Solution:
See Section 10-1.1. DTFT becomes DSFT and impulse response becomes point-spread function.

Solution:
See Section 10-1.1.

Solution:
By inspection, $h[m,n]=h[m]\;h[n]$ where $h[n]=\{1,\underline{2},1\}$. $h[n]$ has DTFT ${\bf H} (e^{j\Omega})=2+2\cos(\Omega)$.
$\displaystyle {\bf H} (e^{j\Omega_1},e^{j\Omega_2})=(2+2\cos(\Omega_1))(2+2\cos(\Omega_2)). $

Solution:
$$ {\bf H} (e^{j\Omega_1},e^{j\Omega_2}) =e^{j\Omega_1}+e^{-j\Omega_1}+e^{j\Omega_2}+e^{-j\Omega_2}-4 =2\cos(\Omega_1)+2\cos(\Omega_2)-4. $$

Solution:
\begin{eqnarray*} {\bf H} (e^{j\Omega_1},e^{j\Omega_2}) &=& 1+e^{-j\Omega_1}+e^{-j\Omega_2}+e^{-j\Omega_1}e^{-j\Omega_2} \\ &=& e^{-j\Omega_1/2}2\cos\left(\frac{\Omega_1}{2}\right) +e^{-j\Omega_2}e^{-j\Omega_1/2}2\cos\left(\frac{\Omega_1}{2}\right) \\ &=& 2\cos\left(\frac{\Omega_1}{2}\right)2 \cos\left(\frac{\Omega_2}{2}\right)e^{-j\Omega_1/2}e^{-j\Omega_2/2}. \\ |{\bf H} (e^{j\Omega_1},e^{j\Omega_2})| &=& 4\left|\cos\left(\frac{\Omega_1}{2}\right)\cos\left(\frac{\Omega_2}{2}\right)\right|. \end{eqnarray*} This passes low wavenumbers and rejects high ones.

Solution:
\begin{eqnarray*} {\bf H} (e^{j\Omega_1},e^{j\Omega_2}) &=& 1-e^{-j\Omega_1}-e^{-j\Omega_2}+e^{-j\Omega_1}e^{-j\Omega_2} \\ &=& e^{-j\Omega_1/2}2j\sin\left(\frac{\Omega_1}{2}\right) -e^{-j\Omega_2}e^{-j\Omega_1/2}2j\sin\left(\frac{\Omega_1}{2}\right) \\ &=& 2j\sin\left(\frac{\Omega_1}{2}\right)2j\sin\left(\frac{\Omega_2}{2}\right) e^{-j\Omega_1/2}e^{-j\Omega_2/2}. \\ |{\bf H}(e^{j\Omega_1},e^{j\Omega_2})| &=& \left|4\sin\left(\frac{\Omega_1}{2}\right)\sin\left(\frac{\Omega_2}{2}\right)\right|. \end{eqnarray*} This passes high wavenumbers and rejects low ones.

Solution:
\begin{eqnarray*} {\bf H} (e^{j\Omega_1},e^{j\Omega_2}) &=& e^{j\Omega_1}e^{j\Omega_2}+e^{j\Omega_1}e^{-j\Omega_2} +e^{-j\Omega_1}e^{j\Omega_2}+e^{-j\Omega_1}e^{-j\Omega_2} \\ &=&(e^{j\Omega_1}+e^{-j\Omega_1})(e^{j\Omega_2} +e^{-j\Omega_2}) \\ &=& 2\cos(\Omega_1) 2\cos(\Omega_2). \end{eqnarray*} This is a bandreject filter, since it rejects midrange frequencies ($\Omega_i\approx\frac{\pi}{2}$).

Solution:
(a) is Eq. (10.29) and (b) is Eq. (10.31).

Solution:
From Eq. (10.10), $$ h[m,n]**\;x[m,n] =\sum_{i=-\infty}^{\infty}\sum_{j=-\infty}^{\infty} x[i,j]\;h[m-i, n-j]. $$ $$ (1)\left[\matrix{1 & 2 & 0\cr 3 & 4 & 0\cr 0 & 0 & 0}\right]+(1)\left[\matrix{0 & 1 & 2\cr 0 & 3 & 4\cr 0 & 0 & 0}\right] +(1)\left[\matrix{0 & 0 & 0\cr 1 & 2 & 0\cr 3 & 4 & 0}\right]+(1)\left[\matrix{0 & 0 & 0\cr 0 & 1 & 2\cr 0 & 3 & 4}\right] =\left[\matrix{1 & 3 & 2\cr 4 & 10 & 6\cr 3 & 7 & 4}\right]. $$

Solution:
From Eq. (10.20a), $$ {\bf X} [k_1,k_2]=\sum_{m=0}^{M-1}\sum_{n=0}^{N-1} x[m,n]\;e^{-j2\pi(k_1m/M+k_2n/N)}. $$ The $(2\times 2)$ 2-D DFT of $\left[\matrix{a & b\cr c & d}\right]$ is \begin{eqnarray*} {\bf X} [k_1,k_2] &=& a+be^{-j\pi k_2}+ce^{-j\pi k_1}+de^{-j\pi(k_1+k_2)} \\ &=& \left[\matrix{a+b+c+d & a-b+c-d\cr a+b-c-d & a-b+c-d}\right]. \end{eqnarray*} The 2-D DFT of $\left[\matrix{3 & 1\cr 4 & 5}\right]$ is $\left[\matrix{13 & 1\cr -5 & 3}\right]$.

Solution:
Keep columns #1 and #3 and rows #1 and #3. See Section 10-4.

Solution:
See Section 10-7.5.

Solution:
The 2-D DFT is given by Eq. (10.20a), $$ {\bf X} [k_1,k_2]=\sum_{m=0}^{M-1}\sum_{n=0}^{N-1}x[m,n] \;e^{-j2\pi(k_1m/M+k_2n/N)}. $$ The $(2\times 2)$ 2-D DFT of $\left[\matrix{a & b\cr c & d}\right]$ is $$ {\bf X} [k_1,k_2]=a+be^{-j\pi k_2}+ce^{-j\pi k_1}+de^{-j\pi(k_1+k_2)} =\left[\matrix{a+b+c+d & a-b+c-d\cr a+b-c-d & a-b+c-d}\right]. $$ The 2-D DFT of $\left[\matrix{42 & 42\cr 28 & 31}\right]$ is $\left[\matrix{143 & -3\cr 25 & 3}\right]$, the 2-D DFT of $\left[\matrix{3 & 1\cr 4 & 5}\right]$ is $\left[\matrix{13 & 1\cr -5 & 3}\right]$. Then $$ {\bf X} [k_1,k_2]=\frac{{\bf Y} [k_1,k_2]}{{\bf H} [k_1,k_2]} =\left[\matrix{11 & -3\cr -5 & 1}\right] $$ and, the 2-D inverse DFT is given by Eq. (10.20b) as $x[m,n]=\left[\matrix{1 & 2\cr 3 & 5}\right]$.

Solution:
From Eq. (10.20a), $$ {\bf X} [k_1,k_2]=\sum_{m=0}^{M-1}\sum_{n=0}^{N-1} x[m,n]\;e^{-j2\pi(k_1m/M+k_2n/N)}. $$ The $(2\times 2)$ 2-D DFT of $\left[\matrix{a & b\cr c & d}\right]$ is $$ {\bf X} [k_1,k_2]=a+be^{-j\pi k_2}+ce^{-j\pi k_1}+de^{-j\pi(k_1+k_2)} =\left[\matrix{a+b+c+d & a-b+c-d\cr a+b-c-d & a-b+c-d}\right]. $$ The 2-D DFT of $\left[\matrix{3 & 1\cr 4 & 6}\right]$ is $\left[\matrix{14 & 0\cr -6 & 4}\right]$. ${\bf H} [1,2]=0$, so ${\bf X} [1,2]$ cannot be computed from $$ {\bf X} [k_1,k_2]=\frac{{\bf Y} [k_1,k_2]}{{\bf H}[k_1,k_2]}. $$

Solution:
See Section 10-7.

Solution:
$$ \{3,1,1,3\}\;©\;\frac1{\sqrt{2}}\{1,1,0,0\} =\frac1{\sqrt{2}}\{6,4,2,4\}. $$ Downsample to $\frac1{\sqrt{2}}\{6,2\}$.

Solution:
$$ \{3,1,1,3\}\;©\;\frac1{\sqrt{2}}\{1,-1,0,0\} =\frac1{\sqrt{2}}\{0,-2,0,2\}. $$ Downsample to $\{0,0\}$. CHECK: Upsample $\frac1{\sqrt{2}}\{6,2\}$ to $\frac1{\sqrt{2}}\{6,0,2,0\}$. $$ \frac1{\sqrt{2}}\{6,0,2,0\}\;©\;\frac1{\sqrt{2}}\{1,0,0,1\}=\{3,1,1,3\} $$ Upsampling $\{0,0\}$ to $\{0,0,0,0\}$ and $$ \{0,0,0,0\}\;©\;\frac1{\sqrt{2}}\{1,0,0,-1\}=\{0,0,0,0\}. $$

Solution:
$\{a,b,c,d\}$ is a QMF pair with $\pm\{d,-c,b,-a\}$.
 
See Eq. 10.109, which is $\displaystyle h[n] = (-1)^n\,g[L-n].$

Solution:
Smith-Barnwell condition: Autocorrelation of $g[n]$ is 0 for even $n\ne 0$. For $g[n]=\{a,b,c,d\}$: $ac+bd=0$ and $a^2+b^2+c^2+d^2=1$. Only ${\bf B}$ satisfes the former. See Section 10-11.4.

Solution:
(a) is Eq. (10.108) and (b) is Eq. (10.113). See Section 10-11.

Solution:
By construction, Daubechies wavelet functions sparsify polynomials.

Solution:
This is the $\ell_1$ norm of $\underline{x}$.

Solution:
Wavelets result in a sparse representation, by minimizing the $\ell_1$ norm. Landweber is just to solve systems of equations. Basis pursuit and IRLS can be used.

Solution:
See Section 10-17.3.

Solution:
See Section 10-14.

Solution:
See Section 10-14.

Solution:
See Sections 10-16 and 10-17.

Solution:
See Eq. (10.166), which is derived in Section 10-14 and is
 
$\displaystyle \hat x[n]= \begin{array}{ll} y[n]-\lambda &{\rm for~}y[n]>+\lambda,\\ y[n]+\lambda &{\rm for~}y[n]<-\lambda,\\ 0 &{\rm for~}|y[n]|<\lambda. \end{array} $

Solution:
Fig. 10-28 shows how to fit the wavelet transform components of various sizes together like a jigsaw puzzle.