# 7.1 MCQs-Laplace Transform
### Laplace Transform MCQs
### Definition and Basic Transforms
1\. The Laplace transform of a function $$f(t)$$ is defined as:
1. $$\int\_{0}^{\infty} e^{-st} f(t) , dt$$
2. $$\int\_{-\infty}^{\infty} e^{-st} f(t) , dt$$
3. $$\int\_{0}^{\infty} e^{st} f(t) , dt$$
4. $$\int\_{-\infty}^{\infty} e^{st} f(t) , dt$$
Show me the answer
**Answer:** 1. $$\int\_{0}^{\infty} e^{-st} f(t) , dt$$
**Explanation:**
* The (unilateral) Laplace transform $$\mathcal{L}{f(t)} = F(s)$$ is defined as: $$F(s) = \int\_{0}^{\infty} e^{-st} f(t) , dt$$
* The integral is taken from $$0$$ to $$\infty$$ (one-sided).
* The parameter $$s$$ is a complex variable: $$s = \sigma + i\omega$$.
***
2\. The Laplace transform of the unit step function $$u(t)$$ (also denoted as $$H(t)$$) is:
1. $$\frac{1}{s}$$ for $$\text{Re}(s) > 0$$
2. $$\frac{1}{s^2}$$
3. 1
4. $$s$$
Show me the answer
**Answer:** 1. $$\frac{1}{s}$$ for $$\text{Re}(s) > 0$$
**Explanation:**
* $$u(t) = \begin{cases} 0, & t < 0 \ 1, & t \ge 0 \end{cases}$$
* $$\mathcal{L}{u(t)} = \int\_{0}^{\infty} e^{-st} \cdot 1 , dt = \left\[ \frac{e^{-st}}{-s} \right]\_{0}^{\infty}$$
* For $$\text{Re}(s) > 0$$, $$e^{-s \cdot \infty} \to 0$$.
* Therefore, $$\mathcal{L}{u(t)} = 0 - \left( \frac{1}{-s} \right) = \frac{1}{s}$$.
***
3\. The Laplace transform of $$f(t) = e^{at}$$ is:
1. $$\frac{1}{s-a}$$ for $$\text{Re}(s) > \text{Re}(a)$$
2. $$\frac{1}{s+a}$$
3. $$\frac{s}{s^2 + a^2}$$
4. $$\frac{a}{s^2 + a^2}$$
Show me the answer
**Answer:** 1. $$\frac{1}{s-a}$$ for $$\text{Re}(s) > \text{Re}(a)$$
**Explanation:**
* $$\mathcal{L}{e^{at}} = \int\_{0}^{\infty} e^{-st} e^{at} , dt = \int\_{0}^{\infty} e^{-(s-a)t} , dt$$
* This integral converges if $$\text{Re}(s-a) > 0$$, i.e., $$\text{Re}(s) > \text{Re}(a)$$.
* $$\int\_{0}^{\infty} e^{-(s-a)t} , dt = \left\[ \frac{e^{-(s-a)t}}{-(s-a)} \right]\_{0}^{\infty} = 0 - \frac{1}{-(s-a)} = \frac{1}{s-a}$$.
***
4\. The Laplace transform of $$f(t) = t^n$$ where n is a positive integer is:
1. $$\frac{n!}{s^{n+1}}$$
2. $$\frac{n!}{s^n}$$
3. $$\frac{s}{s^2 + n^2}$$
4. $$\frac{1}{s^n}$$
Show me the answer
**Answer:** 1. $$\frac{n!}{s^{n+1}}$$
**Explanation:**
* $$\mathcal{L}{t^n} = \int\_{0}^{\infty} e^{-st} t^n , dt$$
* Using integration by parts or the Gamma function: $$\int\_{0}^{\infty} e^{-st} t^n , dt = \frac{\Gamma(n+1)}{s^{n+1}}$$.
* Since n is a positive integer, $$\Gamma(n+1) = n!$$.
* Therefore, $$\mathcal{L}{t^n} = \frac{n!}{s^{n+1}}$$ for $$\text{Re}(s) > 0$$.
* Special cases: $$\mathcal{L}{t} = \frac{1}{s^2}$$, $$\mathcal{L}{t^2} = \frac{2}{s^3}$$.
***
### Transforms of Trigonometric Functions
5\. The Laplace transform of $$\cos(\omega t)$$ is:
1. $$\frac{s}{s^2 + \omega^2}$$
2. $$\frac{\omega}{s^2 + \omega^2}$$
3. $$\frac{1}{s^2 + \omega^2}$$
4. $$\frac{s}{s^2 - \omega^2}$$
Show me the answer
**Answer:** 1. $$\frac{s}{s^2 + \omega^2}$$
**Explanation:**
* Using Euler's formula: $$\cos(\omega t) = \frac{e^{i\omega t} + e^{-i\omega t}}{2}$$.
* Then $$\mathcal{L}{\cos(\omega t)} = \frac{1}{2} \left( \mathcal{L}{e^{i\omega t}} + \mathcal{L}{e^{-i\omega t}} \right) = \frac{1}{2} \left( \frac{1}{s - i\omega} + \frac{1}{s + i\omega} \right)$$.
* Simplify: $$\frac{1}{2} \cdot \frac{(s + i\omega) + (s - i\omega)}{s^2 + \omega^2} = \frac{1}{2} \cdot \frac{2s}{s^2 + \omega^2} = \frac{s}{s^2 + \omega^2}$$.
***
6\. The Laplace transform of $$\sin(\omega t)$$ is:
1. $$\frac{s}{s^2 + \omega^2}$$
2. $$\frac{\omega}{s^2 + \omega^2}$$
3. $$\frac{1}{s^2 + \omega^2}$$
4. $$\frac{s}{s^2 - \omega^2}$$
Show me the answer
**Answer:** 2. $$\frac{\omega}{s^2 + \omega^2}$$
**Explanation:**
* Using Euler's formula: $$\sin(\omega t) = \frac{e^{i\omega t} - e^{-i\omega t}}{2i}$$.
* $$\mathcal{L}{\sin(\omega t)} = \frac{1}{2i} \left( \frac{1}{s - i\omega} - \frac{1}{s + i\omega} \right)$$.
* Simplify: $$\frac{1}{2i} \cdot \frac{(s + i\omega) - (s - i\omega)}{s^2 + \omega^2} = \frac{1}{2i} \cdot \frac{2i\omega}{s^2 + \omega^2} = \frac{\omega}{s^2 + \omega^2}$$.
***
### Properties of Laplace Transform
7\. The linearity property of Laplace transform states that:
1. $$\mathcal{L}{af(t) + bg(t)} = a\mathcal{L}{f(t)} + b\mathcal{L}{g(t)}$$
2. $$\mathcal{L}{f(t)g(t)} = \mathcal{L}{f(t)} \cdot \mathcal{L}{g(t)}$$
3. $$\mathcal{L}{f(t)} = s\mathcal{L}{f(t)} - f(0)$$
4. $$\mathcal{L}{f(at)} = \frac{1}{a}F\left(\frac{s}{a}\right)$$
Show me the answer
**Answer:** 1. $$\mathcal{L}{af(t) + bg(t)} = a\mathcal{L}{f(t)} + b\mathcal{L}{g(t)}$$
**Explanation:**
* Linearity is a fundamental property: The Laplace transform of a linear combination is the same linear combination of the transforms.
* This follows directly from the linearity of the integral: $$\int (af + bg) = a\int f + b\int g$$.
* This property makes the Laplace transform easy to apply to linear differential equations.
***
8\. The First Shifting Theorem (Frequency Shift) states: If $$\mathcal{L}{f(t)} = F(s)$$, then $$\mathcal{L}{e^{at}f(t)} =$$
1. $$F(s-a)$$
2. $$F(s+a)$$
3. $$e^{as}F(s)$$
4. $$e^{-as}F(s)$$
Show me the answer
**Answer:** 1. $$F(s-a)$$
**Explanation:**
* The First Shifting Theorem: $$\mathcal{L}{e^{at}f(t)} = F(s-a)$$.
* Proof: $$\mathcal{L}{e^{at}f(t)} = \int\_{0}^{\infty} e^{-st} e^{at} f(t) , dt = \int\_{0}^{\infty} e^{-(s-a)t} f(t) , dt = F(s-a)$$.
* This is also called the s-shifting property.
* Example: Since $$\mathcal{L}{\cos(\omega t)} = \frac{s}{s^2+\omega^2}$$, then $$\mathcal{L}{e^{at}\cos(\omega t)} = \frac{s-a}{(s-a)^2+\omega^2}$$.
***
9\. The Laplace transform of the derivative $$f'(t)$$ is:
1. $$sF(s)$$
2. $$sF(s) - f(0)$$
3. $$\frac{F(s)}{s}$$
4. $$F(s) - f(0)$$
Show me the answer
**Answer:** 2. $$sF(s) - f(0)$$
**Explanation:**
* This is one of the most important properties for solving differential equations.
* $$\mathcal{L}{f'(t)} = s\mathcal{L}{f(t)} - f(0) = sF(s) - f(0)$$.
* For the second derivative: $$\mathcal{L}{f''(t)} = s^2F(s) - sf(0) - f'(0)$$.
* This property converts differential equations in t to algebraic equations in s.
***
10\. The Laplace transform of the integral $$\int\_0^t f(\tau) , d\tau$$ is:
1. $$\frac{F(s)}{s}$$
2. $$sF(s)$$
3. $$F(s) - f(0)$$
4. $$\frac{F(s)}{s} + \frac{f(0)}{s}$$
Show me the answer
**Answer:** 1. $$\frac{F(s)}{s}$$
**Explanation:**
* Integration in the time domain corresponds to division by s in the frequency domain.
* If $$g(t) = \int\_0^t f(\tau) , d\tau$$, then $$g'(t) = f(t)$$ and $$g(0) = 0$$.
* Using the derivative property: $$\mathcal{L}{g'(t)} = s\mathcal{L}{g(t)} - g(0)$$.
* So $$F(s) = s\mathcal{L}{g(t)} - 0$$, thus $$\mathcal{L}{g(t)} = \frac{F(s)}{s}$$.
***
### Unit Step Function and Time Shifting
11\. The Laplace transform of the unit step function shifted by a, i.e., $$u(t-a)$$ (with $$a > 0$$), is:
1. $$\frac{e^{-as}}{s}$$
2. $$\frac{e^{as}}{s}$$
3. $$\frac{1}{s}$$
4. $$e^{-as}$$
Show me the answer
**Answer:** 1. $$\frac{e^{-as}}{s}$$
**Explanation:**
* $$u(t-a) = \begin{cases} 0, & t < a \ 1, & t \ge a \end{cases}$$
* $$\mathcal{L}{u(t-a)} = \int\_{0}^{\infty} e^{-st} u(t-a) , dt = \int\_{a}^{\infty} e^{-st} , dt$$
* Let $$\tau = t-a$$, then $$dt = d\tau$$, and when $$t=a$$, $$\tau=0$$.
* $$\int\_{a}^{\infty} e^{-st} , dt = \int\_{0}^{\infty} e^{-s(\tau+a)} , d\tau = e^{-as} \int\_{0}^{\infty} e^{-s\tau} , d\tau = e^{-as} \cdot \frac{1}{s}$$.
***
12\. The Second Shifting Theorem (Time Shift) states: If $$\mathcal{L}{f(t)} = F(s)$$, then $$\mathcal{L}{f(t-a)u(t-a)} =$$
1. $$e^{-as}F(s)$$
2. $$e^{as}F(s)$$
3. $$F(s-a)$$
4. $$F(s+a)$$
Show me the answer
**Answer:** 1. $$e^{-as}F(s)$$
**Explanation:**
* Also known as the t-shifting property.
* The function $$f(t-a)u(t-a)$$ represents $$f(t)$$ shifted to the right by a units.
* $$\mathcal{L}{f(t-a)u(t-a)} = e^{-as}F(s)$$.
* Example: To find $$\mathcal{L}{(t-2)^2 u(t-2)}$$, we know $$\mathcal{L}{t^2} = \frac{2}{s^3}$$, so the transform is $$e^{-2s} \cdot \frac{2}{s^3}$$.
***
### Dirac Delta Function
13\. The Laplace transform of the Dirac delta function $$\delta(t)$$ is:
1. 0
2. 1
3. $$\frac{1}{s}$$
4. $$s$$
Show me the answer
**Answer:** 2. 1
**Explanation:**
* The Dirac delta function $$\delta(t)$$ is defined such that $$\int\_{-\infty}^{\infty} \delta(t) , dt = 1$$ and $$\delta(t) = 0$$ for $$t \neq 0$$.
* $$\mathcal{L}{\delta(t)} = \int\_{0}^{\infty} e^{-st} \delta(t) , dt = e^{-s \cdot 0} = 1$$ (using the sifting property).
* For a shifted delta: $$\mathcal{L}{\delta(t-a)} = e^{-as}$$ for $$a > 0$$.
***
14\. The Laplace transform of $$\delta(t-a)$$ for $$a > 0$$ is:
1. 1
2. $$e^{-as}$$
3. $$e^{as}$$
4. $$\frac{e^{-as}}{s}$$
Show me the answer
**Answer:** 2. $$e^{-as}$$
**Explanation:**
* Using the sifting property of the Dirac delta function: $$\int\_{0}^{\infty} e^{-st} \delta(t-a) , dt = e^{-sa}$$ for $$a > 0$$.
* This follows because $$\delta(t-a)$$ "picks out" the value of the integrand at $$t = a$$.
* Therefore, $$\mathcal{L}{\delta(t-a)} = e^{-as}$$.
***
### Inverse Laplace Transform
15\. The inverse Laplace transform of $$\frac{1}{s-a}$$ is:
1. $$e^{at}$$
2. $$e^{-at}$$
3. $$t^n$$
4. $$\cos(at)$$
Show me the answer
**Answer:** 1. $$e^{at}$$
**Explanation:**
* This is the inverse of the basic transform $$\mathcal{L}{e^{at}} = \frac{1}{s-a}$$.
* So, $$\mathcal{L}^{-1}\left{ \frac{1}{s-a} \right} = e^{at}$$ for $$t \ge 0$$.
***
16\. The inverse Laplace transform of $$\frac{s}{s^2 + 4}$$ is:
1. $$\sin(2t)$$
2. $$\cos(2t)$$
3. $$e^{2t}$$
4. $$e^{-2t}$$
Show me the answer
**Answer:** 2. $$\cos(2t)$$
**Explanation:**
* We know $$\mathcal{L}{\cos(\omega t)} = \frac{s}{s^2 + \omega^2}$$.
* Here, $$\frac{s}{s^2 + 4} = \frac{s}{s^2 + 2^2}$$, so $$\omega = 2$$.
* Therefore, $$\mathcal{L}^{-1}\left{ \frac{s}{s^2 + 4} \right} = \cos(2t)$$.
***
### Convolution Theorem
17\. The Convolution Theorem states: The Laplace transform of the convolution $$(f \* g)(t) = \int\_0^t f(\tau)g(t-\tau) , d\tau$$ is:
1. $$F(s) + G(s)$$
2. $$F(s) \cdot G(s)$$
3. $$F(s) - G(s)$$
4. $$\frac{F(s)}{G(s)}$$
Show me the answer
**Answer:** 2. $$F(s) \cdot G(s)$$
**Explanation:**
* The Convolution Theorem: $$\mathcal{L}{(f \* g)(t)} = F(s) \cdot G(s)$$.
* Conversely: $$\mathcal{L}^{-1}{F(s) \cdot G(s)} = (f \* g)(t)$$.
* This is useful for finding inverse transforms of products and for solving integral equations.
* Convolution is commutative: $$f \* g = g \* f$$.
***
### Solving Differential Equations
18\. When solving an initial value problem using Laplace transforms, the differential equation in the time domain is converted to:
1. Another differential equation in s
2. An integral equation
3. An algebraic equation in s
4. A partial differential equation
Show me the answer
**Answer:** 3. An algebraic equation in s
**Explanation:**
* This is the primary advantage of the Laplace transform method for solving linear ordinary differential equations with constant coefficients.
* Derivatives become algebraic terms: $$\mathcal{L}{y'} = sY(s) - y(0)$$, $$\mathcal{L}{y''} = s^2Y(s) - sy(0) - y'(0)$$, etc.
* The ODE transforms into an algebraic equation for $$Y(s)$$, which is then solved algebraically.
* Finally, $$y(t)$$ is obtained by taking the inverse Laplace transform of $$Y(s)$$.
***
19\. The Laplace transform is particularly useful for solving:
1. Nonlinear differential equations
2. Linear differential equations with constant coefficients and initial conditions
3. Partial differential equations without boundary conditions
4. Integral equations only
Show me the answer
**Answer:** 2. Linear differential equations with constant coefficients and initial conditions
**Explanation:**
* The Laplace transform method excels at solving linear ODEs with constant coefficients and given initial conditions (initial value problems).
* It handles discontinuous forcing functions (like step functions) and impulse functions (Dirac delta) naturally.
* The method automatically incorporates the initial conditions.
* For boundary value problems or variable coefficient equations, other methods may be more appropriate.
***
### Final Value Theorem
20\. The Final Value Theorem states: If the limits exist, then $$\lim\_{t \to \infty} f(t) =$$
1. $$\lim\_{s \to 0} sF(s)$$
2. $$\lim\_{s \to \infty} sF(s)$$
3. $$\lim\_{s \to 0} F(s)$$
4. $$\lim\_{s \to \infty} F(s)$$
Show me the answer
**Answer:** 1. $$\lim\_{s \to 0} sF(s)$$
**Explanation:**
* Final Value Theorem: $$\lim\_{t \to \infty} f(t) = \lim\_{s \to 0} sF(s)$$, provided all poles of $$sF(s)$$ have negative real parts (i.e., the limit exists and is finite).
* This theorem gives the steady-state value of a time function without needing the inverse transform.
* Initial Value Theorem (dual): $$\lim\_{t \to 0^+} f(t) = \lim\_{s \to \infty} sF(s)$$.