Homework 3 1. Suppose we have two four-point sequences x[n] and h[n] as follows: (a) Calculate the four-point DFT X[k]. (b) Calculate the four-point DFT.

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Presentation transcript:

Homework 3 1. Suppose we have two four-point sequences x[n] and h[n] as follows: (a) Calculate the four-point DFT X[k]. (b) Calculate the four-point DFT H[k]. (c) Calculate y[n] = x[n] 4 y[n] by doing the circular convolution directly. (d) Calculate y[n] in part (c) by multiplying the DFTs of x[n] and h[n] and performing an inverse DFT.

Homework 3 Solution: The formula of four-point DFT . (a) Substitute x[n] to the formula, then X[0] = 1+0+(-1)+0 = 0, X[1] = 1+(-j)+1+j = 2, X[2] = 1+0+(-1)+0 = 0, X[3] = 1+0+1+0 = 2. (b) Substitute h[n] to the formula, then H[0] = 1+2+4+8 = 15, H[1] = 1+(-2j)+(-4)+8j = -3+6j, H[2] = 1+(-2)+4+(-8) = -5, H[3] = 1+2j+(-4)+(-8j) = -3-6j.

Homework 3 Solution (cont.): (c) y[n] = x[n] 4 h[n] = . (d) Y[k] = X[k]·H[k]. Y[0] = 0, Y[1] = -6+12j, Y[2] = 0, Y[3] = -6-12j. The formula of of four-point IDFT . y[0] = ¼ (0+(-6+12j)+0+(-6-12j)) = -3. y[1] = ¼ (0+(-6j-12)+0+(6j-12)) = -6. y[2] = ¼ (0+(6-12j)+0+(6+12j)) = 3. y[3] = ¼ (0+(6j+12)+0+(-6j+12)) = 6.

Homework 3 2. Suppose that a computer program is available for computing the DFT i.e., the input to the program is the sequence x[n] and the output is the DFT X[k]. Show how the input and/or output sequences may be rearranged such that the program can also be used to compute the inverse DFT

Homework 3 Solution: Given a X[k], k = 0,1,…,N-1. We then obtain a Y[k] as the rearrange of X[k] as follows: Y[0] = X[0], Y[k] = X[N-k], k = 1,2,…,N-1. Taking Y[k] as input of the program, we then obtain Therefore, The IDFT of X[k] can be obtained .

Homework 3 3. Consider the systems shown in the following figure. Suppose that H1(ejw) is fixed and known. Find H2(ejw), the frequency response of an LTI system, such that y2[n] = y1[n] if the inputs to the systems are the same. H1(ejw) 2 2 x[n] y1[n] H2(ejw) x[n] y2[n]

Homework 3 X(e2jw) X(e2jw) H1(ejw) H1(ejw) 2 2 X(ejw) Y1(ejw) Solution: We can analyze the system in the frequency domain: Y1(ejw) is X(e2jw) H1(ejw) downsampled by 2: Y1(ejw) = ½ {X(e2jw/2)H1(ejw/2)+X(e2j(w-2π)/2)H1(ej(w-2π)/2)} = ½ {X(ejw)H1(ejw/2)+X(ej(w-2π))H1(ej(w/2-π))} = ½ {H1(ejw/2)+H1(ej(w/2-π))}X(ejw) = H2(ejw)X(ejw) Therefore, H2(ejw) = ½ {H1(ejw/2)+H1(ej(w/2-π))}. X(e2jw) X(e2jw) H1(ejw) H1(ejw) 2 2 X(ejw) Y1(ejw)