is y(t) = 3 f(t-2) +4 a LTI system?

Question

i just had a pop quiz in one of my classes, i am wondering if i got the answers correct (the solution hasn't been posted online yet).

 

y(t) is the output

f(t) is the input

 

y(t) = 3 f(t-2) +4

 

is the system linear?

 

Is the system time invariant? 

 

 

Although it wasn't asked i was also curious what the answer would be if 

"y(t) is the input
f(t) is the output"

 

 

thanks!

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LTI means linear time invarient (system)

https://en.wikipedia.org/wiki/LTI_system_theory

 

 

here is a similar problem worked out as an example:

 

Problem 1.27
(a) y(t) = x(t − 2) + x(2 − t)
Let us check for linearity.
x1(t) → y1(t) = x1(t − 2) + x1(2 − t)
x2(t) → y2(t) = x2(t − 2) + x2(2 − t)
ax1(t) + bx2(t) = x3(t) → y3(t) = x3(t − 2) + x3(2 − t)
= ax1(t − 2) + bx2(t − 2) + ax1(2 − t) + bx2(2 − t)
= a(x1(t − 2) + x1(2 − t)) + b(x2(t − 2) + x2(2 − t))
= ay1(t) + by2(t)
Hence linear.
Let us check for time-invariance.
x1(t) → y1(t) = x1(t − 2) + x1(2 − t)
x1(t − to) = x2(t) → y2(t) = x2(t − 2) + x2(2 − t)
= x1(t − to − 2) + x2(2 − t − to)
6= y1(t − to)
Note that y1(t − to) = x1(t − to − 2) + x1(2 − t + to). Hence time-variant.
Suppose |x(t)| < B. Then y(t) < B + B = 2B (because |x(t − 2)| < B and |x(2 − t)| < B).
Hence stable.
Not memoryless as the present output at time t depends on t − 2.
Non-Causal because y(-1)=x(-3)+x(3). So depends on future inputs.

Answer 1

This question is vague. In terms of the quantities y and f, the answer depends upon whether or not f(x) is linear. Without knowing this, I cannot be more helpful to you.

 

I also do not understand the term LTI system, so anything I tell you now may be totally worthless.