The outputs of four systems $(S_1, S_2, S_3, S_4)$ corresponding to the input signal $\sin(t)$, for all time $t$, are shown. Based on the given information, which of the four systems is/are definitely NOT LTI (linear and time-invariant)?
Show Hint
Frequency scaling (e.g., $\sin(t) \to \sin(2t)$) and nonlinear operations (e.g., squaring) immediately break LTI behavior.
Given responses to input $\sin(t)$:
\[
S_1 : \sin(t) \rightarrow \sin(-t)
\]
\[
S_2 : \sin(t) \rightarrow \sin(t+1)
\]
\[
S_3 : \sin(t) \rightarrow \sin(2t)
\]
\[
S_4 : \sin(t) \rightarrow \sin^{2}(t)
\]
We test linearity and time invariance for each system based only on the given mapping.
Step 1: Analyze $S_1$.
Output: $\sin(-t)$.
This is simply time-reversal:
\[
y(t)=x(-t).
\]
This operation is linear. Time-reversal is also time-invariant:
If input is delayed,
\[
x(t-t_0) \Rightarrow x(-(t-t_0)) = x(-t+t_0),
\]
which is a shifted version of $x(-t)$.
Thus, $S_1$ is LTI, so it is NOT among the answers.
Step 2: Analyze $S_2$.
Output: $\sin(t+1)$.
This corresponds to:
\[
y(t)=x(t+1).
\]
This is a time shift (advance), but still linear and time-invariant.
Hence, $S_2$ is LTI, so it is NOT among the answers.
Step 3: Analyze $S_3$.
Output: $\sin(2t)$.
To check time invariance:
If input is delayed by $t_0$, input becomes $\sin(t - t_0)$, and a time-invariant system should give
\[
\sin(2(t - t_0)) = \sin(2t - 2t_0).
\]
But the given mapping only tells us that for $\sin(t)$ the output is $\sin(2t)$.
Doubling the argument introduces a scaling of the time axis, which breaks time invariance.
Also, it is not linear:
A linear system would preserve frequency relationships, but
\[
\sin(t) \rightarrow \sin(2t)
\]
shows frequency multiplication, a nonlinear operation.
Hence, $S_3$ is definitely not LTI.
Step 4: Analyze $S_4$.
Output: $\sin^{2}(t)$.
This is unmistakably nonlinear because squaring the input violates superposition:
\[
(\sin(t_1)+\sin(t_2))^2 \neq \sin^2(t_1)+\sin^2(t_2).
\]
Thus $S_4$ is definitely not linear, hence not LTI.
Step 5: Conclusion.
The systems that are definitely NOT LTI are:
\[
\boxed{S_3,\; S_4}
\]
which corresponds to options (C) and (D).
Final Answer: (C), (D)
