LTI systems.

by sahil_time
Tags: systems
sahil_time is offline
Dec16-12, 03:59 AM
P: 108
Hello, my question is that almost all text books say that a linear system will give the output to a weighted sum of impulses which equals the superposition of scaled responses to each of the shifted impulses. But if we apply the same input which is a weighted sum of impulses to a non linear time invariant system, we will get the same output. Because a linear system and a non linear system are two separate systems only when 2 inputs are SIMULTANEOUSLY applied. But here we are just applying a single input REPRESENTED as a weighted sum of impulses. Is that true?
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Runei is offline
Dec16-12, 06:22 AM
P: 134
We can choose to view a signal as a sum of several weighted impulses. Mathematically it is the same.

In a linear system, the input x(t) will give an output y(t). If we decompose x(t) into several impulses, and then send each of these impulses through the system one after each other, we will end up with a response for each impulse.

If we add up the responses, we will find that the total response is the same as it was, when we just sent the signal x(t) through the system.

For a NON-linear system. A signal x(t) will provide a response y(t). If we do as before, and send each impulse through the system they will each provide a response. HOWEVER, if we add these responses together, we cannot be sure that the total response will be the same as for the total input x(t).

So for a non-linear system, we shouldn't think about decomposing a signal into several parts, because it wont work. Only for linear system should we think about this.
sahil_time is offline
Dec16-12, 06:38 AM
P: 108
Thankyou but what is the difference between sending signal x(t) and sending its weighted sum which goes one by one ? I mean its the same right?

Runei is offline
Dec16-12, 07:11 AM
P: 134

LTI systems.

No its not the same. By "one by one" I don't mean we send the weighted sum through. I mean we take ONE of the impulses and use it as an input. Then we record the output. Then we take ANOTHER impulse and use it as an input. Another output is recorded.

If we afterwards take each of our outputs and add them together, we would see that for a linear system, the sum of these outputs are exactly the same, as if we instead of taking one impulse at a time, had taken the signal x(t) which could be represented by a weighted sum of these impulses.

The weighted sum and x(t) are the same. But the impulses that make up the weighted sum, can be treated as individual signals, and we can look at the one-by-one.

For a linear system, the two ways of doing it (ie. using x(t) (aka the weighted sum), or recording each response from each impulse and then adding will give us the same).
For a non-linear system, these two ways do NOT necessarily provide the same result.

Does that help? :)
sahil_time is offline
Dec16-12, 08:35 AM
P: 108
Ok, I think im getting what you are saying. But doesnt an input x(t) when applied, BEHAVE like a weighted sum of impulses?

Lets say


This is clearly a non linear system.
Now if we take an input x(t) = x(t1)δ(t-t1) + x(t2)δ(t-t2)

If we treat
x1(t) = x(t1)δ(t-t1)
x2(t) = x(t2)δ(t-t2)

If we see this will give us the output we DID NOT expect.

x^2(t) = x^2(t1)δ^2(t-t1) + x^2(t2)δ^2(t-t2) + (a term whose value is zero)
= x1^2(t) + x2^2(t)

Can u please give me an example ?

Thanx a lot :)
Runei is offline
Dec16-12, 09:38 AM
P: 134
y(t) = x2(t) + 1

This is also a nonlinear system.

x1(t) = x(t1) δ(t-t1)
x2(t) = x(t2) δ(t-t2)

y1(t) = x2(t1) δ2(t-t1) + 1
y2(t) = x2(t2) δ2(t-t2) + 1

y1(t) + y2(t) = x2(t1) δ2(t-t1) + x2(t2) δ2(t-t2) + 2


x3(t) = x1(t) + x2(t)

y3(t) = = x1(t)2 + x2(t)2 + 1 (as before the third term from the result of the squaring will be zero)

= x2(t1) δ2(t-t1) + x2(t2) δ2(t-t2) + 1

y3 ≠ y1 + y2

In conclusion:

While you might find some systems where it works. Not all will :)
sahil_time is offline
Dec16-12, 10:08 AM
P: 108
Thanx alot Runei :)

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