9.00 Chapter Map - Laplace Transform & s-Domain Analysis | 9.02 Laplace Transforms of Singularity & Elementary Functions | 9.04 Initial and Final Value Theorems: Statements, Proofs, and Bounds
9.03 Mathematical Properties of the Laplace Transform
Core Idea
Mathematical properties of the Unilateral Laplace Transform map complex continuous-time operations {such as differentiation, integration, scaling, and shifting} directly to simple algebraic manipulations in the complex -domain. These properties transform integro-differential systems of equations into linear algebraic equations, allowing for direct system-level analysis, transfer function modeling, and transient circuit solving.
1. Linearity Property
The Laplace transform is a linear operator. If two continuous-time signals and have unilateral Laplace transforms and with regions of convergence and , respectively, then any linear combination of these signals yields:
The resulting Region of Convergence (ROC) is at least the intersection of the individual ROCs:
1.1 Mathematical Proof
By applying the definition of the unilateral Laplace integral:
Since integration is a linear operation, we can distribute the integral and factor out the scalar constants: \mathcal{L}\{a_1 x_1(t) + a_2 x_2(t)\} = a_1 X_1(s) + a_2 X_2(s) \quad lacksquare
2. Time Scaling Property
Time scaling compresses or expands a signal in the time domain. If with ROC , then scaling the independent time variable by a real constant scales the complex frequency variable and its amplitude as:
ight)$$ The scaled **Region of Convergence** becomes: $$ ext{ROC} = rac{R}{a} \quad *\{meaning if the boundary was \sigma > \sigma_0, the new boundary is \sigma > a \sigma_0\}*$$ *Note: For the unilateral Laplace transform, we restrict $a > 0$ because scaling by a negative number ($a < 0$) would reflect the signal into the negative time domain ($t < 0$), moving it outside the causal integration limits $[0^-, \infty)$ of the unilateral transform.* ### 2.1 Mathematical Proof By definition of the unilateral Laplace transform: $$\mathcal{L}\{x(at)\} = \int_{0^-}^{\infty} x(at) e^{-st} \, dt$$ Perform a change of variable. Let $ au = at$. Since $a > 0$, the limits of integration remain unchanged: * At $t = 0^- \implies au = 0^-$ * As $t o \infty \implies au o \infty$ * The differential element is $dt = rac{d au}{a}$ Substituting these into the integral yields: $$\mathcal{L}\{x(at)\} = \int_{0^-}^{\infty} x( au) e^{-s\left( rac{ au}{a} ight)} rac{d au}{a}$$ $$\mathcal{L}\{x(at)\} = rac{1}{a} \int_{0^-}^{\infty} x( au) e^{-\left( rac{s}{a} ight) au} \, d au$$ $$\mathcal{L}\{x(at)\} = rac{1}{a} X\left( rac{s}{a} ight) \quad lacksquare$$ --- ## 3. Time Shifting Property Delaying a signal in time shifts its phase in the frequency domain. In unilateral Laplace analysis, shifting a causal signal by a positive delay $t_0 \ge 0$ requires the signal to be zero-padded for $t < t_0$, which is mathematically enforced by multiplying by a shifted unit step $u(t - t_0)$: $$\mathcal{L}\{x(t - t_0)u(t - t_0)\} = e^{-s t_0} X(s)$$ The **Region of Convergence** remains unchanged: $$ ext{ROC} = R$$ ### 3.1 Mathematical Proof Applying the unilateral definition to the shifted causal signal: $$\mathcal{L}\{x(t - t_0)u(t - t_0)\} = \int_{0^-}^{\infty} x(t - t_0)u(t - t_0) e^{-st} \, dt$$ Since $u(t - t_0) = 0$ for $t < t_0$, the lower integration limit collapses from $0^-$ to $t_0$: $$\mathcal{L}\{x(t - t_0)u(t - t_0)\} = \int_{t_0}^{\infty} x(t - t_0) e^{-st} \, dt$$ Perform a change of variable. Let $ au = t - t_0 \implies t = au + t_0$ and $dt = d au$. The new integration limits become: * At $t = t_0 \implies au = 0^-$ * As $t o \infty \implies au o \infty$ Substituting these variables: $$\mathcal{L}\{x(t - t_0)u(t - t_0)\} = \int_{0^-}^{\infty} x( au) e^{-s( au + t_0)} \, d au$$ Factor out the constant term $e^{-s t_0}$ from the integration over $ au$: $$\mathcal{L}\{x(t - t_0)u(t - t_0)\} = e^{-s t_0} \int_{0^-}^{\infty} x( au) e^{-s au} \, d au$$ $$\mathcal{L}\{x(t - t_0)u(t - t_0)\} = e^{-s t_0} X(s) \quad lacksquare$$ --- ## 4. Shifting in the s-Domain (Frequency Shifting / Modulation) Multiplying a time-domain signal by a complex exponential $e^{s_0 t}$ translates its spectrum in the complex $s$-plane: $$\mathcal{L}\{e^{s_0 t} x(t)\} = X(s - s_0)$$ The resulting **Region of Convergence** is shifted along the real axis: $$ ext{ROC} = R + \Re e(s_0)$$ ### 4.1 Mathematical Proof By definition: $$\mathcal{L}\{e^{s_0 t} x(t)\} = \int_{0^-}^{\infty} [e^{s_0 t} x(t)] e^{-st} \, dt$$ Combine the exponents of the exponential terms: $$\mathcal{L}\{e^{s_0 t} x(t)\} = \int_{0^-}^{\infty} x(t) e^{-(s - s_0)t} \, dt$$ By comparing this directly to the definition of the unilateral transform, we see that the variable $s$ has been replaced by $(s - s_0)$: $$\mathcal{L}\{e^{s_0 t} x(t)\} = X(s - s_0) \quad lacksquare$$ --- ## 5. Differentiation in the Time Domain (With Initial Conditions) One of the most powerful properties of the unilateral Laplace transform is its ability to convert calculus derivatives into algebraic multiplication, while **explicitly incorporating initial state conditions**. This forms the basis for solving electrical circuit transient differential equations. If $x(t) \leftrightarrow X(s)$, then the unilateral transform of its first derivative is: $$\mathcal{L}\left\{ rac{dx(t)}{dt} ight\} = s X(s) - x(0^-)$$ Where $x(0^-)$ represents the initial condition of the signal evaluated at the instant just prior to $t=0$ *{capturing pre-stored energy, such as initial capacitor voltage or inductor current}*. ### 5.1 Mathematical Proof By applying the definition of the unilateral Laplace transform: $$\mathcal{L}\left\{ rac{dx(t)}{dt} ight\} = \int_{0^-}^{\infty} \left( rac{dx(t)}{dt} ight) e^{-st} \, dt$$ Solve this integral using **Integration by Parts**: $$\int u \, dv = u v - \int v \, du$$ Let: * $u = e^{-st} \implies du = -s e^{-st} \, dt$ * $dv = rac{dx(t)}{dt} \, dt = dx(t) \implies v = x(t)$ Applying the limits: $$\mathcal{L}\left\{ rac{dx(t)}{dt} ight\} = \left[ x(t) e^{-st} ight]_{0^-}^{\infty} - \int_{0^-}^{\infty} x(t) (-s e^{-st}) \, dt$$ Evaluating the boundary term $\left[ x(t) e^{-st} ight]_{0^-}^{\infty}$: * At the upper limit $t o \infty$: The exponential term $e^{-st}$ decays to zero within the Region of Convergence ($\Re e(s) > \sigma_0$), forcing $\lim_{t o \infty} x(t) e^{-st} = 0$. * At the lower limit $t = 0^-$: The term evaluates to $x(0^-) e^0 = x(0^-)$. $$\left[ x(t) e^{-st} ight]_{0^-}^{\infty} = 0 - x(0^-) = -x(0^-)$$ Substituting this back and simplifying the remaining integral: $$\mathcal{L}\left\{ rac{dx(t)}{dt} ight\} = -x(0^-) + s \int_{0^-}^{\infty} x(t) e^{-st} \, dt$$ $$\mathcal{L}\left\{ rac{dx(t)}{dt} ight\} = s X(s) - x(0^-) \quad lacksquare$$ ### 5.2 Higher-Order Derivative Generalizations Applying this differentiation property sequentially yields algebraic formulations for higher-order derivatives: #### Second Derivative: $$\mathcal{L}\left\{ rac{d^2x(t)}{dt^2} ight\} = s^2 X(s) - s x(0^-) - x'(0^-)$$ #### Third Derivative: $$\mathcal{L}\left\{ rac{d^3x(t)}{dt^3} ight\} = s^3 X(s) - s^2 x(0^-) - s x'(0^-) - x''(0^-)$$ #### $n$-th Order Derivative: $$\mathcal{L}\left\{ rac{d^n x(t)}{dt^n} ight\} = s^n X(s) - \sum_{k=1}^{n} s^{n-k} x^{(k-1)}(0^-)$$ --- ## 6. Integration in the Time Domain Integrating a signal in the time domain corresponds to division by the complex frequency variable $s$ in the $s$-domain: $$\mathcal{L}\left\{\int_{0^-}^{t} x( au) \, d au ight\} = rac{X(s)}{s}$$ The resulting **Region of Convergence** is bounded by: $$ ext{ROC} \supseteq R \cap \{\Re e(s) > 0\}$$ ### 6.1 Mathematical Proof Let the running integral be defined as a new function $y(t)$: $$y(t) = \int_{0^-}^{t} x( au) \, d au$$ By the fundamental theorem of calculus, the derivative of $y(t)$ is: $$ rac{dy(t)}{dt} = x(t)$$ Take the unilateral Laplace transform of both sides of this equation using the time-differentiation property: $$\mathcal{L}\left\{ rac{dy(t)}{dt} ight\} = \mathcal{L}\{x(t)\}$$ $$s Y(s) - y(0^-) = X(s)$$ By evaluating the running integral at the lower limit: $$y(0^-) = \int_{0^-}^{0^-} x( au) \, d au = 0$$ Therefore, the equation simplifies to: $$s Y(s) = X(s) \implies Y(s) = rac{X(s)}{s}$$ $$\mathcal{L}\left\{\int_{0^-}^{t} x( au) \, d au ight\} = rac{X(s)}{s} \quad lacksquare$$ --- ## 7. Differentiation in the s-Domain (Multiplication by $t$) Multiplying a continuous-time signal by the independent time variable $t$ corresponds to differentiating its Laplace transform with respect to the complex frequency variable $s$: $$\mathcal{L}\{t x(t)\} = - rac{dX(s)}{ds}$$ ### 7.1 Mathematical Proof Start with the definition of the unilateral Laplace transform: $$X(s) = \int_{0^-}^{\infty} x(t) e^{-st} \, dt$$ Differentiate both sides of the equation with respect to the complex variable $s$: $$ rac{dX(s)}{ds} = rac{d}{ds} \left[ \int_{0^-}^{\infty} x(t) e^{-st} \, dt ight]$$ Using **Leibniz's Integral Rule**, we can bring the derivative inside the integral as a partial derivative *{valid because the integrand is continuously differentiable with respect to s within its ROC}*: $$ rac{dX(s)}{ds} = \int_{0^-}^{\infty} rac{\partial}{\partial s} \left[ x(t) e^{-st} ight] \, dt$$ $$ rac{dX(s)}{ds} = \int_{0^-}^{\infty} x(t) (-t e^{-st}) \, dt$$ $$ rac{dX(s)}{ds} = -\int_{0^-}^{\infty} [t x(t)] e^{-st} \, dt$$ Multiplying both sides by $-1$: $$- rac{dX(s)}{ds} = \int_{0^-}^{\infty} [t x(t)] e^{-st} \, dt$$ $$\mathcal{L}\{t x(t)\} = - rac{dX(s)}{ds} \quad lacksquare$$ --- ## 8. Division by $t$ Dividing a time-domain signal by $t$ corresponds to integrating its Laplace transform over the complex variable $s$ from $s$ to infinity: $$\mathcal{L}\left\{ rac{x(t)}{t} ight\} = \int_{s}^{\infty} X(u) \, du$$ *Constraint: This property is valid if and only if the limit $\lim_{t o 0^+} rac{x(t)}{t}$ exists and is finite.* ### 8.1 Mathematical Proof Let $g(t) = rac{x(t)}{t} \implies x(t) = t g(t)$. Take the Laplace transform of both sides of this equation: $$\mathcal{L}\{x(t)\} = \mathcal{L}\{t g(t)\}$$ Apply the differentiation in the $s$-domain property to the right-hand side: $$X(s) = - rac{dG(s)}{ds}$$ Integrate both sides with respect to $s$ from $s$ to $\infty$: $$\int_{s}^{\infty} X(u) \, du = -\int_{s}^{\infty} rac{dG(u)}{du} \, du$$ $$\int_{s}^{\infty} X(u) \, du = -\left[ G(u) ight]_{s}^{\infty}$$ $$\int_{s}^{\infty} X(u) \, du = G(s) - \lim_{u o \infty} G(u)$$ Since $g(t) = rac{x(t)}{t}$ is absolutely integrable, its Laplace transform must vanish at infinity: $\lim_{u o \infty} G(u) = 0$. $$\int_{s}^{\infty} X(u) \, du = G(s)$$ $$\mathcal{L}\left\{ rac{x(t)}{t} ight\} = \int_{s}^{\infty} X(u) \, du \quad lacksquare$$ --- ## 9. Time Convolution Property Convolution in the time domain corresponds to **direct algebraic multiplication** in the complex $s$-domain: $$\mathcal{L}\{x_1(t) * x_2(t)\} = X_1(s) X_2(s)$$ This property is fundamental to LTI system analysis, as it relates the output spectrum $Y(s)$ directly to the input spectrum $X(s)$ and the system transfer function $H(s)$. ### 9.1 Mathematical Proof By definition of the unilateral Laplace transform of a convolution integral: $$\mathcal{L}\{x_1(t) * x_2(t)\} = \int_{0^-}^{\infty} \left[ \int_{0^-}^{\infty} x_1( au) x_2(t - au) \, d au ight] e^{-st} \, dt$$ Interchange the order of integration under Fubini's theorem (since the signals are absolutely integrable): $$\mathcal{L}\{x_1(t) * x_2(t)\} = \int_{0^-}^{\infty} x_1( au) \left[ \int_{0^-}^{\infty} x_2(t - au) e^{-st} \, dt ight] \, d au$$ Since $x_2(t - au)$ is causal, it is zero for $t < au$. Thus, we change the lower integration limit of the inner integral to $ au$: $$\mathcal{L}\{x_1(t) * x_2(t)\} = \int_{0^-}^{\infty} x_1( au) \left[ \int_{ au}^{\infty} x_2(t - au) e^{-st} \, dt ight] \, d au$$ Perform a change of variable on the inner integral. Let $u = t - au \implies t = u + au$ and $dt = du$. * At $t = au \implies u = 0^-$ * As $t o \infty \implies u o \infty$ Substituting these variables: $$\mathcal{L}\{x_1(t) * x_2(t)\} = \int_{0^-}^{\infty} x_1( au) \left[ \int_{0^-}^{\infty} x_2(u) e^{-s(u + au)} \, du ight] \, d au$$ $$\mathcal{L}\{x_1(t) * x_2(t)\} = \int_{0^-}^{\infty} x_1( au) e^{-s au} \left[ \int_{0^-}^{\infty} x_2(u) e^{-su} \, du ight] \, d au$$ Recognize that the inner integral is the definition of $X_2(s)$: $$\mathcal{L}\{x_1(t) * x_2(t)\} = \int_{0^-}^{\infty} x_1( au) e^{-s au} [X_2(s)] \, d au$$ Factor out $X_2(s)$, which is constant with respect to $ au$: $$\mathcal{L}\{x_1(t) * x_2(t)\} = X_2(s) \int_{0^-}^{\infty} x_1( au) e^{-s au} \, d au$$ $$\mathcal{L}\{x_1(t) * x_2(t)\} = X_1(s) X_2(s) \quad lacksquare$$ --- ## 10. Time Correlation Property [PYQ 2015] The cross-correlation of two real-valued signals $x_1(t)$ and $x_2(t)$ is defined as: $$r_{12}(t) = \int_{-\infty}^{\infty} x_1( au) x_2( au - t) \, d au$$ In the Laplace domain, time-domain correlation corresponds to multiplying one transform by the reflected (conjugate) transform of the second: $$\mathcal{L}\{r_{12}(t)\} = X_1(s) X_2(-s)$$ ### 10.1 Mathematical Proof Expressing the Laplace transform of the correlation integral: $$\mathcal{L}\{r_{12}(t)\} = \int_{-\infty}^{\infty} \left[ \int_{-\infty}^{\infty} x_1( au) x_2( au - t) \, d au ight] e^{-st} \, dt$$ Interchange the order of integration: $$\mathcal{L}\{r_{12}(t)\} = \int_{-\infty}^{\infty} x_1( au) \left[ \int_{-\infty}^{\infty} x_2( au - t) e^{-st} \, dt ight] \, d au$$ Perform a change of variable. Let $u = au - t \implies t = au - u$ and $dt = -du$. As $t o -\infty \implies u o \infty$, and as $t o \infty \implies u o -\infty$. Substituting these variables: $$\mathcal{L}\{r_{12}(t)\} = \int_{-\infty}^{\infty} x_1( au) \left[ \int_{-\infty}^{\infty} x_2(u) e^{-s( au - u)} \, du ight] \, d au$$ $$\mathcal{L}\{r_{12}(t)\} = \int_{-\infty}^{\infty} x_1( au) e^{-s au} \left[ \int_{-\infty}^{\infty} x_2(u) e^{su} \, du ight] \, d au$$ The inner integral represents the bilateral Laplace transform of $x_2(t)$ evaluated at $-s$: $$\int_{-\infty}^{\infty} x_2(u) e^{-(-s)u} \, du = X_2(-s)$$ Substituting this back: $$\mathcal{L}\{r_{12}(t)\} = X_2(-s) \int_{-\infty}^{\infty} x_1( au) e^{-s au} \, d au$$ $$\mathcal{L}\{r_{12}(t)\} = X_1(s) X_2(-s) \quad lacksquare$$ --- ## 11. Unified Properties Reference Table The following master table compiles the mathematical properties of the Unilateral Laplace Transform for easy exam reference: | Property | Time Domain $x(t)$ | Laplace Domain $X(s)$ | Region of Convergence (ROC) | | :--- | :--- | :--- | :--- | | **Linearity** | $a_1 x_1(t) + a_2 x_2(t)$ | $a_1 X_1(s) + a_2 X_2(s)$ | At least $R_1 \cap R_2$ | | **Time Scaling** | $x(at), \ a > 0$ | $ rac{1}{a} X\left( rac{s}{a} ight)$ | $R / a$ | | **Time Shifting** | $x(t - t_0) u(t - t_0)$ | $e^{-st_0} X(s)$ | Unchanged ($R$) | | **s-Domain Shifting** | $e^{s_0 t} x(t)$ | $X(s - s_0)$ | $R + \Re e(s_0)$ | | **First Derivative** | $ rac{dx(t)}{dt}$ | $s X(s) - x(0^-)$ | At least $R$ | | **Second Derivative** | $ rac{d^2x(t)}{dt^2}$ | $s^2 X(s) - s x(0^-) - x'(0^-)$ | At least $R$ | | **Time Integration** | $\int_{0^-}^{t} x( au) \, d au$ | $ rac{X(s)}{s}$ | $R \cap \{\Re e(s) > 0\}$ | | **s-Domain Derivative** | $t x(t)$ | $- rac{dX(s)}{ds}$ | Unchanged ($R$) | | **Division by $t$** | $ rac{x(t)}{t}$ | $\int_{s}^{\infty} X(u) \, du$ | Unchanged ($R$) | | **Time Convolution** | $x_1(t) * x_2(t)$ | $X_1(s) X_2(s)$ | At least $R_1 \cap R_2$ | | **Time Correlation** | $r_{12}(t)$ | $X_1(s) X_2(-s)$ | Intersecting Strip | --- ## 12. High-Yield Solved "Exam Killers" ### 12.1 Example 1: The Sawtooth Pulse Wave [PYQ 2018 - 4 Marks] **Question:** Determine the unilateral Laplace transform of the single sawtooth pulse wave shown below with peak amplitude $A$ and duration $T$. ``` Amplitude ^ A | /| | / | | / | | / | 0 o--/--------+------> Time (t) 0 T ``` #### Step-by-Step Singularity Representation: 1. **Formulate the Piecewise Equation:** The signal is defined analytically over one period as: $$x(t) = egin{cases} rac{A}{T} t, & 0 \le t \le T \ 0, & ext{otherwise} \end{cases}$$ 2. **Represent Using Unit Step Functions:** Using unit step windowing to bound the ramp: $$x(t) = \left( rac{A}{T} t ight) [u(t) - u(t-T)]$$ $$x(t) = rac{A}{T} t \, u(t) - rac{A}{T} t \, u(t-T)$$ 3. **Rearrange Terms for Time Shifting:** To apply the time-shifting property $\mathcal{L}\{f(t-T)u(t-T)\} = e^{-sT}F(s)$, we must express the second ramp term in terms of $(t-T)$: $$t = (t - T) + T$$ Substitute this into the expression: $$x(t) = rac{A}{T} t \, u(t) - rac{A}{T} [(t - T) + T] u(t-T)$$ $$x(t) = rac{A}{T} t \, u(t) - rac{A}{T} (t - T) u(t-T) - A u(t-T)$$ 4. **Apply Laplace Transform Properties:** Using the linearity property, take the transform of each individual component: * **Term 1 (Linear Ramp):** $$\mathcal{L}\left\{ rac{A}{T} t \, u(t) ight\} = rac{A}{T} \mathcal{L}\{t \, u(t)\} = rac{A}{T s^2}$$ * **Term 2 (Shifted Ramp):** $$\mathcal{L}\left\{ rac{A}{T} (t - T) u(t-T) ight\} = rac{A}{T} e^{-sT} \mathcal{L}\{t \, u(t)\} = rac{A}{T s^2} e^{-sT}$$ * **Term 3 (Shifted Step):** $$\mathcal{L}\{A u(t-T)\} = A e^{-sT} \mathcal{L}\{u(t)\} = rac{A}{s} e^{-sT}$$ 5. **Assemble the Final Algebraic Expression:** $$X(s) = rac{A}{T s^2} - rac{A}{T s^2} e^{-sT} - rac{A}{s} e^{-sT}$$ Factor out the common term $ rac{A}{T s^2}$ to yield the final simplified representation: $$X(s) = rac{A}{T s^2} \left[ 1 - e^{-sT} - sT e^{-sT} ight] \quad ext{for } \Re e(s) > 0 \quad lacksquare$$ --- ### 12.2 Example 2: Composite Signal Multiplication by $t$ **Question:** Obtain the unilateral Laplace transform of $x(t) = t \sin(\omega_0 t) u(t)$ using the s-domain differentiation property. #### Step-by-Step Mathematical Solution: 1. **Define the Base Signal Transform:** Let $g(t) = \sin(\omega_0 t) u(t)$. Its Laplace transform is: $$G(s) = rac{\omega_0}{s^2 + \omega_0^2}$$ 2. **Apply s-Domain Differentiation:** According to the property, multiplication by $t$ corresponds to the negative derivative with respect to $s$: $$X(s) = \mathcal{L}\{t g(t)\} = - rac{dG(s)}{ds}$$ $$X(s) = - rac{d}{ds} \left[ rac{\omega_0}{s^2 + \omega_0^2} ight]$$ 3. **Perform the Derivative Calculus:** Using the quotient rule: $$X(s) = -\omega_0 \cdot \left[ rac{0 \cdot (s^2 + \omega_0^2) - 1 \cdot (2s)}{(s^2 + \omega_0^2)^2} ight]$$ $$X(s) = -\omega_0 \cdot \left[ rac{-2s}{(s^2 + \omega_0^2)^2} ight]$$ $$X(s) = rac{2 \omega_0 s}{(s^2 + \omega_0^2)^2} \quad ext{for } \Re e(s) > 0 \quad lacksquare$$ --- ### 12.3 Example 3: Higher-Order Trigonometric Power Expansion **Question:** Find the unilateral Laplace transform of $x(t) = \cos^3(3t) u(t)$ using trigonometric identities and Laplace linearity. #### Step-by-Step Mathematical Solution: 1. **Apply the Triple-Angle Identity:** The standard cubic trigonometric expansion is: $$\cos(3 heta) = 4\cos^3( heta) - 3\cos( heta) \implies \cos^3( heta) = rac{1}{4}\cos(3 heta) + rac{3}{4}\cos( heta)$$ 2. **Substitute the Frequency Parameter:** Let $ heta = 3t$: $$x(t) = \cos^3(3t) = rac{1}{4}\cos(9t) + rac{3}{4}\cos(3t)$$ 3. **Apply Laplace Linearity:** Taking the unilateral Laplace transform: $$X(s) = rac{1}{4} \mathcal{L}\{\cos(9t)u(t)\} + rac{3}{4} \mathcal{L}\{\cos(3t)u(t)\}$$ 4. **Substitute Standard Cosine Transform Pairs:** Using $\mathcal{L}\{\cos(\omega t)u(t)\} = rac{s}{s^2+\omega^2}$: $$X(s) = rac{1}{4} \left( rac{s}{s^2 + 81} ight) + rac{3}{4} \left( rac{s}{s^2 + 9} ight)$$ $$X(s) = rac{s(s^2 + 9) + 3s(s^2 + 81)}{4(s^2 + 81)(s^2 + 9)}$$ $$X(s) = rac{4s^3 + 252s}{4(s^2 + 81)(s^2 + 9)} = rac{s(s^2 + 63)}{(s^2 + 81)(s^2 + 9)} \quad ext{for } \Re e(s) > 0 \quad lacksquare$$ --- ## 13. Common Mistakes That Cost Marks > [!danger] **The Time-Scaling vs. Time-Shifting Order Trap** > > A frequent source of lost marks in exams is executing time scaling and time shifting sequentially without accounting for their mutual scaling impact. > * **Correct Approach (Shift First, Then Scale):** > $$ ext{Let } g(t) = x(t - t_0) \leftrightarrow X(s)e^{-s t_0}$$ > $$\mathcal{L}\{x(at - t_0)\} = \mathcal{L}\{g(at)\} = rac{1}{a} G\left( rac{s}{a} ight) = rac{1}{a} X\left( rac{s}{a} ight) e^{-\left( rac{s}{a} ight) t_0}$$ > * **Incorrect Approach (Applying properties independently):** > $$\mathcal{L}\{x(at - t_0)\} eq rac{1}{a} X\left( rac{s}{a} ight) e^{-s t_0}$$ > *Always remember that scaling scales the shift parameter itself if operations are not factored carefully!* > [!warning] **The Unilateral Shift Windowing Omission** > > Under unilateral Laplace conditions, writing $\mathcal{L}\{x(t-t_0)\} = e^{-st_0}X(s)$ is strictly **incorrect** unless the signal is multiplied by $u(t-t_0)$. Without the step multiplier, the integration still begins at $t=0$, causing pre-shifted information in the interval $[0, t_0)$ to be lost or altered, which invalidates the pure exponential shift property. --- ## 14. PYQ Bank — Verbatim Questions & Answer Plans ### 14.1 PYQ 2023 [5 Marks] **Question:** If $x(t)$ is a signal with Laplace transform $X(s)$, then prove $ rac{dX(s)}{ds} = \mathcal{L}\{-t x(t)\}$. * **Answer Plan:** 1. State the definition of the unilateral Laplace transform integral as shown in **Section 7.1**. 2. Perform differentiation with respect to the complex variable $s$ on both sides. 3. Apply Leibniz's rule to bring the derivative operator inside the integration boundaries as a partial derivative. 4. Evaluate the partial derivative of $e^{-st}$ with respect to $s$ to yield $-t e^{-st}$. 5. Factor out the negative sign and identify the resulting integral as the unilateral transform of $-t x(t)$, completing the proof. ### 14.2 PYQ 2018 [4 Marks] **Question:** Find the Laplace transform of a signal sawtooth pulse shown in Fig. 7(c). * **Answer Plan:** 1. Define the piecewise linear equation of a single sawtooth pulse of duration $T$ and amplitude $A$. 2. Write the equivalent singularity representation using step multipliers as shown in **Section 12.1**. 3. Rearrange terms using $(t-T)$ algebraic shifts to allow the use of the time-shifting property. 4. Take individual unilateral Laplace transforms of each term using linearity. 5. Simplify and state the final transform along with its ROC ($\Re e(s) > 0$). ### 14.3 PYQ 2015 [5 Marks] **Question:** Explain the following terms in relations to Laplace transform: i) Linearity, ii) Scaling, iii) Time-shift, iv) Frequency differentiation, v) Time correlation. * **Answer Plan:** 1. **Linearity:** Define and state the equation $\mathcal{L}\{a_1x_1(t)+a_2x_2(t)\} = a_1X_1(s)+a_2X_2(s)$ with its proof (**Section 1**). 2. **Scaling:** State the time-scaling equation $\mathcal{L}\{x(at)\} = rac{1}{a}X(s/a)$ and provide the integration proof (**Section 2**). 3. **Time-Shift:** Define the causal shift $\mathcal{L}\{x(t-t_0)u(t-t_0)\} = e^{-st_0}X(s)$ alongside its proof (**Section 3**). 4. **Frequency Differentiation:** Explain that differentiating the transform corresponds to time multiplication: $ rac{dX(s)}{ds} = \mathcal{L}\{-t x(t)\}$ (**Section 7**). 5. **Time Correlation:** Define cross-correlation in time and prove that it translates to multiplication of conjugate spectra: $\mathcal{L}\{r_{12}(t)\} = X_1(s)X_2(-s)$ (**Section 10**). --- ## 15. Self-Check Before Moving On - [ ] Can you rigorously prove the s-domain differentiation property using Leibniz's rule? [7.1] - [ ] Do you know why a unilateral time shift requires the causal windowing step $u(t-t_0)$? [13.2] - [ ] Can you solve for the Laplace transform of a sawtooth pulse wave starting from its piecewise linear definition? [12.1] - [ ] Have you memorized the time correlation property formula $\mathcal{L}\{r_{12}(t)\} = X_1(s)X_2(-s)$? [10.1] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 4), ECE 2108 Signal & Systems-1 (1).pdf.*