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
- get $X(z)$
- find what is $Y(z)$
- then we can maybe do inverse Z-transform to get the y[k] output sequence in explicit form sequence
- 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
- z-table
- long division
- 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