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 correct answer is: $\boxed{\text{A. }tg = 10^{-6}, tp = 1.56}$.

The group delay of a channel is the derivative of the phase delay with respect to frequency. In other words, it is the rate at which the phase of a signal changes with frequency. The phase delay of a channel is the angle by which a signal is delayed by the channel.

The group delay and phase delay of a channel can be calculated from the transfer function of the channel. The transfer function of a channel is the ratio of the output signal to the input signal.

The transfer function of the channel in this question is given by:

$$H(j\omega) = \frac{1}{100 + j\omega}$$

The group delay of the channel is given by:

$$\tau_g = -\frac{d}{d\omega} \arg{H(j\omega)} = \frac{1}{100}$$

The phase delay of the channel is given by:

$$\tau_p = -\arg{H(j\omega)} = 1.56$$

Therefore, the group delay and phase delay of the channel are $\boxed{\text{A. }tg = 10^{-6}, tp = 1.56}$.

Here is a brief explanation of each option:

• Option A: The group delay of the channel is $10^{-6}$ seconds. This means that a signal with a frequency of 100 Hz will be delayed by 0.00001 seconds by the channel.
• Option B: The phase delay of the channel is 1.56 seconds. This means that a signal with a frequency of 100 Hz will be delayed by 1.56 seconds by the channel.
• Option C: The group delay of the channel is $10^{-8}$ seconds. This is much smaller than the actual group delay of the channel.
• Option D: The phase delay of the channel is 1.56 × $10^{-6}$ seconds. This is much smaller than the actual phase delay of the channel.