Inverse Z-Transform: Partial Fractions vs. Residue Theorem

In summary: I see, I forgot the factor of 3 in the denominator of the second term. Ok, now I agree with everything you have.Now, when you write stuff like this, you should say what u(n) and δ(n) are. I think δ(n) = 1 when n=0 and 0 otherwise. And u(n) is 1 when n >= 0 and 0 otherwise. Once you specify that, then it will be clear that both expressions are the same.
  • #1
peripatein
880
0
Hello,

Homework Statement


I would like to find the inverse Z transform of the following:
F(z)=1-1.25z-1+0.25z-2/[1-(5/6)z-1+(1/6)z-2]
using (a) partial fractions, and (b) residue theorem
I have obtained different results and hence would appreciate some insight on the discrepancy and how it may be resolved.

The Attempt at a Solution


(a) Using partial fractions, I obtained the following:
F(z)=(3/2)[4z2-5z+1/(6z2-5z+1)] = (3/2)[10z/(z-1/3) - 2/(z-1/3) - 6z/(z-1/2)]=
=15z/(z-1/3) - 3/(z-1/3) - 9z/(z-1/2) = 6 + 2/(z-1/3) - (9/2)/(z-1/2)
⇔ f(n) = 6δ(n) + 2(1/3)n-1u(n-1) - (9/2)(1/2)n-1u(n-1) =
6δ(n) + [6(1/3)n - 9(1/2)n]u(n-1)
where u(n) is the step function and δ(n) is the dirac function.
(b) Using residue theorem, I obtained the following:
Residue for z=1/2 is -9, residue for z=1/3 is 6, hence:
f(n) = [-9(1/2)n + 6(1/3)n]u(n)

Where am I erring? I'd appreciate any comments.
 
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  • #2
peripatein said:
Hello,

Homework Statement


I would like to find the inverse Z transform of the following:
F(z)=1-1.25z-1+0.25z-2/[1-(5/6)z-1+(1/6)z-2]
using (a) partial fractions, and (b) residue theorem
I have obtained different results and hence would appreciate some insight on the discrepancy and how it may be resolved.

The Attempt at a Solution


(a) Using partial fractions, I obtained the following:
F(z)=(3/2)[4z2-5z+1/(6z2-5z+1)] = (3/2)[10z/(z-1/3) - 2/(z-1/3) - 6z/(z-1/2)]=
=15z/(z-1/3) - 3/(z-1/3) - 9z/(z-1/2) = 6 + 2/(z-1/3) - (9/2)/(z-1/2)
⇔ f(n) = 6δ(n) + 2(1/3)n-1u(n-1) - (9/2)(1/2)n-1u(n-1) =
6δ(n) + [6(1/3)n - 9(1/2)n]u(n-1)
where u(n) is the step function and δ(n) is the dirac function.
(b) Using residue theorem, I obtained the following:
Residue for z=1/2 is -9, residue for z=1/3 is 6, hence:
f(n) = [-9(1/2)n + 6(1/3)n]u(n)

Where am I erring? I'd appreciate any comments.

What you have written (using standard parsing rules for reading mathematical expressions) is
[tex]F(z) = 1-1.25z^{-1}+\frac{0.25z^{-2}}{1-(5/6)z^{-1}+(1/6)z^{-2}} [/tex]
Is that what you truly indended? If you mean
[tex] \frac{1-1.25z^{-1}+ 0.25z^{-2}}{1-(5/6)z^{-1}+(1/6)z^{-2}} [/tex]
you need to use parentheses around the numerator, like this:
F(z)=[1-1.25z-1+0.25z-2]/[1-(5/6)z-1+(1/6)z-2]

Anyway, your partial-fraction expansion seems wrong; I (or, rather, Maple) get
[tex]F(z) = 1 - \frac{3}{2}\frac{1}{2z-1} - \frac{1}{3z-1} [/tex]
 
  • #3
How could my partial fraction expansion be wrong if upon multiplication I obtain the same F(Z) I started with?? I have just checked myself yet again, and the expansion is absolutely correct, hence:
F(Z)=(3/2)(4z^2-5z+1)/(6z^2-5z+1) = 6 + 2/(z-1/3) - (9/2)/(z-1/2)
I still don't understand why the results are seemingly different, using both methods. I would appreciate some insight on that.
 
  • #4
peripatein said:
How could my partial fraction expansion be wrong if upon multiplication I obtain the same F(Z) I started with?? I have just checked myself yet again, and the expansion is absolutely correct, hence:
F(Z)=(3/2)(4z^2-5z+1)/(6z^2-5z+1) = 6 + 2/(z-1/3) - (9/2)/(z-1/2)
This could not possibly be correct.
Here's why.
$$F(z) = \frac{6z^2 + \text{lower deg. terms}}{6z^2 + \text{lower deg. terms}} = 1 + \text{other rational expressions}$$

Notice that if you do the long division, the constant term is 1, not 6 as you show.
peripatein said:
I still don't understand why the results are seemingly different, using both methods. I would appreciate some insight on that.
 
  • #5
Okay, let's suppose you're right (even though I am still unable to figure out where my mistake is). Long division yields:
1 + 2/(z-1/3) - (9/2)/(z-1/2)
That still leaves me with f(n)=δ(n) + [6(1/3)n - 9(1/2)n]u(n-1) using partial fractions, vs. [6(1/3)n - 9(1/2)n]u(n) using residue theorem.
How would you settle this discrepancy?
 
  • #6
Can you expand on how you applied the residue theorem? I don't get what you did for n=0 when I use the residue theorem.
 
  • #7
peripatein said:
Okay, let's suppose you're right (even though I am still unable to figure out where my mistake is). Long division yields:
1 + 2/(z-1/3) - (9/2)/(z-1/2)
That still leaves me with f(n)=δ(n) + [6(1/3)n - 9(1/2)n]u(n-1) using partial fractions, vs. [6(1/3)n - 9(1/2)n]u(n) using residue theorem.
How would you settle this discrepancy?

Are δ(.) and u(.) defined so that δ(0) = 1 and δ(k) = 0 for k ≥ 1 and u(k) = 1 for k ≥ 0? If so, both results are the same for n ≥ 0; that is, there would be no discrepancy.
 
  • #8
But for n=0:
δ(n) + [6(1/3)^n - 9(1/2)^n]u(n-1) = 1 (since u(-1)=0; u(n-1) only begins at n=1)
whereas
[6(1/3)^n - 9(1/2)^n]u(n) = -3
What am I missing?
 
  • #9
You're not applying the residue theorem correctly.
 
  • #10
I am. The residue of z=0 is zero and contributes nothing to the sum.
 
  • #11
The final result is pretty much that which yields Wolfram. However, I cannot account for the (seeming) discrepancy.
If you think I am not applying it correctly, could you please point out where it is I am faltering?
 
  • #12
I did. You're messing up somewhere. It's hard to be more specific when all you've done is post your answers and ask "what's wrong?" How are we supposed to figure out what you're doing if you don't show us?

All I can say is that when I looked at 2/(z-1/3) and used the complex integral to evaluate what the n=0 term should be, I got 0, not 6(1/3). That's why I'm saying you're not applying the residue method correctly. Maybe you're not using the right integrand.
 
  • #13
Using residue theorem:
F(z)*z^(n-1) = z^n*[6z-(15/2)+(3/2)z^-1]/[(z-1/3)(z-1/2)]
Residue for z=1/2 is -9(1/2)^n, residue for z=1/3 is 6(1/3)^n, hence:
f(n) = [-9(1/2)^n + 6(1/3)^n]u(n)
Is that wrong?
 
  • #14
I'll assume your algebra is correct. When n=0, you still have a singularity at z=0, don't you?, because the numerator is undefined.
 
  • #15
By the way, I get the same partial fraction expansion that Ray posted above, and it doesn't agree with the results of your long division. So you still have other errors to find in your work.
 
  • #16
The expansion is not as Ray wrote it, but rather:
1 + 1/(3z-1) - (3/2)/(2z-1)
I sincerely doubt this is wrong as have checked it over and over again. You may do so yourself.
Alright, now that yields f(n) = δ(n) + [(1/3)n - (3/2)(1/2)n]u(n-1)
Would you agree?
Next, concerning the residue theorem:
f(n)=sum of residues of [zn*(z2-(5/4)z+(1/4))]/[(z)(z-1/2)(z-1/3)]
Residue for z=1/2 : (-3/2)(1/2)n
Residue for z=1/3 : (1/3)n
Residue for z=0 : 0
Hence f(n) = [(1/3)n - (3/2)(1/2)n]u(n)

Here again there appears to be a discrepancy between f(n) = [(1/3)n - (3/2)(1/2)n]u(n) (via residue theorem) and δ(n) + [(1/3)n - (3/2)(1/2)n]u(n-1) (via partial fractions)

Why?
 
  • #17
peripatein said:
The expansion is not as Ray wrote it, but rather:
1 + 1/(3z-1) - (3/2)/(2z-1)
I sincerely doubt this is wrong as have checked it over and over again. You may do so yourself.
Yeah, Ray made a small typo, but you wrote
peripatein said:
Long division yields:
1 + 2/(z-1/3) - (9/2)/(z-1/2)
I was pointing out this result wasn't correct, and it can't be explained away as a simple typo either.

Alright, now that yields f(n) = δ(n) + [(1/3)n - (3/2)(1/2)n]u(n-1)
Would you agree?
Next, concerning the residue theorem:
f(n)=sum of residues of [zn*(z2-(5/4)z+(1/4))]/[(z)(z-1/2)(z-1/3)]
Residue for z=1/2 : (-3/2)(1/2)n
Residue for z=1/3 : (1/3)n
Residue for z=0 : 0
Hence f(n) = [(1/3)n - (3/2)(1/2)n]u(n)

Here again there appears to be a discrepancy between f(n) = [(1/3)n - (3/2)(1/2)n]u(n) (via residue theorem) and δ(n) + [(1/3)n - (3/2)(1/2)n]u(n-1) (via partial fractions)

Why?
You're not calculating the residue for z=0 correctly. Consider the cases n=0 and n≥1 separately.

By the way, you might want to use the partial fractions expansion when calculating the residues. It makes things a bit simpler in my opinion.
 
  • #18
Alright, then please explain to me how to correctly calculate the residue for z=0.
 
  • #19
Did you bother trying what I suggested?
 
  • #20
I am not sure I am supposed to make use of partial fractions here at all as I was asked to specifically solve it using residue theorem after having solved it using partial fractions.
I don't quite understand why z=0 poses a problem. I am interested in solving it correctly for that value so please try to explain to me how it might be attempted without using partial fractions.
 
  • #21
There's no reason to not use the partial fractions expansion to make calculating the residues easier. In any case, I already suggested you do the calculations for n=0 and n>0 separately. Show us what you get (by that, I mean your work, not just the answer) when you do that.
 
  • #22
peripatein said:
The expansion is not as Ray wrote it, but rather:
1 + 1/(3z-1) - (3/2)/(2z-1)
I sincerely doubt this is wrong as have checked it over and over again. You may do so yourself.
Alright, now that yields f(n) = δ(n) + [(1/3)n - (3/2)(1/2)n]u(n-1)
Would you agree?
Next, concerning the residue theorem:
f(n)=sum of residues of [zn*(z2-(5/4)z+(1/4))]/[(z)(z-1/2)(z-1/3)]
Residue for z=1/2 : (-3/2)(1/2)n
Residue for z=1/3 : (1/3)n
Residue for z=0 : 0
Hence f(n) = [(1/3)n - (3/2)(1/2)n]u(n)

Here again there appears to be a discrepancy between f(n) = [(1/3)n - (3/2)(1/2)n]u(n) (via residue theorem) and δ(n) + [(1/3)n - (3/2)(1/2)n]u(n-1) (via partial fractions)

Why?

The partial fraction expansion is exactly as I wrote it. Here is the Maple input (calling your function Fz):
lprint(Fz);
(1-5/4/z+1/4/z^2)/(1-5/6/z+1/6/z^2) <---- the input
Here, the notation a/b/c is Maple's way of writing a/(b*c) in ASCII.
Here is the partial-fraction expansion:
pfz:=convert(Fz,parfrac):
lprint(pfz);
1-3/2/(2*z-1)+1/(3*z-1)

You can re-combine the three terms and get back to the original.
 
  • #23
But that was not what you wrote earlier. Check for yourself. There was a typo.
 
  • #24
Alright, did you mean something like: zn/z + zn/[z(3z-1)] - (3/2)zn/[z(2z-1)]?
I guess it could be written thus.
Obviously there is not difficulty for n>0. However, I am not quite sure how to handle the case where n=0 and would appreciate some guidance.
 
  • #25
Set n equal to 0 in that expression. What do you get?
 
  • #26
I get 1/z + 1/[z(3z-1)] - (3/2)/[z(2z-1)], which will yield 3/2 as the residue for z=0, right?
 
  • #27
Yup.
 
  • #28
Now I seem to be getting strange results for the residues at the other poles (I say strange as they don't quite coincide with the expression obtained using partial fractions). For z=1/2 I get -3(1/2)^n and for z=1/3 I get 3(1/3)^n. I used the same partial fractions expansion as noted above. Hence,
Via residue theorem:
f(n)=[3(1/3)^n - 3(1/2)^n + (3/2)]u(n) for n=0, ergo (3/2)u(n), or simply (3/2); and for n>0, f(n)=[3(1/3)^n - 3(1/2)^n]u(n)
Via partial fractions:
f(n) = δ(n) + [(1/3)^n - (3/2)(1/2)^n]u(n-1), which for n=0 is 1, and for n>0 is [(1/3)^n - (3/2)(1/2)^n]u(n-1)

How come the expressions obtained using the two methods do not match?
 
  • #29
Okay, I have managed. Thank you for your help :).
 

1. What is the inverse Z-transform?

The inverse Z-transform is a mathematical operation that converts a function in the Z-domain back to the time domain. It is the reverse process of the Z-transform, which converts a time domain function into the Z-domain.

2. Why is the inverse Z-transform important?

The inverse Z-transform is important because it allows us to analyze and solve discrete-time systems and signals in the time domain. It is commonly used in fields such as digital signal processing, control systems, and communication systems.

3. How is the inverse Z-transform calculated?

The inverse Z-transform can be calculated using various methods such as partial fraction expansion, power series expansion, and contour integration. The specific method used depends on the form of the Z-transform function.

4. What are some applications of the inverse Z-transform?

The inverse Z-transform has various applications in engineering and science. It is used for analyzing and designing control systems, digital filters, and communication systems. It is also used in time series analysis and signal reconstruction.

5. Are there any limitations of the inverse Z-transform?

One limitation of the inverse Z-transform is that it can only be applied to functions that have a region of convergence (ROC). If the ROC does not include the unit circle in the Z-plane, the inverse Z-transform does not exist. Additionally, the inverse Z-transform may not be applicable to functions with complicated singularities or poles on the unit circle.

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