Laplace Transform MCQ Quiz in मल्याळम - Objective Question with Answer for Laplace Transform - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

L [f(t)] = F(s)

L [sin(at) u(t)] ↔ $\frac{a}{s^2+a^2}$

Application:

With v(t) = sin (10t) u(t), the Laplace transform will be:

$V(s)=\frac{10}{s^2+100}$

Important Points 

Some common Laplace transforms are:

Concept:

$e^{-at}u\left(t\right)\to \frac{1}{s+a}{R}_e\left\{s\right\}>-a$ 

$u\left(t\right)\leftrightarrow \frac{1}{s}{R}_e\left\{s\right\}>-a$

Calculation:

Given:

$F\left(s\right)=\frac{\left(s+2\right)}{s\left(s+3\right)\left(s+4\right)}$ 

$F\left(s\right)=\frac{s+2}{s\left(s+3\right)\left(s+4\right)}$

$F\left(s\right)=\frac{A}{s}+\frac{B}{s+3}+\frac{C}{s+4}$ 

After Partial fraction we’ll get;

$f\left(s\right)=\frac{1}{6s}+\frac{1}{3\left(s+3\right)}-\frac{1}{2\left(s+4\right)}$ 

Now, Taking Laplace inverse of F(s);

$f\left(t\right)=\frac{1}{6}+\frac{1}{3}{e}^{-3t}-\frac{1}{2}{e}^{-4t}$ 

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)

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

Calculation:

Laplace transform of y(t) = e^-3t cos 5t

$L(\cos 4t)=\frac{s}{(s^2+4^2)}$

By applying the property of shifting,

$$\begin{gathered} L\left({\mathbf{e}}^{-5\mathbf{t}}\mathbf{cos}4\mathbf{t}\right)=\frac{(s+5)}{{(s+5)}^2+4^2} \\ =\frac{(s+5)}{{(s+5)}^2+16} \end{gathered}$$

Solution:

The given periodic function can be expressed as the summation of shifted unit step functions as shown:

Similarly x₃(t), x₄(t), … will be shifted unit step functions.

x₁(t) = u(t) - u(t - 1)

x₂(t) = u(t - 2) - u(t - 3)

x(t) can be expressed as:

x(t) = x₁(t) + x₂(t) + x₃(t) + ….

x(t) = u(t) - u(t - 1) + u(t - 2) - u(t - 3)+ …      ---(1)

$u\left(t\right)\stackrel{L.T}{\leftrightarrow }\frac{1}{s}$ 

Time-shifting affects the frequency as:

$u\left(t-t_0\right)\stackrel{L.T.}{\leftrightarrow }\frac{1}{s}\cdot e^{-s{t}_0}$ 

The Laplace transform of x(t) from equation (1), can be written as:

$=\frac{1}{s}-\frac{1}{s}{e}^{-s}+\frac{1}{s}{e}^{-2s}-\frac{1}{s}{e}^{-3s}+\frac{1}{s}{e}^{4s}-\frac{1}{5}{e}^{-5s}+\dots$ 

$=\frac{1}{s}\left[1+e^{-2s}+e^{-4s}+\dots \right]-\frac{1}{s}\left[e^{-s}+e^{-3s}+e^{-5s}+\dots \right]$ 

Simplifying the geometric series, we can write:

$=\frac{1}{s}\left[\frac{1}{1-e^{-2s}}\right]-\frac{1}{s}\left[\frac{e^{-s}}{1-e^{-2s}}\right]$ 

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

$=\frac{1}{s}\left[\frac{1-e^{-s}}{\left(1+e^{-s}\right)\left(1-e^{-s}\right)}\right]$ 

$X\left(s\right)=\frac{1}{s}\left[\frac{1}{1+e^{-s}}\right]$         ---(2)

Given $X\left(s\right)=\frac{1}{s\left(1+e^{ks}\right)}$ 

Comparing this with equation (2) we get:

k = -1

Given the differential equation,

$\frac{d^{2}x}{d{t}^2}=-9x$

Taking the Laplace transform,

s²X(s) – sX(0) – X^’(0) = -9X(s)

or s²X(s) – s(1) – (1) = - 9X(s)

or X(s) (s²+9) = s + 1

$X\left(s\right)=\frac{s+1}{s^2+9}=\frac{s}{s^2+9}+\frac{1}{s^2+9}$

$x\left(t\right)=cos3t+\frac{1}{3}\sin 3t$ (Taking  inverse Laplace.)

Laplace Transform:

The Laplace transform of $t^{n}u(t)=\frac{n}{s^{n+1}}$

1.)  $L[t^0]=\frac{1}{s}$

2.) $L[t]=\frac{1}{s^2}$

3.) $L[sin\omega t]=\frac{\omega }{s^2+{\omega }^2}$

4.) $L[cos\omega t]=\frac{s}{s^2+{\omega }^2}$

Hence, option 2 is correct.

The Fourier transform of a function is equal to its two-sided Laplace transform evaluated on the imaginary axis of the s-plane.

Explanation:

The Fourier transform of x(t) is, denoted by X(jω), is defined as:

$X\left(j\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-j\omega t}dt$

The Laplace transform of x(t), denoted by X(s), is defined as:

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

Where s is a continuous complex variable.

We can also express s as: s = σ + jω

Where σ and ω are the real and imaginary parts of s, respectively

The Laplace transform can be written as:

$X\left(\sigma +j\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-\left(\sigma +j\omega \right)t}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left(x\left(t\right)e^{-\sigma t}\right)e^{-j\omega t}dt$

By comparing the above Laplace and Fourier transform equations, it is clear that Laplace transform of x(t) is equal to the Fourier transform of $x\left(t\right)e^{-\sigma t}$.

When σ = 0 or s = jω, both are identical.

${X\left(s\right)|}_{s=j\omega }=X\left(j\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-j\omega t}dt$

That is, Laplace transform generalizes Fourier transform.

• In the partial fraction method of Laplace transforms, the function F(s) is generally represented as a rational function, which is the ratio of two polynomials:
   $F(S)=\frac{N(S)}{D(S)}=\frac{Numeratorpolynomial}{Denominatorpolynomial}$
• The goal of partial fraction decomposition is to express $F(s)$ as a sum of simpler fractions to facilitate the inverse Laplace transform.
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  F(s).

Concept:

Transfer function:

It is the ratio of the Laplace transform of the output variable to the input variable with initial conditions zero.

TF = $\frac{L\left[output\right]}{L\left[input\right]}|$L[output]L[input]|" id="MathJax-Element-28-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">L[output]L[input]∣∣L[output]L[input]|" id="MathJax-Element-71-Frame" role="presentation" style="position: relative;" tabindex="0">L[output]L[input]∣∣L[output]L[input]|" id="MathJax-Element-19-Frame" role="presentation" style="position: relative;" tabindex="0">L[output]L[input]∣∣L[output]L[input]|" id="MathJax-Element-21-Frame" role="presentation" style="position: relative;" tabindex="0">L[output]L[input]∣∣$\frac{L\left[output\right]}{L\left[input\right]}|$ L[output]L[input]| L[output]L[input]| L[output]L[input]|  

Initial conditions = 0

TF = C(s) / R(s)

If R(s) = 1 i.e impulse input

TF = C(s)

Impulse response (IR) = c(t) = L-1 [C(s)] = L-1[TF]

TF = L-1[IR]

Hence, TF may be also called the impulse response of the system.

Explanation:

From the property of Laplace transform

$LT[e^{-at}]\to \frac{1}{s+a}$

Put a = 2, 

$LT[e^{-2t}]\to \frac{1}{s+2}$

Hence, T.F = 1 / (s+2)

Additional Information

Concept:

The Laplace transform F(s) of a function f(t) is defined by:

$L\text{(}f\left(t\right)\}=F\left(s\right)=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{e}^{-st}f\left(t\right)dt$

From the time-shifting property of Laplace transform:

$L\left\{f(t-a)\right\}=e^{-as}F\left(s\right)$

Application:

L{x(t)} = X(s)

L(x(t-t0)) = $e^{-s{t}_0}X\left(s\right)$