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

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

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?

  • $\begingroup$ why is the "ideal D/C system" running at half the sample rate ($ \frac{1}{T_2} = \frac{1}{2} \cdot \frac{1}{T_1} $ that the subsystems preceding it on the signal chain? how is the portion of the DTFT $Y(\omega)$ for $\frac{\pi}{2} < |\omega| < \pi$ defined? $\endgroup$ Commented Oct 24, 2016 at 23:39
  • 1
    $\begingroup$ BTW, a system that samples a continuous-time system is not LTI. it is LTI if the time variance is limited to integer multiples of the sampling period. $\endgroup$ Commented Oct 24, 2016 at 23:41

1 Answer 1


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?

  • $\begingroup$ My question was poorly worded. As you pointed out, this is not even LTI as the D/C runs at half the sample rate. $\endgroup$
    – ryan80
    Commented Oct 25, 2016 at 11:36

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