Unformatted text preview: he equation we can see that the frequency function is a function of m. As from the lecture notes of lecture 1, we can get that for a real valued x[n] and even integer N, the magnitude of the DFT plot is symmetric and the phase of the DFT plot is anti
symmetric about N/2 point. And the magnitude of all the points except at f1 = m/N and (N
m)/N frequency points are of zero value. The value of !
the DFT at frequency m/N(Hz) is X[m] = ! ∗ ! !" , so the maximum magnitude value is N/2, and the phase at this frequency is !. Also at frequency (N
m)/N !
(Hz), the DFT value is X[N
m] = ! ∗ ! !!" , so the magnitude is N/2, and the phase at this frequency is –!. 4. Answer: The explanation of each statement is as below: b=A; /* T */ (Assign the address of the first element of array A to b.) printf("a: %d...
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 Fall '08
 STAFF
 the00, Passband Ripple, Dpass

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