Steady-State Performance of Sampled Data Systems

What is the steady-state step response of a SD system where the SD system is closed-loop stable and $T>0$ is non-pathological?

Theorem

If $T>0$ is non-pathological and the SD system is closed-loop stable. Let $r(t)=1(t)$. Then $y_{ss}={\lim }_{t\to \infty }=T_{ry}[1]$.

Proof.

• Given: SD system is closed-loop stable, $T>0$ and non-pathological
• WTS: $y_{ss}={\lim }_{t\to \infty }=T_{ry}[1]$

Then:

• DT system is closed-loop stable [theorem from class]
• $T_{ru}[z]$ is BIBO stable [def. of DT closed-loop stability]
• $T_{rx}[z]$ is BIBO stable [lemma from class]
• $T_{ru}[z],T_{ry}[z]$ are stable [theorem from class]
• $u_{ss}={\lim }_{k\to \infty }u[k]=T_{ru}[1]$ [FVT corollary]
• $x_{ss}={\lim }_{k\to \infty }x[k]=T_{rx}[1]$

Recall that the system can be given as:

$$\begin{array}{r}x[k+1]=A_{d}x[k]+B_{d}u[k]\\ y[k]=Cx[k]\end{array}$$

Taking the limit of the first equation as $k\to \infty$:

$$\begin{array}{rl}{x}_{ss}& =A_d{x}_{ss}+B_d{u}_{ss}\\ (I-A_d)x_{ss}& =B_d{u}_{ss}\\ x_{ss}& =(I-A_d{)}^{-1}{B}_d{u}_{ss}\\ & =(I-e^{At}{)}^{-1}{A}^{-1}(e^{AI-I})B{u}_{ss}\\ & =-(I-e^{At}{)}^{-1}\cdot (I-e^{At})\cdot A^{-1}B{u}_{ss}\\ x_{ss}& =-A^{-1}B{u}_{ss}\end{array}$$

Recall that for $t\in [0,T]$, we have $x(kT+t)=e^{At}x[k]+A^{-1}(e^{At}-I)Bu[k]$. Taking the limit as $kT+t\to \infty$ which is equivalent to $k\to \infty$:

$$x(kT+t)\to e^{At}{x}_{ss}+A^{-1}(e^{At}-I)B{u}_{ss}=e^{At}(-A^{-1}B{u}_{ss})+A^{-1}(e^{At}-I)B{u}_{ss}$$

such that

$$x_{ss}=-A^{-1}{e}^{At}B{u}_{ss}+A^{-1}{e}^{At}B{u}_{ss}-A^{-1}B{u}_{ss}$$

so that the limit as continuous time approaches infinity is

$$\underset{\hat{t}\to \infty }{\lim }x(\hat{t})=x_{ss}$$

Now proving for $y$:

$$\begin{array}{rl}{y}_{ss}& =\underset{\hat{t}\to \infty }{\lim }y(\hat{t})=\underset{\hat{t}\to \infty }{\lim }(x(\hat{t}))=C{x}_{ss}\\ y_{ss}& =C{A}^{-1}B{u}_{ss}\\ & =-C{A}^{-1}B{T}_{ru}[1]\end{array}$$

We have:

$$C{A}^{-1}B=-\left[\begin{array}{ccc}{c}_1& \dots & c_n\end{array}\right]\left[\begin{array}{ccc}\frac{1}{{\lambda }_1}& & \\ & \ddots & \\ & & \frac{1}{{\lambda }_n}\end{array}\right]\left[\begin{array}{c}1\\ ⋮\\ 1\end{array}\right]=\sum _{i=1}^n\frac{c_i}{{\lambda }_i}=P(0)=G[1]$$

Thus:

$$y_{ss}=G[1]T_{ru}[1]=T_{ry}[1]$$

Another takeaway:

$$\begin{array}{rl}P(s)& =\dot{x}=Ax+Bu\\ \underset{\hat{t}\to \infty }{\lim }A{x}_{ss}+B{u}_{ss}& =A(-A^{-1}B{u}_{ss})+B{u}_{ss}\\ & =-B{u}_{ss}+B{u}_{ss}\\ & =0\end{array}$$

Thus, $(x_{ss},u_{ss})$ is an equilibrium point for the plant $P(s)$.

Sinusoidal Reference

The above was for a reference $r[k]=1[k]$. Consider now a sinusoidal reference:

$$\begin{array}{rl}r(t)& =\cos ({\omega }_{0}t)\\ r[k]& =\cos ({\omega }_{0}kT)=\cos (k{\omega }_{0}T)=\cos (k\theta )\end{array}$$

In steady-state

$$y[k]⟶\left|T_{ry}[e^{j{\omega }_{0}T}]\right|\cos (k\theta +\angle T_{ry}[e^{j{\omega }_{0}T}])$$

This shows what is happening at the sample points, but what happens between the sample points?

Natural conjecture: in steady state, $y(t)\to \left|T_{ry}[e^{j{\omega }_{0}T}]\right|\cos ({\omega }_{0}t+\angle T_{ry}[e^{j{\omega }_{0}T}])$. But this is incorrect! As $T\to 0$, this becomes closer to correct, but in general this is pretty bad.

Goal: develop a better approximation of the output by making a Fourier-esque expression.

Define $\{{\omega }_n{\}}_{n=0}^{\infty }$ by

$$\begin{array}{rl}{\omega }_0& ={\omega }_0\\ {\omega }_1& ={\omega }_s-{\omega }_0\\ {\omega }_2& ={\omega }_s+{\omega }_0\\ {\omega }_3& =2{\omega }_s-{\omega }_0\\ {\omega }_4& =2{\omega }_s+{\omega }_0\\ & ⋮\\ & =n{\omega }_s-{\omega }_0\\ & =n{\omega }_s+{\omega }_0\end{array}$$

In steady-state:

$$y(t)⟶\frac{1}{T}\sum _{n=0}^{\infty }\left|T_{ru}(e^{j{\omega }_{0}T})\right|\left|P(j{\omega }_n)\right|\left|H(j{\omega }_n)\right|\cos ({\omega }_{n}t+\angle T_{ru}[e^{j{\omega }_{0}T}]+\angle P(j{\omega }_n)+\angle H(j{\omega }_n))$$

This is the true response of a sampled data system to sinusoidal input. The output is an infinite sum, not a true Fourier series, which causes wacky behavior sometimes.

At $N=0$, it gives the output of a true LTI system (exactly at the sample points)

• The frequency content of $y(t)$ is at ${\omega }_0,n{\omega }_s\pm {\omega }_0,\forall n\ge 1$

Sampling Time

We can use this formula to investigate sampling time as well – under what conditions can we collapse this infinite sum down to just the $N=0$ term?

Define the Nyquist Frequency ${\omega }_n=\frac{{\omega }_s}{2}=\frac{2\pi }{T}\frac{1}{2}=\frac{\pi }{T}$.

Assume:

1. The plant is bandwidth-limited with ${\omega }_{bw}\le {\omega }_N$, such that $\left|P(j\omega )\right|\approx 0$ for $\omega >{\omega }_{bw}$ (attenuation is perfect above ${\omega }_{bw}$)
2. ${\omega }_0<{\omega }_n$ (sampling frequency is at least twice the external frequency)

We saw above that the frequency content of $y(t)$ is at ${\omega }_0,n{\omega }_s\pm {\omega }_0$.

For any $n\ge 1$:

$$\begin{array}{rlr}n{\omega }_s\pm {\omega }_0& \ge n{\omega }_s-{\omega }_0& \\ & >n{\omega }_s-{\omega }_n& \text{[via assumption 2]}\\ & \ge {\omega }_s-{\omega }_n& \\ n{\omega }_s\pm {\omega }_0& \ge {\omega }_n& \end{array}$$

Thus, for any $n\ge 1$, $n{\omega }_s\pm {\omega }_0>{\omega }_{bw}$.

Thus:

$$\left|P(j{\omega }_n)\right|\approx 0\forall n\ge 1⟹y(t)=\text{ its n=0 term}$$

as if $y(t)$ were perfectly LTI.

If ${\omega }_n\ge {\omega }_{bw}$, we have ${\omega }_n>0$, such that $y(t)$ is approximately sinusoidal with frequency ${\omega }_0$.

Then:

$$\begin{array}{rl}& {\omega }_n\ge \text{max}({\omega }_{bw},{\omega }_0)\\ & ⟹\frac{1}{2}{\omega }_s\ge \text{max}({\omega }_{bw},{\omega }_0)\\ & ⟹{\omega }_s\ge 2\text{max}({\omega }_{bw},{\omega }_0)\end{array}$$

In practice, $\left|P(j\omega )\right|\ne 0$ for all $\omega >{\omega }_{bw}$, so we add a safety factor and choose

$${\omega }_s\ge (5-10)\times \text{max}\{{\omega }_{bw},{\omega }_0\}$$

• This is where the rule for choosing sampling time came from!