9.01 Foundation of Laplace Transform & s-Plane Representation | 9.03 Mathematical Properties of the Laplace Transform
9.02 Laplace Transforms of Singularity & Elementary Functions
Core Idea
To analyze continuous-time LTI systems in the complex s-domain, we must construct a dictionary of Laplace Transform Pairs for fundamental signal building blocks. By calculating the unilateral Laplace integral for singularity functions {idealized mathematical abstractions containing infinite discontinuities or impulses} and elementary causal functions, we establish a robust algebraic framework. This note mathematically derives each standard transform pair and presents a complete, rigorous proof for the power function ()—a major exam question.
1. Singularity & Linear Growth Functions
Singularity functions are idealized mathematical models used to construct complex waveforms or analyze system responses. Because unilateral Laplace analysis models systems from , these causal functions are implicitly multiplied by the unit step .
1.1 The Dirac Delta (Unit Impulse),
The unit impulse function represents an infinitely narrow, infinitely tall spike at with a total area (weight) of unity.
delta(t) [Area = 1]
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-------o------- t
0
Step-by-Step Derivation:
We evaluate the unilateral integral using the sampling property of the impulse function {which isolates the integrand value at the exact instant the impulse occurs}:
Because the impulse occurs at , we evaluate the exponential at :
- Final Transform:
- Region of Convergence (ROC): The entire s-plane, as the integral is non-zero only at and never diverges.
1.2 The Unit Step Function,
The unit step function represents an instantaneous switch closing at , jumping from to a constant DC value of .
1 +------------ u(t)
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-------o------------ t
0
Step-by-Step Derivation:
Evaluate the definite integral:
ight]_{0}^{\infty} = \lim_{t o \infty} \left( - rac{1}{s} e^{-st} ight) - \left( - rac{1}{s} e^{-s(0)} ight)$$ For the upper limit to converge to $0$, we must require that the real part of $s$ is positive ($\sigma > 0$), making the exponential decay: $$\mathcal{L}\{u(t)\} = 0 - \left( - rac{1}{s} ight) = \mathbf{ rac{1}{s}}$$ * **Final Transform:** $$\mathbf{\mathcal{L}\{u(t)\} = rac{1}{s}}$$ * **Region of Convergence (ROC):** $$\mathbf{\Re e(s) > 0} \quad ext{*{the right-half of the s-plane}*}$$ --- ### 1.3 The Ramp Function, $r(t)$ The **ramp function** grows linearly with a slope of $1$ for $t \ge 0$. It is the running integral of the unit step function. ``` / r(t) = t*u(t) / / ----o--------------- t 0 ``` #### Step-by-Step Derivation: $$r(t) = t \cdot u(t)$$ $$\mathcal{L}\{t \cdot u(t)\} = \int_{0}^{\infty} t e^{-st} \, dt$$ We solve this using **Integration by Parts** ($\int u \, dv = uv - \int v \, du$): * Let $u = t \implies du = dt$ * Let $dv = e^{-st} dt \implies v = - rac{1}{s} e^{-st}$ $$\mathcal{L}\{t \cdot u(t)\} = \left[ t \left(- rac{1}{s} e^{-st} ight) ight]_{0}^{\infty} - \int_{0}^{\infty} \left(- rac{1}{s} e^{-st} ight) dt$$ Evaluate the boundary term: * At $t o \infty$: $\lim_{t o \infty} - rac{t}{s} e^{-st} = 0 \quad$ (provided $\Re e(s) > 0$, as exponential decay dominates linear growth). * At $t = 0$: $0 \cdot \left(- rac{1}{s} ight) = 0$. Thus, the boundary term vanishes entirely. Now, integrate the second term: $$\mathcal{L}\{t \cdot u(t)\} = 0 + rac{1}{s} \int_{0}^{\infty} e^{-st} \, dt = rac{1}{s} \left( rac{1}{s} ight) = \mathbf{ rac{1}{s^2}}$$ * **Final Transform:** $$\mathbf{\mathcal{L}\{t \cdot u(t)\} = rac{1}{s^2}}$$ * **Region of Convergence (ROC):** $$\mathbf{\Re e(s) > 0}$$ --- ## 2. Decaying & Growing Real Exponentials Real exponential signals describe natural circuit decays (such as discharging capacitors) and growing thermal/runaway processes. ### 2.1 Causal Exponential Decay, $e^{-at}u(t)$ #### Step-by-Step Derivation: We evaluate the unilateral integral for a positive real decay constant $a > 0$: $$\mathcal{L}\{e^{-at} u(t)\} = \int_{0}^{\infty} e^{-at} e^{-st} \, dt = \int_{0}^{\infty} e^{-(s+a)t} \, dt$$ $$\mathcal{L}\{e^{-at} u(t)\} = \left[ - rac{1}{s+a} e^{-(s+a)t} ight]_{0}^{\infty}$$ Evaluate the boundary limits: * At the upper limit ($t o \infty$): The term converges to $0$ if and only if the real part of the exponent coefficient is positive, meaning $\Re e(s+a) > 0 \implies \Re e(s) > -a$. * At the lower limit ($t = 0$): The term evaluates to $- rac{1}{s+a}$. $$\mathcal{L}\{e^{-at} u(t)\} = 0 - \left( - rac{1}{s+a} ight) = \mathbf{ rac{1}{s+a}}$$ * **Final Transform:** $$\mathbf{\mathcal{L}\{e^{-at} u(t)\} = rac{1}{s+a}}$$ * **Region of Convergence (ROC):** $$\mathbf{\Re e(s) > -a}$$ --- ### 2.2 Causal Exponential Growth, $e^{at}u(t)$ For an exponentially growing signal with $a > 0$: $$\mathcal{L}\{e^{at} u(t)\} = \int_{0}^{\infty} e^{at} e^{-st} \, dt = \int_{0}^{\infty} e^{-(s-a)t} \, dt$$ Following identical integration steps, the upper limit converges to $0$ only if $\Re e(s - a) > 0 \implies \Re e(s) > a$: $$\mathcal{L}\{e^{at} u(t)\} = \mathbf{ rac{1}{s-a}}$$ * **Final Transform:** $$\mathbf{\mathcal{L}\{e^{at} u(t)\} = rac{1}{s-a}}$$ * **Region of Convergence (ROC):** $$\mathbf{\Re e(s) > a}$$ --- ## 3. Sinusoidal & Hyperbolic Formulations Continuous-time oscillations and bilateral structural decays are analyzed using trigonometric and hyperbolic functions. ### 3.1 Causal Cosine Wave, $\cos(\omega t)u(t)$ #### Step-by-Step Euler Derivation: We express the real cosine wave in terms of complex exponentials using **Euler's identity** *{which decomposes a real oscillation into two counter-rotating complex frequency phasors}*: $$\cos(\omega t) = rac{e^{j\omega t} + e^{-j\omega t}}{2}$$ Substitute this identity into the unilateral Laplace integral: $$\mathcal{L}\{\cos(\omega t) u(t)\} = \mathcal{L}\left\{ rac{e^{j\omega t} + e^{-j\omega t}}{2} u(t) ight\}$$ Apply the **Linearity Property** to separate the transforms: $$\mathcal{L}\{\cos(\omega t) u(t)\} = rac{1}{2} \mathcal{L}\{e^{j\omega t}u(t)\} + rac{1}{2} \mathcal{L}\{e^{-j\omega t}u(t)\}$$ Using our standard real exponential transform with complex constants ($a = \mp j\omega$): $$\mathcal{L}\{\cos(\omega t) u(t)\} = rac{1}{2} \left( rac{1}{s - j\omega} + rac{1}{s + j\omega} ight)$$ Find a common denominator to combine the fractions: $$\mathcal{L}\{\cos(\omega t) u(t)\} = rac{1}{2} \left( rac{(s + j\omega) + (s - j\omega)}{(s - j\omega)(s + j\omega)} ight)$$ $$\mathcal{L}\{\cos(\omega t) u(t)\} = rac{1}{2} \left( rac{2s}{s^2 - (j\omega)^2} ight) = \mathbf{ rac{s}{s^2 + \omega^2}}$$ * **Final Transform:** $$\mathbf{\mathcal{L}\{\cos(\omega t) u(t)\} = rac{s}{s^2 + \omega^2}}$$ * **Region of Convergence (ROC):** $$\mathbf{\Re e(s) > 0} \quad ext{(since $\Re e(s \mp j\omega) > 0 \implies \Re e(s) > 0$)}$$ --- ### 3.2 Causal Sine Wave, $\sin(\omega t)u(t)$ #### Step-by-Step Euler Derivation: Using Euler's identity for the sine wave: $$\sin(\omega t) = rac{e^{j\omega t} - e^{-j\omega t}}{2j}$$ $$\mathcal{L}\{\sin(\omega t) u(t)\} = rac{1}{2j} \left( \mathcal{L}\{e^{j\omega t}u(t)\} - \mathcal{L}\{e^{-j\omega t}u(t)\} ight)$$ $$\mathcal{L}\{\sin(\omega t) u(t)\} = rac{1}{2j} \left( rac{1}{s - j\omega} - rac{1}{s + j\omega} ight)$$ $$\mathcal{L}\{\sin(\omega t) u(t)\} = rac{1}{2j} \left( rac{(s + j\omega) - (s - j\omega)}{s^2 + \omega^2} ight) = rac{1}{2j} \left( rac{2j\omega}{s^2 + \omega^2} ight) = \mathbf{ rac{\omega}{s^2 + \omega^2}}$$ * **Final Transform:** $$\mathbf{\mathcal{L}\{\sin(\omega t) u(t)\} = rac{\omega}{s^2 + \omega^2}}$$ * **Region of Convergence (ROC):** $$\mathbf{\Re e(s) > 0}$$ --- ### 3.3 Hyperbolic Sine ($\sinh$) and Cosine ($\cosh$) Hyperbolic functions describe non-oscillatory exponential growth and decay combinations. * **Hyperbolic Sine:** $$\sinh(at) = rac{e^{at} - e^{-at}}{2}$$ $$\mathcal{L}\{\sinh(at) u(t)\} = rac{1}{2} \left( rac{1}{s - a} - rac{1}{s + a} ight) = \mathbf{ rac{a}{s^2 - a^2}}$$ * **ROC:** $\mathbf{\Re e(s) > |a|}$ (intersection of $\Re e(s) > a$ and $\Re e(s) > -a$). * **Hyperbolic Cosine:** $$\cosh(at) = rac{e^{at} + e^{-at}}{2}$$ $$\mathcal{L}\{\cosh(at) u(t)\} = rac{1}{2} \left( rac{1}{s - a} + rac{1}{s + a} ight) = \mathbf{ rac{s}{s^2 - a^2}}$$ * **ROC:** $\mathbf{\Re e(s) > |a|}$ --- ## 4. The 10-Mark Proof: Laplace of the Power Function, $t^n$ > [!theorem] **Rigorous Proof of $\mathcal{L}\{t^n u(t)\} = rac{n!}{s^{n+1}}$ [PYQ 2024, 2022, 2019]** > > **Theorem Statement:** Prove mathematically that the unilateral Laplace transform of $x(t) = t^n u(t)$ (where $n$ is a positive integer) is given by $ rac{n!}{s^{n+1}}$ with a Region of Convergence $\Re e(s) > 0$. ### Proof by Mathematical Induction: #### Step 0: Setup and Core Definition We define the continuous unilateral integral for the power function: $$I_n(s) = \int_{0}^{\infty} t^n e^{-st} \, dt \quad ext{for } \Re e(s) > 0$$ #### Step 1: Prove the Base Case ($n = 0$) Let $n = 0$. The function is $t^0 u(t) = u(t)$ (the unit step function). $$I_0(s) = \int_{0}^{\infty} t^0 e^{-st} \, dt = \int_{0}^{\infty} e^{-st} \, dt$$ $$I_0(s) = \left[ - rac{1}{s} e^{-st} ight]_{0}^{\infty}$$ Since $\Re e(s) > 0$, the upper limit is $0$: $$I_0(s) = 0 - \left( - rac{1}{s} ight) = rac{1}{s} = rac{0!}{s^{0+1}} \quad ext{(Base case is true!)}$$ #### Step 2: Establish the Recurrence Relation using Integration by Parts We evaluate the integral $I_n(s) = \int_{0}^{\infty} t^n e^{-st} \, dt$ using **Integration by Parts**: $$\int u \, dv = uv - \int v \, du$$ * Let $u = t^n \implies du = n t^{n-1} dt$ * Let $dv = e^{-st} dt \implies v = - rac{1}{s} e^{-st}$ Substitute these parts into the integration formula: $$I_n(s) = \left[ t^n \left( - rac{1}{s} e^{-st} ight) ight]_{0}^{\infty} - \int_{0}^{\infty} \left( - rac{1}{s} e^{-st} ight) \left( n t^{n-1} \, dt ight)$$ $$I_n(s) = \left[ - rac{t^n}{s} e^{-st} ight]_{0}^{\infty} + rac{n}{s} \int_{0}^{\infty} t^{n-1} e^{-st} \, dt$$ #### Step 3: Evaluate the Boundary Term Limits * **Upper Limit ($t o \infty$):** $$\lim_{t o \infty} - rac{t^n}{s} e^{-st} = 0 \quad ext{for } \Re e(s) > 0$$ *{Exponential decay of $e^{-st}$ always dominates the polynomial growth of $t^n$ as $t o \infty$}*. * **Lower Limit ($t = 0$):** Since $n \ge 1$: $$- rac{0^n}{s} e^{-s(0)} = 0$$ Because both boundary evaluations are zero, the entire left term vanishes. We are left with: $$I_n(s) = 0 + rac{n}{s} \int_{0}^{\infty} t^{n-1} e^{-st} \, dt$$ Identify that the remaining integral is exactly the definition of $I_{n-1}(s)$: $$\mathbf{I_n(s) = rac{n}{s} I_{n-1}(s)} \quad ext{--- (Recurrence Relation)}$$ #### Step 4: Apply Induction (Repeated Recurrence Substitution) By substituting the recurrence relation sequentially for each lower index: $$I_n(s) = rac{n}{s} \left( rac{n-1}{s} I_{n-2}(s) ight)$$ $$I_n(s) = rac{n(n-1)}{s^2} I_{n-2}(s)$$ $$I_n(s) = rac{n(n-1)(n-2)\dots(2)(1)}{s^n} I_0(s)$$ Since $n(n-1)(n-2)\dots(1) = n!$: $$I_n(s) = rac{n!}{s^n} I_0(s)$$ #### Step 5: Substitute the Base Case Value Substitute our proven base case value $I_0(s) = rac{1}{s}$ from Step 1: $$I_n(s) = rac{n!}{s^n} \left( rac{1}{s} ight) = \mathbf{ rac{n!}{s^{n+1}}} \quad lacksquare$$ This completes the mathematical proof. The Region of Convergence is strictly restricted to **$\Re e(s) > 0$** because the upper boundary limit $\lim_{t o \infty} t^n e^{-st}$ diverges to infinity if $\Re e(s) \le 0$. --- ## 5. Master Unilateral Laplace Transform Pair Table This table acts as your ultimate, zero-error formula lookup matrix during exams. All signals are causal, meaning $x(t) = 0$ for $t < 0$: | Signal Waveform, $x(t)$ | Laplace Domain Expression, $X(s)$ | Region of Convergence (ROC) | Physical/Circuit Significance | | :--- | :--- | :--- | :--- | | **$\delta(t)$** | $$1$$ | **Entire s-plane** | Flat, infinite-bandwidth impulse excitation. | | **$u(t)$** | $$ rac{1}{s}$$ | **$\Re e(s) > 0$** | DC voltage step source (switch closure). | | **$t \cdot u(t)$** | $$ rac{1}{s^2}$$ | **$\Re e(s) > 0$** | Linear ramp sweep excitation. | | **$t^n \cdot u(t)$** | $$ rac{n!}{s^{n+1}}$$ | **$\Re e(s) > 0$** | High-order transient tracking signals. | | **$e^{-at} u(t)$** | $$ rac{1}{s+a}$$ | **$\Re e(s) > -a$** | First-order source-free RC/RL decay. | | **$e^{at} u(t)$** | $$ rac{1}{s-a}$$ | **$\Re e(s) > a$** | Unstable, exponentially growing thermal runaways. | | **$t^n e^{-at} u(t)$** | $$ rac{n!}{(s+a)^{n+1}}$$ | **$\Re e(s) > -a$** | Multiple-pole response of cascaded systems. | | **$\sin(\omega t) u(t)$** | $$ rac{\omega}{s^2 + \omega^2}$$ | **$\Re e(s) > 0$** | Continuous AC harmonic voltage/current sources. | | **$\cos(\omega t) u(t)$** | $$ rac{s}{s^2 + \omega^2}$$ | **$\Re e(s) > 0$** | Reference AC carrier signals. | | **$e^{-at} \sin(\omega t) u(t)$** | $$ rac{\omega}{(s+a)^2 + \omega^2}$$ | **$\Re e(s) > -a$** | Damped sinusoidal transient (underdamped RLC). | | **$e^{-at} \cos(\omega t) u(t)$** | $$ rac{s+a}{(s+a)^2 + \omega^2}$$ | **$\Re e(s) > -a$** | Reference damped transient oscillation. | | **$\sinh(at) u(t)$** | $$ rac{a}{s^2 - a^2}$$ | **$\Re e(s) > |a|$** | Hyperbolic exponential divergence. | | **$\cosh(at) u(t)$** | $$ rac{s}{s^2 - a^2}$$ | **$\Re e(s) > |a|$** | Symmetrical hyperbolic growth. | --- ## 6. Common Mistakes That Cost Marks > [!danger] **The Unspecified ROC Penalty** > > A Laplace transform is **never** mathematically unique without stating its Region of Convergence. For example, $X(s) = rac{1}{s+a}$ can correspond to the causal right-sided signal $e^{-at}u(t)$ (for $\Re e(s) > -a$) OR the anti-causal left-sided signal $-e^{-at}u(-t)$ (for $\Re e(s) < -a$). Failing to write down the ROC next to your result will cause a **2 to 3-mark penalty**. > [!warning] **The Sine vs. Hyperbolic Sign Confusion** > > A very common algebraic slip under exam pressure is mixing up the denominator signs of trigonometric sines/cosines and hyperbolic sines/cosines: > * Trigonometric: $\mathcal{L}\{\sin(\omega t)\} = rac{\omega}{s^2 \mathbf{+} \omega^2} \quad$ *{contains a plus sign}* > * Hyperbolic: $\mathcal{L}\{\sinh(at)\} = rac{a}{s^2 \mathbf{-} a^2} \quad$ *{contains a minus sign}* > Redoing a long transient derivation with a reversed sign yields incorrect poles, leading to a **zero-mark** evaluation on the circuit solving section. --- ## 7. PYQ Bank — Verbatim Questions & Answer Plans ### 7.1 PYQ 2024/2022/2019 Section B [12/10/08 Marks] **Question:** What is the main difference between Fourier transform and Laplace transform? Find the Laplace transform of $t^n$ function. * **Answer Plan:** 1. **Main Difference:** State that the Fourier transform only converges for absolutely integrable signals ($\int |x(t)| dt < \infty$). The Laplace transform generalizes this by multiplying the signal by an exponential attenuation damping factor $e^{-\sigma t}$, forcing convergence for growing/unstable waveforms (like ramps or growing exponentials) where the Fourier transform fails. Show the mapping $s = \sigma + j\omega$ to relate the two domains mathematically. 2. **Derivation:** Reconstruct the complete **Mathematical Induction Proof** for $t^n$ step-by-step as structured in **Section 4**: * Establish base case $I_0(s) = rac{1}{s}$. * Write KVL-style Integration by Parts. * Prove why the boundary term vanishes for $\Re e(s) > 0$. * Derive the recurrence relation $I_n(s) = rac{n}{s}I_{n-1}(s)$. * Sequence the recurrence to prove $X(s) = rac{n!}{s^{n+1}}$ with ROC $\Re e(s) > 0$. ### 7.2 Foundational Concept PYQ 2017/2015 [6 Marks] **Question:** State and explain Laplace transform and its inverse transform. * **Answer Plan:** 1. **Forward Transform:** Define the Unilateral Laplace integral mapping causal signals to the complex s-plane: $$X(s) = \int_{0^-}^{\infty} x(t) e^{-st} \, dt \quad ext{where } s = \sigma + j\Omega$$ 2. **Inverse Transform:** Define the complex Bromwich contour integral reconstructing the time signal: $$x(t) = rac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} X(s) e^{st} \, ds$$ 3. **Physical Significance:** Discuss how this mapping transforms time-domain differential equations into algebraic equations, allowing easy loop and nodal analysis in circuits. --- ## 8. Self-Check Before Moving On - [ ] Can you rigorously prove the Laplace transform of $t^n$ using integration by parts? [4.0] - [ ] Do you know the exact difference between the denominators of trigonometric sines and hyperbolic sines? [6.0] - [ ] Can you explain why the unilateral Laplace transform limits start at $0^-$ instead of $0$? [1.1] - [ ] Have you memorized the Region of Convergence for a causal decaying exponential $e^{-at}u(t)$? [2.1] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 4), 03 Laplace.pdf, Rabiul sir class note.pdf.*