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# GATE-2012 ECE Q11 (signals)

by on December 27, 2012

Question 11 on signals from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Solution

From the definition of Laplace transform for a function${f(t)}$ defined for all real numbers ${t\ge 0}$ is,

$F(s) = L\{f(t)\}(s) = \int_0^{\infty}e^{-st}f(t)dt$ , where

$s=\sigma + j\omega$ with real numbers $\sigma$ and $\omega$.

To find the Laplace transform of ${tf(t)}$, let us differentiate the above equation by $s$ on both sides,

$\begin{array}{lll}\frac{d}{ds}F(s)&=&\frac{d}{ds}\int_0^{\infty}e^{-st}f(t)dt\\F^'(s)&=&\int_0^{\infty}\frac{d}{ds}e^{-st}f(t)d(t)&\mbox{ (substituting the order of integration and differentiation)}\\&=&\int_0^{\infty}-te^{-st}f(t)d(t)\\&=&-\int_0^{\infty}e^{-st}\underbrace{tf(t)}d(t)\\&=&-L\{tf(t)\}(s)\end{array}$.

Rearranging,

$\Large{L\{tf(t)\}(s) = \int_0^{\infty}te^{-st}f(t)dt=-F^'(s)}$.

Further using the reciprocal rule from calculus, the derivative of $\frac{1}{g(x)}$ is given by,

$\Large{\frac{d}{dx}$$\frac{1}{g(x)}$$ = -\frac{g^'(x)}{$$g(x)$$^2}}$ where$g(x)\ne 0$.

Applying these two aspects to the problem,

$\begin{array}{lll}L\{tf(t)\}(s)=-F^'(s)&=&-\frac{d}{ds}$$\frac{1}{s^2+s+1}$$\\&=&\frac{2s+1}{(s^2+s+1)^2}\end{array}$.

Based on the above, the right choice is (D) $\Large{\frac{2s+1}{(s^2+s+1)^2}$.

Comment : What does the term ‘unilateral’ stand for in the statement unilateral Laplace transform ?

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

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