{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Problem1

# Problem1 - Problem1 Case1 Step#1 N:= 256 fs:= 1000 Defining...

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 0 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 = 0 51 102 153 204 255 6 - 3 - 0 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 0 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 - = 0 100 200 400 - 300 - 200 - 100 - 0 DFT of Input Sequence Discrete Samples Magnitude P1_h n n

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Step # 3 - DFT of WINDOWED SEQUENCE USING HAMMING WINDOW k 0 N 1 - .. = X1_hamm k 0 N 1 - m x m 0.54 0.46 cos 2 π m N 1 - - 0 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 - = 0 100 200 150 - 100 - 50 - 0 Normalized DFT of Hamming Window Discrete frequencies Magnitude in db P1_hamm k k
Comparing two DFT's 0 200 400 600 800 1 10 3 × 400 - 300 - 200 - 100 - 0 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)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
0 100 200 300 400 500 600 700 800 900 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 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 0 N 1 - m x m ( 29 2 = 3.536 = FREQUENCY DOMAIN rms CALCULATION Summing all power coponents over the frequency spectrum 0 N 1 - m Xp m ( 29 2 = 3.536 = ---rms value of signal --- Area under the curve over the frequency spectrum 0 N 1 - m PSD m f ( 29 = 3.536 = ---rms value of signal --- Step # 5 - Estimation of Signal frequency and amplitude PEAK FINDING ALGORITHM !!! STEP A -- Find the max spectrum power using a simple maximum seeking function maximum PSDr in the spectrum MFFT max PSD ( ) = MFFT 1.6 = volts ^ 2 / Hz STEP B -- Program for finding the index of the of all spectral component exceeding LIM

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
LIM MFFT 0.95 = Max i 0 U 0 U i j i i 1 + PSD j LIM if j 0 rows PSD ( ) 2 .. for U = NOTICE THAT I AM ONLY SCANNING THE POSITIVE FREQUENCIES Max 20 ( ) =
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 34

Problem1 - Problem1 Case1 Step#1 N:= 256 fs:= 1000 Defining...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online