The Nyquist filter has impulse response given by
A. $$h\left( t \right) = \frac{{\sin \left( {\frac{{\pi t}}{T}} \right)}}{{\left( {\frac{{\pi t}}{T}} \right)}}$$
B. $$h\left( t \right) = \frac{1}{2}\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi t}}} \right)$$
C. $$h\left( t \right) = \frac{{\sin \left( {\pi t} \right)}}{{\left( {\frac{{\pi t}}{T}} \right)}}$$
D. $$h\left( t \right) = \frac{{\cos \left( {\pi t} \right)}}{{\left( {\frac{{\pi t}}{T}} \right)}}$$
Answer: Option A
This Question Belongs to Electronics And Communications Engineering >> Signal Processing
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