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hwsolB21

# hwsolB21 - x2[n =(1/10*exp(j*4*pi*n/5(iii X3 = 0 0 0 0 j...

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Sheet1 Page 1 HW 21 SOLUTION ______________ (i) X1 = [ 0 1 0 0 0 0 0 0 0 1 ].' The Fourier frequencies present in x1 are: w = (2*pi/10) = pi/5 and w = 9*(2*pi/10) = 9*pi/5 The corresponding Fourier sinusoids are v1[n] = exp(j*pi*n/5) and v9[n] = exp(j*9*pi*n/5) = exp(-j*pi*n/5) (n = 0:9). v9 is the complex conjugate of v1 since it has frequency w = 9*pi*n/5 (same as -pi/5). Thus x1 = (1/10)*V*X1 = (1/5)*cos(pi*n/5) (ii) X2 = [ 0 0 0 0 1 0 0 0 0 0 ].' x2 = (1/10)*V*X2 = v4/10 where v4 is the k=4th Fourier sinusoid, at frequency w = 4*(2*pi/10) = 4*pi/5. Thus
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Unformatted text preview: x2[n] = (1/10)*exp(j*4*pi*n/5) (iii) X3 = [ 0 0 0 0 j 0 -j 0 0 0 ].' x3 = (1/10)*V*X3 = (v4 - v6)/(10*j) v6 is the complex conjugate of v4 since it has frequency w = 6*(2*pi/10) = 6*pi/5 (same as -4*pi/5). Thus x3[n] = (-j/10)*(exp(j*4*pi*n/5) - exp(-j*4*pi*n/5)) = (1/5)*sin(4*pi*n/5) (iv) X4 = [ 0 1 0 0 5*j 0 -5*j 0 0 1 ].' By linearity: x4[n] = (1/5)*sin(pi*n/5) + sin(4*pi*n/5) (v) 2cos(2*pi*n/5) is the sum of v2 and v8. 2*cos(3*pi*n/5) is the sum of v3 and v7. Therefore X5 = [ 40 0 -5 -35 0 0 0 -35 -5 0].'...
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