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# GATE-2012 ECE Q34 (signals)

by on December 29, 2012

Question 34 on signals from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Solution

Let us Laplace transform to find $y(t)$ and later

$F(s) = L\{f(t)\}(s) = \int_{0^-}^{\infty}e^{-st}f(t)dt$,  where$s=\sigma + j\omega$ with real numbers $\sigma$ and $\omega$.

Using integration by parts,

$\begin{array}{lll}F(s)&=&\int_{0^-}^{\infty}e^{-st}f(t)dt&\\&=&$f(t)\frac{e^{-st}}{-s}$_{0^-}^{\infty}\mbox{ }-\int_{0^-}^{\infty}\frac{df(t)}{dt}\frac{e^{-st}}{-s}dt\\&=&\frac{f(0^-)}{s}\mbox{ }+\frac{1}{s}\underbrace{\int_{0^-}^{\infty}e^{-st}f'(t)dt}_{L\{f'(t)\}}\end{array}$.

Rearranging,

$\Large{\begin{array}{lll}L\{f'(t)\}&=&sF(s)-f$$0^-$$\end{array}}$.

Extending this to find the Laplace Transform of the second derivative of the function,

$\Large{{\begin{array}{lll}L\{f''(t)\}&=&sL\{f'(t)\}-f'$$0^-$$\\&=&s$$F(s)-f\(0^-$$\)-f'$$0^-$$\\&=&sF(s)-sf$$0^-$$-f'$$0^-$$\end{array}}$.

Coming back to the problem,

${\frac{d^2y(t)}{dt}+2\frac{dy(t)}{dt}+y(t) = \delta(t)}$

Taking Laplace transform,

$\begin{array}{lll}s^2Y(s)-sy(0^-)-y'(0^-)+2$sY(s)-y(0^-)$+Y(s)&=&1\$s^2+2s+1$Y(s)-$s+2$y(0^-)-y'(0^-)&=&1\$s^2+2s+1$Y(s)-($s+2$*-2)-0&=&1\$s^2+2s+1$Y(s)+2s+4&=&1\\Y(s)&=&\frac{-2s-3}{s^2+2s+1}\end{array}$.

To find the inverse Laplace transform, let us revisit the Laplace transform for some simple functions.

For $f(t) = e^{-t}$, the Laplace transform is,
$\begin{array}{lll}F(s)&=&\int_0^{\infty}e^{-st}e^{-t}u(t)dt\\&=&\int_0^{\infty}e^{-(s+1)t}dt\\&=&\frac{1}{s+1}\end{array}$.

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

Also from the earlier discussion in this post,

${\begin{array}{lll}L\{\frac{d}{dt}te^{-t}\}(s)&=&sF(s)-f$$0^-$$\\&=&\frac{s}{(s+1)^2}\end{array}}$

Applying the above equations to find the inverse Laplace transform

$\begin{array}{lll}Y(s)&=&\frac{-2s}{(s+1)^2}+\frac{-3}{(s+1)^2}\\y(t)&=&-2\frac{d}{dt}$te^{-t}$ + -3te^{-t}\\&=&+2te^{-t}-2e^{-t}-3te^{-t}\\&=&-2e^{-t}-te^{-t}\end{array}$.

Taking the differential,

$\begin{array}{lll}\frac{dy(t)}{dt}&=&\frac{d}{dt}\(-2e^{-t}-te^{-t})\\&=&2e^{-t}+te^{-t}-e^{-t}\\&=&e^{-t}+te^{-t}\end{array}$.

Plugging in ${t=0^{+}}$ ,

$\begin{array}{lll}\frac{dy}{dt}|_{t=0^{+}}&=&e^{-t}+te^{-t}|_{t=0^+}&=&1\end{array}$

Based on the above, the right choice is (D) 1

## References

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

[2] Wiki entry on Laplace transform of function’s derivative

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