Question

(M) Example 3 in Section 4.8 displayed a low-pass linear filter that changed a signal \(\left\{ {{{\bf{y}}_k}} \right\}\) into \(\left\{ {{{\bf{y}}_{k + 1}}} \right\}\) and changed a higher-frequency signal \(\left\{ {{{\bf{u}}_k}} \right\}\) into the zero signal, where \({{\bf{y}}_k} = \cos \left( {\frac{{\pi k}}{4}} \right)\) and \({w_k} = \cos \left( {\frac{{3\pi k}}{4}} \right)\). The following calculations will design a filter with approximately those properties. The filter equation is

\({a_0}{{\bf{y}}_{k + 2}} + {a_1}{{\bf{y}}_{k + 1}} + {a_2}{{\bf{y}}_k} = {z_k}\)for all \(k\) (8)

Because the signals are periodic, with period 8, it suffices to study equation (8) for \(k = 0, \ldots ,7\). The action on the two signals described above translates into two sets of eight equations, shown below:

\(\begin{aligned}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{y_{k + 2}}\,\,\,\,\,{y_{k + 1}}\,\,\,\,\,{y_k}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{y_{k + 1}}\\\begin{aligned}{{}}{k = 0}\\{k = 1}\\ \vdots \\{}\\{}\\{}\\{}\\{K = 7}\end{aligned}\left( {\begin{aligned}{{}}0&{.7}&1\\{ - .7}&0&{.7}\\{ - 1}&{ - .7}&0\\{ - .7}&{ - 1}&{ - .7}\\0&{ - .7}&{ - 1}\\{.7}&0&{ - .7}\\1&{.7}&0\\{.7}&1&{.7}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{a_0}}\\{{a_1}}\\{{a_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}}{.7}\\0\\{ - .7}\\{ - 1}\\{ - .7}\\0\\{.7}\\1\end{aligned}} \right)\end{aligned}\)

\(\begin{aligned}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{w_{k + 2}}\,\,\,\,\,{w_{k + 1}}\,\,\,\,\,{w_k}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\begin{aligned}{{}}{k = 0}\\{k = 1}\\ \vdots \\{}\\{}\\{}\\{}\\{K = 7}\end{aligned}\left( {\begin{aligned}{{}}0&{ - .7}&1\\{.7}&0&{.7}\\{ - 1}&{.7}&0\\{.7}&{ - 1}&{.7}\\0&{.7}&{ - 1}\\{ - .7}&0&{.7}\\1&{ - .7}&0\\{ - .7}&1&{ - .7}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{a_0}}\\{{a_1}}\\{{a_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}}0\\0\\0\\0\\0\\0\\0\\0\end{aligned}} \right)\end{aligned}\)

Write an equation \(A{\bf{x}} = {\mathop{\rm b}\nolimits} \), where A is a \(16 \times 3\) matrix formed from the two coefficient matrices above and where b in \({\mathbb{R}^{16}}\) is formed from the two right sides of the equations. Find \({a_0},{a_1},\) and \({a_2}\) given by the least-squares solution of \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \). (The .7 in the data above was used as an approximation for \(\frac{{\sqrt 2 }}{2},\) to illustrate how a typical computation in an applied problem might proceed. If .707 were used instead, the resulting filter coefficients would agree to at least seven decimal places with \(\frac{{\sqrt 2 }}{4},\frac{1}{2},\) and \(\frac{{\sqrt 2 }}{4},\) the values produced by exact arithmetic calculations.)

Short answer

The values are \({a_0} = {a_2} \approx .353535\), and \({a_1} = .5\).

Step-by-step solution

Least-square solution

When \(A\) is a \(m \times n\) matrix and \({\bf{b}}\) in \({\mathbb{R}^m}\), then \(\widehat {\bf{x}}\) in \({\mathbb{R}^n}\) is aleast-squares solutionof \(A{\bf{x}} = {\bf{b}}\) such that

\(\left\| {{\bf{b}} - A\widehat {\bf{x}}} \right\| \le \left\| {{\bf{b}} - A{\bf{x}}} \right\|\) for every \({\bf{x}}\) in \({\mathbb{R}^n}\).

Determine \({a_0},{a_1},\) and \({a_2}\) given by the least-squares solution of \(A{\mathop{\rm x}\nolimits}  = {\mathop{\rm b}\nolimits} \)

The given filter equation is \({a_0}{y_{k + 2}} + {a_1}{y_{k + 1}} + {a_2}{y_k} = {z_k}\) for all \(k\).

Take \(.7\) as an approximation of \(\frac{{\sqrt 2 }}{2}\), therefore \({a_0} = {a_2} \approx .353535\), and \({a_1} = .5\). Take \(.707\) as an approximation of \(\frac{{\sqrt 2 }}{2}\), therefore, \({a_0} = {a_2} \approx .35355339,\) and \({a_1} = .5\).

Thus, the values are \({a_0} = {a_2} \approx .353535\), and \({a_1} = .5\).