Hilbert transform of the signal u(t) – u(t – t 0) is
A. <span class="math-tex">\(\frac{1}{\pi }\ln \left( {\frac{t}{{t + {t_0}}}} \right)\)</span>
B. <span class="math-tex">\(\frac{1}{\pi }\ln \left( {\frac{t}{{t - {t_0}}}} \right)\)</span>
C. <span class="math-tex">\(\frac{1}{\pi }\ln \left( {\frac{{t - {t_0}}}{t}} \right)\)</span>
D. <span class="math-tex">\(\frac{1}{\pi }{\rm{sin\;c}}\left( {\frac{t}{{{t_0}}}} \right)\)</span>
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Right Answer is: B
SOLUTION
x(t) = u(t) –u (t – t0)
\(h\left( t \right) = \frac{1}{{\pi t}}\)
the Hilbert transform of the signal is defined as
\(\hat x\left( t \right) = \mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( \tau \right)h\left( {t - \tau } \right)d\tau \)
\(= \mathop \smallint \limits_{ - \infty }^{ + \infty } \left[ {u\left( \tau \right) - u\left( {t - \tau } \right)} \right]\frac{1}{{\pi \left( {t - \tau } \right)}}d\tau \)
\(= \frac{1}{\pi }\mathop \smallint \limits_0^{{t_0}} \frac{1}{{\left( {t - \tau } \right)}}d\tau \)
\(\Rightarrow \frac{{ - 1}}{\pi }\left[ {\ln \left( {t - \tau } \right)} \right]_0^{{t_0}}\)
\(\frac{1}{\pi }\ln \left( {\frac{t}{{t - {t_0}}}} \right)\)
