Discrete-Time Systems

shteii01

Joined Feb 19, 2010
4,644
And the definition is about the two properties?

I am not satisfied with this definition. I understand that it works but I want another one.
I have a feeling that if you can do that, you might get a Nobel Prize or something similar.

I am still surprised you have not opened your textbook.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
I have a feeling that if you can do that, you might get a Nobel Prize or something similar.

I am still surprised you have not opened your textbook.
What if we say that the input x(t) = x*f(t), where x is the physical variable like voltage, and f(t) is a varying function. For example: x(t)=Vcos(ft), where x = V, f(t) = cos(ft).
Then, we could define a linear system like this:
T[x(t)] = x*g(t).

A non-linear system would look like this:
T[x(t)] = x*g(t)+h(t), or other forms.

For example, the RC series circuit is linear and we have that the output to unit step function is: x*g1(t), or to the sinewave: x*g2(t).
If we have voltage on the capacitor, then h(t) appears.

And from this, we see that the system is a special case of a function, where we just spread the values of x in time with the help of f(t). In other words, this x is the same x as in f(x)=ax.
For a given t, we have that T[x] = x*constant... so we have the function. The function is a special case of a system when we look to a particular moment in time.

I haven't seen this in textbooks. I thought about this. So take it with a pinch of salt. Could this be a good definition?
Or is there a linear system that does not satisfy this definition?

Or put in other words. Would a linear ODE always give as answer T[x(t)] = x*g(t), where g(t) is particular on x(t)?
 
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WBahn

Joined Mar 31, 2012
26,398
V1 = A*u(t)
V2 = B*u(t)^2

I do not get it where you are trying to go with this.

The first system is linear and we can verify this by checking the scaling property and additivity property. Because it is defined like it is defined, then for every input, the output will have the form A*Vin.

The second system is not linear.

But I am talking about something else. I do not know a priori how the system is defined. I just have a system, a black box, to which I input the unit step function and I observe the response. And I see that the response is such that the system is linear. Can I conclude based on this that the system is linear for any input? In other words, is the linearity of the system independent on what input I apply?
You are definitely missing the entire point. Do the math like I suggested.

If I take the two systems I described and put each in a black box and you applied a unit step function to the input, you will observe the same response from BOTH black boxes! Namely, the output will be a unit step function. So, based on that, how can you claim to be able to conclude that a system is linear at all by applying a unit step input and observing the output?
 

WBahn

Joined Mar 31, 2012
26,398
My black box is a system described by a differential equation relating the input to the output.
So, if I change the input waveform, then the differential equation, i.e. the model, is the same. Only the output changes.

If the differential equation is a linear ODE, then the system is linear, no matter what input I apply.

And suppose that I do not know that the system is linear or not. I apply the unit step function and see the response. Having the input and the output in hand, surely I can deduce how the differential equation looks and if it is a linear ODE, and thus deduce if the system is linear. Right?

I do not think that for the same input, two different differential equations can produce the same output.
Sure they can! And a linear and non-linear system can produce the same output for a particular input. For a specific input, you can design a literally infinite number of systems that will produce exactly the same output.

f1(x) = 40x(x - 0.5)
f2(x) = 20x
f3(x) = x² + 19x

Apply a unit step input, x = 1·u(t), to any of these and you will get a result of 20·u(t) as the output.
 

WBahn

Joined Mar 31, 2012
26,398
The usual RC series circuit where the input is a voltage source and the initial voltage on the capacitor is 0... this circuit is definitely linear. I am not saying this. Is on the internet. It is linear in the sense that it satisfies the additivity and scaling properties.

If the initial voltage on the capacitor is not 0, then it is not a linear circuit anymore.

I am not trying to develop a new language. I am talking about the linear from linear, time-invariant systems.

Are there LTIs where the gain is not linear?
If yes, I can see that the "linear" word is used differently depending on the context.
A system is linear if and only if it obeys superposition. Namely

Vout(A·V1 + B·V2) = A·Vout(V1) + B·Vout(V2)

Any (practical) system, no matter how nonlinear, can be approximated as a linear system for sufficiently small signal excursions. That is the entire basis for small-signal analysis.

What do you even by the gain being linear?

If the gain is proportional to the input signal, does that constitute a "linear gain"?

Consider the gain of a simply RC voltage divider -- the gain of the circuit is frequency dependent. Yet it is very much linear circuit because it obeys superposition.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Ok. I understand.

The thing that bugs me is that in this edx course, they say that the response of a system to the unit step function is a(n)*u(n).
And then they say that if the system is LTI then i know the response of the system to any linear combination of unit step functions.

And ok, I agree.

But if the response was a(n)*u(n) + constant, then I could not have said that.. ''hey, lets say that this system is LTI.''

And this constant they add separately after the LTI.

So if a system is LTI or not depends on how the output waveform looks like.

So there must be a general form for the output waveform of an LTI.

(I say 'observe' but i'm not referring to viewing the output on the oscilloscope, but how its formula looks like. So Au(t) is different than Bu(t)^2 )
 
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WBahn

Joined Mar 31, 2012
26,398
Ok. I understand.

The thing that bugs me is that in this edx course, they say that the response of a system to the unit step function is a(n)*u(n).
And then they say that if the system is LTI then i know the response of the system to any linear combination of unit step functions.
Notice what they are saying and what they are not saying.

They are saying that IF the system is LTI, then knowing the response to a unit step means that you know the response to any linear combination of unit step functions.

They are NOT saying that if you know the response to a unit step input that you know whether or not the system is LTI.

HUGE difference!

So if a system is LTI or not depends on how the output waveform looks like.
Not based on a single input.

So there must be a general form for the output waveform of an LTI.
Nope. Because a linear system's behavior depends on how it responds to different inputs. A single input tells you pretty much nothing.

(I say 'observe' but i'm not referring to viewing the output on the oscilloscope, but how its formula looks like. So Au(t) is different than Bu(t)^2 )
You need to make up your mind -- one instant you are talking about black boxes in which you don't know what the system looks like and now you are talking specifically about the case where you are relying on having full knowledge of what the system looks like. Which is it?
 

Thread Starter

RdAdr

Joined May 19, 2013
214
So how do I reconcile the fact that if the response had this form: a(n)*u(n) + constant, then i could not have said ''if the system is LTI...''?

I can only say 'if' if the response is a(n)*u(n).
 

WBahn

Joined Mar 31, 2012
26,398
What is there to reconcile? There are certainly responses that cannot be consistent with the system being (large-signal) LTI. Having ANY output for no input is one such case. But the reverse is not true -- knowing that the output is consistent with the system being LTI does not tell you where it actually IS an LTI system.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
It seems paradoxical to me.

If there are certainly responses that cannot be consistent with the system being LTI, then surely there must be some general form for the output.

For the linear function, the general form is y=ax.
And depending on what a is, we have different linear functions and to each such function we can apply different inputs like 2,3,4, etc.
And y=x and y=x^2 produce the same output for the same input 1, but they are still different.

Surely there must be something for the LTI too.
 

WBahn

Joined Mar 31, 2012
26,398
There is nothing paradoxical about it. If an animal does not have feathers you know it can't be a duck. That does NOT mean that if an animal DOES have feathers that is MUST be a duck.

You need to learn what is a necessary condition, what is a sufficient condition, and what the difference is between the two.

Furthermore, the general form for a linear function is NOT y = ax.

\(
y(x) \: = \: A \frac{dy(x)}{dx}
\)

is also a linear function and is not of the form y = ax.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
That is a linear first order ODE and its solution for y(x) is an exponential. Is a system where the x is time, the output is y(x) and the input is 0. So y(x) can be written equivalently as an exponential which is not linear.

The system is indeed linear (if the initial y(0) = 0), but its solution is not.

I am referring to straight lines on a plane. The general form is y = ax. I make distinction between the model and the function that arises from it as the output.

I do not think that people are searching for a theory of everything and in the same time there is no general form for the output of LTIs. Maybe not a general form as in the form of a single formula, maybe different formulas according to a classification of LTIs. I do not know.

I think is safe to say that the solution of any linear ODE has the form:
x*f(t), where x is the physical variable, for example V.
If the system is not linear then it has the form x*f(t)+initial condition

If we apply multiple inputs then it is sum of xi*fi(t).

That is what I always observe. I do not have the math background to demonstrate but I think is something trivial.

And in the edx course, because the output is a(n)*u(n), we can suppose that there is an LTI that has at output this function, or we can build one. Because it looks like x*f(t).
If, on the other hand, it was a(n)*u(n)+offset+noise(n), then this is something else and we cannot say that this can be the output of an LTI. It can be the output of a system which has some initial conditions imposed on it. Or, in other words, the output of an LTI in series with some kind of adder.


Regarding the black box. Maybe I was not clear from the beginning. You can see the output waveform on an oscilloscope and can't tell if it is the output of an LTI or not. I agree.
But I was talking in terms of differential equations (being the black boxes) and their solutions. So if I model the output of a system with a solution which can be the output of an LTI, then that system is an LTI.
With the oscilloscope, yes, I think I would need at least two inputs and their scaled versions too to test if the system is linear.
 
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WBahn

Joined Mar 31, 2012
26,398
What do straight lines in a plane have to do with linear, time-invariant systems?

It's pretty clear that you aren't able to keep straight the distinction between a linear system and a signal that, when plotted, is a straight line. They are two very different things.

You claim that the output of the system I gave you is an exponential if the input is 0. But the output is 0 if the input is 0. If the input is sinusoidal then the output is sinusoidal. That's essentially the differential equation governing the V-I characteristic for a capacitor or an inductor.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
What do straight lines have to do with LTIs? In my opinion, everything. It must be some kind of link between them. Maybe the straight line is a special case of the LTI when the straight line is viewed in time.

Both of them respect the superposition principle.

I know they are very different things, but maybe they are different point of views of the same thing. Just my belief. I did not see a proof or disproof in the textbooks so I will just believe this because it seems like a nice idea. Maybe in year 3000, they will say that the straight line is a special case of the LTI.

The LTI has as input and output functions. The straight line has as input and output single numbers.
But the functions are just numbers when extended in time.

The LTI is a linear map between functions, the linear function is a linear map between single numbers.

Just my belief. I started to have this belief because in the edx course, they talk a little about linear functions and then they make the transition to LTI. They do not specifically say how the transition is made, they just present the facts. But I think there is a link somewhere there.

Yes, the output is constant*exponential. If this constant which is y(0) = 0, then the output is 0. In other words, the output is a familiy of functions depending on the values y(0). But every function from the family is an exponential.
For the sin input is still has the form x*f(t) + initial condition. It is not an exponential, but it is not linear too.
For the input 0, it is 0. Which still has the form x*f(t).

We see that it is linear in terms of functions. It is mapping linearly a function to another function if the initial condition is 0.

And if the system is just linear, not time-invariant, then it is mapping linearly functions to changing functions. So the mapping is changing in time.
 
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WBahn

Joined Mar 31, 2012
26,398
You are making claims based on what you would like to be true, without much concern for whether it actually IS true.

You are confusing signals with systems. This is not uncommon -- I recall having a very similar confusion back when I was a junior. The stuff I had supposedly learned in my first signals and systems course hadn't fully gelled and so when I started getting a bit deeper into it my confusion came to the surface. Go back and read a good sophomore-level introductory signals and systems text.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
I am certainly not confusing them.
One is the model which might be a differential equation relating the input to the output. Another is the output function that arises from the model.
Different functions applied to the model give rise to different output functions.
(the confusion one has sometimes is because sometimes the model is of the form T[x(n)] = n*y(n). It is not a differential equation. And so sometimes there is confusion.)

Making claims based on what I would like to be true. Maybe, yes. But that is how discoveries are made. It is not always a logical process. Sometimes a little intuition can take someone far. And you go on this intuition and prove logical whatever you want to prove.

Making claims are good, in my opinion, as long as you do not present them as 100% true facts.
I am not saying I discovered something here. I think it is already known. But is a discovery for me based on observation.

And I said belief after all. I did not say it is 100% true and proven.
 
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MrAl

Joined Jun 17, 2014
7,952
Hi,

An equation is linear if the ratio of the difference between any two x values and the difference between the corresponding two y values is the same for any two points, or:
(x2-x1)/(y2-y1)=K

for any two points (x1,y1) and (x2,y2), and K being a constant.

The function A*sin(w*t) is non linear in time, but it could be considered linear in A.

A linear system would satisfy:
f(a)+f(b)=f(a+b) (superposition)

and also:
a*f(b)=f(a*b) (homogenous)

In words:
The sum of the function of the individual inputs equals the function of the sum of the individual inputs, and
Multiplication of the input by a factor X results in a multiplication of the output by the same factor X.

This makes it clear that A*sin(w*t) cant be linear in time but it can be linear in A. It might also be called linear in sin(w*t) even though sin(w*t) is not considered linear itself.

So besides other things, we have to know what the independent variable is.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Hi,

An equation is linear if the ratio of the difference between any two x values and the difference between the corresponding two y values is the same for any two points, or:
(x2-x1)/(y2-y1)=K

for any two points (x1,y1) and (x2,y2), and K being a constant.

The function A*sin(w*t) is non linear in time, but it could be considered linear in A.

A linear system would satisfy:
f(a)+f(b)=f(a+b) (superposition)

and also:
a*f(b)=f(a*b) (homogenous)

In words:
The sum of the function of the individual inputs equals the function of the sum of the individual inputs, and
Multiplication of the input by a factor X results in a multiplication of the output by the same factor X.

This makes it clear that A*sin(w*t) cant be linear in time but it can be linear in A. It might also be called linear in sin(w*t) even though sin(w*t) is not considered linear itself.

So besides other things, we have to know what the independent variable is.
That is correct. But why do you say this?
 
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