Problem1 - Problem1 Case1 Step 1 Defining the signal...

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Unformatted text preview: Problem1 Case1 : Step # 1- Defining the signal N-Points N 256 = N-Points N 256 = f s 1000 = Sampling Frequency - Hz f s 1 10 3 × s Hz ⋅ = t 1 f s = f0 20 f s N ⋅ = Analog Input Frequency - Hz f0 78.125 = A p 5 = n N 1- .. = x n A p sin 2 π ⋅ f0 f s ⋅ n π 3 + ⋅ = Input Signal f0 f s N 20 = BIN frequency !! Theoretical rms value !! rms_a A p 2 = rms_a 3.536 = Theoretical total power (rms ^2) value !! rms_a 2 12.5 = 51 102 153 204 255 6- 3- 3 6 INPUT SEQUENCE Time [ Sample ] Amplitude [ Volts ] x n n Step # 2- Using the new definition of the N-Point DFT of x[n] f n f s N n ⋅ = These are the BIN frequencies [ Hz ] DFT X n N 1- m x m e i- 2 π ⋅ N ⋅ m ⋅ n ⋅ ⋅ ∑ = = P1 n 20 log X n ( 29 ⋅ = P_r max P1 ( ) = P2 n P1 n = P1_h n P1 n P_r- = 100 200 400- 300- 200- 100- DFT of Input Sequence Discrete Samples Magnitude P1_h n n Step # 3- DFT of WINDOWED SEQUENCE USING HAMMING WINDOW k N 1- .. = X1_hamm k N 1- m x m 0.54 0.46 cos 2 π ⋅ m N 1- ⋅ ⋅- m ≤ N 1- ≤ if 0 otherwise ⋅ e i- 2 ⋅ π ⋅ m ⋅ k N ⋅ ⋅ ∑ = = P1_hamm k 20 log X1_hamm k ( 29 ⋅ = P_r_hamm max P1_hamm ( ) = P2_hamm k P1_hamm k = P_r_hamm 50.743 = P1_hamm k P1_hamm k P_r_hamm- = 100 200 150- 100- 50- Normalized DFT of Hamming Window Discrete frequencies Magnitude in db P1_hamm k k Comparing two DFT's 200 400 600 800 1 10 3 × 400- 300- 200- 100- Comparision of DFT sequences Analog Frequency -Hz Power in dbW P1_hamm k P1_h k k f s N ⋅ Step 5 : Power Spectral Density Xp n 1 N X n ⋅ = Normalized DFT ∆ f f s N = ∆ f 3.906s Hz ⋅ = P n Xp n ( 29 2 = Power Spectrum (volts ^2) PSD n Xp n ( 29 2 ∆ f = Power Spectral Density (volts ^2 / Hz) 100 200 300 400 500 600 700 800 900 1 10 3 × 1 10 5- × 1 10 4- × 1 10 3- × 0.01 0.1 1 10 Power Spectrum Frequency [ Hz ] Power Spectrum [ volts^2 ] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 10 6- × 1 10 5- × 1 10 4- × 1 10 3- × 0.01 0.1 1 Power Spectral Density Normalized frequency [ unitless ] Power Spectral Desnity [ volts^2 / Hz ] TIME DOMAIN rms CALCULATION---rms value of signal --- 1 N N 1- m x m ( 29 2 ∑ = ⋅ 3.536 = FREQUENCY DOMAIN rms CALCULATION Summing all power coponents over the frequency spectrum N 1- m Xp m ( 29 2 ∑ = 3.536 =---rms value of signal --- Area under the curve over the frequency spectrum N 1- m PSD m ∆ f ⋅ ( 29 ∑ = 3.536 =---rms value of signal --- Step # 5- Estimation of Signal frequency and amplitude PEAK FINDING ALGORITHM !!!...
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Problem1 - Problem1 Case1 Step 1 Defining the signal...

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