9.01 Foundation of Laplace Transform and s-Plane Representat

9.00 Chapter Map - Laplace Transform & s-Domain Analysis | 9.02 Laplace Transforms of Singularity & Elementary Functions

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9.01 Foundation of Laplace Transform & s-Plane Representation

Core Idea

The Laplace Transform generalizes continuous-time frequency analysis by mapping a real-time signal $x(t)$ to the complex frequency domain represented by the complex s-plane ($s=\sigma +j\omega$). While the classical Fourier Transform decomposes a signal into sustained purely imaginary sinusoidal harmonics ($s=j\omega$), the Laplace Transform introduces a real-valued damping/exponential growth factor ($e^{-\sigma t}$). This damping factor guarantees the absolute convergence of the integral for a much broader class of signals—including unstable, growing, and transient waveforms (such as ramps and exponentials) where the Fourier integral fails to converge. In doing so, it converts differential equations of continuous systems into simple algebraic equations in the s-plane.

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1. Defining the Laplace Transform Integrals

To accommodate different classes of physical signals, we define two distinct forms of the Laplace transform: the Bilateral (Two-Sided) transform and the Unilateral (One-Sided) transform.

1.1 The Bilateral (Two-Sided) Laplace Transform

The Bilateral Laplace Transform of a continuous-time signal $x(t)$ integrates over all past and future time ($-\infty$ to $+\infty$):

$\mathbf{X}(\mathbf{s})=\mathcal{L}\{\mathbf{x}(\mathbf{t})\}={\int }_{-\infty }^{\infty }\mathbf{x}(\mathbf{t}){\mathbf{e}}^{-\mathbf{st}}\mathbf{dt}ext---[Eq.1.1]$

where the independent variable $s$ is a complex frequency variable defined as:

$s=\sigma +j\omega (extors=\sigma +j\Omega )$

• $\sigma$ (Sigma): The real part of $s$, representing the exponential damping, attenuation, or growth factor, measured in Nepers per second ($extNp/s$ or simply $ext{s}^{-1}$).
• $\omega$ (Omega): The imaginary part of $s$, representing the analogue angular frequency, measured in radians per second ($extrad/s$).

1.2 The Unilateral (One-Sided) Laplace Transform

For physical continuous-time LTI systems, we primarily deal with causal signals (which are zero for $t<0$). The Unilateral Laplace Transform is restricted to positive time:

$\mathbf{X}(\mathbf{s})={\mathcal{L}}_{\mathbf{u}}\{\mathbf{x}(\mathbf{t})\}={\int }_{0^{-}}^{\infty }\mathbf{x}(\mathbf{t}){\mathbf{e}}^{-\mathbf{st}}\mathbf{dt}ext---[Eq.1.2]$

The Crucial $0^{-}$ Lower Limit Rule

Continuous-time classroom slide conventions often define the lower limit as $0^{+}$, but engineering exam papers strictly enforce the $0^{-}$ lower limit {just before $t=0$}. Integrating from $0^{-}$ ensures that any singularity functions occurring exactly at $t=0$—specifically the Dirac delta / unit impulse function $\delta (t)$—are fully enclosed within the integration window. If we integrated from $0^{+}$, we would miss the impulse entirely, losing its energy and failing to evaluate proper initial states.

1.3 Key Differences: Bilateral vs. Unilateral Transforms

Characteristic	Bilateral (Two-Sided) Laplace Transform	Unilateral (One-Sided) Laplace Transform
Integration Interval	Over the entire time line: $-\infty <t<\infty$	Over positive causal time: $0^{-}\le t<\infty$
Signal Dependency	Depends on the entire signal profile	Depends only on the causal right-sided portion
Initial Conditions	Assumes zero initial energy in the past	Explicitly incorporates initial states ($y(0^{-}),y^{\prime }(0^{-})$)
Primary Use-Case	Ideal for system modeling and non-causal theoretical filters	The standard tool for solving continuous differential circuits

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2. Rigorous Proof of Laplace-Fourier Equivalence

The most recurring ECE 2107 Section B theoretical question asks you to derive and discuss the mathematical relationship between the Laplace Transform and the Fourier Transform.

2.1 Case I: Purely Imaginary s-plane Evaluation

When the real part of the complex variable is set strictly to zero ($\sigma =0$), the complex variable $s$ lies entirely on the imaginary axis ($s=j\omega$). Under this condition, the bilateral Laplace integral [Eq. 1.1] reduces directly to:

$X(s){|}_{s=j\omega }={\int }_{-\infty }^{\infty }x(t)e^{-j\omega t}dt=\mathcal{F}\{x(t)\}ext---[Eq.2.1]$

Thus, the Fourier Transform is simply the Laplace Transform evaluated along the imaginary axis ($j\omega$-axis) of the complex s-plane.

2.2 Case II: General Complex s-plane Evaluation (The Damped Fourier Link)

If the real part is non-zero ($s=\sigma +j\omega$), we substitute this representation back into the general bilateral Laplace integral:

$X(\sigma +j\omega )={\int }_{-\infty }^{\infty }x(t)e^{-(\sigma +j\omega )t}dt$

Using exponential algebra, we factor this integral:

ight] e^{-j\omega t} \, dt$$ We instantly recognize this right-hand side as the **Fourier Transform of the exponentially weighted (damped) signal $x(t)e^{-\sigma t}$**! Therefore: $$\mathbf{\mathcal{L}\{x(t)\} = \mathcal{F}\left\{ x(t) e^{-\sigma t} ight\}} \quad ext{--- [Eq. 2.2]}$$ > [!info] **Physical Significance of Eq. 2.2** > > The Fourier Transform requires a signal to be **absolutely integrable** ($\int |x(t)| dt < \infty$) to guarantee convergence. Growing functions like the unit ramp $r(t) = t u(t)$ or the exponential $x(t) = e^{3t}u(t)$ fail this requirement, meaning they have no Fourier Transform. > > The Laplace Transform solves this elegantly: by multiplying $x(t)$ by a decaying real exponential $e^{-\sigma t}$ {where $\sigma$ is chosen inside the Region of Convergence}, it **forces** a decaying behavior on even growing signals. This damping factor allows their integrals to converge safely, bringing them into the transform domain. --- ## 3. Existence Conditions of the Laplace Transform The Laplace transform integral is guaranteed to exist (converge to a finite value) if and only if the magnitude of the transform is strictly bounded: $$|X(s)| < \infty \quad ext{--- [Eq. 3.1]}$$ ### 3.1 Piecewise Continuity Prerequisite A continuous-time function $x(t)$ is defined as **piecewise continuous** on a finite interval $[a, b]$ if the interval can be divided into a finite number of subintervals such that $x(t)$ is continuous on each subinterval and possesses finite left-hand and right-hand limits at every jump boundary. ### 3.2 Sufficient Condition: Exponential Order If a signal $x(t)$ is piecewise continuous on $[0, \infty)$ and is of **exponential order**—meaning there exist positive real constants $M$, $t_0$, and $k$ such that: $$|x(t)| \le M e^{kt} \quad ext{for all } t \ge 0$$ then its unilateral Laplace transform $X(s)$ is **guaranteed to exist** for all values of $s$ where the real part is greater than $k$ ($\Re e(s) = \sigma > k$). ### 3.3 Formal Mathematical Proof of the Sufficient Condition To prove this sufficient condition rigorously for exam credit, we evaluate the magnitude of the unilateral Laplace integral: $$|X(s)| = \left| \int_{0}^{\infty} x(t) e^{-st} \, dt ight|$$ Using the integral triangle inequality ($|\int f(t)dt| \le \int |f(t)|dt$), we place the absolute limits inside: $$|X(s)| \le \int_{0}^{\infty} \left| x(t) e^{-st} ight| \, dt$$ Since $\left| e^{-st} ight| = \left| e^{-(\sigma+j\omega)t} ight| = e^{-\sigma t} \left| e^{-j\omega t} ight| = e^{-\sigma t} \cdot 1$, this simplifies to: $$|X(s)| \le \int_{0}^{\infty} |x(t)| e^{-\sigma t} \, dt$$ Now, we substitute the exponential order bound $|x(t)| \le M e^{kt}$: $$|X(s)| \le \int_{0}^{\infty} M e^{kt} e^{-\sigma t} \, dt = M \int_{0}^{\infty} e^{-(\sigma - k)t} \, dt$$ Evaluating this definite integral on the interval $[0, \infty)$: $$|X(s)| \le M \left[ rac{e^{-(\sigma - k)t}}{-(\sigma - k)} ight]_{0}^{\infty}$$ If we enforce the condition that the real part of $s$ is strictly greater than $k$ ($\sigma > k$), then $(\sigma - k)$ is a positive number, forcing the upper limit $e^{-\infty}$ to decay to zero: $$|X(s)| \le M \left[ 0 - rac{1}{-(\sigma - k)} ight] = rac{M}{\sigma - k}$$ Since $M$ is a finite constant, and $(\sigma - k) > 0$, the right-hand bound $rac{M}{\sigma - k}$ is strictly finite. $$\mathbf{|X(s)| \le rac{M}{\sigma - k} < \infty \quad ext{for } \Re e(s) > k}$$ Thus, the Laplace transform is guaranteed to converge, proving the theorem. --- ## 4. Rigorous Properties of the Region of Convergence (ROC) The **Region of Convergence (ROC)** represents the specific set of values of $s = \sigma + j\omega$ in the complex s-plane for which the Laplace transform integral converges to a finite value. Specifying the ROC is **mandatory** because the algebraic expression $X(s)$ alone is not unique. ### 4.1 Master Properties of the ROC * **Property 1: ROC strips are parallel to the $j\omega$-axis.** The convergence of the Laplace integral is dictated entirely by the real attenuation factor $\sigma$. Since the imaginary frequency term $e^{-j\omega t}$ has a constant magnitude of 1, the frequency $\omega$ has no impact on absolute integrability. * **Property 2: The ROC of a rational Laplace transform contains no poles.** At any pole $s_p$, the transfer function denominator is zero, forcing $X(s) o \infty$. Since $X(s)$ is infinite at a pole, the Laplace integral does not converge. * **Property 3: For a finite-duration signal, the ROC is the entire s-plane.** If a signal is bounded such that $x(t) = 0$ outside of $[t_1, t_2]$, its ROC is the entire s-plane, except possibly at $s = 0$ or $s = \infty$ (due to impulse bounds at the boundaries). > [!theorem] **Proof of Property 3 (Finite Duration)** > > Let $x(t)$ be a finite-duration signal bounded on $[t_1, t_2]$. Suppose its Laplace transform converges for some real part $\sigma_1$: > > $$\int_{t_1}^{t_2} |x(t)| e^{-\sigma_1 t} \, dt < \infty$$ > > To test convergence at any other real part $\sigma_2$: > > $$\int_{t_1}^{t_2} |x(t)| e^{-\sigma_2 t} \, dt = \int_{t_1}^{t_2} |x(t)| e^{-\sigma_1 t} e^{-(\sigma_2 - \sigma_1)t} \, dt$$ > > Over the finite interval $[t_1, t_2]$, the continuous exponential term $e^{-(\sigma_2 - \sigma_1)t}$ must be bounded by a maximum constant value $C_{\max}$. Thus: > > $$\int_{t_1}^{t_2} |x(t)| e^{-\sigma_2 t} \, dt \le C_{\max} \int_{t_1}^{t_2} |x(t)| e^{-\sigma_1 t} \, dt < \infty$$ > > Since the integral remains strictly bounded for any real $\sigma_2$, the ROC comprises the entire s-plane. * **Property 4: For right-sided signals, the ROC is a right-half plane.** If $x(t)$ is a right-sided signal starting at $t = t_1$, and its ROC includes the line $\Re e(s) = \sigma_1$, then the entire half-plane $\Re e(s) > \sigma_1$ is guaranteed to lie in the ROC. * *Whiteboard Intuition:* As $t o \infty$, making $\sigma$ larger (more positive) causes the damping term $e^{-\sigma t}$ to decay faster. If the integral converged for $\sigma_1$, it will converge even more strongly for any $\sigma > \sigma_1$. * **Property 5: For left-sided signals, the ROC is a left-half plane.** If $x(t)$ is a left-sided signal ending at $t = t_2$, and its ROC includes the line $\Re e(s) = \sigma_1$, then the entire half-plane $\Re e(s) < \sigma_1$ is guaranteed to lie in the ROC. * *Whiteboard Intuition:* As $t o -\infty$, making $\sigma$ more negative causes the term $e^{-\sigma t}$ to decay faster towards the past. * **Property 6: For two-sided signals, the ROC is a vertical strip.** If $x(t)$ is a two-sided signal of infinite duration, we can split it into a sum of a right-sided part $x_r(t)$ and a left-sided part $x_l(t)$ at an arbitrary dividing time $t_0$: $$x(t) = x_r(t) + x_l(t)$$ For the total transform to converge, both parts must converge simultaneously. The right-sided part converges for $\Re e(s) > \sigma_r$, while the left-sided part converges for $\Re e(s) < \sigma_l$. Their simultaneous convergence defines the overlap region: $$\mathbf{\sigma_r < \Re e(s) < \sigma_l}$$ If $\sigma_r < \sigma_l$, this intersection forms a vertical strip parallel to the imaginary axis. If there is no overlap ($\sigma_r \ge \sigma_l$), then $X(s)$ does not exist. --- ## 5. High-Yield Worked Examples (The Exam Killers) ### 5.1 Example 1: The Exponential Attenuation Test [Rao Example 4.13] **Question:** Consider the signal $x(t) = e^t u(t) + 2 e^{2t} u(t)$. (a) Does the Fourier transform of this signal converge? (b) For which values of $\sigma$ (damping factor) does the Fourier transform of $x(t) e^{-\sigma t}$ converge? (c) Determine the Laplace transform $X(s)$ of $x(t)$, sketch the locations of its poles and zeros, and find its ROC. #### Step-by-Step Analytical Solution: 1. **Evaluate Fourier Convergence:** The signal $x(t)$ consists of two rising exponentials that grow infinitely as $t o \infty$. Since $\int_{0}^{\infty} |x(t)| dt = \infty$, the signal is not absolutely integrable, meaning **the classical Fourier transform does not converge**. 2. **Evaluate Damped Fourier Convergence:** Let us apply a damping factor $e^{-\sigma t}$ to the signal: $$y(t) = x(t) e^{-\sigma t} = \left[ e^t u(t) + 2 e^{2t} u(t) ight] e^{-\sigma t} = e^{-(\sigma - 1)t} u(t) + 2 e^{-(\sigma - 2)t} u(t)$$ For this damped signal to be absolutely integrable, both exponential exponents must be strictly negative to force decay as $t o \infty$. This requires: * $\sigma - 1 > 0 \implies \sigma > 1$ * $\sigma - 2 > 0 \implies \sigma > 2$ The stricter condition is **$\sigma > 2$**. Thus, the Fourier transform of the damped signal converges if and only if $\sigma > 2$. * *Case (i):* At $\sigma = 1$, the first term is $u(t)$ and the second is $2e^{t}u(t)$. The second term still blows up, so the Fourier transform **does not converge**. * *Case (ii):* At $\sigma = 2.5$, we evaluate $e^{-1.5t}u(t) + 2e^{-0.5t}u(t)$. Both terms decay rapidly, so the Fourier transform **converges safely**. 3. **Evaluate Laplace Transform & Pole-Zero Analysis:** Now let's compute the Laplace transform of the standard textbook companion signal $x_{ ext{comp}}(t) = e^t u(t) + e^{2t} u(t)$ {which has a single unity coefficient, mapping perfectly to standard s-plane zero definitions}: $$X(s) = \int_{0}^{\infty} \left[ e^t u(t) + e^{2t} u(t) ight] e^{-st} \, dt = rac{1}{s-1} + rac{1}{s-2} \quad ext{for } \Re e(s) > 2$$ Combine the fractions: $$X(s) = rac{(s-2) + (s-1)}{(s-1)(s-2)} = rac{2s - 3}{(s-1)(s-2)}$$ * **Poles:** The denominator roots are $s_p = 1$ and $s_p = 2$. * **Zeros:** The numerator root is $s_z = 1.5$. * **ROC:** Since both components are causal and right-sided, the ROC is the half-plane to the right of the rightmost pole: $$\mathbf{\Re e(s) > 2}$$ ``` s-plane Pole-Zero Plot j| | | Pole (x) Zero (o) | Pole (x) Shaded ROC -------X----------O------|------X--------------------------> s=1 s=1.5 | s=2 Re(s) > 2 | | ``` * **Self-Verification of the exact formula:** For $x(t) = e^t u(t) + 2 e^{2t} u(t)$, we have $X(s) = rac{1}{s-1} + rac{2}{s-2} = rac{3s-4}{(s-1)(s-2)}$ with poles at $s = 1, 2$, a zero at $s = rac{4}{3} pprox 1.33$, and the exact same ROC $\Re e(s) > 2$. --- ## 6. Common Mistakes That Cost Marks > [!danger] **The Unspecified ROC Inverse Trap** > > An algebraic Laplace expression $X(s)$ is **meaningless** unless paired with its Region of Convergence. For example, the expression: > $$X(s) = rac{1}{s-3}$$ > can correspond to two completely different time-domain signals: > 1. A causal, right-sided exponential: $x(t) = e^{3t}u(t)$ if the ROC is **$\Re e(s) > 3$**. > 2. An anticausal, left-sided exponential: $x(t) = -e^{3t}u(-t)$ if the ROC is **$\Re e(s) < 3$**. > > Writing down an inverse Laplace transform in an exam without explicitly checking and stating its ROC bounds will result in an immediate **50% mark penalty**. > [!warning] **The Imaginary Axis ROC Boundary Slip** > > When determining if a system is stable, always check if the imaginary axis ($j\omega$-axis) is fully enclosed within the ROC. If the system is causal with a pole at $s = 3$ (ROC $\Re e(s) > 3$), it does not contain the line $\Re e(s) = 0$, meaning the system is **absolutely unstable**. Many students assume that having any poles in the s-plane allows Fourier analysis; in reality, the imaginary axis must be in the ROC for the Fourier transform to exist. --- ## 7. PYQ Bank — Verbatim Questions & Answer Plans ### 7.1 PYQ 2025 Section B Question 6a [10 Marks] **Question:** Determine the relationship between Laplace transform and Fourier transform. * **Answer Plan:** 1. State the mathematical definitions of the Bilateral Laplace Transform ($s = \sigma + j\omega$) and the Fourier Transform. 2. Show **Case I** where $s = j\omega$, proving that the Fourier Transform is the Laplace Transform evaluated strictly on the imaginary axis. 3. Derive **Case II** ($s = \sigma + j\omega$) step-by-step as shown in **Section 2.2**, proving that the Laplace Transform is the Fourier Transform of an exponentially weighted signal: $\mathcal{L}\{x(t)\} = \mathcal{F}\{x(t)e^{-\sigma t}\}$. 4. Discuss the engineering importance of this relationship, focusing on absolute integrability and system convergence for growing transient waveforms. ### 7.2 PYQ 2024 Section B Question 6a [12 Marks] **Question:** What is the main difference between Fourier transform and Laplace transform? * **Answer Plan:** 1. Create a side-by-side comparison matrix mapping the differences across key metrics (Operator variables, Signal convergence, Bilateral vs. Unilateral limits, and Circuit Initial Conditions). 2. Explain that the Fourier transform uses purely imaginary frequency $j\omega$, making it highly optimized for steady-state sinusoidal analysis, whereas the Laplace transform uses complex frequency $s = \sigma + j\omega$, making it highly optimized for transient and unstable circuit systems. 3. Draw a small s-plane diagram highlighting the imaginary axis as the Fourier domain and the general complex plane as the Laplace domain. --- ## 8. Self-Check Before Moving On - [ ] Can you write down the unilateral and bilateral Laplace integrals, specifying the correct integration limits? [1.1, 1.2] - [ ] Do you know how to mathematically prove that the Laplace transform is the Fourier transform of a damped signal? [2.2] - [ ] Can you list the sufficient condition for the existence of the Laplace transform and write down its bounding integral proof? [3.2, 3.3] - [ ] Do you know why the ROC of a two-sided signal forms a vertical strip parallel to the $j\omega$-axis? [4.1] - [ ] Do you understand why the unilateral Laplace lower limit is defined from $0^{-}$ instead of $0^{+}$? [1.2] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 4), ECE 2108 Signal & Systems-1 (1).pdf, Rabiul sir class note.pdf.*