Final answer:
The system y(t)=cos(2t)x(t) is linear, not time-invariant, not memoryless, causal, and BIBO-stable.
Step-by-step explanation:
The system y(t)=cos(2t)x(t) can be analyzed to determine if it is linear, time-invariant, memoryless, causal, and BIBO-stable:
(a) To determine linearity, we need to check if the system satisfies the superposition property. Let's assume x1(t) and x2(t) are input signals and their corresponding outputs are y1(t) and y2(t). If we apply the input x(t) = Ax1(t) + Bx2(t), the output will be y(t) = Acos(2t)x1(t) + Bcos(2t)x2(t). Since this satisfies the superposition property, the system is linear.
(b) To determine if it is time-invariant, we need to check if a time shift in the input signal results in the same time shift in the output signal. If we apply a time shift T to the input x(t), the output will be y(t) = cos(2(t-T))x(t-T). Since the output is not the same as cos(2t)x(t), the system is not time-invariant.
(c) To determine if it is memoryless, we need to check if the output at any time t only depends on the value of the input at the same time t. In this case, the output depends on both the current input value x(t) and the current time t, so the system is not memoryless.
(d) To determine causality, we need to check if the output at any time t only depends on the current and past values of the input. In this case, since the output depends on the current input value x(t) and the past value of cos(2t), the system is causal.
(e) To determine BIBO-stability, we need to check if the system produces bounded output for bounded input. Since the output is cos(2t)x(t), which is bounded if the input x(t) is bounded, the system is BIBO-stable.