Final answer:
The tested system y[n]=nx[n]+2x[n+4] does not meet the criterion for time invariance, as the output changes in a non-equivalent manner when the input signal is time-shifted. It does not meet the criterion for linearity as well since the output response to a scaled input is not a scaled output.
Step-by-step explanation:
To determine whether the system represented by y[n]=nx[n]+2x[n+4] is linear and time-invariant, two properties must be tested: superposition (linearity) and time invariance. For linearity, the principle of superposition must hold, which means that if two signals x1[n] and x2[n] are input into the system to produce outputs y1[n] and y2[n] respectively, then the system is linear if for any constants A and B, Ay1[n] + By2[n] is the output for the input Ax1[n] + Bx2[n].
To check time invariance, the system must produce the same output if the input signal is shifted in time. So, if x[n - no] leads to y[n - no], where no is a fixed number, then the system is time-invariant. Let's apply a time shift to the input x[n] and see if the output will be equivalent to y[n] shifted by the same amount:
Shifting the input by no, we have:
x1[n] = x[n - no]
The output becomes y1[n] = n * x1[n] + 2 * x1[n + 4] = n * x[n - no] + 2 * x[(n + 4) - no]
If we replace n with (n + no), the output should ideally be:
y[n + no] = (n + no)x[n] + 2x[n + 4]
Comparing the shifted output y1[n] and y[n + no], we see they are not equal because of the multiplicative term n in the expression. This indicates the system is not time-invariant.