TUGAS PENGOLAHAN SINYAL DIGITALKELOMPOK 1Oleh :Nikko Haendratnio H1C007061Henky Agung W.H1C008066Muhamad Abduh H1C009003Suwitno H1C009039KEMENTERIAN PENDIDIKAN NASIONALUNIVERSITAS JENDRAL SOEDIRMANFAKULTAS SAINS DAN TEKNIKJURUSAN TEKNIKPROGRAM STUDI TEKNIK ELEKTROPURBALINGGA2011/2012TUGAS PSD CHAPTER 41. given a sequence x(n) for3 where x(0) = 1, x(1) = 1, x(2) = -1 and x(3) = 0, compute its DFT X(k).Answer:X(0) = = ( x(0) + x(1) + x(2) + x(3) ) =(1+1-1+0) = 1X(1) ==(++ )=x(0) - jx(1) x(2) + jx(3) = 1 j + 1 + 0 =2-jX(2) == (++ )= x(0) - x(1) + x(2) x(3)= 1 1 1 + 0= - 1 X(3)== (++ )= x(0) + jx(1) - x(2) jx(3)= 1 + j + 1 - 0 = 2 + j2. Given a sequence x(n) for 0 > X = fft ([4 3 2 1])X =10.0000 2.0000 - 2.0000i 2.0000 2.0000 + 2.0000iDFT Using matlab with zero padding :>> X = fft ([4 3 2 1 0 0])X =10.00003.5000-4.3301i2.5000-0.8660i 2.00002.5000+0.8660i 3.5000+4.3301i5. Using the DFT sequence X(k) forcomputed in problem 4.4, evaluate the inverse DFTx(0) and x(4)Answer:Dari soal 4 didapatkan: x(0) = 10X(1) = 3,5 4,3301jX(2) = 2,5 0,8660jX(3) = 2X(4) = 2,5 + 0,8660jX(5) = 3,5 + 4,3301jDitanyakan x(0) dan x(4)X(0)=(x(0) + x(1) + x(2) + x(3) + x(4) + x(5)=( 10 + 3,5 - 4,3301j + 2,5 0,8660j + 2 + 2,5 + 0,8660j + 3,5 + 4,3301j)=(10 + 3,5 + 2,5 + 2 + 2,5 + 3,5)=(24)=4X(4)= =( ++++)=( 10 (3,5 - 4,3301j) (2,5 0,8660j) + 2 (2,5 + 0,8660j) (3,5 + 4,3301j))=(0)= 06. Consider a digital sequence sampled at the rate of 20,000 Hz. If we usethe 8,000-point DFT to compute the spectrum, determine : a. the frequency resolutionb. the folding frequency in the spectrum.Answer:a.A f = fsN= 20.0008000= 2.5 Hzb. f max = N2A f = 10.000 Hz7. We use the DFT to compute the amplitude spectrum of a sampled data sequence with a sampling rate fs = 2,000 Hz. It requires the frequency resolution to be less than 0.5 Hz. Determine the number of data points used by theFFTalgorithm and actual frequencyresolution inHz, assuming that the data samples are available for selecting the number of data points.Jawab:A f =0.5 HzN=fsA f =20000.5 =4000Since we use the FFT to compute the spectrum, the number of the data points must be a powerof 2, thatis,N=215=32768And the resulting frequency resolution can be recalculated asA f = fsN=200032768=0.061 Hz8. Penyelesaian :Diketahui:N = 4fs = 100 HzW4 = e jn2x(0) = -1; x(1)= 2; x(2) = 1; x(3) = 4Ditanyakan: amplitude spectrum, phase spectrum and power spectrum . . .?Jawab: using matlab function we can get DFT X(k)>> X=fft ([-1 2 1 4])X = 6.0000-2.0000 + 2.0000i-6.0000-2.0000 2.0000iwe get X(0)= 6; X(1)= -2+2j = 2.828