# Discrete Time Processing of Continuous Time Signals LTI Concept

I wonder if the input signal (CT) violates Shannon-Nyquist Theorem for a given sampling rate, is there any chance for the overall system not to be LTI although discrete time system is LTI?

Thanks.

• That reads like a homework question. VERY MUCH. The answer is yes. If you want to know why, you might want to start by explaining what you've figured out so far. – Marcus Müller Mar 20 '17 at 20:59
• Do you mean a system like this: input -> ADC -> DSP -> DAC -> output? Such a system can't be LTI, even if the DSP system is. – MBaz Mar 20 '17 at 21:31
• @MarcusMüller Actually it's not a homework question. I was solving problems from Discrete-Time Signal Processing by Oppenheim, and then I encountered with such a statement:"It is important to emphasize that the linear and time invariant behavior of Figure 10 depends on two factors. First, the discrete-time system must be linear and time invariant. Second, the input signal must be band limited, and the sampling rate must be high enough so that any aliased components are removed by discrete-time system."(p178). Figure 10 represent a system as follows; x(t)->C/D->x[n]->DT System->y[n]->D/C->y(t). – Canberk Mar 20 '17 at 22:27
• I cannot figure out why do we need these two conditions. First one seems reasonable, but i did not understand how the violation of the second condition affects the linearity and time invariance. – Canberk Mar 20 '17 at 22:28
• @Canberk make a drawing of a signal that violates Nyquist, then mark the times with red dots at which the ADC samples. Now, shift signal in time a little bit, mark sampling points with green dots. "reconstruct" signal as good as possible by connecting red dots with red line that is as low-frequency as possible, do the same with green. Both lines are totally different. Hence, your system is time-variant. End of Proof. – Marcus Müller Mar 20 '17 at 22:34

x(t) -> C/D -> DSP -> D/C -> y(t)

where $x(t)$ and $y(t)$ are sufficiently bandlimited and the C/D and D/C blocks are ideal, then the system behaves as if it were LTI. Note that the system is not truly LTI, since it is not LTI for all inputs.
In particular, if the process of discretizing $x(t)$ produces aliases, then the output $y(t)$ can potentially contain frequencies not present in the input. Since an LTI system cannot create new frequencies, then in conclusion the system is not LTI.
• in short; "the output response $y(t)$ of the above system is identical to that of an LTI continuous-time system (whose impulse response is determined by the DSP block) for the class of input signals $x(t)$ which are bandlimited to $\pi /T_s$ where $T_s$ is the sampling period used in both the ideal C/D and D/C converters..." – Fat32 Mar 20 '17 at 23:44