Question

In the table shown below, List-I are the system responses to input \(x(t)\) and List-II are the system properties. The correct matching is ______.

List-I	List-II
(M) \(x^2(t)\)	(I) Linear and Time-invariant
(N) \(x(-t)\)	(II) Non-linear and Time-varying
(O) \(|x(3-t)|\)	(III) Linear and Time-varying
(P) \(x(t-3)\)	(IV) Non-linear and Time-invariant

• A. M-IV, N-III, O-II, P-I
• B. M-II, N-I, O-IV, P-III
• C. M-III, N-IV, O-II, P-I
• D. M-I, N-IV, O-III, P-II

Hint:

Check superposition (scaling and adding inputs) for linearity, and check whether a delayed input just delays the output for time-invariance.

Answer 1

The Correct Option is A

Step 1: Recall the two tests we need.
A system is linear if, for any inputs \(x_1(t)\), \(x_2(t)\) and constants \(a\), \(b\), the response to \(ax_1(t)+bx_2(t)\) equals \(a\) times the response to \(x_1(t)\) plus \(b\) times the response to \(x_2(t)\) (superposition). A system is time-invariant if delaying the input by \(t_0\) simply delays the output by the same \(t_0\), with no other change in shape.

Step 2: Test (M) \(y(t)=x^2(t)\).
For input \(ax_1+bx_2\), the output is \((ax_1+bx_2)^2=a^2x_1^2+b^2x_2^2+2abx_1x_2\), which is not \(a\,x_1^2+b\,x_2^2\) in general, since the cross term \(2abx_1x_2\) spoils it. So (M) is non-linear. For a delayed input \(x(t-t_0)\), the output is \(x^2(t-t_0)\), which is exactly the original output delayed by \(t_0\). So (M) is time-invariant. This matches (IV) Non-linear and Time-invariant.

Step 3: Test (N) \(y(t)=x(-t)\).
For input \(ax_1+bx_2\), the output is \(ax_1(-t)+bx_2(-t)\), which is exactly \(a\) times the first output plus \(b\) times the second, so (N) is linear. For time-invariance, delay the input to get \(x(t-t_0)\); the output of this new input is \(x(-t-t_0)\). But delaying the original output by \(t_0\) gives \(x(-(t-t_0))=x(-t+t_0)\). These two expressions agree only when \(t_0=0\), so (N) is time-varying. This matches (III) Linear and Time-varying.

Step 4: Test (O) \(y(t)=|x(3-t)|\).
The absolute value bars make superposition fail in general, since if \(x_1\) and \(x_2\) have opposite signs, \(|x_1+x_2|\ne|x_1|+|x_2|\), so (O) is non-linear. The argument \(3-t\) both reflects and shifts time, and by the same reasoning as Step 3 this reflection makes the system time-varying. So (O) matches (II) Non-linear and Time-varying.

Step 5: Test (P) \(y(t)=x(t-3)\).
This is a pure delay by \(3\) units. Superposition clearly holds, since the response to \(ax_1+bx_2\) is \(ax_1(t-3)+bx_2(t-3)\). It is also time-invariant, since delaying the input by \(t_0\) just adds another \(t_0\) to the existing delay, giving exactly the output delayed by \(t_0\). So (P) matches (I) Linear and Time-invariant.

Step 6: Collect the matches and check the options.
We have M-IV, N-III, O-II, P-I. Comparing with the given choices, this combination is exactly option (A).
\[ \boxed{\text{M-IV, N-III, O-II, P-I}} \]