Question

The system $y(t) = tx(t) + 4$ is_______.

• A. Non-linear, time-varying and unstable
• B. Linear, time-varying and unstable
• C. Non-linear, time-invariant and unstable
• D. Non-linear, time-varying and stable

Hint:

Check linearity first by scaling, then time-invariance by shifting, and finally stability by testing bounded input/output.

Answer 1

The Correct Option is A

Step 1: Test for linearity.
A system is linear if it satisfies:
1. Additivity: $x_1 + x_2 \rightarrow y_1 + y_2$
2. Homogeneity: $kx(t) \rightarrow ky(t)$
Let’s apply scaling: \[ x(t) \rightarrow kx(t) \Rightarrow y(t) = t \cdot kx(t) + 4 \Rightarrow ky(t) = kt \cdot x(t) + 4k \neq y(t) \] So the "+4" term makes it non-linear. 
Step 2: Test for time invariance.
Shift input: $x(t - t_0) \Rightarrow y(t) = t \cdot x(t - t_0) + 4$
Compare with shifted output:
$y(t - t_0) = (t - t_0) \cdot x(t - t_0) + 4$
They are not equal. Hence, it is time-varying.
Step 3: Stability.
If input is bounded, e.g., $x(t) = \sin(t)$, still $y(t) = t \sin(t) + 4$, which grows unbounded as $t \rightarrow \infty$. Hence, unstable.
Conclusion: $\boxed{\text{Non-linear, time-varying and unstable}}$