1) y(t)=dx(t)/dt 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 :)
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: [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. 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 y_{1}(t) = x(t)+x(t-2) then the input αx(t) will give y_{2}(t) = αx(t)+αx(t-2) = α(x(t)+x(t-2)) = αy_{1}(t). This a scaled input will produce a scaled output. What about superposition? Well here we run into problems. Because if x_{1}(t) < 0 and x_{2}(t) < 0 for all t, but x_{1}(t) + x_{2}(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.
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! Thankyou :)
y(t) = [x(t) + x(t-2)] u(x(t)) x_{1}(t) → y_{1}(t) x_{2}(t) → y_{2}(t) x_{3}(t) = αx_{1}(t) + βx_{2}(t) y_{3}(t) = [x_{3}(t) + x_{3}(t-2)] u(x_{3}(t)) = [αx_{1}(t) + βx_{2}(t) + αx_{1}(t-2) + βx_{2}(t-2)] u(x_{3}(t)) = [αx_{1}(t) + αx_{1}(t-2) + βx_{2}(t) + βx_{2}(t-2)] u(x_{3}(t)) = [α(x_{1}(t) + x_{1}(t-2)) + β(x_{2}(t-2) + x_{2}(t))] u(x_{3}(t)) If we need it to be y_{3}(t) = αy_{1}(t) + βy_{2}(t) we would need to somehow decompose u(x_{3}(t)). But there is no way we can do this. If x_{1}(t) ≥ 0, for all t AND x_{2}(t) ≥ 0, for all t Then we can remove the unit step function. And the thing behaves like a linear system. If x_{1}(t) + x_{2}(t) ≤ 0, for all t 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_{1}(t_{0}) < 0 x_{2}(t_{0}) < 0 x_{1}(t_{0}) + x_{2}(t_{0}) ≥ 0 If this happens then y_{1}(t_{0}) = 0; y_{2}(t_{0}) = 0; However y_{3}(t_{0}) = [x_{3}(t_{0}) + x_{3}(t_{0}-2)] u(x_{3}(t_{0})) = x_{3}(t_{0}) + x_{3}(t_{0}-2) = α (x_{1}(t_{0}) + x_{1}(t_{0}-2)) + β (x_{2}(t_{0}) + x_{2}(t_{0}-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.