Z transforms Two questions

can somebody please help me with the two questions

I have the answers but I not able to prove it



see large image here

https://obrazki.elektroda.pl/78_1179740083.jpg

find the inverse z transform by partial fraction
and then find the index 'n' for which u get a negative value of x
Hope its the answer of the question.

the inverse z-transform is to difficult to solve analytically since the powers of the z transform are very high
please help me out

Don't kill me, I'll try and help, but could you tell what's right-sided sequence? I know plenty of English, but the special Z transform apparently not....

Thanks... I'm trying to work something up in the meantime...

Added after 44 minutes:

If you still want an answer...Here's how to do it
your Z transform is like a transmittance Y(z)/U(z), U is input, Y is output. You need to find the output when input is a Dirac, a Kronecker impulse, or however you may want to call it. The idea is that its value is 1 only in 0

What you do is that you put:

Y(z)/ D(z)= X(z) . You multiply Y(z) by denominator and D(z) by nominator of the fraction. Divide the equality by z-3 on both sides and then go into temporal by simply : Y(z)*z-5 --> y(k-5) where y(k) is actually y(kT)...

then separate the y(k) term on one side and leave the rest on the other side:

you get:

4y(k)=2d(k)-2d(k-5)-d(k-3)-4d(k+3) + y(k-5)+y(k-2)

This relationship, for k<0, will give you a non-zero value for y(k) only at k=-3 because it is the only negative value than makes any of the "k-5, k-3...k+3.." in the d 0: d(k+3)=0 only in k=-3 and 4y(-3)=-4d(k+3)=-4 => y(-3)=-1....

I just realised that y(k-5) and y(k-2) need to eb zero, so in hope that maybe you can reason that, I hope I haven't given you a wrong idea...

tzushky great idea. problem solved.

Many thanks to you.
help button pressed
15 points given

bye . keep it up

jim

thanks I like it here

Enjoy your Z transforms

Hi
well;there is a simple way,BUT result is different(!)


as you know :
N(z)/D(z)=sum(ai*z-i,i=-infinite:infinite)
i.e. dividing numerator to denominator will give you the coefficients.

Z^8 x (2Z^-8+...)/[Z^7 * (Z^-8+...)]=
(4Z^8 - 2Z^5 +...)/(-4Z^5 + 2Z^3)

4 Z^8 - 2 Z^5 +... |__ -4 Z^5 + 2 Z^3
-Z^3 - .5 Z ...

i.e. for n<0 : x(-3)=-1 & x(-1)=-.5

also I used MATLAB ,where I fiund the same result!!?
any ideas which answer is correct???

regards
Armin

Hi, I also know that you can get the results by dividing the two polynomials, but I also knew that the rule was to have the denominator of the form 1+ sum(coeff_i*z^-i, i=0:infinity) , so I fuigured it should be the same here, but in order to do that, you can't simplify the numerator by the necessary z^-3 to get a constant at the denominator, and you should separate the 4 at the nominator, or something like that... I let it go... because I decided it didn't respect that rule. I may be wrong...

Also, if Matlab is allowed, a decomposition in simple fractions would also be possible, but still long and too complicated by hand so I thought of this method as the simplest, no calculation involved...

I honestly don't know if what you applied is perfectly allowed, I have Z experience, but not that much ... It's good though that we get to review all methods...

Regards,
Tzushky

Hi
thanks to Tzushky for his/her deep insight about the problem

I think I have to clarify my last post.
I used the following well known procedure:

1)multipled both numerator & denominator by z^8

2)ordered new numerator & new denominator

3)used only first 2 coefficients of numerator & first 2 coefficients of denumerator in division procedure.
(since after two steps of division other n<0 coefficients are zero)


and finally i checked my results with results i obtained in MATLAB to see if they are correct.

the point is that my result is different from Jim's instructor result(!!)
best wishes
Armin