Part (a)\[ x\left ( n\right ) =\alpha ^{\left \vert n\right \vert }\,,\ \ \ \ \ 0<\left \vert \alpha \right \vert <1 \] When \(n<0\), then \(x\left ( n\right ) =\alpha ^{-n}\) and when \(n\geq 0\,,\) then \(x\left ( n\right ) =\alpha ^{n}\), hence we split the sum \begin{align} X\left ( z\right ) & ={\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n\right ) z^{-n}\nonumber \\ & ={\displaystyle \sum \limits _{n=-\infty }^{-1}} \alpha ^{-n}z^{-n}+{\displaystyle \sum \limits _{n=0}^{\infty }} \alpha ^{n}z^{-n}\nonumber \\ & ={\displaystyle \sum \limits _{1}^{\infty }} \alpha ^{n}z^{n}+{\displaystyle \sum \limits _{n=0}^{\infty }} \alpha ^{n}z^{-n}\nonumber \\ & =-1+{\displaystyle \sum \limits _{0}^{\infty }} \alpha ^{n}z^{n}+{\displaystyle \sum \limits _{n=0}^{\infty }} \alpha ^{n}z^{-n}\nonumber \\ & =-1+{\displaystyle \sum \limits _{0}^{\infty }} \left ( \alpha z\right ) ^{n}+{\displaystyle \sum \limits _{n=0}^{\infty }} \left ( \frac{\alpha }{z}\right ) ^{n}\nonumber \\ & =-1+\frac{1}{1-\alpha z}+\frac{1}{1-\alpha z^{-1}}\tag{1} \end{align}
Where for the first sum \({\displaystyle \sum \limits _{0}^{\infty }} \left ( \alpha z\right ) ^{n}\,\), we need \(\left \vert \alpha z\right \vert <1\) or \(\left \vert z\right \vert <\left \vert \frac{1}{\alpha }\right \vert \) for convergence, and for the second sum \({\displaystyle \sum \limits _{n=0}^{\infty }} \left ( \frac{\alpha }{z}\right ) ^{n}\), we need \(\left \vert \frac{\alpha }{z}\right \vert <1\) for convergence, or \(\left \vert z\right \vert >\left \vert \alpha \right \vert \) Hence, since \(\ 0<\left \vert \alpha \right \vert <1\), we have the ROC as \[ \left \vert \alpha \right \vert <\left \vert z\right \vert <\left \vert \frac{1}{\alpha }\right \vert \] To help see where the poles and zeros are, expression (1) is simplfied to \[ X\left ( z\right ) =\frac{z\left ( 1-\alpha ^{2}\right ) }{\left ( 1-\alpha z\right ) \left ( z-\alpha \right ) }\] We now see that a pole exist at \(z=\frac{1}{\alpha }\) and at \(z=\alpha \) and a zero at \(z=0\)
Part (2)\[ x\left ( n\right ) =Ar^{n}\cos \left ( \omega _{0}n+\phi \right ) u\left ( n\right ) ,\ \ \ \ \ 0<r<1 \]\begin{align*} \Im \left ( x\left ( n\right ) \right ) & =X\left ( z\right ) ={\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n\right ) z^{-n}\\ & ={\displaystyle \sum \limits _{n=0}^{\infty }} Ar^{n}\cos \left ( \omega _{0}n+\phi \right ) z^{-n}\\ & =\frac{A}{2}{\displaystyle \sum \limits _{n=0}^{\infty }} r^{n}\left ( e^{i\left ( \omega _{0}n+\phi \right ) }+e^{-i\left ( \omega _{0}n+\phi \right ) }\right ) z^{-n}\\ & =\frac{A}{2}{\displaystyle \sum \limits _{n=0}^{\infty }} r^{n}\left ( e^{i\omega _{0}n}e^{i\phi }+e^{-i\omega _{0}n}e^{-i\phi }\right ) z^{-n}\\ & =\frac{A}{2}{\displaystyle \sum \limits _{n=0}^{\infty }} r^{n}e^{i\omega _{0}n}e^{i\phi }z^{-n}+\frac{A}{2}{\displaystyle \sum \limits _{n=0}^{\infty }} r^{n}e^{-i\omega _{0}n}e^{-i\phi }z^{-n}\\ & =\frac{A}{2}e^{i\phi }{\displaystyle \sum \limits _{n=0}^{\infty }} \left ( \frac{re^{i\omega _{0}}}{z}\right ) ^{n}+\frac{A}{2}e^{-i\phi }{\displaystyle \sum \limits _{n=0}^{\infty }} \left ( \frac{r}{e^{i\omega _{0}}z}\right ) ^{n}\\ & =\frac{A}{2}e^{i\phi }\frac{1}{1-re^{i\omega _{0}}z^{-1}}+\frac{A}{2}e^{-i\phi }\frac{1}{1-r\left ( e^{i\omega _{0}}z\right ) ^{-1}} \end{align*}
Where the first sum \({\displaystyle \sum \limits _{n=0}^{\infty }} \left ( \frac{re^{i\omega _{0}}}{z}\right ) ^{n}\) requires that \(\left \vert \frac{re^{^{i\omega _{0}}}}{z}\right \vert <1\) or \(\left \vert z\right \vert >r\) (since \(\left \vert e^{^{i\omega _{0}}}\right \vert =1)\), and the second sum \({\displaystyle \sum \limits _{n=0}^{\infty }} \left ( \frac{r}{e^{i\omega _{0}}z}\right ) ^{n}\) requires that \(\left \vert \frac{r}{e^{i\omega _{0}}z}\right \vert <1\) or \(\left \vert z\right \vert >\left \vert r\right \vert \), hence the ROC is \(\left \vert z\right \vert >\left \vert r\right \vert \), and since \(\left \vert r\right \vert <1\), then the ROC contains the unit circle, i.e. the sequence \(x\left ( n\right ) \) has a DTFT transform as well (it is BIBO stable).
To find poles and zeros: Since the ROC is surrounded by poles, we conclude that \(z=r\) is a pole, and since this is a causal signal, the zero is at \(z=\infty \)
Part(3)\[ x\left ( n\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0\leq n\leq N-1\\ 0 & & N\leq n\\ 0 & & n<0 \end{array} \right . \]\begin{align*} X\left ( z\right ) & ={\displaystyle \sum \limits _{n=0}^{N-1}} z^{-n}\\ & =\frac{1-z^{-N}}{1-z^{-1}} \end{align*}
Where the only condition is that \(\left \vert \frac{1}{z^{n}}\right \vert <\infty \) for any \(n\) in the above range. Since \(n\geq 0\), then this implies that \(z\neq 0\)
Hence the ROC is the complete z plane, except for \(z=0\)
To find poles and zero, easier to rewrite \(X\left ( z\right ) \) in powers of \(z\)
\[ X\left ( z\right ) =\frac{z^{N}-1}{z^{N}-z^{N-1}}=\frac{z^{N}-1}{z^{N-1}\left ( z-1\right ) }\] So we have \(N\) zeros around the unit circle (roots of unity) and one pole at \(z=1\) and \(N-1\) poles at \(z=0.\) notice pole at \(z=1\) would cancel the zero at \(z=0\)
part (4)\[ x\left ( n\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0\leq n\leq N\\ 2N-n & & N+1\leq n\leq 2N\\ 0 & & 2N\leq 0\\ 0 & & 0>n \end{array} \right . \]\begin{align} X\left ( z\right ) & ={\displaystyle \sum \limits _{n=0}^{N}} z^{-n}+{\displaystyle \sum \limits _{n=N+1}^{2N}} \left ( 2N-n\right ) z^{-n}\nonumber \\ & ={\displaystyle \sum \limits _{n=0}^{N}} z^{-n}+2N{\displaystyle \sum \limits _{n=N+1}^{2N}} z^{-n}-{\displaystyle \sum \limits _{n=N+1}^{2N}} nz^{-n}\nonumber \\ & ={\displaystyle \sum \limits _{n=0}^{N}} z^{-n}+2N\left ({\displaystyle \sum \limits _{n=0}^{2N}} z^{-n}-{\displaystyle \sum \limits _{n=0}^{N}} z^{-n}\right ) -\left ({\displaystyle \sum \limits _{n=0}^{2N}} nz^{-n}-{\displaystyle \sum \limits _{n=0}^{N}} nz^{-n}\right ) \nonumber \\ & =\frac{1-z^{-\left ( N+1\right ) }}{1-z}+2N\left ( \frac{1-z^{-\left ( 2N+1\right ) }}{1-z^{-1}}-\frac{1-z^{-\left ( N+1\right ) }}{1-z^{-1}}\right ) -{\displaystyle \sum \limits _{n=0}^{2N}} nz^{-n}+{\displaystyle \sum \limits _{n=0}^{N}} nz^{-n} \tag{1} \end{align}
To find \({\displaystyle \sum \limits _{n=0}^{2N}} nz^{-n}\,,\) I will use the relation that if the z-transform of \(x\left ( n\right ) \) is \(X\left ( z\right ) \), then the z-transform of \(nx\left ( n\right ) \) is \(-zX^{\prime }\left ( z\right ) \). Hence, let \[ x\left ( n\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0\leq n\leq 2N\\ 0 & & N\leq n\\ 0 & & n<0 \end{array} \right . \] Then \begin{align*} X\left ( z\right ) & ={\displaystyle \sum \limits _{n=0}^{2N}} z^{-n}\\ & =\frac{1-\left ( z^{-1}\right ) ^{2N+1}}{1-z^{-1}}=\frac{1-z^{-2N-1}}{1-z^{-1}} \end{align*}
Hence \begin{align*} -zX^{\prime }\left ( z\right ) & =-z\frac{d}{dz}\left ( \frac{1-z^{-2N-1}}{1-z^{-1}}\right ) \\ & =-z\left [ \frac{-\left ( -2N-1\right ) z^{-2N-2}}{\left ( 1-z^{-1}\right ) }-\frac{\left ( 1-z^{-2N-1}\right ) }{\left ( 1-z^{-1}\right ) ^{2}z^{2}}\right ] \\ & =\left [ \frac{z^{-2N}\left ( -2N\left ( z-1\right ) +z\left ( z^{2N}-1\right ) \right ) }{\left ( z-1\right ) ^{2}}\right ] \end{align*}
Hence\[{\displaystyle \sum \limits _{n=0}^{2N}} nz^{-n}=\left [ \frac{z^{-2N}\left ( -2N\left ( z-1\right ) +z\left ( z^{2N}-1\right ) \right ) }{\left ( z-1\right ) ^{2}}\right ] \] Similarly, for \({\displaystyle \sum \limits _{n=0}^{N}} nz^{-n},\) Let \[ x\left ( n\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0\leq n\leq N\\ 0 & & N\leq n\\ 0 & & n<0 \end{array} \right . \] then \begin{align*} X\left ( z\right ) & ={\displaystyle \sum \limits _{n=0}^{N}} z^{-n}\\ & =\frac{1-\left ( z^{-1}\right ) ^{N+1}}{1-z^{-1}}=\frac{1-z^{-N-1}}{1-z^{-1}} \end{align*}
Hence \begin{align*} -zX^{\prime }\left ( z\right ) & =-z\frac{d}{dz}\left ( \frac{1-z^{-N-1}}{1-z^{-1}}\right ) \\ & =\frac{z^{-N}\left ( N-Nz+z\left ( z^{N}-1\right ) \right ) }{\left ( z-1\right ) ^{2}} \end{align*}
Therefore \[{\displaystyle \sum \limits _{n=0}^{N}} nz^{-n}=\frac{z^{-N}\left ( N-Nz+z\left ( z^{N}-1\right ) \right ) }{\left ( z-1\right ) ^{2}}\] Going back to (1) and substitute the above results, we obtain\begin{align*} X\left ( z\right ) & =\frac{1-z^{-\left ( N+1\right ) }}{1-z}+2N\left ( \frac{1-z^{-\left ( 2N+1\right ) }}{1-z^{-1}}-\frac{1-z^{-\left ( N+1\right ) }}{1-z^{-1}}\right ) -{\displaystyle \sum \limits _{n=0}^{2N}} nz^{-n}+{\displaystyle \sum \limits _{n=0}^{N}} nz^{-n}\\ & =\frac{1-z^{-\left ( N+1\right ) }}{1-z}+2N\left ( \frac{1-z^{-\left ( 2N+1\right ) }}{1-z^{-1}}-\frac{1-z^{-\left ( N+1\right ) }}{1-z^{-1}}\right ) \\ & -\left [ \frac{z^{-2N}\left ( -2N\left ( z-1\right ) +z\left ( z^{2N}-1\right ) \right ) }{\left ( z-1\right ) ^{2}}\right ] \\ & +\frac{z^{-N}\left ( N-Nz+z\left ( z^{N}-1\right ) \right ) }{\left ( z-1\right ) ^{2}} \end{align*}
This can be simplified little more to be\begin{align*} X\left ( z\right ) & =\frac{1-z^{-\left ( N+1\right ) }}{1-z}+2N\left ( \frac{1-z^{-\left ( 2N+1\right ) }}{1-z^{-1}}-\frac{1-z^{-\left ( N+1\right ) }}{1-z^{-1}}\right ) \\ & +\frac{-z^{-2N}\left ( -2N\left ( z-1\right ) +z\left ( z^{2N}-1\right ) \right ) +z^{-N}\left ( N-Nz+z\left ( z^{N}-1\right ) \right ) }{\left ( z-1\right ) ^{2}} \end{align*}
That is as closed form as I can get it to be.
Since this is finite sequence, and \(n>0\), then the sum converges as long as \(z\neq 0.\) Hence the ROC is all the z-plane except for \(z=0.\)
To find poles and zero, easier to rewrite \(X\left ( z\right ) \) in powers of \(z\), but too complicated to do. I think my solution here is not what I should have done, but I am not sure now what else to do.
Determine whether or not the function \(F\left ( z\right ) =z^{\ast }\) can correspond to the z transform of a sequence.
Answer:
I assume we are only to consider a real sequence \(x\left ( n\right ) \) and not complex sequence.
Consider \(z\) at the unit circle, hence \(z=e^{j\omega }\), therefore \(z^{\ast }=e^{-j\omega }=z^{-1}\), So, we want to find a sequence \(x\left ( n\right ) \,\), such that \[{\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}=z^{-1}\] But if \(x\left ( n\right ) =\delta \left ( n-1\right ) \), then\[{\displaystyle \sum \limits _{n}} \delta \left ( n-1\right ) z^{-n}={\displaystyle \sum \limits _{n}} \delta \left ( n-1\right ) z^{-1}=z^{-1}{\displaystyle \sum \limits _{n}} \delta \left ( n-1\right ) =z^{-1}\] Hence, we found such a sequence.
Show that if \(X\left ( z\right ) \) is the z transform of a sequence \(x\left ( n\right ) \), then
(a) \(z^{n_{0}}X\left ( z\right ) \) is the z-transform of \(x\left ( n+n_{0}\right ) \)
(b) \(X\left ( a^{-1}z\right ) \) is the z-transform of \(a^{n}x\left ( n\right ) \)
(c)\(-zX^{\prime }\left ( z\right ) \) is the z-transform of \(nx\left ( n\right ) \)
Part(a)\begin{align*} \Im \left ( x\left ( n\right ) \right ) & =X\left ( z\right ) ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}\\ z^{n_{0}}X\left ( z\right ) & =z^{n_{0}}{\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n\right ) z^{-n}={\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n\right ) z^{-n+n_{0}} \end{align*}
Let \(m=n-n_{0}\), hence \(n=m+n_{0}\) and when \(n=-\infty \), \(=-\infty \), and then \(n=\infty \), \(m=\infty \), then the above becomes\[ z^{n_{0}}X\left ( z\right ) ={\displaystyle \sum \limits _{m=-\infty }^{\infty }} x\left ( m+n_{0}\right ) z^{-m}\] But \(m\) is a dummy variable, rename it back to \(n\), we have\[ z^{n_{0}}X\left ( z\right ) ={\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n+n_{0}\right ) z^{-n}\] But RHS above is just the z-transform of \(x\left ( n+n_{0}\right ) \). QED Part(b)\begin{align*} \Im \left ( x\left ( n\right ) \right ) & =X\left ( z\right ) ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}\\ X\left ( \frac{z}{a}\right ) & ={\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n\right ) \left ( \frac{z}{a}\right ) ^{-n}\\ & ={\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n\right ) a^{n}z^{-n}\\ & ={\displaystyle \sum \limits _{n=-\infty }^{\infty }} \left [ a^{n}x\left ( n\right ) \right ] z^{-n} \end{align*}
Hence, RHS is \(\Im \left ( a^{n}x\left ( n\right ) \right ) \), hence \[ \Im \left ( a^{n}x\left ( n\right ) \right ) =X\left ( \frac{z}{a}\right ) \] Part(c)\begin{align*} \Im \left ( x\left ( n\right ) \right ) & =X\left ( z\right ) ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}\\ X^{\prime }\left ( z\right ) & ={\displaystyle \sum \limits _{n}} x\left ( n\right ) \left ( -nz^{-n-1}\right ) \\ & =-{\displaystyle \sum \limits _{n}} x\left ( n\right ) n\frac{z^{-n}}{z}\\ & =-\frac{1}{z}{\displaystyle \sum \limits _{n}} \left [ nx\left ( n\right ) \right ] z^{-n} \end{align*}
Hence \[ -zX^{\prime }\left ( z\right ) ={\displaystyle \sum \limits _{n}} \left [ nx\left ( n\right ) \right ] z^{-n}\] Therefore\[ \Im \left ( nx\left ( n\right ) \right ) =-zX^{\prime }\left ( z\right ) \] QED
part (1) show that \(\Im \left ( x^{\ast }\left ( n\right ) \right ) =X^{\ast }\left ( z^{\ast }\right ) \). By definition, \[ \Im \left ( x\left ( n\right ) \right ) =X\left ( z\right ) ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}\] For the sequence \(x^{\ast }\left ( n\right ) \)\begin{equation} \Im \left ( x^{\ast }\left ( n\right ) \right ) ={\displaystyle \sum \limits _{n}} x^{\ast }\left ( n\right ) z^{-n}\tag{1} \end{equation} But \begin{align*} X^{\ast }\left ( z^{\ast }\right ) & =\left ({\displaystyle \sum \limits _{n}} x\left ( n\right ) \left ( z^{\ast }\right ) ^{-n}\right ) ^{\ast }\\ & =\left ({\displaystyle \sum \limits _{n}} x\left ( n\right ) \left ( z^{-n}\right ) ^{\ast }\right ) ^{\ast } \end{align*}
Move the outside conjugate operation to inside the sum results in\begin{align} X^{\ast }\left ( z^{\ast }\right ) & ={\displaystyle \sum \limits _{n}} x^{\ast }\left ( n\right ) \left ( \left ( z^{-n}\right ) ^{\ast }\right ) ^{\ast }\nonumber \\ & ={\displaystyle \sum \limits _{n}} x^{\ast }\left ( n\right ) z^{-n}\tag{2} \end{align}
Compare (1) and (2), they are the same. Hence \(\Im \left ( x^{\ast }\left ( n\right ) \right ) =X^{\ast }\left ( z^{\ast }\right ) \) Part (2)
Show that \(\Im \left ( x\left ( -n\right ) \right ) =X\left ( \frac{1}{z}\right ) \)
By definition, \begin{equation} \Im \left ( x\left ( n\right ) \right ) =X\left ( z\right ) ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n} \tag{1} \end{equation} For the sequence \(x\left ( -n\right ) \), we have\[ \Im \left ( x\left ( -n\right ) \right ) ={\displaystyle \sum \limits _{n}} x\left ( -n\right ) z^{-n}\] Let \(m=-n\), hence when \(n=-\infty ,m=\infty \), and when \(n=\infty ,m=\infty \), so the above becomes\[ \Im \left ( x\left ( -n\right ) \right ) ={\displaystyle \sum \limits _{m=\infty }^{-\infty }} x\left ( m\right ) z^{m}\] But \(m\) is a dummy variable, hence rename to \(n\), we have\begin{equation} \Im \left ( x\left ( -n\right ) \right ) ={\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n\right ) z^{n} \tag{2} \end{equation} Now, in (1), Let \(z=\frac{1}{z}\)on both sides, we obtain\begin{align} X\left ( \frac{1}{z}\right ) & ={\displaystyle \sum \limits _{n}} x\left ( n\right ) \left ( \frac{1}{z}\right ) ^{-n}\nonumber \\ & ={\displaystyle \sum \limits _{n=-\infty }^{\infty }} x\left ( n\right ) z^{n} \tag{3} \end{align}
Compare (2) and (3), we see they are the same, hence\[ \Im \left ( x\left ( -n\right ) \right ) =X\left ( \frac{1}{z}\right ) \] Part (3)
Show that \(\Im \left ( \operatorname{Re}\left [ x\left ( n\right ) \right ] \right ) =\frac{1}{2}\left [ X\left ( z\right ) +X^{\ast }\left ( z^{\ast }\right ) \right ] \)
By definition, \[ \Im \left ( x\left ( n\right ) \right ) =X\left ( z\right ) ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}\] Let \[ x\left ( n\right ) =\operatorname{Re}\left ( x\left ( n\right ) \right ) +j\operatorname{Im}\left ( x(n)\right ) \] Now \[ X^{\ast }\left ( z\right ) =\left ({\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}\right ) ^{\ast }={\displaystyle \sum \limits _{n}} x^{\ast }\left ( n\right ) \left ( z^{-n}\right ) ^{\ast }\] Then\[ X^{\ast }\left ( z^{\ast }\right ) ={\displaystyle \sum \limits _{n}} x^{\ast }\left ( n\right ) z^{-n}\] Hence\begin{align*} X\left ( z\right ) +X^{\ast }\left ( z^{\ast }\right ) & ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}+{\displaystyle \sum \limits _{n}} x^{\ast }\left ( n\right ) z^{-n}\\ & ={\displaystyle \sum \limits _{n}} \left [ x\left ( n\right ) +x^{\ast }\left ( n\right ) \right ] z^{-n}\\ & ={\displaystyle \sum \limits _{n}} \left [ \operatorname{Re}\left ( x\left ( n\right ) \right ) +j\operatorname{Im}\left ( x(n)\right ) +\operatorname{Re}\left ( x\left ( n\right ) \right ) -j\operatorname{Im}\left ( x(n)\right ) \right ] z^{-n}\\ & ={\displaystyle \sum \limits _{n}} 2\operatorname{Re}\left ( x\left ( n\right ) \right ) \ z^{-n} \end{align*}
Hence \[ \frac{1}{2}\left ( X\left ( z\right ) +X^{\ast }\left ( z^{\ast }\right ) \right ) ={\displaystyle \sum \limits _{n}} \operatorname{Re}\left ( x\left ( n\right ) \right ) \ z^{-n}\] But RHS\(\ \)above is \(\Im \left ( \operatorname{Re}\left ( x\left ( n\right ) \right ) \right ) \), hence\[ \Im \left ( \operatorname{Re}\left ( x\left ( n\right ) \right ) \right ) =\frac{1}{2}\left ( X\left ( z\right ) +X^{\ast }\left ( z^{\ast }\right ) \right ) \] Part (4)
Show that \(\Im \left ( \operatorname{Im}\left [ x\left ( n\right ) \right ] \right ) =\frac{1}{2j}\left [ X\left ( z\right ) -X^{\ast }\left ( z^{\ast }\right ) \right ] \)
By definition, \[ \Im \left ( x\left ( n\right ) \right ) =X\left ( z\right ) ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}\] Let \[ x\left ( n\right ) =\operatorname{Re}\left ( x\left ( n\right ) \right ) +j\operatorname{Im}\left ( x(n)\right ) \] Hence\begin{align*} X\left ( z\right ) -X^{\ast }\left ( z^{\ast }\right ) & ={\displaystyle \sum \limits _{n}} x\left ( n\right ) z^{-n}-{\displaystyle \sum \limits _{n}} x^{\ast }\left ( n\right ) z^{-n}\\ & ={\displaystyle \sum \limits _{n}} \left ( x\left ( n\right ) -x^{\ast }\left ( n\right ) \right ) z^{-n}\\ & ={\displaystyle \sum \limits _{n}} \left ( \operatorname{Re}\left ( x\left ( n\right ) \right ) +j\operatorname{Im}\left ( x(n)\right ) -\left [ \operatorname{Re}\left ( x\left ( n\right ) \right ) -j\operatorname{Im}\left ( x(n)\right ) \right ] \right ) z^{-n}\\ & ={\displaystyle \sum \limits _{n}} 2j\operatorname{Im}\left ( x(n)\right ) z^{-n} \end{align*}
Hence\[ \frac{1}{2j}\left [ X\left ( z\right ) -X^{\ast }\left ( z^{\ast }\right ) \right ] ={\displaystyle \sum \limits _{n}} \operatorname{Im}\left ( x(n)\right ) \ z^{-n}\] But RHS above is \(\Im \left ( \operatorname{Im}\left ( x(n)\right ) \right ) \), hence\[ \Im \left ( \operatorname{Im}\left ( x(n)\right ) \right ) =\frac{1}{2j}\left [ X\left ( z\right ) -X^{\ast }\left ( z^{\ast }\right ) \right ] \]
Consider a finite impulse response filter with unit-sample response \(h\left ( n\right ) \) of length (\(2N+1)\). If \(h\left ( n\right ) \) is real and even, show that the zeros of the system function occur in mirror image pairs about the unit circle. i.e. if \(H\left ( z\right ) =0\) for \(z=\rho e^{j\theta }\) then \(H\left ( z\right ) =0\) also for \(z=\left ( \frac{1}{\rho }\right ) e^{j\theta }\)
Answer:
Since \(h\left ( n\right ) \) is real and even, then \(\Im \left ( h\left ( n\right ) \right ) \) is real and even. This can be seen as follows\begin{align*} \Im \left ( h\left ( n\right ) \right ) & =H\left ( z\right ) ={\displaystyle \sum \limits _{n=-N}^{N}} h\left ( n\right ) z^{-n}\\ & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=-N}^{0}} h\left ( n\right ) z^{-n}+{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) z^{-n}\\ & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=0}^{N}} h\left ( -n\right ) z^{n}+{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) z^{-n} \end{align*}
Since \(h\left ( n\right ) \) is real and even, then \(h\left ( -n\right ) =h\left ( n\right ) \), so the above becomes\begin{align} H\left ( z\right ) & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) z^{n}+{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) z^{-n}\nonumber \\ & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) \left ( z^{n}+z^{-n}\right ) \tag{1} \end{align}
Now, let \(z=\rho e^{j\theta }\), so\begin{align} H\left ( z\right ) & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) \left ( \rho e^{j\theta }+\frac{1}{\rho }e^{-j\theta }\right ) ^{n}\nonumber \\ & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) \left ( \rho e^{j\theta }+\frac{1}{\rho }e^{-j\theta }\right ) ^{n} \tag{2} \end{align}
And let \(z=\frac{1}{\rho }e^{j\theta }\) again in (1), we obtain\begin{align} H\left ( z\right ) & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) \left ( \left ( \frac{1}{\rho }e^{j\theta }\right ) ^{n}+\left ( \frac{1}{\rho }e^{j\theta }\right ) ^{-n}\right ) \nonumber \\ & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) \left ( \left ( \frac{1}{\rho }e^{j\theta }\right ) ^{n}+\left ( \rho e^{-j\theta }\right ) ^{n}\right ) \nonumber \\ & =-h\left ( 0\right ) +{\displaystyle \sum \limits _{n=0}^{N}} h\left ( n\right ) \left ( \frac{1}{\rho }e^{j\theta }+\rho e^{-j\theta }\right ) ^{n} \tag{3} \end{align}
Compare (2) and (3), they are the same. Hence \(H\left ( z\right ) \) is the same at \(z=\frac{1}{\rho }e^{j\theta }\) and at \(z=\rho e^{j\theta }\), therefore \(H\left ( z\right ) \) is even function w.r.t. unit circle. Hence if a zero occurs outside a unit circle, there will be a mirror image of this zero inside the unit circle (since \(H\left ( z\right ) \) will have the same value, which is zero in this case) at both location.
This problem is similar to looking at the DTFT \(F\left ( e^{j\omega }\right ) \), which is real and even when \(x\left ( n\right ) \) is real and even. The difference is that \(F\left ( e^{j\omega }\right ) \) will be even about the \(y-axis\) while \(H\left ( z\right ) \) is even w.r.t. unit circle.