Does this equation have a closed form

Question

I am on chapter 6 of "Understanding digital signal processing", 3rd Edition by Richard G. Lyons. But I am stuck on question 6.15. Here is the network in question.

I wrote down the difference equation as $$y(n)=x(n)+y(n-1)-Qy(n-1),$$ where $x(n)=D$(i.e. constant for all $n$), $Q$ is positive and less than $1$ and network is at rest at $n=0$ (i.e $y(n-1)=0$ when $n=0.$) In order to answer the question, I need to develop the series equation for $y(n)$ and find its closed form.

I have developed the series expression for $$y(n)=D\sum^n_{k=0}(-Q)^k {{(n+1)}!\over{(k+1)!(n-k)!}}.$$

And the terms without simplifying the factorials are $$y(n)=D{(n+1)!\over n!} - DQ{(n+1)!\over 2(n-1)!} +DQ^2{(n+1)!\over 6(n-2)!} -DQ^3{(n+1)!\over 24(n-3)!} . .. +(-Q)^nD$$.

My issue is that I cannot find where I have gone wrong because this series for $y(n)$ does not seem to have a common ratio which makes it hard for me to write it in the closed form as in $$\sum^{N-1}_{n=p}r^n={r^p-r^N\over{1-r}}.$$

My question is, is my developed $y(n)$ suitable? If it is suitable, does it have a closed form?

Thanks

Answer 1

$$y[n] = x[n] + (1-Q) y[n-1]$$

where $y[-1] = 0$ and otherwise $n \ge 0$.

this is a difference equation with both a "natural response" (when the input is always zero, but you might have initial conditions) and a "forced response". a.k.a. the "transient response" and the "steady-state response" the math guys might call these the "homogeneous solution" and the "particular solution".

this is LTI so then the exponential function is an eigenfunction for the natural response.

for the natural solution, try $$ y_\text{n}[n] = y_\text{n}[0] a^n $$

then $$\begin{align} y[n] &= x[n] + (1-Q) y[n-1] \\ y_\text{n}[0] a^n &= 0 + (1-Q) y_\text{n}[0] a^{n-1} \\ a^n &= (1-Q) a^{n-1} \\ &= (1-Q) a^n a^{-1} \\ 1 &= (1-Q) a^{-1} \\ \end{align}$$

so i guess $a = 1-Q$.

for steady state, you posit a solution that is like the input plus all of its finite differences. since the input is constant and all finite differences of a discrete constant function are zero, then the steady state output is a constant. you sum the scaled steady state with the scaled transient response for the complete response.

$$\begin{align} y[n] &= x[n] + (1-Q) y[n-1] \\ C + y_\text{n}[0] a^n &= D + (1-Q)(C + y_\text{n}[0] a^{n-1}) \\ C + y_\text{n}[0] (1-Q)^n &= D + (1-Q)(C + y_\text{n}[0] (1-Q)^{n-1}) \\ &= D + (1-Q)C + y_\text{n}[0] (1-Q)^n) \\ C &= D + (1-Q)C \\ \end{align}$$

i guess that means that $QC = D$ and $C = \tfrac{D}{Q}$.

now you add your natural response to your forced response.

$$ y[n] = y_\text{n}[n] + C $$

lastly, for your initial condition

$$\begin{align} y[n] &= x[n] + (1-Q) y[n-1] \\ y[0] &= x[0] + (1-Q) y[-1] \\ C + y_\text{n}[0] (1-Q)^0 &= D + (1-Q)\cdot 0 \\ C + y_\text{n}[0] &= D \\ \end{align}$$

i guess that means that $ y_\text{n}[0] = D-C = D - \tfrac{D}{Q} $.

$$\begin{align} y[n] &= y_\text{n}[n] + C \\ &= y_\text{n}[0]a^n + C \\ &= (D - \tfrac{D}{Q})(1-Q)^n + \tfrac{D}{Q} \\ \end{align}$$

for $n \ge 0$.