|Dec16-12, 03:52 AM||#1|
Signals and Systems.
Is this system memoryless?
2) y(t) = 0 if x(t)<0
x(t) + x(t-2) if x(t)>=0
Is this system linear?
|Dec16-12, 06:09 AM||#2|
1) The system is not memoryless. The definition of the derivative can be expressed in two ways. What it was before, and what it will be in the future, like the following:
The two definitions will give the same result. However, the first one is has memory (in that it uses a past value of x(t)) and the second one is non-causal (in that it uses a future value of x(t)).
If you had a system that needed to compute the derivative, the first definition is the only one you would be able to use, and that would give you a system with memory.
2) If the system is linear it must satisfy superposition and homogeneity.
Homogeneity: It's easy to see that the system is homogeneous. If an input x(t) gives y1(t) = x(t)+x(t-2) then the input αx(t) will give y2(t) = αx(t)+αx(t-2) = α(x(t)+x(t-2)) = αy1(t).
This a scaled input will produce a scaled output.
What about superposition? Well here we run into problems. Because if x1(t) < 0 and x2(t) < 0 for all t, but x1(t) + x2(t) > 0 for all t, then the response for each of the inputs will be 0, but if you add them together, they will provide an output.
Thus, the system is not linear.
|Dec16-12, 06:30 AM||#3|
Runei you are right about the first one. But the answer to the second one is that the system is Linear. This is how i tried to evaluate but unfortunately could not prove Linearity.
For the second question y(t)= [x(t)+x(t-2)]u[x(t)] which condenses the signal in one equation. Now we give two inputs to test superposition,
x1(t)→ y1(t)= [x1(t)+ x1(t-2)]u[x1(t)]
x2(t)→ y2(t)= [x2(t)+ x2(t-2)]u[x2(t)]
Now let x3(t) = ax1(t) + bx2(t) To test homogeniety
therefore y3(t) = [x3(t)+x3(t-2)]u[x3(t)]
= [ax1(t) + bx2(t) + ax1(t-2) + bx2(t-2)]u[ax1(t) + bx2(t)]
We have to prove that y3(t) = ay1(t) + by2(t).
I got stuck here!
|Dec16-12, 07:59 AM||#4|
Signals and Systems.
y(t) = [x(t) + x(t-2)] u(x(t))
x1(t) → y1(t)
x2(t) → y2(t)
x3(t) = αx1(t) + βx2(t)
y3(t) = [x3(t) + x3(t-2)] u(x3(t))
= [αx1(t) + βx2(t) + αx1(t-2) + βx2(t-2)] u(x3(t))
= [αx1(t) + αx1(t-2) + βx2(t) + βx2(t-2)] u(x3(t))
= [α(x1(t) + x1(t-2)) + β(x2(t-2) + x2(t))] u(x3(t))
If we need it to be y3(t) = αy1(t) + βy2(t) we would need to somehow decompose u(x3(t)). But there is no way we can do this.
x1(t) ≥ 0, for all t AND
x2(t) ≥ 0, for all t
Then we can remove the unit step function. And the thing behaves like a linear system.
x1(t) + x2(t) ≤ 0, for all t
Then the system also behaves linearly (though the output will always be zero).
The two equations above are not satisfied, that means that for SOME t, we might have the following.
x1(t0) < 0
x2(t0) < 0
x1(t0) + x2(t0) ≥ 0
If this happens then
y1(t0) = 0;
y2(t0) = 0;
y3(t0) = [x3(t0) + x3(t0-2)] u(x3(t0))
= x3(t0) + x3(t0-2)
= α (x1(t0) + x1(t0-2)) + β (x2(t0) + x2(t0-2))
This equation may or may not give us 0. But we can't be sure. Thus the combined signal x1+x2 does not necesarrily provide the combined output from each of them.
|Dec16-12, 08:41 AM||#5|
Your solution is elegant. I think the answer in the solution manual is wrong.
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