Solved: [M] Example 3 in Section 4.8 displayed a low-pass

Chapter 6, Problem 26E

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Problem 26E

[M] Example 3 in Section 4.8 displayed a low-pass linear filter that changed a signal  and changed a higher-frequency signal  into the zero signal, where

 The following calculations will design a filter with approximately those properties. The filter equation is

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

Write an equation Ax = b, where A is a 16 × 3 matrix formed from the two coefficient matrices above and where b in R16 is formed from the two right sides of the equations. Find  given by the least-squares solution of Ax = b. (The .7 in the data above was used as an approximation for , 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 , the values produced by exact arithmetic calculations.)

Reference Example 3 in 4.8: