GATE 2024 EE Q.33: solution and answer

Official answer: 0.0Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Check the symmetry of the signal. $e^{-t^4}$ is even, because $(-t{)}^4=t^4$, and $\cos t$ is even. So $x(t)=e^{-t^4}\cos t$ is real and even: $x(-t)=x(t)$.

2. Symmetry property of the Fourier transform. A real, even signal has a real, even transform: $X(-\omega )=X(\omega )$.

3. The derivative exists. ${\int }_{-\infty }^{\infty }|tx(t)|dt\le {\int }_{-\infty }^{\infty }|t|e^{-t^4}dt<\infty$, so $X(\omega )$ is differentiable and we can differentiate under the integral sign.

4. The derivative of an even function is odd. Differentiate $X(-\omega )=X(\omega )$ to get $-X^{\prime }(-\omega )=X^{\prime }(\omega )$. At $\omega =0$ this reads $X^{\prime }(0)=-X^{\prime }(0)$, so $X^{\prime }(0)=0$.

5. Check with the frequency-differentiation property. $\frac{dX}{d\omega }={\int }_{-\infty }^{\infty }(-jt)x(t)e^{-j\omega t}dt$, so
   $$X^{\prime }(0)=-j{\int }_{-\infty }^{\infty }t{e}^{-t^4}\cos tdt.$$

6. Odd integrand, symmetric limits. $t{e}^{-t^4}\cos t$ is odd (odd $\times$ even $\times$ even). The integral converges absolutely because $e^{-t^4}$ decays so fast. An odd function integrated over $(-\infty ,\infty )$ gives $0$, so $X^{\prime }(0)=0$.

7. Same answer from modulation. Let $G(\omega )=\mathcal{F}\{e^{-t^4}\}$, which is real and even. Then $X(\omega )=\frac{1}{2}[G(\omega -1)+G(\omega +1)]$, and $X^{\prime }(0)=\frac{1}{2}[G^{\prime }(-1)+G^{\prime }(1)]=0$ because $G^{\prime }$ is odd.

8. Rounded to 1 decimal place: $X^{\prime }(0)=0.0$.