10.2.2 Linear Time-Invariant (LTI) Systems with Random Inputs

Linear Time-Invariant (LTI) Systems:

A linear time-invariant (LTI) system can be represented by its impulse response (Figure 10.6). More specifically, if X(t) is the input signal to the system, the output, Y(t), can be written as
Y(t)=h(α)X(tα)dα=X(α)h(tα)dα.
The above integral is called the convolution of h and X, and we write
Y(t)=h(t)X(t)=X(t)h(t).
Note that as the name suggests, the impulse response can be obtained if the input to the system is chosen to be the unit impulse function (delta function) x(t)=δ(t). For discrete-time systems, the output can be written as (Figure 10.6)
Y(n)=h(n)X(n)=X(n)h(n)=k=h(k)X(nk)=k=X(k)h(nk).
The discrete-time unit impulse function is defined as
δ(n)={1n=00otherwise
For the rest of this chapter, we mainly focus on continuous-time signals.
LTI
Figure 10.6 - LTI systems.

LTI Systems with Random Inputs:

Consider an LTI system with impulse response h(t). Let X(t) be a WSS random process. If X(t) is the input of the system, then the output, Y(t), is also a random process. More specifically, we can write
Y(t)=h(t)X(t)=h(α)X(tα)dα.
Here, our goal is to show that X(t) and Y(t) are jointly WSS processes. Let's first start by calculating the mean function of Y(t), μY(t). We have
μY(t)=E[Y(t)]=E[h(α)X(tα)dα]=h(α)E[X(tα)]dα=h(α)μXdα=μXh(α)dα.
We note that μY(t) is not a function of t, so we can write
μY(t)=μY=μXh(α)dα.
Let's next find the cross-correlation function, RXY(t1,t2). We have
RXY(t1,t2)=E[X(t1)Y(t2)]=E[X(t1)h(α)X(t2α)dα]=E[h(α)X(t1)X(t2α)dα]=h(α)E[X(t1)X(t2α)]dα=h(α)RX(t1,t2α)dα=h(α)RX(t1t2+α)dα(since X(t) is WSS).
We note that RXY(t1,t2) is only a function of τ=t1t2, so we may write
RXY(τ)=h(α)RX(τ+α)dα=h(τ)RX(τ)=h(τ)RX(τ).
Similarly, you can show that
RY(τ)=h(τ)h(τ)RX(τ).
This has been shown in the Solved Problems section. From the above results we conclude that X(t) and Y(t) are jointly WSS. The following theorem summarizes the results.

Theorem
Let X(t) be a WSS random process and Y(t) be given by
Y(t)=h(t)X(t),
where h(t) is the impulse response of the system. Then X(t) and Y(t) are jointly WSS. Moreover,
  1. μY(t)=μY=μXh(α)dα;
  2. RXY(τ)=h(τ)RX(τ)=h(α)RX(tα)dα;
  3. RY(τ)=h(τ)h(τ)RX(τ).

Frequency Domain Analysis:

Let's now rewrite the statement of Theorem 10.2 in the frequency domain. Let H(f) be the Fourier transform of h(t),
H(f)=F{h(t)}=h(t)e2jπftdt.
H(f) is called the transfer function of the system. We can rewrite
μY=μXh(α)dα
as
μY=μXH(0)
Since h(t) is assumed to be a real signal, we have
F{h(t)}=H(f)=H(f),
where shows the complex conjugate. By taking the Fourier transform from both sides of RXY(τ)=RX(τ)h(τ), we conclude
SXY(f)=SX(f)H(f)=SX(f)H(f).
Finally, by taking the Fourier transform from both sides of RY(τ)=h(τ)h(τ)RX(τ), we conclude
SY(f)=SX(f)H(f)H(f)=SX(f)|H(f)|2.
SY(f)=SX(f)|H(f)|2


Example
Let X(t) be a zero-mean WSS process with RX(τ)=e|τ|. X(t) is input to an LTI system with
|H(f)|={1+4π2f2|f|<20otherwise
Let Y(t) be the output.
  1. Find μY(t)=E[Y(t)].
  2. Find RY(τ).
  3. Find E[Y(t)2].
  • Solution
    • Note that since X(t) is WSS, X(t) and Y(t) are jointly WSS, and therefore Y(t) is WSS.
      1. To find μY(t), we can write
        μY=μXH(0)=01=0.
      2. To find RY(τ), we first find SY(f).
        SY(f)=SX(f)|H(f)|2.
        From Fourier transform tables, we can see that
        SX(f)=F{e|τ|}=21+(2πf)2.
        Then, we can find SY(f) as
        SY(f)=SX(f)|H(f)|2={2|f|<20otherwise
        We can now find RY(τ) by taking the inverse Fourier transform of SY(f).
        RY(τ)=8sinc(4τ),
        where
        sinc(f)=sin(πf)πf.
      3. We have
        E[Y(t)2]=RY(0)=8.




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