Final answer:
The system y[n]=x²[n-1] is not time-invariant, not linear, but it is causal. It is not BIBO stable and it is a system with memory, as the current output depends on the past input.
Step-by-step explanation:
When analyzing the system y[n]=x²[n-1], we need to determine various characteristics, including time invariance, linearity, causality, BIBO stability, and whether it is a system with memory. To determine if a system is time-invariant, we apply a time shift to the input and check if this results in the same time shift in the output. In this case, applying a time shift t, we have y[n+t]=(x[n+t-1])² ≠ x²[n-1+t], showing that the system is not time-invariant.
For linearity, we check if the system adheres to the principles of superposition (additivity and homogeneity). In this case, if we have two inputs, x1[n] and x2[n], the corresponding outputs would be y1[n]=(x1[n-1])² and y2[n]=(x2[n-1])², respectively. The sum of the inputs squared, (x1[n]+x2[n])² ≠ y1[n]+y2[n], does not equal the sum of the outputs, indicating nonlinearity. Similarly, for homogeneity, λx[n-1]² ≠ λy[n] when λ is a scaling constant.
Regarding causality, a system is causal if the output at any time n depends only on the present and past inputs, not future inputs. Since y[n] depends only on x[n-1], the system is causal. To determine BIBO (Bounded Input, Bounded Output) stability, we see if bounded inputs always produce bounded outputs. Since the square of a bounded input can result in an unbounded output, this system is not BIBO stable. Lastly, this system is with memory because the current output y[n] depends on the past input x[n-1], not just the current input x[n].