Continuous-time signals are real functions of t∈R with t in seconds, and discrete-time…

Continuous-time signals are real functions of \(t\in\mathbb{R}\) with \(t\) in seconds, and discrete-time signals are real sequences indexed by \(n\in\mathbb{Z}\). The unit step satisfies \(u(t)=1\) for \(t\ge 0\) and \(u(t)=0\) for \(t<0\). Shifts are \((S_a x)(t)=x(t-a)\) and \((S_k x)[n]=x[n-k]\). Inputs and outputs of \(N\), \(D\), and \(G\) are voltages in volts. Classify each system as linear or not, time-invariant or not, causal or not, memoryless or not, and BIBO stable or not. Provide an explicit witness for every negative claim. - \(N\): \(y(t)=\bigl(x(t)\bigr)^3\), with the cube acting pointwise on the numerical value of the voltage in volts and producing an output in \(\mathrm{V}^3\); treat the input and output as real numerical signals for the purpose of the five classifications. - \(D\): \(y(t)=\dfrac{x(t)-x(t-0.010)}{0.010}\), where the factor \(0.010\) has SI unit seconds. - \(G\): \(y(t)=(2t+1)x(t)\), where the factor \((2t+1)\) is dimensionless. - \(R\): \(y[n]=x[-n]\). For \(D\), compute the output when \(x=u\), giving a piecewise formula for all \(t\in\mathbb{R}\). For \(N\), compute the output at \(t=0\) when \(x(t)=2\cos(10\pi t)\,\mathrm{V}\). For \(G\), compute the response to \(x(t)=e^{-t}u(t)\,\mathrm{V}\) at \(t=1\).

A pointwise power is still a function of the present sample only. A two-tap difference uses a delayed sample. A gain that grows with \(t\) can turn a constant into an unbounded output. Reversal changes the time index.