Question

Consider a system with input x(t) and output y(t) = x(e^t). The system is

• A. Causal and time invariant.
• B. Non-causal and time varying.
• C. Causal and time varying.
• D. Non-causal and time invariant.

Hint:

If the output uses $x$ at $g(t)$ with $g(t)>t$ for some $t$, the system is non-causal. For time invariance, compare $T\{x(t-\tau)\}$ with $T\{x(t)\}$ shifted: any mismatch implies time-varying.

Answer 1

The Correct Option is B

Causality test: For any time instant \(t_0\), the system output is \[ y(t_0) = x(e^{t_0}). \] Notice that \(e^{t_0} > t_0\) for all real \(t_0\) (since the function \(f(t) = e^t - t\) has a global minimum at \(t=0\), with value \(f(0)=1 > 0\)). This means the output at present time \(t_0\) depends on the input at a future time \(e^{t_0}\). Therefore, the system is non-causal.

Time invariance test: Consider a shifted input \(x_1(t) = x(t-\tau)\). Then the corresponding output is \[ y_1(t) = x_1(e^t) = x(e^t - \tau). \] On the other hand, if we take the original output and shift it in time, we get \[ y(t-\tau) = x(e^{\,t-\tau}). \] In general, \(e^t - \tau \neq e^{\,t-\tau}\), so \[ y_1(t) \neq y(t-\tau). \] Hence, the system is time-varying.

Final Answer:
\[ \boxed{\text{The system is non-causal and time-varying (Option B).}} \]