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Linearity of y(t) = x²(t)

Short answer

Squaring gives y(ax1+bx2)=a²x1²+b²x2²+2ab x1x2, which is not a y1+b y2. Superposition fails, so the system y(t)=x²(t) is non-linear.

Determine whether the signal y(t) = x²(t) is linear or not.

A system is linear if it satisfies superposition: ax1(t)+bx2(t)→ay1(t)+by2(t)a x_1(t) + b x_2(t) \to a y_1(t) + b y_2(t).

Given: y(t)=x2(t)y(t) = x^2(t)

For individual inputs: y1(t)=x12(t)y_1(t) = x_1^2(t), y2(t)=x22(t)y_2(t) = x_2^2(t).

For the combined input x3(t)=ax1(t)+bx2(t)x_3(t) = a x_1(t) + b x_2(t):

y3(t)=[ax1(t)+bx2(t)]2=a2x12(t)+b2x22(t)+2ab x1(t)x2(t)\begin{aligned} y_3(t) &= \left[a x_1(t) + b x_2(t)\right]^2 \\ &= a^2x_1^2(t) + b^2x_2^2(t) + 2ab\,x_1(t)x_2(t) \end{aligned}

Required for linearity: ay1(t)+by2(t)=ax12(t)+bx22(t)a y_1(t) + b y_2(t) = a x_1^2(t) + b x_2^2(t).

Clearly y3(t)≠ay1(t)+by2(t)y_3(t) \ne a y_1(t) + b y_2(t) (the coefficients become a2a^2, b2b^2 and a cross term 2ab x1x22ab\,x_1x_2 appears).

Simple check with homogeneity: doubling the input, x→2xx \to 2x, gives y=4x2y = 4x^2, i.e. the output becomes four times, not twice.

Answer: y(t)=x2(t)y(t) = x^2(t) is a non-linear system (a square-law device, such as that used in square-law modulators and detectors).

✓ Answer checked by Claude Opus 5.5 (high effort)

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