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LTI system

Started by Sharan123 4 years ago6 replieslatest reply 4 years ago108 views

I would like to know if there an intuitive way, without going through rigorous maths, to check if a given system is LTI system or not.

I have the following examples in mind,

1) y = a + b (a and b are inputs)

2) y = a*b

3) y = x + x delayed version

Thanks,

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Reply by Tim WescottNovember 2, 2016

Dunno what's too rigorous for you, but if it obeys superposition then it's linear, and you can always test that.  If the characteristics of the system don't vary by time, then it's time invariant.

If the system is a collection of linear operations, then it's linear.

A system state or input times a constant is a linear operation, as is one of those things delayed by a constant amount.  So (1) is linear because addition is linear, and (3) is linear (assuming the delay is constant) because constant delay is linear and addition is linear.

(2) isn't linear, and without just declaring it so, I don't know how to show that without using superposition.

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Reply by jithinrjNovember 2, 2016
I think there are 2 ways to check if a system is LTI. First it has to be linear. There are 2 conditions for that -a) For 0 input, the output should be 0. b) If the input is $\alpha x$, output should be $\alpha y$. In order to check if it is time invariant (TI), a delayed input should give an output delayed by the same amount. That is $x[n-k]$ input should give $y[n-k]$ as output. Now you can apply this criterea and see which systems are LTI.
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Reply by Y(J)SNovember 2, 2016

We normally speak of LTI regarding a single input (x) - single output (y) system, while 2 of your examples have 2 inputs (a and b). A system with 2 inputs can be linear in one input and not in the other.

For the single input version there are easy ways to check:

1. the time (t or n) must not appear in any explicit way - no multiplying by a function of time, no if time>t, etc.

2. the output must be a weighted sum of inputs and outputs (ARMA) with constant weights, such as a0 x_n + a1 x_n-1 or a0 x_n + b1 y_n-1.

Thus your third example (y = x_n + x_n-m) is clearly LTI.

Y(J)S

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Reply by Tim WescottNovember 2, 2016

There's no reason that the concept of a linear system can't be extended to multiple-input, multiple-output systems.

In fact, one of the required classes for my undergraduate degree was state-space control systems, the book for which was Kalaith's "Linear Systems".  The math used was only applicable to linear systems, so there you are.

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Reply by Y(J)SNovember 3, 2016

Sure, the concept is easily extendible (multiple dimensional signal processing is a very well established field). All I was saying is that you can no longer say that the system is linear, you have to say linear in THIS input.

Regarding TI, it is harder to say whether a system is TI or isn't when there are multiple inputs. For example, assume a system that for n<0 is

y = a0 x1_n + a1 x1_n-1 + b0 x2_n + b1 x2_n-1

and for n>0 is

y = a0 x1_n + a1 x1_n-1 + c0 x2_n + c1 x2_n-1 

This system is not TI since it is explicitly dependent on the time n. However, this dependence only effects x2, so you could say that it is LTI in x1.

However, the whole reason for ever defining LTI is the fact that LTI systems are "filters" in the sense that Y(w) = H(w) X(w) where H(w) is the frequency response. In particular, Y(w) has no frequencies that were not in X(w) (or in the additive 2-input case, no frequencies that were not present in either X1(w) or X2(w)). This doesn't hold for the case given above, so the concept of LTI is meaningless.

Y(J)S

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Reply by Tim WescottNovember 2, 2016

Here's another thought: if you can put the system description into the form of a linear state-space system without making any approximations, then it has to be linear.

I.e., 

$$x_n = A x_{n-1} + B u_n$$ $$y_n = C x_{n-1} + D u_n$$

where \(u_n\) is the input (possibly a vector), \( x_n \) is the state (almost always a vector), \(y_n\) is the output (possibly a vector), and \(A\), \(B\), \(C\) and \(D\) are all matrices of the correct dimensions to make the math work out.  If the matrices are constant, then the thing is also time-invariant.