For a function g(t), it is given that \(\mathop \smallint \limits
If, \(y\left( t \right) = \mathop \smallint \limits_{ - \infty }^t g\left( \tau \right)d\tau \), then \(\mathop \smallint \limits_{ - \infty }^\infty y\left( t \right)dt\) is:
A. 0
B. <span class="math-tex">\(-j\)</span>
C. <span class="math-tex">\(-\frac{j}{2}\)</span>
D. <span class="math-tex">\(\frac{j}{2}\)</span>
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Right Answer is: B
SOLUTION
\(G\left( \omega \right)\) =\(\mathop \smallint \limits_{ - \infty }^{ + \infty } g\left( t \right){e^{ - j\omega t}}dt = \omega {e^{ - 2{\omega ^2}}}\)
As \(G\left( \omega \right)\)= \(-G\left( -\omega \right)\)
\(G\left( \omega \right)\) is odd symmetry
\(\therefore \;\mathop \smallint \limits_{ - \infty }^\infty g\left( t \right)dt = 0\)
Now,
\(y\left( t \right) = \mathop \smallint \limits_{ - \infty }^t g\left( \tau \right)d\tau = g\left( t \right)*u\left( t \right)\)
\(\therefore Y\left( {j\omega } \right) = U\left( {j\omega } \right)G\left( {j\omega } \right)\)
\(Y\left( {j\omega } \right) = \omega {e^{ - 2{\omega ^2}}}\left( {\frac{1}{{j\omega }} + \pi \delta \left( \omega \right)} \right)\)
\(Y\left( {j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty y\left( t \right){e^{ - j\omega t}}dt \)
\(= \;{e^{ - 2{\omega ^2}}}\left( {\frac{1}{j} + \omega \pi \delta \left( \omega \right)} \right)\)
\( = {e^{ - 2{\omega ^2}}}\left( {\frac{1}{j} + 0.\pi \delta \left( \omega \right)} \right)\)
\( Y\left( {j0} \right) = \mathop \smallint \limits_{ - \infty }^\infty y\left( t \right)dt \)
\(= \frac{1}{j} = - j\)
\(Y\left( 0 \right) = \;\mathop \smallint \limits_{ - \infty }^{ + \infty } g\left( t \right)dt = - j\)
So, (b) is correct