TVSC_Lecture_3

124 112 here and in the following a as an entry

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Unformatted text preview: hat for large systems with n ∼ 10, 000 and given that r << n, the complexity of implementation could be reduced to O(8 · r · n) compared to O((1/2)n2 ) for a straight-forward implementation. However, the question is how can we u1 u3 u2 x2 1/2 1/3 Z 1/3 y1 x3 u4 1/4 1/4 Z x4 Z y3 y2 y4 Abbildung 4: Simplified implementation of the matrix T in Equation 8 tell if implementation complexity reduction is possible for a given system, and how to determine the minimal-complexity implementation? To answer this, we take a closer look at the implementations. For the system in Figure 4 we can read off the individual realization matrices · 1/2 ·1 Σ1 = Σ2 = 1/3 1 1/3 1 1/4 1 Σ3 = 1/4 1 Σ4 = · 1 · 1 , which we can decompose and reassemble to form the time-varying realization matrix A C Σ= B D with the corresponding block-diagonal matrices [·] [ A1 ] [1/3] [A2 ] = A = [A3 ] [A4 ] B = [B1 ] [B2 ] [B3 ] [B4 ] = [1/4] [·] [1/2] , [1/3] [1/4] [·] , (14) 11 Lecture 3 C = [C1 ] [C2 ] [C3 ] [C4 ] [D1 ] D = [·] = = [D2 ] [D3 ] [D4 ] , [1] [1] [1] [1] . [1] [1] [1] Realization of Inverse Operator Based on the realization given in the previous section, we can determine a realization Γ, which implements the inverse operator T −1 . We write the realization of the inverse operator as ˆˆ AB ˆD Cˆ Γ= We can determine ˆ Ak Γk = ˆk C . the individual realization matrices Γk as − − ˆ Bk Ak − Bk · Dk 1 · Ck Bk · Dk 1 = −1 −1 ˆ −Dk · Ck Dk Dk . (15) For the in system depicted in Equation 14, the inverse realization can be done by using Equation 15, and is given as Γ1 = ˆ A= ˆ B= ˆ C= ˆ D= · · 1/2 1 ˆ [A1 ] ˆ [B1 ] ˆ [C1 ] ˆ [D1 ] Γ2 = ˆ [A2 ] ˆ [B2 ] ˆ [C2 ] ˆ [D2 ] ˆ [A3 ] ˆ [B3 ] ˆ [C3 ] ˆ [D3 ] 0 −1 ˆ [A4 ] ˆ [B4 ] ˆ [C4 ] 1/3 0 1/4 Γ3 = 1 −1 1 [·] [1/3] = [1/4] = = Γ4 = [·] [1/2] [1/3] [1/4] [·] = [1] [1] [1] [1] [1] [·] · −1 · 1 (16) [1] ˆ [1] [D4 ] The realization is depicted in Figure 5. The inverse transfer function can be read-off as 1 0 −1 −1/2 1 . ˆ ˆ ˆ ˆ T −1 = D − C 1 − Z A ZB = 0 −1/3 1 0 0 −1/4 1 (17) 12 Lecture 3 y1 1/3 1/2 Z y4 1/4 Z −1 u1 y3 y2 Z −1 u2 −1 u3 Abbildung 5: Inverse realization of the system u4 13 Lecture 3 Efficient Methods for Solving Linear Systems of Equations In a more general setting we consider the problem of solving the set of linear equations T · u = y , for the variable u where here the matrix T is not assumed to have any special structure. The variable u takes on the interpretation of an input signal applied to the input of a linear time-variant system, which we describe by means of the matrix T . The signal y is the corresponding output signal computed by a linear matrix-vector multiplication. In many technical situations we may need to invert this map. We can solve the inversion problem by determining the inverse of T , which we may compute using standard linear algebra tools such as Gaussian elimination or similar. However, for a general matrix T this inversion process costs O(n3 ) operations, if n denotes the size of the matrix n. Similar to the strategy of using Fast Fourier techniques...
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This note was uploaded on 07/07/2013 for the course EI 2012 taught by Professor Tum during the Winter '12 term at TU München.

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