4.5 HW4, March. 10,2010

  4.5.1 Problem 1 (chapter 2, 2, page 78)
  4.5.2 Problem 2 (problem 7 in text book, chapter 2, page 79)
  4.5.3 Problem (3) (problem 10, chapter 2, page 80)
  4.5.4 Problem (4) (Problem 11, chapter 2, page 80)
  4.5.5 Problem (5) problem 15, chapter 2, page 81
  4.5.6 key solution
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4.5.1 Problem 1 (chapter 2, 2, page 78)

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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 \(z=0\)

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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 \(z=\infty \)

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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.

4.5.2 Problem 2 (problem 7 in text book, chapter 2, page 79)

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.

4.5.3 Problem (3)  (problem 10, chapter 2, page 80)

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

4.5.4 Problem (4) (Problem 11, chapter 2, page 80)

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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\[ \Im \left ( \operatorname{Im}\left ( x(n)\right ) \right ) =\frac{1}{2j}\left [ X\left ( z\right ) -X^{\ast }\left ( z^{\ast }\right ) \right ] \]

4.5.5 Problem (5) problem 15, chapter 2, page 81

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.

4.5.6 key solution

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