Laplace transforms vs Z transforms

[Yuseph]

Hey guys,

I know in continuous systems laplace is the norm. But what about sampled systems. Did anyone use laplace transforms to determine difference equations from sampled systems.

 

[Deleted member 4993]

Hey guys,

I know in continuous systems laplace is the norm. But what about sampled systems. Did anyone use laplace transforms to determine difference equations from sampled systems.

You say:

I know in continuous systems laplace is the norm.​

I have no idea what you are talking about. Please define your statement.

 

[Yuseph]

Sorry i thought you guys knew about Z transforms

 

[QuantAbu]

Hi there, I am a new to learning real analysis. I would like to ask why they set epsilon = (x-y^2)/4(x+1), I get the numerator but how did they come up with 4(x+1), like x+1 is y but why multiply by 4?
Second Question: how did they reduce the third term to (x-y^2)/4 * x/4x in the fourth equality and again from that to just x-y^2. is it a just increasing the numerator and decreasing the denominator in a clever way?

Thank You for your time!

 

[Steven G]

1st of all (x-y^2)/4(x+1) = [math]\dfrac{x-y^2}{4}*(x+1), \ not \ \dfrac{x-y^2}{4*(x+1)}[/math]
2nd, where does it say or even imply that y=x+1? So why say that y=x+1. y is the least upper bound while x+1 is an upper bound. The best you can say is that y < x+1.

You keep making statements about inequalities as if they are equalities. You can't do that!
3rd, (x-y^2)/4 * x/4x = (x-y^2)/16. Now (x-y^2)/16 < (x-y^2)/2 so when you add (x-y^2)/2 you get less that x-y^2! Now when you add y^2 to that you get x, which is what we wanted! So why did we choose epilson = (x-y^2)/[4(x+1)]? Because it worked, that is we got the initial expression to be less than x. Can you think of another expression which epilson could have equaled?[/math]

 

[QuantAbu]

Thank you so much for the things you have taught me today.
If you don't mind may I ask one more question? in the third inequality, the third term (x-y^2)/(4*(x-1)) * (x-y^2)/(4*(x-1)) was reduced to ((x-y^2)/4)* (x/(4*x)), how was it done?

 

[HallsofIvy]

Again, this is an inequality, not an equation! One was NOT "reduced" to the other, they are not equal. One is less than the other. if x< y then a+ b+ x< a+ b+ y.

 

[QuantAbu]

Thank you very much! So the question would be why is [ (x-y^2)/(4*(x-1)) * (x-y^2)/(4*(x-1)) ]< [ ((x-y^2)/4)* (x/(4*x)) ]

 

[Steven G]

Thank you very much! So the question would be why is [ (x-y^2)/(4*(x-1)) * (x-y^2)/(4*(x-1)) ]< [ ((x-y^2)/4)* (x/(4*x)) ]

No I will not answer your question, you are in real analysis and should have some number sense so I want you to figure it out.

Here is a hint: If A/(BC)< A/(BD) and all variables are positive then what can you conclude about some (two of them) of the variables?

 

[QuantAbu]

Correct me if my thinking is wrong here, So in the numerator we have two positive numbers( here number is defined as (x-y^2) ) we remove the y^2 from the second number, since x>y^2 this increases the numerator. making it (x-y^2)*x. We make the denominator smaller by taking 1 instead of x+1, and x instead of x+1. hence we have ((x-y^2).x )/ (4*x) which is greater than the previous term.

As per your question C have to be greater than D to make A/(BD) greater than A/(BC)

 

[Steven G]

I will write e instead of the symbol for epsilon.

You are given that y^2<x (so x>y^2 and x-y^2>0), e = (x-y^2)/[4*(x+1)] and y<x+1 (since the lub < ub)

Now (y+e)^2 = y^2 + 2ye + e^2 (just foiled out)
= y^2 + 2y(x-y^2)/[4*(x+1)] +(x-y^2)/[4*(x+1)]*(x-y^2)/[4*(x+1)] (replaced e with (x-y^2)/[4*(x+1)]
< y^2 + 2y(x-y^2)/[4*y] +(x-y^2)/[4*(x+1)]*(x-y^2)/[4*(x+1)] (replaced something in the denominator with some smaller, y<x+1)
< y^2 + (x-y^2)/2 + (x-y^2)/[4]*(x)/[4*(x)] (replaced 4(x+1) with 4 which makes the denominator smaller which makes the positive fraction larger. replaced the factor x+1 with something smaller, namely x, which makes the positive fraction larger)replaced x-y^2 with x.
=y^2 + (x-y^2)/2 + (x-y^2)/16 (arithmetic)
= y^2 + 9(x-y^2)/16 (arithmetic)
< y^2 + (x-y^2) (since x-y^2>0 and (x-y^2)> (9/16)(x-y^2)
= x (arithmetic)

 

[QuantAbu]

Thank You Sir, I like the way you teach. You made me think about the problem first which helped me a lot indeed. I will come here again if I stumble upon something which I can't figure out.