Determine whether the systems are linear, time invariant, causal and stable.

 Determine whether the systems are linear, time invariant, causal and stable

1. y(n) = n x(n) 

2. y(t) = x(t) + x(t-2) for t = 0 , 0 for t < 0

 

1. y(n) = n x(n) 

y1(n) = n x1(n) and y2(n) = n x2(n) and

y3(n) = T [a1x1(n) + a2x2(n)] = n[a1x1(n) + a2x2(n)] = a1nx1(n)+a2nx2(n) 

y’3(n) = a1 y1(n) + a2y2(n) 

         = a1nx1(n) + a2nx2(n)

 Since y3(n) = y’3 (n), the system is linear. 

    Since, y(n) is function of ‘n’, the system is time variant. Hence the output is not bounded even if x(n) is bounded. Hence the system is unstable. 

    Output is function of present input only, hence the system is causal.

2. y(t) = x(t) + x(t-2) for t = 0 , 0 for t < 0 

y1(t) = x1(t) + x1(t-2) 

y2(t) = x2(t) + x2(t-2) and

y3(t) = T[a1x1(t) + a2x2(t)] = T[a1x1(t)]+T[a2x2(t)] 

        = a1x1(t) + a1x1(t-2) + a2x2(t2)y’3(t)

        = a1y1(t) + a2y2(t) 

        = a1x1(t) + a1x1(t-2)+a2x2(t) + a2x2(t-2) 

Since, y3(t) = y’3(t), the system is linear. y(t) is not the function of time directly. Hence the system is time invariant. 

    Output depends upon present and past input. Hence this is causal system. As long as x(t) is bounded x(t-2) will be bounded and hence y(t) will be bounded. Hence this is stable system. Since the term x(t-2) is present, the system requires memory.