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The input to a channel is a bandpass signal. It is obtained by linearly modulating a sinusoidal carrier with a single-tone signal. The output of the channel due to this input is given by
$$y\left( t \right) = \left( {{1 \over {100}}} \right)\cos \left( {100t - {{10}^{ - 6}}} \right)\cos \left( {{{10}^6}t - 1.56} \right)$$
The group delay (tg) and the phase delay (tp) in seconds, of the channel are

The input to a channel is a bandpass signal. It is obtained by linearly modulating a sinusoidal carrier with a single-tone signal. The output of the channel due to this input is given by
$$y\left( t \right) = \left( {{1 \over {100}}} \right)\cos \left( {100t - {{10}^{ - 6}}} \right)\cos \left( {{{10}^6}t - 1.56} \right)$$
The group delay (tg) and the phase delay (tp) in seconds, of the channel are Correct Answer t<sub>g</sub> = 10<sup>8</sup>, t<sub>p</sub> = 1.56

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