I have a F system and a input of this system is e x(n) = e^iπn/4 and output is y(n)=cos(πn/4)
So is the F system is LTI or it would be ?
I try to using addivity and Homogeneity but I cant do it
İs this system is a LTI or not
signalsystemtime
Related Solutions
This question belongs better to dsp.stackexchange.
The input signal in part (b) has 2 terms, and is the sum of 2 exponentials, one with frequency $$\omega_1 = 0$$ and the other with frequency $$\omega_2 = \pi/2$$
Since the system is linear, then the output is the sum of the output to each individual input. That is: $$y[n] = y_1[n] + y_2[n]$$
Also, since complex exponentials are eigenvectors of LTI systems, then the output will be the same exponential multiplied by the frequency response (i.e. the eigenvalue).
So,
$$ y_1[n] = 1 \times H(e^{j 0}) $$
and
$$ y_2[n] = e^{j n \pi/2} \times H(e^{j \pi/2}) $$
Where $$H(e^{j0}) = 1$$ and $$H(e^{j\pi/2})= 2+j$$.
First of all lets correct your mistake: a particular solution of the equation is not the one for which you need the initial conditions.
Now lets see if I can give you some intuition...
The symbol \$x\$, as well as constant term, (usually) stand for the inputs of the system (the term "input signals" is more appropriate, but we are not seeking to be rigorous here, right?). What the inputs are? Depends on the system in question. For mechanical system the inputs may represent external forces, for electrical system the inputs may represent the currents flowing into a component, etc. You can see inputs as external factors - the system is an entity by itself, and the inputs affect the system in some way when applied to it.
The symbol \$y\$ in the equation is the most interesting part - it is an output of the system. Usually, the "solution" of the DE will be a functional description of \$y\$ as a function of time (\$t\$). Mathematically, the solution, is a function that satisfies the DE when substituted for \$y\$.
The good thing about DEs is that once you have an answer it is very simple to test whether it is correct. The bad thing is that it is difficult to derive the answer (sometimes it is even impossible to get an exact answer).
After we saw what the symbols in the equation stand for, let's see how to solve these mother f***s.
A homogeneous solution of DE:
This is the function which satisfies the DE when you reduce it to become homogeneous. What does homogeneous mean? Well, just throw out of the equation all the terms which do not contain \$y\$ or its derivatives.
From an intuitive point of view a homogeneous equation represents an output of the system when there are no inputs (or all the inputs are zero, which is the same thing). The question arises: how comes a system can have an output when there are no inputs? Well, in general, it can't (don't take this statement too literally). However, the system may have initial conditions. Think of a mass on a spring as an example - if the string was initially stretched, even if there are no other forces on the mass - it will oscillate. The homogeneous solution in this case is the description of these "free" oscillations.
There is one single solution which is common to all homogeneous DEs. Try to think by yourself what it is. Don't know? Read the previous paragraph and think again. This solution is zero. If you think of it, it makes perfect sense - each system with zero initial conditions will remain at rest. "Remain at rest" is a normal way of saying: \$y=0\$. This solution is called "trivial" for obvious reasons, and it is not the solution you're interested in.
What you do want to know is how the system will behave when there are no inputs, but some non-zero initial conditions. The non-trivial homogeneous solution will provide this information.
A particular solution of DE:
A particular solution is a function which satisfies the whole DE. The usual way to obtain these solutions is to "guess". No joke, really. I have a friend who always complaint: "but we did not take a course on guessing!". Of course this approach works for the simplest cases only. If you can't guess a particular solution, you still have some mathematical tricks which can help you, but in the real world the usual approach would be to say "well, f*** it!" and perform a numerical analysis.
While the homogeneous solution represents a "free" or "natural" response of the system, the particular solution is its "driven" or "forced" response - the response to the inputs.
A general solution of DE:
A general solution of DE is a function which describes the general form of the output of the system given its inputs. It is just the sum of the homogeneous solution and the particular solution.
Usually, this function contains unknown parameters (constants). The number of these parameters equals the number of degrees of freedom in the system (forget this statement if you don't know what does "degree of freedom" mean). The presence of these parameters is the reason this solution is called "general" - it is not exact, but represents the general form of all possible solutions.
The existence of general solution is astonishingly non-trivial, but we take it as granted. Think of it: we can describe the system (potentially very complicated) based on its inputs with a single mathematical function, and the only things which change with initial conditions are few parameters.
An exact solution of DE (solution):
Once you got the initial conditions for the system you can find the exact solution. Just substitute the initial conditions into the equation, and solve the resulting equations together.
Note: if you got \$n\$ parameters in your general solution, you'll have to provide \$n\$ independent initial conditions to solve the DE exactly.
Note2: the word "initial" is a bit misleading - these conditions may be specified for any time, not necessarily the initial time.
Notes:
There are two points which are not that obvious by themselves, and it is hard to explain them intuitively.
- If the particular solution is the solution of the whole equation, then why do we add to it the homogeneous solution to obtain the general solution?
- If the homogeneous solution is the solution of the undriven system which has non-zero initial conditions, then why do we substitute the initial conditions into the general solution?
Take these two as granted for now. If you'll come back to LTI systems a few times in the future - these will become very intuitive to you, even though you'll find it hard to explain why they are intuitive :)
Hope this helps and not too long - I must stop posting books to the forum.
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Best Answer
You don't have enough information to know if the system is LTI or not.
To know a system is linear, you have to know that for any input \$x(n)\$ producing an output \$y(n)\$, then a scaled input \$A\cdot{}x(n)\$ produces the output \$A\cdot{}y(n)\$, no matter what input was chosen for \$x(n)\$. Since you only know the output for one particular input, you don't know if your system is linear or not.
However, the point is largely moot. We normally don't try to prove a system is LTI or not. We just assume a system is (approximately) LTI in order to model its behavior. Real systems are usually nonlinear (because a large enough input will produce saturation or damage the system or cause numerical overflow) and time-variant (because they are turned on and off from time to time). So LTI-ness is just an approximation to the real behavior, or a given of a pedagogical problem, to allow modeling the system.