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${v}_{o}[1]=\alpha $ ,
${v}_{o}[2]=(1-\alpha )\alpha +\alpha $ ,
${v}_{o}[3]=(1-\alpha {)}^{2}\alpha +(1-\alpha )\alpha +\alpha $ ,
${v}_{o}[k]=\alpha \sum _{k=0}^{n-1}(1-\alpha {)}^{k}=1-(1-\alpha {)}^{n}$
[We have made use of an important formula for the summation of a geometric series which we will prove shortly.] Hence, the solution is
${v}_{o}[n]=(1-(1-\alpha {)}^{n})u[n]$
The solution
${v}_{o}[n]=(1-(1-\alpha {)}^{n})u[n]$
is shown plotted below for $\alpha $ = 0.25.
4/ Important side issue — the summation of finite geometric series
Suppose
$S=\sum _{n=l}^{k}{a}^{n}={a}^{l}+{a}^{l+1}+\text{.}\text{.}\text{.}+{a}^{k}$
Then
$\mathrm{\alpha S}=\alpha \sum _{n=l}^{k}{a}^{n}={a}^{l+1}+{a}^{l+2}+\text{.}\text{.}\text{.}+{a}^{k+1}$
Hence,
$(1-\alpha )S=(1-\alpha )\sum _{n=l}^{k}{a}^{n}={a}^{l}-{a}^{k+1}$
and
$S=\sum _{n=l}^{k}{a}^{n}=\frac{{a}^{l}-{a}^{k+1}}{(1-\alpha )}$
Conclusion
X. GENERAL LINEAR DIFFERENCE EQUATION
For a DT system, such as the ones shown previously, we can write the relation between an input variable x[n] and an output variable y[n]by a general difference equation of the form
$\sum _{k=0}^{K}{a}_{k}y[n+k]=\sum _{l=0}^{L}{b}_{l}x[n+1]$
We seek a solution of this system for an input that is of the form
$x[n]={\text{Xz}}^{n}u[n]$
where X and z are in general complex quantities and u[n] is the unit step function. We will also assume that the system is at rest for n<0, so that y[n] = 0 for n<0.
We will seek the solution for n>0 by finding the homogeneous (unforced solution) and then a particular solution (forced solution).
1/ Homogeneous solution
a/ Geometric (exponential) solution
Let the homogeneous solution be ${y}_{h}[n]$ , then the homogeneous equation is
$\sum _{k=0}^{K}{a}_{k}y[n+k]=0$
To solve this equation we assume a solution of the form
${y}_{h}[n]={\mathrm{A\lambda}}^{n}$
Since
${y}_{h}[n+k]={\mathrm{A\lambda}}^{n+k}$
we have
$\sum _{k=0}^{K}{a}_{k}{\mathrm{A\lambda}}^{n+k}=0$
b/ Characteristic polynomial
The equation can be factored to yield
$(\sum _{k=0}^{K}{a}_{k}{\lambda}^{k}){\mathrm{A\lambda}}^{n}=0$
We are not interested in the trivial solution
${y}_{h}[n]={\mathrm{A\lambda}}^{n}=0$
Therefore, we can divide by ${\mathrm{A\lambda}}^{n}$ to obtain the characteristic polynomial
$\sum _{k=0}^{K}{a}_{k}{\lambda}^{k}=0$
This polynomial of order K has K roots which can be exposed by writing the polynomial in factored form
$\prod _{k=1}^{K}(\lambda -{\lambda}_{k})=0$
c/ Natural frequencies
The roots of the characteristic polynomial $\{{\lambda}_{1},{\lambda}_{2},\text{.}\text{.}\text{.}{\lambda}_{K}\}$ are called natural frequencies. These are frequencies for which there is an output in the absence of an input.
If the natural frequencies are distinct, i.e., if ${\lambda}_{i}\ne {\lambda}_{k}$ for $i\ne k$ the most general homogeneous solution has the form
${y}_{h}[n]=\sum _{k=1}^{K}{A}_{k}{\lambda}_{{k}^{n}}$
Two-minute miniquiz problem
Problem 10-1 — Natural frequencies of the ladder network
The electric ladder network shown below has the following difference equation
$-{v}_{o}[n+1]+(2+\text{rg}){v}_{o}[n]-{v}_{o}[n-1]={\text{rgv}}_{i}[n]$
Find the natural frequencies.
Solution
The natural frequencies are determined by the homogeneous solution
$-{v}_{o}[n+1]+(2+\text{rg}){v}_{o}[n]-{v}_{o}[n-1]=0$
Substituting a solution of the form ${v}_{0}[n]={\mathrm{A\lambda}}^{n}$ results in
$$ $-{\mathrm{A\lambda}}^{n+1}+(2+\text{rg}){\mathrm{A\lambda}}^{n}-{\mathrm{A\lambda}}^{n-1}-1=0$
which yields the characteristic polynomial
${\lambda}^{2}-(2+\text{rg})\lambda +1=0$
whose roots are the characteristic frequencies
${\lambda}_{\mathrm{1,2}}=(1+\frac{\text{rg}}{2})\pm \sqrt{(1+\frac{\text{rg}}{2}{)}^{2}-1}$
d/ Natural frequencies of the ladder network
The electric ladder network shown below has the following difference equation
$-{v}_{o}[n+1]+(2+\text{rg}){v}_{o}[n]-{v}_{o}[n-1]={\text{rgv}}_{i}[n]$
The natural frequencies are:
${\lambda}_{\mathrm{1,2}}=(1+\frac{\text{rg}}{2})\pm \sqrt{(1+\frac{\text{rg}}{2}{)}^{2}-1}$
The natural frequencies depend on rg,
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