inverse z Transform Hw

Thread Starter

Ramiel

Joined Feb 19, 2018
74
Hello everyone.
I have an inverse z transform hw.
I solved them but I am not sure. May I have some feedback please for verification.
Thanks in advance
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MrAl

Joined Jun 17, 2014
7,586
Hi,

I am in a bit of a hurry today i can look this over later today if no one else does first.
But, in the mean time why dont you apply the Z transform to your results and see if you get the original question expression?
If you take the Z transform of an expression y and you get Y, then when you take the inverse transform of Y you should get y back again. Likewise if you take the inverse transform of Y and get y then when you take the Z transform of y you should get Y back again.
This is really the best way to handle this anyway so you can always have a way to check your results yourself in the future.
 

Thread Starter

Ramiel

Joined Feb 19, 2018
74
Hi,

I am in a bit of a hurry today i can look this over later today if no one else does first.
But, in the mean time why dont you apply the Z transform to your results and see if you get the original question expression?
If you take the Z transform of an expression y and you get Y, then when you take the inverse transform of Y you should get y back again. Likewise if you take the inverse transform of Y and get y then when you take the Z transform of y you should get Y back again.
This is really the best way to handle this anyway so you can always have a way to check your results yourself in the future.
Yes I am surprised no one actually gave any response. The thing is i am not sure if Z^2 gives delta(n+2), since my prof never gave such an example. I hope you can take a look at my work later today and let me know what you think.
 

MrAl

Joined Jun 17, 2014
7,586
Yes I am surprised no one actually gave any response. The thing is i am not sure if Z^2 gives delta(n+2), since my prof never gave such an example. I hope you can take a look at my work later today and let me know what you think.
Hello again,

I am sorry i cant read the solutions there is a dark shadow over both images.
Can you take a better shot without shadows?
 

MrAl

Joined Jun 17, 2014
7,586
Hi,

Sorry but maybe you better type it out so i can read it ok. That second set is almost worse than the first unfortunately :)
Sorry but we have to have something that is not too hard to read or else it makes it much more difficult to help.
 

Thread Starter

Ramiel

Joined Feb 19, 2018
74
Hi,

Sorry but maybe you better type it out so i can read it ok. That second set is almost worse than the first unfortunately :)
Sorry but we have to have something that is not too hard to read or else it makes it much more difficult to help.
I know my handwriting is small, so I rewrote it in a bigger font, under the sun. Is it okay or do you want me to type it?51C0A04A-4879-4B0B-AA4B-31A339B8F0DA.jpeg540116A8-7B0F-4450-A392-F6EDDCC78F4B.jpeg
 

MrAl

Joined Jun 17, 2014
7,586
Hi,

That looks better, and yes one of the problems was the tiny 'font' in the previous set.
Give me some time to take a look.
 

Thread Starter

Ramiel

Joined Feb 19, 2018
74
Hi,

That looks better, and yes one of the problems was the tiny 'font' in the previous set.
Give me some time to take a look.
Sure take your time, but take into consideration the ROC given in the question in the beginning of the thread
 

Thread Starter

Ramiel

Joined Feb 19, 2018
74
Hello,

DId you make a mistake in line 4 of this one?

Also, what table of transforms are you using?
Yes you are right i just copied it wrongly from my previous work it should be like this. I just missed a Z. The rest of the steps are the same. And I attached the tables I usedimage.jpgimage.jpgimage.jpg
 

MrAl

Joined Jun 17, 2014
7,586
Hello again,

Thanks for the table that's a pretty good one actually.

What did you do to get the inverse transform of z^2 ?
I ask because if we try to use entry #4 in the table which is z^(-m) we would have to make m=-2 and that would seem to imply that the result would be delta[n+2].
Oh that is what you got ok :)
Is that the part you had the problem with for that question number?

Using that table this seems pretty simple.
What is amazing is that Wolfram can not find the solution. It does not know what to do with z^2.
 

MrAl

Joined Jun 17, 2014
7,586
Hello,

@MrAl ,
This seems to be an other version of the book:
https://archive.org/details/discrete-time-signal-processing/mode/2up

There are the same tables in chapter 4:

View attachment 206535

View attachment 206536

Bertus
Hello there,

Very sorry i could not give you 10 likes for that link as we are only allowed 1 :)

That's a great reference i actually have the 1975 hard cover version by the same two guys one from MIT the other was with Bell Labs right here in New Jersey. Back then it was called "Digital Signal Processing" published by Prentice Hall where i got a lot of my technical library from the 1970's to around the 1990's sometime.

I took a quick look and i see they expanded the Z Transform section a lot. Back then there were less pages in that section and the table was very very short, although they did give lots of test problems.

Thanks a lot and i bet the OP is thankful too for that link.
 

Thread Starter

Ramiel

Joined Feb 19, 2018
74
Hello again,

Thanks for the table that's a pretty good one actually.

What did you do to get the inverse transform of z^2 ?
I ask because if we try to use entry #4 in the table which is z^(-m) we would have to make m=-2 and that would seem to imply that the result would be delta[n+2].
Oh that is what you got ok :)
Is that the part you had the problem with for that question number?

Using that table this seems pretty simple.
What is amazing is that Wolfram can not find the solution. It does not know what to do with z^2.
I actually have no problem. I just want someone to verify if my answers are correct or not
 

MrAl

Joined Jun 17, 2014
7,586
Ok well i thought i would let you know that i got a different result than you did so i will check my result tomorrow to make sure it is right.
 

MrAl

Joined Jun 17, 2014
7,586
Hello again,

Here is what i got for the simpler quiestion (where the 'd' represents the delta symbol):
d[n+2]+2*d[n+1]+2*d[n]+2^(n-1)*u[n-1]

The way i got this was to use partial fraction decomposition and then just transform one term at a time. When i check it i get something that looks right however this is tricky.

What i would suggest trying next is taking both your solution and my solution and generating the two sequences, then using long division divide the numerator of the original problem by the denominator and generate the stream of coefficients of powers of z, then try to correlate to either sequence and see what looks like it is working.
When i do this i get at least in part coefficients 1, 2, 4, 8, 16, 32, etc., etc., which are the coefficients of z^-1, z^-2, z^-3, etc., etc.
I believe we should get the same results for both the correct sequence and the coefficients.

What makes this a little tricky is the impulses. I am not sure i remember how to handle them that well i might have to do some more reading.
 

MrAl

Joined Jun 17, 2014
7,586
Hello again,

Sorry i cant be of more help just yet as i am having a bit of a problem finding some of my reference materials and am drawing a blank with this particular function (the second one).
Most of these functions are proper fractions and so you can create a difference equation quite easily and check it with long division.
 
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