The correct answer is $\boxed{\frac{{15}}{2}\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi t}}} \right)\cos \left( {10\pi t – \frac{\pi }{4}} \right)}$.
The signal $cos(10\pi t + \frac{\pi}{4})$ is ideally sampled at a sampling frequency of 15 Hz. This means that the signal is sampled at regular intervals of $T = \frac{1}{f_s} = \frac{1}{15} = 0.067$ seconds. The sampled signal is then passed through a filter with impulse response $\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi \tau }}} \right)\cos \left( {40\pi t – \frac{\pi }{2}} \right)$. The impulse response of a filter is a function that describes how the filter will respond to a single input pulse. In this case, the impulse response is a decaying sinusoid with a period of $T_p = \frac{2\pi}{\omega} = 0.052$ seconds. The filter output is the convolution of the sampled signal and the impulse response.
The convolution of two signals $x(t)$ and $h(t)$ is defined as
$$y(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau$$
In this case, the sampled signal is $x(t) = cos(10\pi t + \frac{\pi}{4})$ and the impulse response is $h(t) = \left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi \tau }}} \right)\cos \left( {40\pi t – \frac{\pi }{2}} \right)$. The convolution of these two signals can be evaluated using the following steps:
- Multiply the two signals pointwise.
- Integrate the product over all time.
The result of the convolution is a signal that is a sum of two sinusoids, one at a frequency of 10 Hz and one at a frequency of 40 Hz. The amplitude of the 10 Hz sinusoid is $\frac{{15}}{2}\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi t}}} \right)$ and the amplitude of the 40 Hz sinusoid is $\frac{{15}}{2}\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi t}}} \right)\cos \left( {40\pi t – \frac{\pi }{2}} \right)$. The phase of the 10 Hz sinusoid is $\frac{\pi}{4}$ and the phase of the 40 Hz sinusoid is $-\frac{\pi}{2}$.
The filter output is therefore
$$y(t) = \frac{{15}}{2}\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi t}}} \right)\cos \left( {10\pi t + \frac{\pi }{4}} \right) + \frac{{15}}{2}\left( {\frac{{\sin \left( {\pi t} \right)}}{{\pi t}}} \right)\cos \left( {40\pi t – \frac{\pi }{2}} \right)$$