- #1

- 108

- 0

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?

Thankyou :)

- Thread starter sahil_time
- Start date

- #1

- 108

- 0

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?

Thankyou :)

- #2

- 193

- 17

[itex]\frac{dx}{dt}=\lim{Δt\rightarrow0}\frac{x(t)-x(t-Δt)}{Δt}[/itex]

or

[itex]\frac{dx}{dt}=\lim{Δt\rightarrow0}\frac{x(t+Δt)-x(t)}{Δt}[/itex]

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.

Homogeneity: It's easy to see that the system is homogeneous. If an input x(t) gives y

This a scaled input will produce a scaled output.

What about superposition? Well here we run into problems. Because if x

Thus, the system is not linear.

- #3

- 108

- 0

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!

Thankyou :)

- #4

- 193

- 17

x

x

x

y

= [αx

= [αx

= [α(x

If we need it to be y

If

x

x

Then we can remove the unit step function. And the thing behaves like a linear system.

If

x

Then the system also behaves linearly (though the output will always be zero).

However, if

The two equations above are not satisfied, that means that for SOME t, we might have the following.

x

x

x

If this happens then

y

y

However

y

= x

= α (x

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.

- #5

- 108

- 0

Your solution is elegant. I think the answer in the solution manual is wrong.

Thankyou :)

Thankyou :)

- Last Post

- Replies
- 7

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 17K

- Last Post

- Replies
- 5

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 10

- Views
- 3K

- Last Post

- Replies
- 8

- Views
- 7K

- Last Post

- Replies
- 6

- Views
- 3K