# Going from output of D/C to input of D/C

$$y_c[n] {\longrightarrow} \boxed{\text{D/C}} {\longrightarrow} y(t)$$

If the following equation describes the IDEAL D/C converter:

$$y_c(t) = \sum \limits_{n=-\infty}^{\infty} y_d[n]~ \text{sinc}\left(\frac{\pi (t - nT)}{T}\right)\tag{1}$$

$$Y_c(\Omega) = \begin{cases} T~ Y_d(\omega) \Big|_{\omega=\Omega T} & |\Omega| < \left(\Omega_r = \frac{\pi}{T_s}\right) \\0 & \text{otherwise}\end{cases}$$

How to prove that we can find $$y_c[n]$$ from $$y(t)$$ using this formula:

$$y_d[n] = y_c(nT)\tag{2}$$

It doesn't seem entirely obvious to me by rearrange equation $$(1)$$, that it equals $$(2)$$.

conceptually, if function $$f$$ represents a D/C converter, and function $$g$$ represents a C/D converter, and $$f$$ and $$g$$ are inverse functions of each other. then:

$$y(t) = f\Big( y_n[n]\Big)$$

$$g\Big( y(t) \Big) = g\bigg(f\Big(y_n[n]\Big)\bigg)$$

$$g\Big( y(t) \Big) = y_n[n]$$

You won't get the discrete $$y_d[m]$$ by re-arranging the equations, but by sampling the $$y_c(t)$$ at integral multiples of $$T$$. When you sample the RHS in equation1 exactly at $$t=mT$$ instances, you get $$y_d[m]$$. You need to remember that $$sinc(m-n) = 1$$ for $$m=n$$ and $$0$$ otherwise, and (BTW in equation 1, in $$sinc$$ argument $$\pi$$ won't be there). Anyway, take RHS and evaluate the expression at $$t = mT$$ where $$m$$ are integers.
$$y_c(t) \big|_{t=mT} = \sum^{\infty}_{n=-\infty}y_d[n] \, \mbox{sinc} \left( \frac{t-nT}{T} \right) \Big|_{t=mT} = y_d[n]$$