Laplace Transform Analysis PDF

# Laplace Transform Analysis ## Unit 3: Laplace Transform Analysis ### Contents - **Part-1: Review of Laplace Transform** (3-2D to 3-5D) - **Part-2: Properties of Laplace Transform** (3-5D to 3-8D) - **Part-3: Initial and Final Value Theorems** (3-8D to 3-11D) - **Part-4: Inverse Laplace Transfor...

# Laplace Transform Analysis ## Unit 3: Laplace Transform Analysis ### Contents - **Part-1: Review of Laplace Transform** (3-2D to 3-5D) - **Part-2: Properties of Laplace Transform** (3-5D to 3-8D) - **Part-3: Initial and Final Value Theorems** (3-8D to 3-11D) - **Part-4: Inverse Laplace Transform** (3-11D to 3-13D) - **Part-5: Convolution Theorem, Impulse Response** (3-13D to 3-18D) - **Part-6: Application of Laplace Transform to Analysis of Networks** (3-18D to 3-24D) - **Part-7: Waveform Synthesis and Laplace Transform to Complex Waveforms** (3-24D to 3-27D) ### Part-1: Review of Laplace Transform **Que 3.1.** What is Laplace transform? Also explain the different types of Laplace transform. **Answer** **Laplace Transform:** Let x(t) be defined for t > 0 and has zero value for t ≤ 0. Then the Laplace Transform of x(t) is denoted as L[x(t)] = X(s) and is given as: $$ L[x(t)] = X(s) = \int_{0}^{\infty} x(t)e^{-(σ+jω)t} dt = \int_{0}^{\infty} x(t)e^{-st} dt $$ Where, *s* is a complex variable and is equal to σ + jω. **Types of Laplace Transform:** 1. **Unilateral Laplace Transform:** When the integral is taken from '0' to '∞', the Laplace Transform is called one-sided, or unilateral Laplace Transform. This is mainly used for the analysis of causal signals. 2. **Right-sided:** When the integral is taken for positive side i.e. limits are taken from '0' to '∞'. $$ X(s) = \int_{0}^{\infty} x(t)e^{-st} dt $$ 3. **Left-sided:** When the integral is taken for negative side, i.e., limits are taken from '-∞' to '0'. $$ X(s) = \int_{-\infty}^{0} x(t)e^{-st} dt $$ 4. **Bilateral (two-sided) Laplace Transform:** When the integral is taken from -∞ to ∞, the Laplace Transform is then called bilateral or two-sided Laplace Transform. $$ X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt $$ ### Part-2: Properties of Laplace Transform **Que 3.5.** What are the different properties of Laplace Transform? **Answer** 1. **Linearity:** Let *x1(t)* and *x2(t)* be two signals with Laplace Transforms *X1(s)* and *X2(s)* respectively. Then: $$ L[a_1x_1(t) + a_2x_2(t)] = a_1X_1(s) + a_2X_2(s) $$ Where *a1* and *a2* are constants. 2. **Time Shifting:** Let x(t) be a signal. Then: $$ L[x(t - t_0)] = e^{-st_0}X(s) $$ Where *t0* is a constant. 3. **Frequency Shifting:** Let *x(t)* be a signal. Then: $$ L[e^{at} x(t)] = X(s - a) $$ 4. **Differentiation Theorem:** Let *x(t)* be a signal. Then the Differentiation Theorem gives: $$ L[\frac{dx(t)}{dt}] = sX(s) - x(0-) $$ where: - *x(0-)* indicates the value of *x(t)* at the time right before *t=0* 5. **Differentiation by *s*:** The differentiation in complex frequency domain corresponds to multiplication by time *t* in the time domain. Let *x(t)* be a signal. Then: $$ L[tx(t)] = -\frac{dX(s)}{ds} $$ Then: $$ L[t^nx(t)] = (-1)^n\frac{d^nX(s)}{ds_n} $$ 6. **Integration Theorem:** This theorem states: Let *x(t)* be a signal. Then: $$ L[\int_{0}^{t} x(\tau) d\tau] = \frac{X(s)}{s} $$ 7. **Scaling Property:** This theorem states that: Let *x(t)* be a signal. Then: $$ L[x(at)] = \frac{1}{a}X(\frac{s}{a}) $$ ### Part-3: Initial and Final Value Theorems **Que 3.9.** State and prove initial and final value theorems of Laplace Transform. **Answer** **i. Initial Value Theorem:** This theorem states that if F(s) is the Laplace transform of f(t) then: $$ f(0^+) = \lim_{t \to 0^+} f(t) = \lim_{s \to \infty} sF(s) $$ **Proof:** 1. Consider the Laplace transform of the real differentiation: $$ L[\frac{df(t)}{dt}] = sF(s) - f(0-) $$ 2. Taking the limit as *s* approaches ∞ on both sides: $$ \lim_{s \to \infty} L[\frac{df(t)}{dt}] = \lim_{s \to \infty} [sF(s) - f(0-)] $$ 3. Consider the left hand side of the equation: $$ \lim_{s \to \infty} L[\frac{df(t)}{dt}] = \lim_{s \to \infty} \int_{0}^{\infty} \frac{df(t)}{dt}e^{-st} dt = 0 $$ This is because lim_{s \to \infty} e^{-st} = 0, by the definition of Laplace transform 4. Putting the result from step 3 into the equation in step 2: $$ 0 = \lim_{s \to \infty} [sF(s) -f(0-)] $$ Which simplifies to: $$ f(0-) = \lim_{s \to \infty} sF(s) $$ 5. But as the function *f(t)* is continuous, *f(0-) = f(0+)* , i.e., the initial value of *f(t)* Therefore, the initial value theorem is proven: $$ f(0^+) = \lim_{t \to 0^+} f(t) = \lim_{s \to \infty} sF(s) $$ **ii. Final Value Theorem:** This theorem states that if *F(s)* is the Laplace transform of *f(t)*, then the final value of the function *f(t)* is given by: $$ f(\infty) = \lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s) $$ **Proof:** 1. Consider the Laplace transform of the real differentiation: $$ L[\frac{df(t)}{dt}] = sF(s) - f(0-) $$ 2. Taking the limit as *s* approaches 0 on both sides: $$ \lim_{s \to 0} L[\frac{df(t)}{dt}] = \lim_{s \to 0} [sF(s) - f(0-)] $$ 3. Consider the left hand side of the equation: $$ \lim_{s \to 0} L[\frac{df(t)}{dt}] = \lim_{s \to 0} \int_{0}^{\infty} \frac{df(t)}{dt}e^{-st} dt = \lim_{s \to 0} \int_{0}^{\infty} e^{-st} df(t) $$ 4. Applying integration by parts: $$ \lim_{s \to 0} \int_{0}^{\infty} e^{-st} df(t) = \lim_{s \to 0} [e^{-st}f(t)]_{0}^{\infty} - \lim_{s \to 0} \int_{0}^{\infty} -se^{-st}f(t) dt $$ This simplifies to: $$ \lim_{s \to 0} [e^{-st}f(t)]_{0}^{\infty} + \lim_{s \to 0} \int_{0}^{\infty} se^{-st}f(t) dt $$ 5. Evaluating the limits: $$ \lim_{s \to 0} [e^{-st}f(t)]_{0}^{\infty} = f(\infty) - f(0) $$ And assuming that the Laplace transform exists, this is equivalent to 0. Additionally, we can rewrite the right side of the equation in step 4 as: $$ \lim_{s \to 0} \int_{0}^{\infty} se^{-st}f(t) dt = \lim_{s \to 0} sF(s) $$ 6. Putting the results back into the equation from step 2: $$ f(\infty) - f(0-) = \lim_{s \to 0} sF(s) -f(0-) $$ Simplifying, we get: $$ f(\infty) = \lim_{s \to 0} sF(s) $$ Therefore, the final value theorem is proven: $$ f(\infty) = \lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s) $$ ### Part-4: Inverse Laplace Transform **Que 3.12.** What is inverse Laplace transform? **Answer** Inverse Laplace transform converts a function in the frequency domain, like *F(s)*, to a function in the time domain, like *f(t)*. It is represented as: $$ f(t) = L^{-1}[F(s)] $$ Or more formally, using the inverse Laplace transform integral: $$ f(t) = \frac{1}{2\pi j} \int_{σ-j∞}^{σ+j∞} F(s)e^{st} ds $$ This integral is evaluated along a vertical line in the complex plane, where *σ* is a real number chosen such that the path of integration lies to the right of all singularities of *F(s)*. ### Part-5: Convolution Theorem, Impulse Response **Que 3.15.** State and prove convolution theorem. **Answer** **Convolution Theorem:** This theorem states that if *f1(t)* and *f2(t)* are two functions which are equal to *0* for *t < 0*, then *f1(t) * f2(t)* (the convolution of *f1(t)* and *f2(t)*) has a Laplace transform that is the product of the Laplace transforms of *f1(t)* and *f2(t)*. **Proof:** 1. Let *F1(s)* and *F2(s)* be the Laplace transforms of *f1(t)* and *f2(t)* respectively. $$ F₁(s) = L\{f₁(t)\} = \int_{0}^{\infty} e^{-st} f₁(t) dt $$ and $$ F₂(s) = L\{f₂(t)\} = \int_{0}^{\infty} e^{-st} f₂(t) dt $$ 2. Define the convolution of *f1(t)* and *f2(t)* as: $$ f₁(t) * f₂(t) = \int_{0}^{\infty} f₁(τ)f₂(t - τ)dτ $$ 3. The Laplace Transform of the convolution can be written as: $$ L\{f₁(t) * f₂(t) \} = \int_{0}^{\infty} e^{-st} \left[ \int_{0}^{\infty} f₁(τ)f₂(t - τ)dτ \right] dt $$ 4. Using Fubini's theorem to change the order of integration, and then substituting *p = t - τ*: $$ L\{f₁(t) * f₂(t) \} = \int_{0}^{\infty}f₁(τ) \left[ \int_{0}^{\infty} e^{-st} f₂(t - τ) dt \right] dτ $$ $$ L\{f₁(t) * f₂(t) \} = \int_{0}^{\infty}f₁(τ) \left[ \int_{τ}^{\infty} e^{-sp} f₂(p) dp \right] dτ $$ 5. Recognizing the inner integral as the Laplace transform of *f2(p)*, we can rewrite: $$ L\{f₁(t) * f₂(t) \} = \int_{Ο}^{\infty}f₁(τ)F₂(s)e^{-sτ} dτ $$ 6. Finally, recognizing the integral as the Laplace transform of *f1(t)*: $$ L\{f₁(t) * f₂(t)\} = F₂(s) * \int_{0}^{\infty}f₁(τ)e^{-sτ} dτ = F₂(s)F₁(s) $$ Therefore, we have proven the convolution theorem: $$ L\{f₁(t) * f₂(t)\} = F₁(s)F₂(s) $$ **Properties of Convolution:** 1. **Commutative:** Convolution is commutative: $$ x₁(t) * x₂(t) = x₂(t) * x₁(t) $$ 2. **Distributive:** Convolution is distributive over addition: $$ x₁(t) * [x₂(t) + x₃(t)] = [x₁(t) * x₂(t)] + [x₁(t) * x₃(t)] $$ 3. **Associative:** Convolution is associative: $$ x₁(t) * [x₂(t) * x₃(t)] = [x₁(t) * x₂(t)] * x₃(t) $$ 4. **Shift Property:** - If *z(t) = x₁(t) * x₂(t)*, then *z(t - T) = x₁(t) * x₂(t - T)* - If *z(t) = x₁(t) * x₂(t)*, then *z(t - T) = x₁(t - T) * x₂(t)* 5. **Convolution with an Impulse:** Convolution of a signal *x(t)* with a unit impulse is the signal itself: $$ x(t) * δ(t) = x(t) $$ 6. **Width Property:** Let the duration of *x1(t)* and *x2(t)* be *T1* and *T2* respectively. Then the duration of the signal obtained by convolving *x1(t)* and *x2(t)* is *T1+T2*. 7. **Differentiation Property:** $$ \frac{d}{dt}[x(t) * h(t)] = \frac{dx(t)}{dt} * h(t) = x(t) * \frac{dh(t)}{dt} $$ 8. **Time-Scaling Property:** $$ x(at) * h(at) = \frac{1}{|a|} y(at) $$ ### **Part-6: Application of Laplace Transform to Analysis of Networks** **Que 3.20**. Draw the transformed circuit components representation in Laplace form. **Answer** **Independent Sources:** - **Voltage Source:** A voltage source *v(t)* in the time domain is transformed into a voltage source *V(s)* in the Laplace domain:  - **Current Source:** A current source *i(t)* in the time domain is transformed into a current source *I(s)* in the Laplace domain:  **Dependent Sources:** - **Voltage-Controlled Voltage Source (VCVS):** A VCVS with a gain *k* is represented as a voltage source *kV(s)* in the Laplace domain. - **Voltage-Controlled Current Source (VCCS):** A VCCS with a transconductance *g* is represented as a current source *gV(s)* in the Laplace domain. - **Current-Controlled Voltage Source (CCVS):** A CCVS with a transresistance *r* is represented as a voltage source *rI(s)* in the Laplace domain. - **Current-Controlled Current Source (CCCS):** A CCCS with a current gain *β* is represented as a current source *βI(s)* in the Laplace domain. **Passive Components:** - **Resistor:** A resistor *R* is represented by its resistance *R* in the Laplace domain. The KVL equation across the resistor becomes *VR(s) = R*I(s)*, and the KCL equation becomes *I(s) = (1/R)*VR(s)*  - **Capacitor:** A capacitor *C* is represented by its impedance *1/(sC)* in the Laplace domain. - The KVL equation across the capacitor becomes *VC(s) = (1/(sC))*I(s) + (1/s)VC(0)*, where *VC(0)* is initial voltage across the capacitor. - The KCL equation becomes *I(s) = sC*VC(s) - CVC(0)*.  - **Inductor:** An inductor *L* is represented by its impedance *sL* in the Laplace domain. - The KVL equation across the capacitor becomes *VL(s) = sL*I(s) - L*I(0)*, where *I(0)* is the initial current through the inductor. - The KCL equation becomes *I(s) =(1/(sL))*VL(s) + (1/s)*I(0)*.  ### Part-7: **Waveform Synthesis and Laplace Transform to Complex Waveforms** **Que 3.26.** Find the Laplace transform of the following waveforms shown in Fig. 3.26.1(a) and 3.26.1(b). **Answer** **(a)**  1. The waveform can be represented as a combination of two functions: - *x1(t) = 3t* for *0 ≤ t ≤ 2* - *x2(t) = 3t + c* for *2 ≤ t ≤ 4*, where *c* is a constant. 2. Finding the value of *c*: When *t = 2*, *x1(t) = 6*. So, *x2(t) = 6* for *t = 2*, which gives: $$ 6 = 3(2) + c $$ Solving for *c*, we get: $$ c = 12 $$ Therefore: $$ x2(t) = 3t + 12 $$ 3. Applying the Laplace transform to *x1(t)*: $$ X₁(s) = \int_{0}^{2} 3te^{-st} dt = \frac{3}{s^2} (1 - e^{-2s} - 2se^{-2s}) $$ 4. Applying the Laplace transform to *x2(t)*: $$ X₂(s) = \int_{2}^{4} (3t + 12) e^{-st} dt = \frac{3}{s^2} (e^{-2s} - e^{-4s} - 4se^{-4s} + 4se^{-2s}) $$ 5. The Laplace transform of the entire waveform *x(t)* is the sum of the Laplace transforms of *x1(t)* and *x2(t)*: $$ X(s) = X₁(s) + X₂(s) = \frac{3}{s^2} (1 - 4e^{-4s} + 3e^{-2s} - 4se^{-4s} + 6se^{-2s}) $$ **(b)**  1. Let's write the waveform mathematically: $$ x(t) = 3u(t) + 1.5t u(t-2) - 1.5t u(t-4) - 6u(t - 4) $$ 2. Taking the Laplace transform on both sides, we get: $$ X(s) = \frac{3}{s} + \frac{1.5}{s^2} (e^{-2s} - e^{-4s}) - \frac{6}{s} e^{-4s} $$ **Que 3.27.** Determine the Laplace transform of the non-sinusoidal waveform in Fig. 3.27.1. **Answer**  1. The mathematical description of the periodic wave with period *2a* is: $$ f₁(t) = 1; 0 < t ≤ a $$ $$ f₁(t) = -1; a ≤ t ≤ 2a $$ 2. Let *F₁(s)* be the Laplace transform of *f₁(t)* for the time *0 ≤ t ≤ 2a* : $$ F₁(s) = \int_{0}^{a} e^{-st} dt + \int_{a}^{2a} (-1)e^{-st} dt = \frac{1}{s}(1 - e^{-as})^2 $$ 3. The Laplace transform of the periodic function *f(t)* can be expressed in terms of *F₁(s)*: $$ F(s) = \frac{F₁(s)}{1 - e^{-2as}} = \frac{\frac{1}{s}(1 - e^{-as})^2}{1 - e^{-2as}} = \frac{1}{s}(1 - e^{-as}) $$ ### **Very Important Questions** 1. **Calculate the Laplace transform for the function F(t) = e^-at sin(hbt).** - The answer is provided in **Que 3.2** on page 3-3D. 2. **Find the Laplace transform of signal u(t).** - The answer is provided on page 3-4D. 3. **What are the different properties of Laplace transform?** - The answer is provided in **Que 3.5** on page 3-6D. 4. **If the Laplace transform of x(t) is (s + 2)/(s² + 4s + 5), determine the Laplace transform of y(t) = x(2t - 1) u(2t - 1).** - The answer is provided in **Que 3.6** on page 3-7D. 5. **A signal has Laplace transform X(s) = (s + 2)/(s² + 4s + 5). Find the Laplace transform Y(s) of the following signals:** - **i. y(t) = tx(t)** - **ii. y(t) = e^t x(t)** - The answer is provided in **Que 3.8** on page 3-8D. 6. **Calculate the inverse Laplace transform of right-sided sequences with the following transfer functions:** - **X₁ (s) = (s + 3)/(s(s + 1)(s + 2))** - The answer is provided in **Que 3.13** on page 3-12D. 7. **State and prove convolution property of Laplace transform and then using this property find Laplace inverse transform of 8/(s + 1)(s + 2).** - The answer is provided in **Que 3.17** on page 3-15D. 8. **For a transfer function H(s) = (s + 10)(s² + 3s + 2). Find the response due to input x(t) = sin(2t) u(t).** - The answer is provided in **Que 3.19** on page 3-18D. 9. **Initially switch is closed for a long time and steady-state condition has reached. At t = 0 switch is opened. Find the expression of current through inductor.** - The answer is provided in **Que 3.21** on page 3-20D. 10. **Using Laplace transform, solve differential equation: 2x^'' + 7x^' + 6x = 0, where, x(0) = 0, x^'(0) = 1.** - The answer is provided in **Que 3.25** on page 3-24D.