Cooper Doyle

2022-07-15

Let $a\in (-1,1)$ and $n=0,1,2,\cdots $. By decomposing a rational function into simple fractions we have derived the following identity:

${\frac{1}{n!}\frac{{d}^{n}}{d{z}^{n}}\left(\frac{Az+B}{{z}^{2}-(1+a)z+1}\right)|}_{z=0}=\frac{A\mathrm{sin}(\varphi \cdot n)+B\mathrm{sin}(\varphi \cdot (n+1))}{\mathrm{sin}(\varphi )}$

where $\varphi :=\mathrm{arccos}((1+a)/2)$. In other words we have shown that the inverse Z-transform of the rational function in parentheses on the left hand side is a waveform with both time independent frequency and amplitude. Now the question would be what happens if the second order polynomial in the denominator is replaced by some integer power of it. What kind of waveform are we going to get in this case? In other words we ask for the following:

${\frac{1}{n!}\frac{{d}^{n}}{d{z}^{n}}\left(\frac{Az+B}{({z}^{2}-(1+a)z+1{)}^{p}}\right)|}_{z=0}=?$

where $p=1,2,3,\cdots $

${\frac{1}{n!}\frac{{d}^{n}}{d{z}^{n}}\left(\frac{Az+B}{{z}^{2}-(1+a)z+1}\right)|}_{z=0}=\frac{A\mathrm{sin}(\varphi \cdot n)+B\mathrm{sin}(\varphi \cdot (n+1))}{\mathrm{sin}(\varphi )}$

where $\varphi :=\mathrm{arccos}((1+a)/2)$. In other words we have shown that the inverse Z-transform of the rational function in parentheses on the left hand side is a waveform with both time independent frequency and amplitude. Now the question would be what happens if the second order polynomial in the denominator is replaced by some integer power of it. What kind of waveform are we going to get in this case? In other words we ask for the following:

${\frac{1}{n!}\frac{{d}^{n}}{d{z}^{n}}\left(\frac{Az+B}{({z}^{2}-(1+a)z+1{)}^{p}}\right)|}_{z=0}=?$

where $p=1,2,3,\cdots $

Sanaa Hinton

Beginner2022-07-16Added 15 answers

We start from the following identity:

$\frac{1}{{[{z}^{2}-(1+a)z+1]}^{p}}=2Re\left[\sum _{l=0}^{p-1}{\textstyle (}\genfrac{}{}{0ex}{}{2p-2-l}{p-1}{\textstyle )}\frac{(-1{)}^{p}}{(z-\mathrm{exp}(\u0131\varphi ){)}^{l+1}(-2\u0131\mathrm{sin}(\varphi ){)}^{2p-l-1}}\right]$

which is just a simple fraction decomposition of the left hand side. Now since

$Az+B=A\cdot (z-\mathrm{exp}(\u0131\varphi ))+A\mathrm{exp}(\u0131\varphi )+B$ we can express the desired rational function through inverse powers of monomials of z only, i.e. through linear combinations of terms of the kind $(z-\mathrm{exp}(\u0131\varphi ){)}^{-l}$ and their complex conjugates only. Those monomials are very simple to differentiate and therefore we arrive at the result without any substantial problems. Having said all this we just state the final result:

$\begin{array}{rcl}& & {\frac{1}{n!}\frac{{d}^{n}}{d{z}^{n}}\left(\frac{Az+B}{[{z}^{2}-(1+a)z+1{]}^{p}}\right)|}_{z=0}=\\ & & \sum _{l=0}^{p-1}\frac{2(-1{)}^{l+p+1}}{(2\mathrm{sin}(\varphi ){)}^{-l+2p-1}}{\textstyle (}\genfrac{}{}{0ex}{}{l+n}{n}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{2p-l-2}{p-1}{\textstyle )}\frac{1}{l+n}\\ & & (An\mathrm{cos}(\varphi (l+n)-\frac{1}{2}\pi (-l+2p-1))+B(l+n)\mathrm{cos}(\varphi (l+n+1)-\frac{1}{2}\pi (-l+2p-1)))\end{array}$

$\frac{1}{{[{z}^{2}-(1+a)z+1]}^{p}}=2Re\left[\sum _{l=0}^{p-1}{\textstyle (}\genfrac{}{}{0ex}{}{2p-2-l}{p-1}{\textstyle )}\frac{(-1{)}^{p}}{(z-\mathrm{exp}(\u0131\varphi ){)}^{l+1}(-2\u0131\mathrm{sin}(\varphi ){)}^{2p-l-1}}\right]$

which is just a simple fraction decomposition of the left hand side. Now since

$Az+B=A\cdot (z-\mathrm{exp}(\u0131\varphi ))+A\mathrm{exp}(\u0131\varphi )+B$ we can express the desired rational function through inverse powers of monomials of z only, i.e. through linear combinations of terms of the kind $(z-\mathrm{exp}(\u0131\varphi ){)}^{-l}$ and their complex conjugates only. Those monomials are very simple to differentiate and therefore we arrive at the result without any substantial problems. Having said all this we just state the final result:

$\begin{array}{rcl}& & {\frac{1}{n!}\frac{{d}^{n}}{d{z}^{n}}\left(\frac{Az+B}{[{z}^{2}-(1+a)z+1{]}^{p}}\right)|}_{z=0}=\\ & & \sum _{l=0}^{p-1}\frac{2(-1{)}^{l+p+1}}{(2\mathrm{sin}(\varphi ){)}^{-l+2p-1}}{\textstyle (}\genfrac{}{}{0ex}{}{l+n}{n}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{2p-l-2}{p-1}{\textstyle )}\frac{1}{l+n}\\ & & (An\mathrm{cos}(\varphi (l+n)-\frac{1}{2}\pi (-l+2p-1))+B(l+n)\mathrm{cos}(\varphi (l+n+1)-\frac{1}{2}\pi (-l+2p-1)))\end{array}$

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