Asynchronous Linear Systems with Exponential Inputs

FREQUENCY RESPONSE AND MEAN-SQUARED STABILITY

4.2 Asynchronous Linear Systems with Exponential Inputs

Consider a discrete time-invariant system with 𝑅 inputs, 𝑃 outputs, and 𝑁 state variables, whose state-space description is given as follows:

x𝑘+1 =A x𝑘 +B u𝑘 +w𝑘, (4.1)

y𝑘 =C x𝑘 +D u𝑘, (4.2)

wherex₀∈C^𝑁is the initial state vector (initial condition), andw𝑘 ∈C^𝑁 is the noise term with the following statistics:

E[w𝑘] =0, E w_𝑘 w^H𝑠

=𝛿(𝑘 −𝑠) 𝚪, (4.3) where 𝛿(·) denotes the discrete Dirac delta function, and𝚪 is allowed to be non- diagonal.

The matrices in the state-space model in (4.1) have the following dimensions:

A∈C^𝑁×𝑁, B ∈C^𝑁×𝑅, C∈C^𝑃×𝑁, D ∈C^𝑃×𝑅, (4.4) whereAis referred to as thestate-transition matrix, and the columns of the matrices BandDwill be denoted as follows:

B=[B₁ · · · B𝑅], D=[D₁ · · · D𝑅]. (4.5) We further assume that the input signalu𝑘 consists of 𝑅exponential signals in the following form:

u𝑘 = 𝛼^𝑘

1 · · · 𝛼^𝑘

𝑅

T

, (4.6)

where𝛼_𝑖’s are assumed to be distinct without loss of generality. Furthermore, we always assume that

|𝛼_𝑖| ≤ 1, ∀1≤ 𝑖 ≤ 𝑅, (4.7)

so that u𝑘 stays bounded throughout the iterations. While exponential inputs may seem restrictive, they form the basis for more general practical signals, making this study useful.

In the noise free case, i.e.,𝚪=0, it is well-known from linear system theory that the output vectory𝑘 ∈C^𝑃in (4.2) can be written as follows:

y𝑘 =y^ss_𝑘 +y^tr_𝑘, (4.8)

wherey^ss_𝑘 denotes the steady-state component, andy^tr_𝑘 denotes the transient compo- nent that are given as follows:

y^ss_𝑘 =

𝑅

Õ

𝑖=1

H𝑖(𝛼_𝑖) 𝛼^𝑘

𝑖, y^tr_𝑘 =C A^𝑘 x

0−x^ss

0

. (4.9)

whereH𝑖(𝑧) ∈C^𝑃denotes the transfer function that relates the𝑖^{𝑡 ℎ}input to the output, which is given as follows:

H𝑖(𝑧) =D𝑖+C 𝑧I−A_-1

B𝑖. (4.10)

We also note that the termx^ss

0 in (4.9) is given as follows:

x^ss₀ =

𝑅

Õ

𝑖=1

𝛼_𝑖I−A_-1

B𝑖. (4.11)

It is clear from (4.9) that when the state-transition matrixAis a stable matrix, i.e., the following holds true:

𝜌(A) <1, (4.12)

then the transient componenty^tr_𝑘 converges to zero as the iterations progress leaving only the steady-state componenty^ss_𝑘 in the output signal. In fact, stability ofAis also necessary for the transient part to converge to zero, which is a well-known result from linear system theory [95, 200].

In this chapter we consider the following randomized model:

(x𝑘+1)𝑖 =











A x𝑘 +B u𝑘+w𝑘

𝑖, 𝑖 ∈ T𝑘+1, (x𝑘)𝑖, 𝑖 ∉T𝑘+1,

(4.13)

y𝑘 =C x𝑘 +D u𝑘, (4.14)

whereT𝑘 denotes the set of indices updated at the 𝑘^{𝑡 ℎ} iteration. The model (4.13) is very similar to the standard synchronous recursions in (4.2) except the fact that only a random subset of indices (denoted byT𝑘) are updated in every iteration, and the remaining indices stay the same. The specific stochastic model regarding the selection of the indices will be elaborated next.

4.2.1 Random Selection of the Update Sets

In the asynchronous model considered in (4.13), we assume that the𝑖^{𝑡 ℎ} index (state variable) is updated independently with probability𝑝_𝑖in every iteration. That is,

P[𝑖 ∈ T𝑘] = 𝑝_𝑖 ∀𝑘 , (4.15)

where𝑝_𝑖will be referred to asthe update probability of the𝑖^{𝑡 ℎ} index. We will useP to denote the diagonal matrix consisting of the index selection probabilities. More precisely,

P=diag

[𝑝

1 𝑝

2 · · · 𝑝_𝑁]

. (4.16)

It is assumed thatPsatisfies0≺ P I, where the positive definiteness follows from the fact that no index should be left out permanently during the updates of (4.13).

See Section 4.4 for further details. Additionally, tr(P) =E[|T𝑘|]denotes the number of indices updated per iteration on average.

4.2.2 Frequency Response in the Mean

Due to the random updates of the state variables it is clear from (4.13) that the state vectorx𝑘 is a random vector. So, the state vector will not have an exponential behavior exactly even when the input is a simple exponential (𝑅 = 1 in (4.6)).

Nevertheless, we will show thatx𝑘 still behaves like a sum of exponential signalsin a statistical sense:

Theorem 4.1. Assume that the randomized asynchronous state recursions in(4.13) are initialized independently and randomly. Then, the expectation of the state vector in(4.13)is as follows:

E[x𝑘] =x^ss_𝑘 +x^tr_𝑘, (4.17) where

x^ss_𝑘 =

𝑅

Õ

𝑖=1

(𝛼_𝑖I−A)¯ ⁻¹B¯𝑖𝛼^𝑘

𝑖, x^tr_𝑘 =A¯^𝑘 E[x

0] −x^ss

0

, (4.18)

and

A¯ =I+P (A−I), B¯𝑖 =P B𝑖. (4.19)

Proof. See Section 4.8.2.

Here, ¯Awill be referred to asthe average state-transition matrix, and ¯Bthe average input matrix. We can also representx^ss_𝑘 in (4.18) as follows:

x^ss_𝑘 =

𝑅

Õ

𝑖=1

x^ss

0,𝑖 𝛼^𝑘

𝑖 where x^ss

0,𝑖 =(𝛼_𝑖I−A)¯ ⁻¹B¯𝑖, (4.20) which will be useful later in the chapter.

As an immediate corollary to Theorem 4.1, we present the following result regarding the expected behavior of the output:

Corollary 4.1. Assume that the randomized asynchronous state recursions in(4.13) are initialized independently and randomly. Then, the expectation of the output in (4.14)is as follows:

E[y𝑘] =y^ss_𝑘 +y^tr_𝑘, (4.21) where

y^ss_𝑘 =

𝑅

Õ

𝑖=1

H¯𝑖(𝛼_𝑖) 𝛼^𝑘

𝑖, y^tr_𝑘 =CA¯^𝑘 E[x₀] −x^ss₀

, (4.22)

and

H¯𝑖(𝑧) =D𝑖+C 𝑧I−A¯-1B¯𝑖. (4.23) This can, therefore, be regarded as the “transfer function in the mean” from the𝑖^{𝑡 ℎ} input node to the output.

Proof. Due to (4.6), (4.14), and Theorem 4.1, the expectation ofy𝑘 can be written as follows:

E[y𝑘] =CE[x𝑘] +

𝑅

Õ

𝑖=1

D𝑖𝛼^𝑘

𝑖 =y^ss_𝑘 +y^tr_𝑘, (4.24)

wherey^ss_𝑘 andy^tr_𝑘 are as in (4.22).

Regarding the form in (4.21) we first note that the termsy^ss_𝑘 andy^tr_𝑘 are deterministic quantities, and the expectation is with respect to the random selection of the indices, the input noise, and the random selection of the initial condition.

Corollary 4.1 shows that the random output vector y𝑘 behaves the same as its deterministic counterpart (4.8) in expectation. That is,E[y𝑘] can be decomposed into steady-state and transient parts similar to (4.8). Therefore, the quantity ¯H𝑖(𝑧) given in (4.23) can be regarded as the “transfer function” from the𝑖^{𝑡 ℎ} input to the output in theexpectation sense.

It is clear from (4.22) that as long as the average state transition matrix ¯Ais stable, i.e., the following holds true:

𝜌(A)¯ <1, (4.25)

the component y^tr_𝑘 converges to zero irrespective of the observation matrixC and the statistics of the initial condition. Thus, the condition (4.25) is both necessary and sufficient forE[y𝑘]to behave like a sum of exponentials, that is,

𝑘lim→∞E

y𝑘 −y^ss_𝑘

=0. (4.26)

This shows that when (4.25) is satisfied an exponential input results in an exponential outputin expectationeven with the randomized asynchronous state recursions.

4.2.3 A Running Numerical Example

In order to demonstrate the behavior of the random vector y𝑘, we consider the following state-space model with 𝑁 =4 state variables, 𝑅=1 input, and 𝑃=1 output:

A= 1 10





















-4 -1 2 -6 4 -6 -5 3

2 -2 7 2

5 9 -3 1





















, B=



















 1 4 2 3





















, C=



















 1 1 1 1





















T

, D=0, (4.27)

where we point out that the matrix A is not stable since 𝜌(A) ≈1.0441. As we shall discuss later in Section 4.3, stability of the randomized asynchronous state recursions does not require stability of the state-transition matrix in general.

In the following numerical example we assume thatP=𝑝I, i.e., all nodes have the update probability𝑝and assume that𝚪=0. Furthermore, we assume that the input signal has the following complex exponential form:

𝛼=𝑒^{𝑗2𝜋/100} =⇒ u𝑘 =𝑒^{𝑗2𝜋 𝑘/100}. (4.28) In Figure 4.1 we visualize a realization of the output signal y𝑘 together with the steady-state componenty^ss_𝑘 as well as the input signal𝑢_𝑘 for three different update probabilities, namely, 𝑝 ∈ {0.1, 0.3, 0.6}. The figure shows only the real part of the signals for convenience.

From Figure 4.1 it is clear that the random vectory𝑘is not a complex exponential in a strict sense, yet it “behaves like” one. We also note thaty𝑘has the same “frequency”

as the input signal irrespective of the update probabilities.

Figure 4.1 shows also that the response of the random asynchronous system depends on the update probabilities, which is also apparent from the expression in (4.23). In fact, the response of the random asynchronous updates running on a system denoted with (A,B,C,D) can be represented in an average sense as the response of a syn- chronous system(A¯,B¯,C,D). As a result, even when a single state variable updates its value with a different probability, the response of the overall system changes. In short, different update probabilities result in different “frequency responses” while leaving the output frequency unchanged.