# solving recurrence relation with Z-transform (LTI and causal)

I just wanted to doublecheck answers for my sanity's sake (exam next week)

problem statement

recurrence relation, solve it

$$y[n+1]= 35 + y[n]*0.5$$

according to my teacher it will be such that the input sequence must be causal, so that we have re-arranged form such as

$$y[n+1]- y[n]*0.5= 35u[n]$$ ,where $$y[0]=20$$, and causal input

My understanding is that we should go as follows

1. get $$X(z)$$
2. find what is $$Y(z)$$
3. then we can maybe do inverse Z-transform to get the y[k] output sequence in explicit form sequence
4. typically the most difficult phase seems to be the inverse Z transform so it can be tricky, but we have a table, and our teacher taught us to either use
1. z-table
2. long division
3. partial fraction expansion and z-table

Firstly, Z-transform both sides

$$Z(y[n+1])-0.5Z(y[n])=35*Z(u[n])$$

according to the z-table we receive

$$z*Y(z)-z*y[0]-0.5*Y(z)=35*\frac{z}{z-1}$$

combine terms and re-factor and emplace y[0]=20

$$Y(z)(z-0.5)=35*\frac{z}{z-1}+z*20$$

$$Y(z)=35*\frac{z}{(z-1)(z-0.5)}+20*\frac{z}{z-0.5}$$

We designate R(z) for the purposes of partial fraction expansion, because it is not found in the z-table

$$R(z)=\frac{z}{(z-1)(z-0.5)}$$

however the rightsided term is found directly in the z-table $$Z^{-1}(20*\frac{z}{z-0.5}) = 20* (0.5^k)$$

start partial fraction expansions $$R(z)/z = \frac{1}{(z-1)(z-0.5)}=\frac{a}{(z-1)}+\frac{b}{(z-0.5)}$$, then we find that, $$a=2,b=-2$$

$$R(z)=2*\frac{z}{(z-1)}-2*\frac{z}{(z-0.5)}$$

$$Y(z)=35*(2*\frac{z}{z-1}-2*\frac{z}{z-0.5})+20*(\frac{z}{z-0.5})$$

apply inverse Z transform

$$y[k]=20*(0.5)^k+70*u[k]-70*0.5^k$$

$$y[k]=u[k]*(70-50*0.5^k)$$

tabulating some results

• k=-1; y[-1]=0
• k=0; y[0]= 20
• k=1; y1= 45
• k=2; y[2]= 115/2

With such simple recursions it's often easiest (and instructive) to actually go through the recursion and find a pattern. With $$y[0]=20$$ and using $$c=35$$ we get

\begin{align}y[1]&=c+\frac{y[0]}{2}\\y[2]&=c+\frac{c}{2}+\frac{y[0]}{4}\\y[3]&=c+\frac{c}{2}+\frac{c}{4}+\frac{y[0]}{8}\\\vdots\end{align}

So we can conclude

\begin{align}y[n]&=c\left[1+\frac12+\frac14+\ldots+\left(\frac12\right)^{n-1}\right]+y[0]\left(\frac12\right)^n\\&=c\sum_{k=0}^{n-1}\left(\frac12\right)^k+y[0]\left(\frac12\right)^n\\&=c\cdot\frac{1-\left(\frac12\right)^{n}}{1-\frac12}+y[0]\left(\frac12\right)^n\\&=2c+\big(y[0]-2c\big)\left(\frac12\right)^n,\qquad n>0\end{align}

which is the same expression you obtained.

• I dont think that is the right approach. I think you made mistake where you assumed y[0]=35. I think I said in the beginning that y[0]=20. Anyway, I inputted the recurrence relation into my casio calculator recursive mode (that mode can also calculate newton-raphson and other recursive relations) It seems that you can easily compute the values recursively with computer. I got: k=0, y[0]=20; k=1, y[1]=45; k=2, y[2]=57,5 etc. etc. The problem apparently wants to have explicit defined sequence – Late347 Nov 18 '18 at 16:21
• @Late347: yes, I missed the line in your equation where $y[0]=20$ was specified. I'll update my answer. – Matt L. Nov 18 '18 at 16:28
• I used excel to compute recursive relation / recurrence relation, so this problem was funny because the answer is "known" so you can recursively compute always results, but explicit formula was apparently desired... – Late347 Nov 18 '18 at 16:31