Laplace Transform MCQ Quiz - Objective Question with Answer for Laplace Transform - Download Free PDF

Explanation:

Laplace Transform of a Unit Parabolic Function

Definition: The Laplace transform is a widely used integral transform in engineering and physics that converts a function of time (t) into a function of complex frequency (s). The Laplace transform of a function f(t) is defined as:

$$\mathcal{L}\{f(t)\}=F(s)={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)dt$$

Where:

• $\mathcal{L}$ denotes the Laplace transform operator
• $f(t)$ is the time-domain function
• $F(s)$ is the Laplace-transformed function in the frequency domain
• $s$ is a complex number frequency parameter

Unit Parabolic Function: The unit parabolic function, also known as the second-order ramp function, is given by:

$$f(t)=\frac{t^2}{2}$$

for $t\ge 0$. This function represents a parabolic increase starting from $t=0$.

Laplace Transform of the Unit Parabolic Function:

To find the Laplace transform of the unit parabolic function $f(t)=\frac{t^2}{2}$, we use the definition of the Laplace transform:

$$\mathcal{L}\left\{\frac{t^2}{2}\right\}={\int }_0^{\mathrm{\infty }}{e}^{-st}\cdot \frac{t^2}{2}dt$$

We can simplify this by factoring out the constant $\frac{1}{2}$:

$$\mathcal{L}\left\{\frac{t^2}{2}\right\}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-st}{t}^{2}dt$$

To solve this integral, we use the standard Laplace transform formula for $t^n$:

$$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$

For $n=2$, we have:

$$\mathcal{L}\{t^2\}=\frac{2!}{s^{2+1}}=\frac{2}{s^3}$$

Therefore, the Laplace transform of $\frac{t^2}{2}$ is:

$$\mathcal{L}\left\{\frac{t^2}{2}\right\}=\frac{1}{2}\cdot \frac{2}{s^3}=\frac{1}{s^3}$$

Thus, the Laplace transform of a unit parabolic function occurring at $t=0$ is $\frac{1}{s^3}$.

Correct Option Analysis:

The correct option is:

Option 1: $\frac{1}{s^3}$

This option correctly represents the Laplace transform of the unit parabolic function $\frac{t^2}{2}$, as derived above.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: 1

This option is incorrect because the Laplace transform of the unit parabolic function is not a constant value. The correct Laplace transform has a dependency on the complex frequency parameter $s$.

Option 3: $\frac{1}{s^2}$

This option is incorrect as it represents the Laplace transform of a unit ramp function $t$, not the unit parabolic function $\frac{t^2}{2}$. The unit ramp function's Laplace transform is indeed $\frac{1}{s^2}$, but for the parabolic function, it is $\frac{1}{s^3}$.

Option 4: $\frac{1}{s}$

This option is incorrect because it represents the Laplace transform of a unit step function, not the unit parabolic function. The unit step function (Heaviside function) has a Laplace transform of $\frac{1}{s}$.

Conclusion:

Understanding the Laplace transform of different functions is essential for solving differential equations and analyzing systems in the frequency domain. The unit parabolic function, represented by $\frac{t^2}{2}$, has a Laplace transform of $\frac{1}{s^3}$. This transform is particularly useful in control systems and signal processing, where such functions are commonly encountered. By comparing the correct option with the incorrect ones, we can clearly see the importance of accurately applying the Laplace transform formulas to various functions.

Explanation:

Laplace Transform of $e^{at}\cos {\omega }_{0}t\cdot u(t)$

Definition: The Laplace transform is an integral transform that converts a function of time (usually denoted as $t$) into a function of a complex variable (usually denoted as $s$). It is particularly useful for solving differential equations and analyzing linear time-invariant systems.

The Laplace transform of a function $f(t)$, where $t\ge 0$, is defined as:

$$\mathcal{L}\{f(t)\}=F(s)={\int }_0^{\mathrm{\infty }}f(t)e^{-st}dt$$

Laplace Transform of $e^{at}\cos {\omega }_{0}t\cdot u(t)$:

Given the function $f(t)=e^{at}\cos {\omega }_{0}t\cdot u(t)$, we need to find its Laplace transform. Here, $u(t)$ is the unit step function which ensures that the function is defined for $t\ge 0$.

The Laplace transform of $e^{at}\cos {\omega }_{0}t\cdot u(t)$ can be derived using the known Laplace transform pair:

$$\mathcal{L}\{e^{at}\cos {\omega }_{0}t\}=\frac{s+a}{(s+a{)}^2+{\omega }_0^2}$$

This formula is derived from the standard Laplace transform properties and the use of the formula for the cosine function combined with the exponential function.

Derivation:

Let us derive the Laplace transform step-by-step:

1. Consider the function $f(t)=e^{at}\cos {\omega }_{0}t$.
2. The Laplace transform is defined as:

$$\mathcal{L}\{f(t)\}={\int }_0^{\mathrm{\infty }}{e}^{at}\cos {\omega }_{0}t\cdot e^{-st}dt$$

3. Combine the exponential terms:

$$\mathcal{L}\{f(t)\}={\int }_0^{\mathrm{\infty }}{e}^{(a-s)t}\cos {\omega }_{0}tdt$$

4. Use the Laplace transform of $\cos {\omega }_{0}t$, which is $\frac{s}{s^2+{\omega }_0^2}$, and shift it by $a$:

$$\mathcal{L}\{e^{at}\cos {\omega }_{0}t\}=\frac{s}{s^2+{\omega }_0^2}{|}_{s\to s-a}$$

5. Apply the shift property:

$$\mathcal{L}\{e^{at}\cos {\omega }_{0}t\}=\frac{s-(a)}{(s-(a){)}^2+{\omega }_0^2}$$

6. Simplify the expression:

$$\mathcal{L}\{e^{at}\cos {\omega }_{0}t\}=\frac{s+a}{(s+a{)}^2+{\omega }_0^2}$$

Conclusion:

Thus, the Laplace transform of $e^{at}\cos {\omega }_{0}t\cdot u(t)$ is:

$$\boxed{{\displaystyle \frac{s+a}{(s+a{)}^2+{\omega }_0^2}}}$$

This matches Option 3 in the given problem, confirming it as the correct answer.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: $\frac{a}{(s+a{)}^2+{\omega }_0^2}$

This option is incorrect because the numerator should contain $s+a$ rather than just $a$. The Laplace transform involves both the $s$ and $a$ terms in the numerator as derived above.

Option 2: $\frac{s}{(s+a{)}^2+{\omega }_0^2}$

This option is also incorrect as it does not include the $a$ term in the numerator. The correct transform incorporates both $s$ and $a$ in the numerator.

Option 4: $\frac{{\omega }_0}{(s+a{)}^2+{\omega }_0^2}$

This option is incorrect as well since the numerator should be $s+a$, not ${\omega }_0$. The correct expression must include the $s$ term from the original function's Laplace transform.

Conclusion:

Understanding the Laplace transform of exponential and trigonometric functions is critical for solving differential equations and analyzing linear systems. The correct Laplace transform of $e^{at}\cos {\omega }_{0}t\cdot u(t)$ is $\frac{s+a}{(s+a{)}^2+{\omega }_0^2}$, as it accurately represents the combined effects of the exponential growth/decay and the cosine function in the $s$-domain.

${\mathrm{L}}^{-1}\left[\frac{1}{{\left(\mathrm{S}+\mathrm{a}\right)}^2+{\mathrm{b}}^2}\right]=\frac{1}{\mathrm{b}}{\mathrm{e}}^{-\mathrm{a}\mathrm{t}}\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{b}\mathrm{t}$

a = 2

b = 9

${\mathrm{L}}^{-1}\left[\frac{1}{{\left(\mathrm{S}+\mathrm{a}\right)}^2+{\mathrm{b}}^2}\right]=\frac{1}{2}{\mathrm{e}}^{-2\mathrm{t}}\mathrm{S}\mathrm{i}\mathrm{n}2\mathrm{t}$

Concept:

The initial value theorem gives:

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

Calculation:

Using the initial value theorem,

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}.\frac{(2\mathrm{s}+3)}{(\mathrm{s}+1)(\mathrm{s}+3)}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}{\mathrm{s}}^2.\frac{(2+\frac{3}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{1}{\mathrm{s}})(1+\frac{3}{\mathrm{s}})}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}.\frac{(2+\frac{3}{\mathrm{s}})}{(1+\frac{1}{\mathrm{s}})(1+\frac{3}{\mathrm{s}})}}$

∴ x(0+) = 2

Calculation: To find the Laplace transform of sinh(at), we use the definition of the Laplace transform: $$\mathcal{L}\{f(t)\}={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)dt$$ Given $f(t)=\sinh (at)$, we have: $$\mathcal{L}\{\sinh (at)\}={\int }_0^{\mathrm{\infty }}{e}^{-st}\sinh (at)dt$$ Recall that $\sinh (at)=\frac{e^{at}-e^{-at}}{2}$, so: $$\mathcal{L}\{\sinh (at)\}={\int }_0^{\mathrm{\infty }}{e}^{-st}\left(\frac{e^{at}-e^{-at}}{2}\right)dt$$ This can be split into two integrals: $$\mathcal{L}\{\sinh (at)\}=\frac{1}{2}({\int }_0^{\mathrm{\infty }}{e}^{-st}{e}^{at}dt-{\int }_0^{\mathrm{\infty }}{e}^{-st}{e}^{-at}dt)$$ Simplifying the exponents, we get: $$\mathcal{L}\{\sinh (at)\}=\frac{1}{2}({\int }_0^{\mathrm{\infty }}{e}^{-(s-a)t}dt-{\int }_0^{\mathrm{\infty }}{e}^{-(s+a)t}dt)$$ Evaluating each integral, we find: $${\int }_0^{\mathrm{\infty }}{e}^{-(s-a)t}dt=\frac{1}{s-a}\text{for}s>a$$ $${\int }_0^{\mathrm{\infty }}{e}^{-(s+a)t}dt=\frac{1}{s+a}\text{for}s>-a$$ So, we have: $$\mathcal{L}\{\sinh (at)\}=\frac{1}{2}(\frac{1}{s-a}-\frac{1}{s+a})$$ Combining the fractions: $$\mathcal{L}\{\sinh (at)\}=\frac{1}{2}\left(\frac{(s+a)-(s-a)}{(s-a)(s+a)}\right)$$ Simplifying the numerator: $$\mathcal{L}\{\sinh (at)\}=\frac{1}{2}\left(\frac{2a}{s^2-a^2}\right)$$ Therefore: $$\mathcal{L}\{\sinh (at)\}=\frac{a}{s^2-a^2}$$ Final Answer: The Laplace transform of $\sinh (at)$ is $\frac{a}{s^2-a^2}$. The correct answer is option 3.

Concept:

Bilateral Laplace transform:

$L\left[x\left(t\right)\right]=x\left(s\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-st}dt$

Unilateral Laplace transform:

$L\left[x\left(t\right)\right]=x\left(s\right)=\underset{0}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-st}dt$

Some important Laplace transforms:

	f(t)	f(s)	ROC
1.	δ(t)	1	Entire s-plane
2.	e-at u(t)	$\frac{1}{s+a}$	s > - a
3.	e-at u(-t)	$\frac{1}{s+a}$	s < - a
4.	cos ω0 t u(t)	$\frac{s}{s^2+{\omega }_0^2}$	s > 0
5.	te-at u(t)	$\frac{1}{{\left(s+a\right)}^2}$	s > - a
6.	sin ω0t u(t)	$\frac{{\omega }_0}{s^2+{\omega }_0^2}$	s > 0
7.	u(t)	1/s	s > 0

 

Calculation:

$\sin \omega t.u(t)\leftrightarrow \frac{\omega }{s^2+{\omega }^2}$

By applying frequency differentiation property,

$e^{-at}\sin \omega t.u(t)\leftrightarrow \frac{\omega }{{\left(s+a\right)}^2+{\omega }^2}$

Concept:

Final value theorem:

A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression

Final value theorem states that the final value of a system can be calculated by

$x\left(\mathrm{\infty }\right)=\underset{s\to 0}{lim}sX\left(s\right)$

 Where X(s) is the Laplace transform of the function.

For the final value theorem to be applicable system should be stable in steady-state and for that real part of poles should lie in the left side of s plane.

Initial value theorem:

$x\left(0\right)=\underset{t\to 0}{lim}x\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sX\left(s\right)$

It is applicable only when the number of poles of X(s) is more than the number of zeros of X(s).

Calculation:

Given that, $X\left(s\right)=\frac{4s+1}{s^2+6s+3}$

Initial value,

 $$\begin{gathered} x\left(0\right)=\underset{s\to \mathrm{\infty }}{lim}s\frac{4s+1}{s^2+6s+3} \\ =\underset{\frac{1}{s}\to 0}{lim}\frac{4+\frac{1}{s}}{1+\frac{6}{s}+\frac{3}{s^2}}=4 \end{gathered}$$

Concept:

 

Initial value theorem:

The initial value theorem is one of the basic properties of the Laplace transform used to find the response of the system at the initial state (t = 0) in the Laplace domain. Mathematically it is given by

$\mathbf{f}\left(0^{+}\right)=\underset{t\to 0}{lim}\mathbf{f}\left(\mathbf{t}\right)=\underset{s\to \mathrm{\infty }}{lim}sF\left(s\right)$

Where

f(t) is system function

F(s) is Laplace transform of system function f(t)

f(0+) is the initial value of the system

NOTE:

• For the final value theorem to be applicable system should be stable in a steady-state and for that real part of the poles should lie on the left side of the s plane.
• In the given problem exactly the final value theorem is not applied but just X(0+) is calculated.
   

Calculation:

Given that, 

$X(s)=\frac{3s+5}{s^2+10s+21}$

$\mathbf{x}\left(0^{+}\right)=\underset{x\to 0}{lim}\mathbf{x}\left(\mathbf{t}\right)=\underset{s\to \mathrm{\infty }}{lim}sX\left(s\right)$

$\mathbf{x}\left(0^{+}\right)=\underset{x\to 0}{lim}\mathbf{f}\left(\mathbf{t}\right)=\underset{s\to \mathrm{\infty }}{lim}s.\frac{3s+5}{s^2+10s+21}$

$=\underset{s\to \mathrm{\infty }}{lim}\frac{3s^2+5s}{s^2+10s+21}$

= 3

The given Laplace transform of f(t) as:

$F\left(s\right)=\frac{5s+3}{s(s+1)}$

Performing partial fraction:

$\frac{5s+3}{s(s+1)}=\frac{A}{s}+\frac{B}{s+1}$

⇒ 5s + 3 = A(s + 1) + B(s)

⇒ Putting s = 0

5(0) + 3 = A

A = 3

Putting s = -1

⇒ 5(-1) + 3 = -B

-2 = -B

B = 2

F(s) can now be written as:

$F\left(s\right)=\frac{3}{s}+\frac{2}{s+1}$

Taking the Inverse Laplace transform, we get:

f(t) = 3 u(t) + 2e^-t u(t)

Concept:

If x(t) ↔ X(s)

Then $\frac{x\left(t\right)}{t}\leftrightarrow \underset{s}{\overset{\mathrm{\infty }}{\int }}X\left(s\right)ds$

Final value theorem states that:

$\underset{t\to \mathrm{\infty }}{lim}f\left(t\right)=\underset{s\to 0}{lim}sf\left(s\right)$

Calculation:

Given $x\left(t\right)=\left(\frac{\sin t}{\pi t}\right)u\left(t\right)$

$\sin t\leftrightarrow \frac{1}{s^2+1}$

So, $\frac{\sin t}{t}\leftrightarrow \underset{s}{\overset{\mathrm{\infty }}{\int }}\frac{1}{s^2+1}ds$

= tan^-1 (∞) – tan^-1 (s)

$=\frac{\pi }{2}-{\tan }^{-1}\left(s\right)$

$\frac{1}{\pi }\left(\frac{\sin t}{t}\right)\leftrightarrow \frac{1}{\pi }\left[\frac{\pi }{2}-{\tan }^{-1}\left(s\right)\right]$

The above signal x(t) is passed through causal LTI system with:

$H\left(s\right)=\frac{1}{s}$ ,

Output Y(s) = X(s).H(s)

$=\frac{1}{s}\left[\frac{1}{\pi }\left(\frac{\pi }{2}-{\tan }^{-1}\left(s\right)\right)\right]$

$\underset{\mathrm{t}\to \mathrm{\infty }}{lim}\mathrm{y}\left(\mathrm{t}\right)=\underset{\mathrm{s}\to 0}{lim}\mathrm{s}\mathrm{Y}\left(\mathrm{s}\right)$

$\underset{t\to \mathrm{\infty }}{\mathrm{y}\left(\mathrm{t}\right)}={}_{\underset{s\to 0}{lim}\frac{s}{s}\left(\frac{1}{\pi }\right)\left(\frac{\pi }{2}-{\tan }^{-1}\left(s\right)\right)}$

$\Rightarrow \mathrm{l}\mathrm{i}{\mathrm{m}}_{s\to 0}\left(\frac{1}{\pi }.\frac{\pi }{2}-\frac{1}{\pi }.{\tan }^{-1}\left(s\right)\right)=\frac{1}{2}=0.5$

Concept:

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

Calculation:

Using the initial value theorem,

$\mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}\mathrm{s}\mathrm{X}(\mathrm{s})}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\mathrm{s}.\frac{(3\mathrm{s}+5)}{({\mathrm{s}}^2+10\mathrm{s}+21)}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{{\mathrm{s}}^2(3+\frac{5}{\mathrm{s}})}{{\mathrm{s}}^2(1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2})}}$

$\Rightarrow \mathrm{x}(0^{+})={\displaystyle \underset{\mathrm{s}\to \mathrm{\infty }}{lim}\frac{3+\frac{5}{\mathrm{S}}}{1+\frac{10}{\mathrm{S}}+\frac{21}{{\mathrm{S}}^2}}}$

∴ x(0+) = 3

Alternate method:

using Inverse Laplace Transform method,

we have,

$\mathrm{X}(\mathrm{s})=\frac{3\mathrm{S}+5}{{\mathrm{S}}^2+10\mathrm{S}+21}=\frac{3\mathrm{S}+5}{{\mathrm{S}}^2+10\mathrm{S}+21+4-4}$

$\mathrm{X}(\mathrm{s})=\frac{3\mathrm{S}+5}{(\mathrm{S}+5{)}^2-2^2}=\frac{3\mathrm{S}+15-10}{(\mathrm{S}+5{)}^2-2^2}$

$\mathrm{X}(\mathrm{S})=\frac{3(\mathrm{S}+5)}{(\mathrm{S}+5{)}^2-2^2}-\frac{10}{(\mathrm{S}+5{)}^2-2^2}$

Taking inverse Laplace transform

x(t) = 3e-5t cosh2t - 5e-5t sinh2t

x(t) = e-5t(3 cosh2t - 5 sinh2t)

At t = 0+,

x(0+) = e0(3cosh 0 - 5 sinh0)

x(0+) = 1(3 - 0)

x(0+) = 3

Concept:

Let the Laplace transform of function of f(t) is L [f(t)] = F (s)

By using first shifting rule

 If L [f(t)] = F (s), then

L [e^at f(t)] = F (s – a)

$L\left(t^n\right)=\frac{n!}{\left(s^{n+1}\right)}$

$L(\cos at)=\frac{s}{(s^2+a^2)}$

Calculation:

Laplace transform of y(t) = 3 t⁴

L [y(t)] = L (3 t⁴)

$L\left(3t^4\right)=\frac{3\times 4!}{\left(s^{4+1}\right)}=\frac{3\times (4\times 3\times 2\times 1)}{\left(s^5\right)}$

$L\left(3t^4\right)=\frac{72}{\left(s^5\right)}$

Important Points

Some common Laplace transforms are:

f(t)	F(s)	ROC
δ (t)	1	All s
u(t)	$\frac{1}{s}$	Re (s) > 0
t	$\frac{1}{s^2}$	Re (s) > 0
tn	$\frac{n!}{s^{n+1}}$	Re (s) > 0
e-at	$\frac{1}{s+a}$	Re (s) > -a
t e-at	$\frac{1}{{\left(s+a\right)}^2}$	Re (s) > -a
tn e-at	$\frac{n!}{{\left(s+a\right)}^n}$	Re (s) > -a
Sin at	$\frac{a}{s^2+a^2}$	Re (s) > 0
Cos at	$\frac{s}{s^2+a^2}$	Re (s) > 0

Given that,

Poles = (2, -1) and (-2, 1)

Zeros = (2, 1), (-2, -1)

Now, drawing the Poles and Zeros to the Real and Imaginary axis-

The transfer function can be written as

$H\left(s\right)=\frac{\left[s-\left(2+j\right)\right]\left[s-\left(-2-j\right)\right]}{\left[s-\left(2+j\right)\right]\left[s-\left(-2+j\right)\right]}$

$H\left(s\right)=\frac{s^2-3-4j}{s^2-3+4j}$

Observations:

• Magnitude of H(jω) = 1. Therefore, the given system is an all-pass filter.
• Since one pole on the RHS, the system is unstable.
• Since the pole does not have complex conjugate poles and zeros present, the system is complex.

$\frac{Y\left(s\right)}{X\left(s\right)}=\frac{s-2}{s+3}$

S Y(s) + 3 Y(s) = S X(s) – 2 X(s)

If initial conditions are included then

S Y(s) – Y(0) + 3 Y(s) = S X(s) – X(0) – 2 X(s)

Y(0) = -2,         X(0^-) = 0

S Y(s) + 2 + 3 Y(s) = S X(s) – 2 X(s)

S Y(s) + 3 Y(s) + 2 = (s – 2) ⋅ X(s)

Put $X\left(s\right)=\frac{-3}{s-2}$

S Y(s) + 3 Y(s) = -5

$Y\left(s\right)=\frac{-5}{s+3}$

y(t) = – 5 e^-3t u(t)

y(∞) = 0

or         Y(s) = X(s) ⋅ H(s)

$=\left(\frac{-3}{s-2}\right).\left(\frac{s-2}{s+3}\right)=\frac{-3}{s+3}$

If initial conditions included

S Y(s) – Y(0) + 3 Y(s) = – 3

$Y\left(s\right)=\frac{-5}{s+3}$

y(t) = – 5 e^-3t u(t)

Concept:

Some of the common Laplace Transform is as shown:

Signal	Transform	ROC
δ(t)	1	All s
u(t)	1/s	Re (s) > 0
cos ω0t u(t)	$\frac{s}{s^2+{\omega }_0^2}$	Re (s) > 0
sin ω0t u(t)	$\frac{{\omega }_0}{s^2+{\omega }_0^2}$	Re (s) > 0