linear time invariant system

Table of Contents

LTI system S fulfills the properties

The output y of an LTI system is defined as the linear convolution of the impulse response and the input signal.

For a discrete system,

S(x[n])=ΣNi=0x[i]h[in]=(xh)[n]

1. Exponentials as eigenfunctions for the continuous case

We show that the exponential functions, z(t)=est where sC.

Consider a system S mapping a signal x(t)=z(t) to y(t), now

y(t)=(xh)(t)=τ=z(tτ)h(τ)=τ=es(nm)h(τ)=τ=estesτh(τ)Exponent laws=estτ=esτh(τ)Term does not depend on n=z(n)τ=z(m)h(m)Substitute z back

Where H(s)=τ=esτh(τ)

The whole summation on the right is a constant scaling factor. This means that after putting in the exponential to the system, we got back the same exponential scaled by amplitude or phase, but never frequency.

Note that H(s) is obtained by the two sided laplace transform of the impulse response in general.

This also implies that if s=jω, i.e z[n] representing a sinusoid, then the eigenvalue is given by the fourier transform.

In either case, H(s) is known as the transfer function.

2. Roots of unity as eigenvectors in discrete time

From the above, it appears that every vector given by any exponential would be an eigenvector, i.e satisfies Cx=λx for a circulant matrix C.

This isn't the case for the discrete scenario. Because the circulant matrix is diagonalizable in the Fourier basis (a result stemming from the convolution theorem), we can only have n eigenvectors for an n point DFT.

where specifically, the eigenvectors are exactly

vi=[1,ωi,ω2i,,ω(N1)i]

For all indices 0iNi.

Where, from the fourier matrix, ω=ej2πn. So, ω2i=ej2πn2i. Each eigenvector is a row of the dft matrix, but that specificity is only for convention because the matrix is symmetric.

We can interpret each row of the dft matrix (of the form ej2πnkn) where k is the row index and n is the column index as really being a sequence of complex phasors rotating at frequency k. This is why for an N point dft, there are going to be N complex exponentials with different frequencies that are eigenvectors.

The larger the DFT matrix, the more exponentials we can fit into the basis set of the eigenspace, but note that the since roots of unity are modular, the actual exponentials can't keep growing to infinity, since there is a fixed space (2π radians) around the complex plane at a fixed magnitude.

As we approach an ∞-point DFT, it turns out that from the above result, all complex exponentials, (meaning, every rotating phasors regardless of frequency), are eigenvectors.

  1. Proof outline

    Consider a simple 3×3 circulant matrix:

    [h(0)h(1)h(2)h(2)h(0)h(1)h(1)h(2)h(0)][ω0ω1ω2]

    The product is then

    [ω0h(0)+ω1h(1)+ω2h(2)ω0h(2)+ω1h(0)+ω2h(1)]=[ω0h(0)+ω1h(1)+ω2h(2)ω(ω1h(2)+ω0h(0)+ωh(1)]Factor out root=[ω0h(0)+ω1h(1)+ω2h(2)ω(ω2h(2)+ω0h(0)+ωh(1))]3rd root of unity modular symmetry=[ω0h(0)+ω1h(1)+ω2h(2)ω(ω0h(0)+ω1h(1)+ω2h(2))]Rearranging terms=(h(0)+ωh(1)+ω2h(2))[ω0ω1ω2]

    From this it is clearly visible that the second row is really ω times the entry above it.

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