GATE 2018 EE Q.40: solution and answer

Official answer: 9.5 to 10.5Source: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Standard pair: $e^{-at}u(t)⟷\frac{1}{a+j\omega }$ for $a>0$. Here $a=10$.

2. Frequency-differentiation property: $tx(t)⟷j\frac{dX(\omega )}{d\omega }$. For $X_1(\omega )=\frac{1}{10+j\omega }$, $\frac{d{X}_1}{d\omega }=\frac{-j}{(10+j\omega {)}^2}$, so $j\frac{d{X}_1}{d\omega }=j\cdot \frac{-j}{(10+j\omega {)}^2}=\frac{1}{(10+j\omega {)}^2}$.

3. Hence $$x(t)=t{e}^{-10t}u(t).$$ The same result follows from convolving $e^{-10t}u(t)$ with itself, since a product (here a square) in frequency is a convolution in time: ${\int }_0^t{e}^{-10\tau }{e}^{-10(t-\tau )}d\tau =t{e}^{-10t}$ for $t>0$.

4. At $t=1$: $x(1)=1\cdot e^{-10}=e^{-10}$, which is positive, so its logarithm is defined.

5. $\ln x(1)=\ln e^{-10}=-10$, so $\left|\ln x(1)\right|=10.0$.