$) begins overlapping $x(\tau )$. $y(t)={\int }_{t-1}^3(1)d\tau +{\int }_3^{t+1}(-1)d\tau =(4-t)-(t-2)=4-2t$

• Interval 4 ($4\le t<6$): Pure Negative Overlap The positive lobe has exited the boundaries of $x(\tau )$ completely. Only the negative lobe overlaps, spanning from $t-1$ to $3$: $y(t)={\int }_{t-1}^3(-1)d\tau =-[3-(t-1)]=t-4$

• Interval 5 ($t\ge 6$): No Overlap The trailing edge of $h(t-\tau )$ (at $\tau =t-3$) has passed the trailing edge of $x(\tau )$ (at $\tau =3$). $t-3\ge 3⟹t\ge 6⟹y(t)=0$

[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]

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

Substitute the active step functions $u(\tau )$ and $u(t-\tau )$ into the limits of integration. The product $u(\tau )u(t-\tau )$ is non-zero if and only if $0\le \tau \le t$: $y(t)={\int }_0^t\left(e^{-\tau /RC}\right)\left(\frac{1}{RC}{e}^{-(t-\tau )/RC}\right)d\tau ,t\ge 0$

Factor out the terms that do not depend on the dummy integration variable $\tau$: $y(t)=\frac{1}{RC}{e}^{-t/RC}{\int }_0^t{e}^{-\tau /RC}{e}^{\tau /RC}d\tau$ $y(t)=\frac{1}{RC}{e}^{-t/RC}{\int }_0^t(1)d\tau$ $y(t)=\frac{t}{RC}{e}^{-t/RC}u(t)\text{(Q.E.D.)}$

[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]

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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 $u(\tau )$ and $u(t-\tau )$.
• The Correction: Always rewrite the step functions as limits of integration before doing calculus: $u(\tau )⟹\text{Lower limit becomes }0$ $u(t-\tau )⟹\text{Upper limit becomes }t$
• 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.

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6. Verbatim PYQ Bank

[PYQ 2025, 2022]

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

• Answer Plan: Refer to Section 2.1 of this note. Use the coordinate transformation substitution $\tau =2\lambda$ 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 $t$: $y(t)={\int }_0^t\tau e^{-\tau /RC}\cdot \frac{1}{RC}{e}^{-(t-\tau )/RC}d\tau =\frac{e^{-t/RC}}{RC}{\int }_0^t\tau d\tau =\frac{t^2}{2RC}{e}^{-t/RC}u(t)$

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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?

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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).