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# Discrete-time Filter as a Matrix Operation

Module by: Ivan Selesnick. E-mail the author

Consider a linear time-invariant system with impulse response hn h n . (Is assumed in these notes that all impulses response hn h n are real-valued.) (Figure 1)

The block diagram is represented in matrix form as y0y1y2y3y4y5y6=( h0000 h1h000 h2h1h00 h3h2h1h0 0h3h2h1 00h3h2 000h3 )x0x1x2x3x4x5x6 y0 y1 y2 y3 y4 y5 y6 h0 000 h1 h0 00 h2 h1 h0 0 h3 h2 h1 h0 0 h3 h2 h1 00 h3 h2 000 h3 x0 x1 x2 x3 x4 x5 x6 where Q=( h0000 h1h000 h2h1h00 h3h2h1h0 0h3h2h1 00h3h2 000h3 ) Q h0 000 h1 h0 00 h2 h1 h0 0 h3 h2 h1 h0 0 h3 h2 h1 00 h3 h2 000 h3

The vectors are of infinite length. The matrix has infinitely many rows and infinitely many columns. We can write convolution as a matrix-vector multiplication: yn=hn*xn y=Qx y n h n x n y Q x where QQ is the convolution matrix. The matrix QQ is also called the Toeplitz matrix - a Toeplitz matrix is constant along its diagonals. Note that the impulse response hnh n appears in each column of the matrix QQ.

What is the transpose of the linear time-invariant system with impulse response hn h n ? Usually, we do not think about the transpose of a system. However, we can derive the transpose of the system by using the matrix representation. We just need to look at the transpose of the matrix QQ. The columns of QQ are the rows of QT Q , so QT Q has the following form: QT=( h3000 h2h300 h1h2h30 h0h1h2h3 0h0h1h2 00h0h1 000h0 ) Q h3 000 h2 h3 00 h1 h2 h3 0 h0 h1 h2 h3 0 h0 h1 h2 00 h0 h1 000 h0 We can observe that QT Q is again a convolution matrix (it is constant along the diagonals). So QT Q represents a linear time-invariant system. The impulse response of this system can be found by looking at the columns of QT Q , which we see is the flipped version of hn h n . Therefore, QT Q represents convolution by hn h n . (Figure 2)

y=Qxyn=hn*xn y Q x y n h n x n z=QTyzn=hn*yn z Q y z n h n y n Can the filter hn h n represent an orthogonal transformation? Equivalently, is it possible that QT Q is the inverse of QQ? If QTQ=I Q Q I , then z=QTy=QTQx=Ix=x z Q y Q Q x I x x or zn=xn z n x n . But

zn=hn*yn=hn*hn*xn z n h n y n h n h n x n
(1)
so zn=xn z n x n only if hn*hn=δn h n h n δ n

Note that hn*hn=δnH1zHz=1HeiωHeiω=1|Heiω|2=1 h n h n δ n H1z Hz 1 H ω H ω 1 H ω 2 1 That means Hz Hz must be an allpass system. The only orthonormal LTI systems are allpass systems. But they provide no frequency selective filtering - the allpass filter does not provide a subband decomposition of the signal xn x n .

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