Respuesta :
Answer:
The following system is not linear.
The following system is time-invariant
Step-by-step explanation:
To determine whether a system is linear, the following condition must be satisfied:
[tex]f(a) + f(b) = f(a+b)[/tex]
For [tex]y(t) = cos(3t)[/tex], we have
[tex]y(a) = cos(3at)[/tex]
[tex]y(b) = cos(3bt)[/tex]
[tex]y(a+b) = cos(3(a+b)t) = cos(3at + 3bt)[/tex]
In trigonometry, we have that:
[tex]cos(a+b) = cos(a)cos(b) - sin(a)sin(b)[/tex]
So
[tex]cos(3at + 3bt) = cos(3at)cos(3bt) - sin(3at)sin(3bt)[/tex]
[tex]y(a) + y(b) = cos(3at) + cos(3bt)[/tex]
[tex]y(a+b) = cos(3(a+b)t) = cos(3at + 3bt) = cos(3at)cos(3bt) - sin(3at)sin(3bt)[/tex]
Since [tex]y(a) + y(b) \neq y(a+b)[/tex], the system [tex]y(t) = cos(3t)[/tex] is not linear.
If the signal is not multiplied by time, it is time-invariant. So [tex]y(t) = cos(3t)[/tex]. Now, for example, if we had [tex]y(t) = t*cos(3t)[/tex] it would not be time invariant.
