I have system for filtering a continuous-time signal using a discrete-time filter where sampling is performed by multiplication with the impulse train.
The frequency response of the discrete-time system is given as $$H_d(e^{j\omega})=\dfrac{1}{1-\frac{1}{2}e^{-j\omega}+3e^{-3j\omega}}$$
And I want to find an equation relating input $x(t)$ and output $y(t)$ of my overall system. It is guaranteed that input is a band-limited signal and sampling frequency is high enough. In my Signals and Systems book frequency response of the continuous system is given as: $$H_c(j\omega)=H_d(e^{j\omega T})\left(u(\omega+\tfrac{1}{2}\omega_s)-u(\omega-\tfrac{1}{2}\omega_s)\right)\quad,$$ $T$ is the sampling period, $\ \omega_s = \frac{2 \pi}{T}$ is the (angular) sampling frequency (radian/second), $\ u(\omega)$ denotes the unit step function (taken from Oppenheim and Willsky, Signals and Systems Second Edition, p. 540, eq. 7.25).
I have found the difference equation which stands for the discrete-time system but I couldn't understand how can I relate it to continuous time system. I will appreciate any help. $$y[n] − \tfrac{1}{2}y[n − 1] + 3y[n − 3] = x[n]$$