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${\lambda}_{\mathrm{1,2}}=(1+\frac{\text{rg}}{2})\pm \sqrt{(1+\frac{\text{rg}}{2}{)}^{2}-1}$
The following plot shows the locus of natural frequencies as rg increases.
Both natural frequencies lie along the positive, real λ-axis. When rg = 0, ${\lambda}_{\mathrm{1,2}}$ = 1. As rg increases, one natural frequency decreases toward λ = 0 and the other natural frequency increases. What is the physical significance of this pattern?
Note that the product of the two natural frequencies is 1,
${\lambda}_{1}{\lambda}_{2}=(1+\frac{\text{rg}}{2}+\sqrt{(1+\frac{\text{rg}}{2}{)}^{2}-1})\times (1+\frac{\text{rg}}{2}-\sqrt{(1+\frac{\text{rg}}{2}{)}^{2}-1})=1$
Therefore, ${\lambda}_{1}=\frac{1}{{\lambda}_{2}}$ and the form of the homogeneous solution is
${v}_{o}[n]={A}_{1}{\lambda}_{{1}^{n}}+{A}_{2}{\lambda}_{{1}^{-n}}$
The two terms of the homogeneous solution are shown below.
Thus, the homogeneous solution consists of a linear combination of decaying and growing geometric functions whose rates of decrease and increase are the same. Increasing the quantity rg increases the magnitude of the rate of change of voltage.
2/ Particular solution
Next we find a particular solution to the difference equation
$\sum _{k=0}^{K}{a}_{k}{y}_{p}[n+k]=\sum _{l=0}^{L}{b}_{l}x[n+1]$
for n>0 and
$x[n]={\text{Xz}}^{n}$
where z does not equal one of the natural frequencies. We assume that
${y}_{p}[n]={\text{Yz}}^{n}$
and we solve for Y . Substitution for both x[n] and y[n]in the difference equation yields, after factoring,
$(\sum _{k=0}^{K}{a}_{k}{z}^{k}){\text{Yz}}^{n}=(\sum _{l=0}^{L}{b}_{l}{x}^{l}){\text{Xz}}^{n}$
After dividing both sides of the equation by ${z}^{n}$ we can solve for Y which has the form
$Y=\tilde{H}(z)X$
where
$\tilde{H}(z)=\frac{\sum _{l=0}^{L}{b}_{l}{z}^{l}}{\sum _{k=0}^{K}{a}_{k}{z}^{k}}$
3/ System function
$\tilde{H}(z)=\frac{\sum _{l=0}^{L}{b}_{l}{z}^{l}}{\sum _{k=0}^{K}{a}_{k}{z}^{k}}\frac{\Leftarrow \text{zeros}}{\Leftarrow \text{poles}}$
a/ Example—reconstruction of difference equation from $\tilde{H}(z)$
Suppose
$\tilde{H}(z)=\frac{Y}{X}=\frac{z}{z+1}$
what is the difference equation that relates y[n] to x[n]? Crossmultiply the equation and multiply both sides by ${z}^{n}$ to obtain
$(z+1)Y{z}^{n}=zX{z}^{n}$
which yields
$({z}^{n+1}Y+{z}^{n}Y={z}^{n+1}X$
from which we can obtain the difference equation
y[n+1] + y[n]= x[n+1]
b/ Pole-zero diagram
$\stackrel{\text{~}}{H}$ characterizes the difference equation and $\stackrel{\text{~}}{H}$ is characterized by K + L + 1 numbers: K poles, L zeros, and one gain constant. Except for the gain constant, $\stackrel{\text{~}}{H}$ is characterized by a pole-zero diagram which is a plot of the locations of poles and zeros in the complex-z plane.
4/ Total solution
The general solution is
$y[n]=\sum _{k=1}^{K}{A}_{k}{\lambda}_{{k}^{n}}+X\tilde{H}(z){z}^{n}\text{for}n>0$
and
y[n] = 0 for n<0.
The general solution can be written compactly as follows
$y[n]=(\sum _{k=1}^{K}{A}_{k}{\lambda}_{{k}^{n}}+X\tilde{H}(z){z}^{n})u[n]$
5/ Initial conditions
To completely determine the total solution we need to determine the K coefficients { ${A}_{1},{A}_{2},\text{.}\text{.}\text{.}{A}_{K}$ }. These are determined from K initial conditions which must be specified. These conditions result in a set of K algebraic equations that need to be solved to obtain the initial conditions so that the total solution can be specified. We shall find another, and simpler, method to determine the total solution later.
Example — discretized CT system
We have previously considered the discretized approximation to a lowpass filter.
The equilibrium equation is
$\frac{{\text{dv}}_{o}(t)}{\text{dt}}=-\frac{1}{\text{RC}}{v}_{o}(t)+\frac{1}{\text{RC}}{v}_{i}(t)$
We know that the unit step response of this network starting from initial rest is
${v}_{o}(t)=(1-{e}^{-t/\text{RC}})u(t)$
We showed that a discretized approximation to this system yields the difference equation
${v}_{o}[n+1]=(1-\beta ){v}_{o}+{\mathrm{\alpha v}}_{i}[n]$
where α = T/(RC). We will determine the solution by finding the homogeneous and particular solution. But all the information we need is contained in the system function which we can obtain by substituting ${v}_{i}[n]={V}_{i}{z}^{n}$ and ${v}_{o}[n]={V}_{o}{z}^{n}$ into the difference equation to obtain
so that
$\tilde{H}(z)=\frac{{\tilde{V}}_{o}(z)}{{\tilde{V}}_{i}(z)}=\frac{\alpha}{z-(1-\alpha )}$
The natural frequency is $\lambda =1-\alpha $ so that the solution has the form
${v}_{o}[n]=(A(1-\alpha {)}^{n}+\tilde{H}(1)(1{)}^{n})u[n]$
where we have made use of the fact that $u[n]={\text{1}}^{n}u[n]$ . Since $\stackrel{\text{~}}{H}$ (1) = 1,
${v}_{o}[n]=(A(1-\alpha {)}^{n}+1)u[n]$
Finally, the initial condition, ${v}_{o}[0]=\text{0}$ , implies that A = −1 so that the solution is
${v}_{o}[n]=(1-1-\alpha {)}^{n})u[n]$
which is the same result we obtained earlier by solving the difference equation iteratively.
We compare the step response of the CT system with the DT approximation
${v}_{o}(t)=(1-{e}^{-t/\text{RC}})u(t)$ and ${v}_{o}[n]=(1-(1-\frac{T}{\text{RC}}{)}^{n})u(t)$
The solutions are shown below for RC = 1 and T/(RC) = 0.1.
XI. CONCLUSIONS
Logic for an analysis method for LTI systems
1/ Structural versus functional descriptions
Just as with CT systems, DT systems can be described either structurally, with a block diagram or a network diagram, or functionally by a system function.
$\stackrel{\text{~}}{H}(z)$ characterizes system
2/ Steady-state response to zn is particularly simple
Since the steady-state response to a complex geometric (exponential) is so simple, it is desirable to represent arbitrary DT functions as sums (integrals) of building-block complex geometric (exponential) functions chosen so that steady-state dominates. The steady-state response to each complex geometric (exponential) is readily computed. For a DT LTI system, the response to an arbitrary input can be computed by superposition.
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