The expression of the filter can be written as H z 1 2 z 2 z 4 1 z 2 2 1 z 1 2

# The expression of the filter can be written as h z 1

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This preview shows page 3 - 6 out of 11 pages. ECE 513 Exam #2 Instructor: Dr. Cranos Williams 4 3.Frequency Sampling Method: (20 Pts)We want to design a filterh(n) whose magnitude response is 0.5 at|ω|= 4π/7, at-tenuates frequencies (zero response) for|ω|<4π/7, and has unity gain at frequencies|ω|>4π/7, where-πωπ.(a) Define the values of anN= 7 point DFT,H(k), for all values ofk, which matchesthe response described above.(b) Given that we desire alinearphase andcausalfilter, write a formula forH(k),defined for all values ofk= 0, . . . , N-1.(c) Using the definition of the inverse DFT (h(n) =1NN-1k=0H(k)ej2πknN) and theH(k) defined in part b.), write a formula forh(n) whereh(n) is composed ofONLY real components. HINT:h(n) will be of the formh(n) =A1cos(ω1(n-n1)) +A2cos(ω2(n-n2))(7) Solution: ECE 513 Exam #2 Instructor: Dr. Cranos Williams 5 (c) (EXTRA CREDIT 5 pts) Now, we can solve for h ( n ) using the inverse DFT. h ( n ) = 1 N N - 1 X k =0 H ( k ) e j 2 πkn N = 1 7 6 X k =0 | H ( k ) | e - j 2 πk 3 7 e j 2 πkn 7 = 1 7 6 X k =0 | H ( k ) | e j 2 πk ( n - 3) 7 = 1 7 | H (2) | e j 2 π 2( n - 3) 7 + | H (3) | e j 2 π 3( n - 3) 7 + | H (4) | e j 2 π 4( n - 3) 7 + | H (5) | e j 2 π 5( n - 3) 7 = 1 7 1 2 e j 2 π 2( n - 3) 7 + e j 2 π 3( n - 3) 7 + e j 2 π 4( n - 3) 7 + 1 2 e j 2 π 5( n - 3) 7 (11) You can multiply any term by e - j 2 π 7( n - 3) 7  • • • 