{"id":51201,"date":"2024-04-15T23:21:26","date_gmt":"2024-04-15T23:21:26","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=51201"},"modified":"2024-04-15T23:21:26","modified_gmt":"2024-04-15T23:21:26","slug":"for-a-function-gt-it-is-given-that-intlimits_-infty-infty-gleft-t-righte-jomega-tdt-omega-e-2omega-2-for-any-real-value-omega-if","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/for-a-function-gt-it-is-given-that-intlimits_-infty-infty-gleft-t-righte-jomega-tdt-omega-e-2omega-2-for-any-real-value-omega-if\/","title":{"rendered":"For a function g(t), it is given that $$\\int\\limits_{ – \\infty }^{ + \\infty } {g\\left( t \\right){e^{ – j\\omega t}}dt = \\omega {e^{ – 2{\\omega ^2}}}} $$ for any real value $$\\omega $$ . If $$y\\left( t \\right) = \\int\\limits_{ – \\infty }^t {g\\left( \\tau \\right)} d\\tau ,\\,{\\rm{then}}\\,\\int\\limits_{ – \\infty }^{ + \\infty } {y\\left( t \\right)} dt$$ is. . . . . . . ."},"content":{"rendered":"

[amp_mcq option1=”0″ option2=”#NAME?” option3=”$$ – {j \\over 2}$$” option4=”$${j \\over 2}$$” correct=”option3″]<\/p>\n

The correct answer is $\\boxed{{j \\over 2}}$.<\/p>\n

To solve this, we can use the following property of the Fourier transform:<\/p>\n

$$\\int_{-\\infty}^{\\infty} y(t) e^{-j\\omega t} dt = \\pi \\left[ \\text{Re} \\left( \\hat{y}(\\omega) \\right) + j \\text{Im} \\left( \\hat{y}(\\omega) \\right) \\right]$$<\/p>\n

where $\\hat{y}(\\omega)$ is the Fourier transform of $y(t)$.<\/p>\n

In this case, we are given that<\/p>\n

$$\\hat{g}(\\omega) = \\omega e^{-2\\omega^2}$$<\/p>\n

Therefore, the Fourier transform of $y(t)$ is<\/p>\n

$$\\hat{y}(\\omega) = \\pi \\omega e^{-2\\omega^2}$$<\/p>\n

The inverse Fourier transform of $\\hat{y}(\\omega)$ is<\/p>\n

$$y(t) = \\frac{\\pi}{2} \\int_{-\\infty}^{\\infty} \\omega e^{-2\\omega^2} e^{j\\omega t} d\\omega = \\frac{j}{2}$$<\/p>\n

Therefore,<\/p>\n

$$\\int_{-\\infty}^{\\infty} y(t) dt = \\frac{j}{2}$$<\/p>\n

\n

Here is a brief explanation of each option:<\/p>\n