Input output relations in signals and systems

[Jncik]

Homework Statement

I have a trouble understanding what would be the output when for example we say let's input the signal x(k*n) or x(n-n0) to the system...

this has given me problems when having to solve systems for which I have to check the property of time invariance

suppose we have a discrete system

y[n] = x[2n]

then, find what the system would be for the following inputs

i. x[2n]
ii. x[n-2]
iii. x[n/3]
iv. x[-n]
v. x[-n+3]
vi. x[-n/4 + 1]

Homework Equations

The Attempt at a Solution

well the thing that makes it confusing is that x[n-3] or what ever is not y[n-3]... but rather something else..

my solutions would be

i. y[n] = x[4n]
ii. y[n] = x[2n - 2]
iii.y[n] = x[2/3 n]
iv. y[n] = x[-n]
v. y[n] = x[-2n + 3]
vi. y[n] = x[-2*n/4 + 1]

my questions:

1. are my answers correct?
2. is there any method on finding the answer using some formula?
3. would these answers be the same for continuous time?

thanks in advance

 

[lanedance]

not 100% i understand the question but i would simplify rearranging with the x expression to find the related y index

lets call
y[n] = x[2n]

first one is the same as the definition
x[2n] = y[n]

for the 2nd
x[n-2] = x[2(n/2-1)]= y[(n/2-1)]

what do you think?

 

[Jncik]

I'm not sure about it

for the second one, my book has the same answer I gave.. but I can't understand why this is the answer..

the y[n] = x[2n]

is the definition of the system...

now if I input a x[2n] won't it be y[n] = x[4n]? since, the system will take the input and make the x[2n] signal to be in a region which is 2 times less than the region of x[2n]

 

[lanedance]

yeah, sorry I'm not totally getting it...

can you explain exactly what the system, notation and indicies represent? also what other answers the book has, we may be able to work backwards

otherwise i may have to pass this one onto someone else

 

[Jncik]

basically the system is

y(t) = x(2t) where y(t) is the output and x(2t) is the input..

and I'm trying to understand what the relation between the output would be for shifted inputs

for example

if the system is y(t) = x(t)

then the relation in this system for input x(2t) would be y(t) = x(2t)

while for x(t-1) it would be y(t) = x(t-1)

now I saw an example of time invariance

where the system y(t) = x(2t) is NOT time invariant...

the explanation for this is that if we put an input of x(t-t₀) we get an output of

y(t) = x(2t - t₀)

the system would be time invariant if the same shifting for y would give us the same system

but in this case

y(t-t₀) = x(2t - 2t₀) <> x(2t - t₀) hence the system is not time invariant...

now what I don't get is, WHY, for input x(t-t₀) we get the system y(t) = x(2t - t0) and not the system y(t) = x(2t - 2t₀)...

there is obviously a difference between "shifting the time for y" and "assuming a shifted time for input x" and this is the difference that I'm trying to understand..

note, that for the examples above, I know the answer only for y[n] = x[2n] which is essentially what I just explained, and the same result is written in the book, the other examples are just what I thought of, when trying to figure out the result for different systems

 

[SammyS]

Some of your statements appear to be contradictory - although I'm not necessarily sure I understand all of the notation.

Do you have a few clear and fairly simple examples from your notes or book --- or a decent website to look at?

 

[Jncik]

Some of your statements appear to be contradictory - although I'm not necessarily sure I understand all of the notation.

Do you have a few clear and fairly simple examples from your notes or book --- or a decent website to look at?

I asked a question some days ago

https://www.physicsforums.com/showthread.php?t=512231

and the user said

"y2(t)=x2(2t)=x1(2t-t0)=y1(t-t0/2)
hence it is time variant :P
"
now, this is where it all started, and I figured out that maybe he was correct, but these gave me some more questions that had nothing to do with time invariance and that's why I posted this new question

as you can see my question had to do with proving that the system y(t) = x(2t) is time variant

I said

what I tried to do is

suppose we apply a signal

x1(t) and we get an output of y1(t) = x1(2t)

now, suppose we apply a signal x2(t) = x1(t-t0)

we get an output of y2(t) = x2(t) = x1(2t - 2t0) = y1(t-t0)

we get an output of y2(t) = x2(t) = x1(2t - 2t0) = y1(t-t0)


now

the output y2(t) is wrong, it should be y2(t) = x1(2t-t0)

and this is what I don't understand, why it's wrong, why it is like that and not what I said before

if you still have trouble understanding what I mean, it's ok, I will read my notes again and maybe I ll read something that will clear my thoughts

 

[Jncik]

ok I think after reading again the replies + notes I have this

we have a system y(t) = x(2t)

if we want to shift y by 1. for example we want y(t-1)

it means that we want to shift it by 1 in the x axis..

but in order to shift it by 1 in the x axis, we need to shift x(2t) too by 1 in the x axis..

now, since the range of x(2t) in the x axis, is 2 time less than the range of x(t) in the x axis, it means that, if we want to shift the x(2t) by 1, we will need to shift x(2t) by 2..

now this 2, is not the same range as in y, because y is just y(t) and x is x(2t)..

it refers to the range of each step in x(2t).. which is obviously 2 times less than x(t) itself

hence

y(t-1) = x(2(t-1)) = x(2t - 2)

now..

having the system y(t) = x(2t)

and if I say "let's input x(t-1) to this system

we have that for input x₂(t) = X(t-1)

I get a y₂(t) = X₂(2t) = X(2t-1)

and that's why it's not x(2t-2)...

sorry for being stupid.. I think this is the correct explanation..

hence for

i. x[2n]
ii. x[n-2]
iii. x[n/3]
iv. x[-n]
v. x[-n+3]
vi. x[-n/4 + 1]

i.

x₁[n] = x[2n]

hence y₁[n] = x₁[2n] = x[4n]

ii.

x₁[n] = x[n-2]

hence y₁[n] = x₁[2n] = x[2n-2]

iii.x₁[n] = x[n/3]

hence y₁[n] = x₁[2n] = x[2n/3]

iv.

x₁[n] = x[-n]

hence y₁[n] = x₁[2n] = x[-2n]

v.

x₁[n] = x[-n+3]

hence y₁[n] = x₁[2n] = x[-2n+3]

vi.

x₁[n] = x[-n/4 + 1]

hence y₁[n] = x₁[2n] = x[-2n/4 + 1]

 

[SammyS]

I think you can simplify that last one to x[-n/2 + 1] .

I found a somewhat helpful example in http://en.wikipedia.org/wiki/Time_invariant#Simple_example".

Basically it said, if you have the time variable outside of the argument, then it's time-variant -- as in: y(t) = t·x(2t).

Then an input of x[2n] would give: y[n] = (2n)·x[4n] .

I'll think about this spme more.

 

[Jncik]

I think you can simplify that last one to x[-n/2 + 1] .

I found a somewhat helpful example in http://en.wikipedia.org/wiki/Time_invariant#Simple_example".

Basically it said, if you have the time variable outside of the argument, then it's time-variant -- as in: y(t) = t·x(2t).

Then an input of x[2n] would give: y[n] = (2n)·x[4n] .

I'll think about this spme more.

that was a good example thanks