# finding inverse Z transform with usage of tables (LTI and causal sequences)

problem is as follows

try to find the inverse Z

$$Z^{-1} (\frac{3}{z+2}) = ???$$

with the usage of z-transform tables

Ok, so in order to find something from the table, I thought that we expand with z and try to see if we get closer to any kind of expression in the tables

$$Z^{-1} (\frac{3z}{z*(z+2)}) = Z^{-1} (3*\frac{1}{z}\frac{z}{z+2})$$

then we get constant out

$$=3*Z^{-1}(z^{-1}*\frac{z}{z+2})$$

then we see that right side fraction can be found in the table, but the combined sum expression can also be found with further manipulation (as can be seen later)

So, I think we declare temp variable here to clean it up (this portion is found in the table green portion)

$$F(z) = \frac{z}{z-(-2)} \iff f[k]=(-2)^k$$

then we can make it so

$$3*Z^{-1}(z^{-1}*F(z))$$

then we find the direct expression in the table (we have delay of one unit because of $$z^{-1}$$ term, it is the orange portion)

$$3*Z^{-1}(z^{-1}*F(z)) = 3* u[n-1]*f[n-1]$$

so it looks like answer should be

$$3* u[n-1]*f[n-1] = 3*u[n-1]*(-2)^{n-1}$$

my teacher hasnt provided the solutions guide yet, despite that he promised to do so, and I reminded him about it, so I was just checking if this is the right solution and thinking...

I did the same problem, but I used polynomial long-division style to get the following series representation

$$3*z^{-1}-6*z^{-2}+12*z^{-3}-24*z^{-4}...$$

so based on that I guess that the $$x[n]=3*(-1)^n*2^n$$

where $$n>=0$$

so, I was just a little bit confused does there actually exist the delay of one unit in the original input sequence (result of inverse Z), sometimes I get confused about the causal sequence when you multiply with the unit step sequence, but then, it seems that sometimes also it can just be stated that the original sequence must have been causal sequence, so you just define n>=0 or something like that???

Your solution is correct. This is the way I see it (assuming that the sequence is causal/right-sided):

$$\mathcal{Z}^{-1}\left\{\frac{3}{z+2}\right\}[n]=3\mathcal{Z}^{-1}\left\{z^{-1}\frac{z}{z+2}\right\}[n]=\\=3\mathcal{Z}^{-1}\left\{\frac{z}{z+2}\right\}[n-1]=3(-2)^{n-1}u[n-1]$$

• thank you for checking my answers, please take a look at my other homework posts if you are inclined to doublecheck them and if you aren't too busy! I'm trying to prepare for my dsp exam next week, and we don't have any rehersal exercises, so I'm just reviewing our old exercises, and redoing them... – Late347 Nov 18 '18 at 6:07
• I don't really have a study group at my school, because my friends are a little bit behind in their homework exercises, so they won't be so useful in helping me, so I typically tend to post some of my homework to be checked at dsp stack exchange or math stack exchange... – Late347 Nov 18 '18 at 6:09