Continuous-time frequency response question Suppose the CTFT of continuous-time input $x_c(t)$ to an LTI system is $X_c(j\Omega)$ and that of its continuous-time output $y_c(t)$ is $Y_c(j\Omega)$. We have,
$$X_c(j\Omega) = 0,\phantom{1}\text{for } |\Omega| \ge \frac{\pi}{T_1}$$.

The LTI system samples the signal (C/D operation at sampling rate $1/T_1$) and passes it through the discrete-time system with response $H(e^{j\omega})$. The result of this DT system is then passed through an ideal D/C system (operating at rate $\frac{1}{T_2} = \frac{1}{2T_1}$) to yield $Y_c(j\Omega)$. 

I want to find out the overall frequency response of the system. I can write

$$Y_c(j\Omega) = \begin{cases}\frac{T_2}{T_1} H(e^{j\Omega T_2})X_c\left(j\Omega \frac{T_2}{T_1}\right),\phantom{1}\text{for } |\Omega| &lt; \frac{\pi}{T_2}\\0,\phantom{1}\text{for } |\Omega| \ge \frac{\pi}{T_2}\end{cases}$$

Since I get $X_c\left(j\Omega \frac{T_2}{T_1}\right)$ in the above expression and not $X_c\left(j\Omega \right)$, I am not sure how to find the overall frequency response. Is it possible to express $X_c\left(j\Omega \frac{T_2}{T_1}\right)$ in terms of $X_c\left(j\Omega \right)$ using some continuous-time filter?
Is it possible to express $X_c\left(j\Omega \frac{T_2}{T_1}\right)$ in terms of $X_c\left(j\Omega \right)$ using some continuous-time filter?


an LTI system cannot do that.  continuous-time "filters" need not be LTI but usually we mean LTI when we consider "filters" in analog or digital filters.

you can change the spectrum of $X_c\left(j\Omega \right)$ to $X_c\left(j\Omega \frac{T_2}{T_1}\right)$ by speeding up (or slowing down) the time index of $x_c(t)$ to $x_c\left(t \frac{T_1}{T_2}\right)$.

it appears you want a pitch shifter or similar.  is that it?