One of the most notable difference between the continuous-time Fourier transform and discrete-time Fourier transform is that the discrete-time Fourier transform domain is periodic. As a result, for the continuous-time Fourier transform, multiplication in time is proportional to convolution in frequency. In contrast, for the discrete-time Fourier transform, multiplication in time is proportional to circular convolution in frequency.
Circular convolution is convolution for two periodic signals. It causes signals to "wrap-around" in the frequency domain. Mathematically, the circular convolution between $X(\omega)$ and $H(\omega)$ is expressed as $$ Y(\omega) = \frac{1}{2 \pi} \int_{2\pi} X(\theta) H(\omega - \theta) \, d\theta \; , $$ where the integral notation represents the integral over any one period of length $2 \pi$.
There are two approaches you can use to compute a circular convolution of two signals $$ Y(\omega) = X(\omega) \circledast H(\omega) \; . $$ Both approaches can easier in different scenarios.
In the first approach, the circular convolution is equivalent to normal convolution between a single period of one signal (either $X(\omega)$ or $H(\omega)$) and the other periodic signal.
In the second approach, we consider a period of either signal over the same frequency range (e.g., in $0 \leq \omega \leq 2 \pi$). Then when computing the output signal, we assume that delayed signals "wrap" around period. We then make that result periodic again.
Consider two periodic signals (note: these do not correspond to real-time signals since they are not symmetric) $$ \begin{align} X(\omega) &= \sum_{m=-\infty}^{\infty} \delta(\omega - \pi - 2 \pi m) \\ H(\omega) &= \sum_{m=-\infty}^{\infty} \delta(\omega + 3 \pi / 2 - 2 \pi m) \end{align} $$ To compute $Y(\omega) = X(\omega) \circledast H(\omega)$, we can use either one of the two above approaches.
If we use the first approach, we can use a single period of $X(\omega)$ and keep $H(\omega)$ periodic. Then we have $$ \begin{align} X_{2 \pi}(\omega) &= \delta(\omega - \pi) \\ H(\omega) &= \sum_{m=-\infty}^{\infty} \delta(\omega + 3 \pi / 2 - 2 \pi m) \; . \end{align} $$ Then we compute $$ \begin{align} Y(\omega) &= X(\omega) \circledast H(\omega) \\ &= X_{2\pi}(\omega) * H(\omega) \\ &= H(\omega - \pi) \\ &= \sum_{m=-\infty}^{\infty} \delta(\omega + \pi / 2 - 2 \pi m) \\ &= \sum_{m=-\infty}^{\infty} \delta(\omega - 3 \pi / 2 - 2 \pi m) \\ \end{align} $$
If we use the second approach, we can use a single period of both $X(\omega)$ and $H(\omega)$ between $0 \leq \omega \leq 2 \pi$. Then we have $$ \begin{align} X_{2 \pi}(\omega) &= \delta(\omega - \pi) \\ H_{2 \pi}(\omega) &= \delta(\omega + 3 \pi / 2) \; . \end{align} $$ Then we compute $$ \begin{align} X_{2\pi}(\omega) * H_{2\pi}(\omega) &= \delta(\omega + \pi / 2) \\ \end{align} $$ However, $\delta(\omega + \pi / 2)$ is outside of our range ($0 \leq \omega \leq 2 \pi$). Instead, we wrap around to the signal. This is equivalent to adding a delay of $- 2 \pi$ to the result to get $$ \begin{align} \delta(\omega + \pi / 2 - 2 \pi) &= \delta(\omega - 3 \pi / 2) \\ \end{align} $$ If we make that periodic again, we get $$ \begin{align} Y(\omega) &= \sum_{m=-\infty}^{\infty} \delta(\omega - 3 \pi / 2 - 2 \pi m) \\ \end{align} $$