# ISU EE 524 - High-resolution Parametric (36 pages)

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- School:
- Iowa State University
- Course:
- Ee 524 - Dig Signal Process

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High resolution Parametric Subspace Methods The first parametric subspace based method was the Pisarenko method which was further modified leading to the MUltiple SIgnal Classification MUSIC method MUSIC Method Recall the model x t As t n t where A is a Vandermonde matrix A 1 1 e j 1 e j 2 e j 1 N 1 e j 2 N 1 1 e j L e j L N 1 R E x t xH t AE s t sH t AH 2I ASAH 2I and S E s t sH t is a full rank diagonal matrix EE 524 Fall 2003 9 1 From the eigenanalysis of R we see that the eigenvalues of R are k 2 k 1 L signal subspace k 2 k L 1 N noise subspace the noise subspace eigenvectors are orthogonal to the column space of A Let the signal and noise subspace eigenvectors be given by E u1 u2 uL signal subspace G uL 1 uL 2 uN noise subspace Then RG 2G On the other hand RG ASAH 2I G ASAH G 2G Comparing the above two equations we conclude AH G 0 EE 524 Fall 2003 9 2 The noise subspace eigenvectors of R are orthogonal to the columns of A In turn the signal subspace eigenvectors span the same subspace as the column space of A The true frequencies L l 1 are the solutions to aH GGH a N aH EE H a 0 where GGH PA EE H I GGH PA MUSIC Spectral Estimate PMUSIC 1 1 H H H H a GG a N a EE a In practice we do not know R so PbMUSIC 1 1 H H H H b b b b a GG a N a E E a where b R K 1 X x k xH k K k 1 bG b H PbA G EE 524 Fall 2003 9 bE bH I G bG b H PbA E 3 Remarks the number of signals L must be known or estimated MUSIC involves eigendecomposition computational cost can be quite intensive if the MUSIC estimate is evaluated with a fine grid EE 524 Fall 2003 9 4 Root MUSIC Algorithm Instead of computing the spectral MUSIC estimate root the polynomial bG b H a z 0 aT 1 z G where a z 1 z 1 z N 1 Remarks MUSIC polynomial has the order 2N 2 and therefore 2N 2 roots the roots form N 1 pairs where one root is the conjugate reciprocal of another i e if z is a root then 1 z will be a root as well bG b H a z aT 1 z G bG b H a z H 0 aT 1 z G bG b H a 1 z aT z G 1 H b b a 1 e z GG a e z with ze z T If z root then 1 z root too EE 524 Fall 2003 9 5 Example Let z 0 8 ej 4 be a root of the MUSIC polynomial Then the conjugate reciprocal root is 1 1 j 4 1 25 e z 0 8e j 4 Thus the angle of the root does not change but it lies at the opposite side of the unit circle Select only the roots that lie inside the unit circle Estimates of signal frequencies b zl l 1 2 L using L roots closest to the unit circle so called signal roots with z 1 EE 524 Fall 2003 9 6 Root MUSIC has better performance at low SNR or small number of samples than spectral MUSIC because it is insensitive to radial errors more precisely sensitive only to errors that cause subspace swapping Root MUSIC has much simpler implementation than spectral MUSIC because it does not require any search over frequency EE 524 Fall 2003 9 7 Minimum norm Method PbMN 1 H 2 a w where the vector w has the first element equal to 1 and minimum norm w belongs to the sample noise subspace Optimization problem min wH w w subject to wH e1 1 bG b H w w G Substituting the second constraint into the objective function and first constraint yields H b bH bG bH b bH wH w wH G zG G w w GG w I bG b H e1 1 wH e1 wH G With these expressions our optimization problem transforms to bG b H w subject to wH G bG b H e1 1 min wH G w EE 524 Fall 2003 9 8 Hence H bG b H w 1 wH G bG b H e1 1 eH b b Q w wH G G G w 1 bG b H w G bG b H e1 Q G bG b H w G bG b H e1 G bG b H w G bG b H e1 to the constraint Substituting the solution G bG b H e1 1 we get equation wH G bG b H e1 1 eH G 1 1 H H bG b e1 e G 1 Finally bG b H e1 w G 1 bG b H e1 G bG b H e1 eH G 1 where we used the second constraint Substituting this solution into the expression for PbMN we get H b bH 2 1 e G G e 1 1 PbMN H bG b H e1 2 a w 2 aH G The constant in the numerator does not alter the shape of the spectrum and as a rule is omitted PbMN EE 524 Fall 2003 9 1 1 H H 2 H H 2 b b b b a GG e1 1 a E E e1 9 ESPRIT Method Consider A 1 1 e j 1 e j 2 e j 1 N 1 e j 2 N 1 1 e j L e j L N 1 Let A and A be the matrices with eliminated first and last row respectively It can be readily shown that AD A D diag ej 1 ej L Let E and E be formed from the signal eigenvector matrix E in the same way as A and A from A EE 524 Fall 2003 9 10 Recall that E and A span the same signal subspace Therefore E AC E AC ADC E AC 1 1 D C AC E EC 1D 1C A DCC z I Equivalently E EC 1DC E E where C 1DC In practice both C and D are unknown LS solution for H 1 E E H E E From it follows that the diagonal elements of D are the eigenvalues of EE 524 Fall 2003 9 11 ESPRIT Step 1 Compute the eigendecomposition of the sample b and obtain the sample signal subspace covariance matrix R b E b b and E Step 2 Form the matrices E Step 3 Compute H H b b b 1 b E E E E b b and Step 4 Form the eigenvalues bl l 1 2 L of obtain the frequency estimates as follows bl bl EE 524 Fall 2003 …

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