15.2: Input- Output Stability

At this point, it is important to make a connection between the stability of a system and its input-output behavior. The most important notion is that of `\(l_{p}\)-stability (\(p\)-stability).

Definition 15.1

A system with input signal \(u\) and output signal \(y\) that is obtained from \(u\) through the action of an arbitrary operator \(H\), so \(y = H(u)\), is `\(l_{p}\)-stable or \(p\)-stable \((p = 1, 2, \infty)\) if there exists a finite \(C \in \mathbb{R}\) such that

\[\|y\|_{p} \leq C\|u\|_{p} \ \tag{15.16}\]

for every input \(u\).

A \(p\)-stable system is therefore characterized by the requirement that every input of finite \(p\)-norm gives rise to an output of finite \(p\)-norm. For the case \(p =\infty\), this notion is known as Bounded-Input Bounded-Output (BIBO) stability. We will see that BIBO stability is equivalent to \(p\)-stability for finite-dimensional LTI state-space systems, but not necessarily in other cases.

Example 15.4

The system described by one integrator:

\[\dot{y}=u\nonumber\]

is not BIBO stable. A step input is mapped to a ramp which is unbounded. It is not hard to see that this system is not \(p\)-stable for any \(p\).

15.3.1 BIBO Stability of LTI Systems

A continuous-time LTI system may be characterized by its impulse response matrix, \(\mathcal{H}(\cdot)\), whose \((i, j)\)th entry \(h_{ij} ( \cdot )\) is the impulse response from the \(j\)th input to the \(i\)th output. In other words the input-output relation is given by

\[y(t)=\int \mathcal{H}(t-\tau) u(\tau) d \tau\nonumber\]

Theorem 15.1

A CT LTI system with \(m\) inputs, \(p\) outputs, and impulse response matrix \(\mathcal{H}(t)\) is BIBO stable if and only if

\[\max _{1 \leq i \leq p} \sum_{j=1}^{m} \int\left|h_{i j}(t)\right| d t<\infty\nonumber\]

Proof

The proof of sufficiency involves a straightforward computation of bounds. If \(u\) is an input signal that satisfies \(\|u\|_{\infty}<\infty\), i.e. a bounded signal, then we have

\[y(t)=\int \mathcal{H}(t-\tau) u(\tau) d \tau\nonumber\]

and

\[\begin{aligned}
\max _{1 \leq i \leq p}\left|y_{i}(t)\right| &=\max _{i}\left|\int \sum_{j=1}^{m} h_{i j}(t-\tau) u_{j}(\tau) d \tau\right| \\
& \leq\left[\max _{i} \int \sum_{j}\left|h_{i j}(t-\tau)\right| d \tau\right] \max _{j} \sup _{t}\left|u_{j}(t)\right|
\end{aligned}\nonumber\]

It follows that

\[\|y\|_{\infty}=\sup _{t} \max _{i}\left|y_{i}(t)\right| \leq\left[\max _{i} \sum_{j} \int\left|h_{i j}(t)\right| d t\right]\|u\|_{\infty}<\infty\nonumber\]

In order to prove the converse of the theorem, we show that if the above integral is infinite then there exists a bounded input that will be mapped to an unbounded output. Let us consider the case when \(p = m = 1\), for notational simplicity (in the general case, we can still narrow the focus to a single entry of the impulse response matrix). Denote the impulse response by \(h(t)\) for this scalar case. If the integral

\[\int|h(t)| d t\nonumber\]

is unbounded then given any (large) \(M\) there exists an interval of length \(2T\) such that

\[\int_{-T}^{T}|h(t)| d t>M\nonumber\]

Now by taking the input \(u_{M}(t)\) as

\[u_{M}(t)=\left\{\begin{array}{ll}
\operatorname{sgn}(h(-t)) & -T \leq t \leq T \\
0 & |t|>T
\end{array}\right.\nonumber\]

we obtain an output \(y_{M}(t)\) that satisfies

\[\begin{aligned}
\sup _{t}\left|y_{M}(t)\right| \geq y_{M}(0) &=\int_{-T}^{T} h(0-\tau) u_{M}(\tau) d \tau \\
&=\int_{-T}^{T}|h(0-\tau)| d \tau \\
&>M
\end{aligned}\nonumber\]

In other words, for any \(M > 0\), we can have an input whose maximum magnitude is 1 and whose corresponding output is larger than \(M\). Therefore, there is no finite constant \(C\) such that the inequality (24.3) holds.

Further reflection on the proof of Theorem 15.1 reveals that the constant \(\|\mathcal{H}\|_{1}\) defined by

\[\|\mathcal{H}\|_{1}=\max _{i} \sum_{j} \int\left|h_{i j}(t)\right| d t\nonumber\]

is the smallest constant \(C\) that satisfies the inequalty (24.3) when \(p = \infty\). This number is called the `\(l_{1}\)-norm of \(\mathcal{H}(t)\). In the scalar case, this number is just the `\(l_{1}\)-norm of \(h( \cdot )\), regarded as a signal.

The discrete-time case is quite similar to continuous-time where we start with a pulse response matrix, \(\mathcal{H}(\cdot)\), whose \((i, j)\)th entry \(h_{ij} ( \cdot )\) is the pulse response from the \(j\)th input to the \(i\)th output. The input-output relation is given by

\[y(t)=\sum_{\tau} \mathcal{H}(t-\tau) u(\tau)\nonumber\]

Theorem 15.2

A DT LTI system with \(m\) inputs, \(p\) outputs, and pulse response matrix \(\mathcal{H}(t)\) is BIBO stable if and only if

\[\max _{1 \leq i \leq p} \sum_{j=1}^{m} \sum_{t}\left|h_{i j}(t)\right|<\infty\nonumber\]

In addition, the constant \(\|\mathcal{H}\|_{1}\) defined by

\[\|\mathcal{H}\|_{1}=\max _{i} \sum_{j} \sum_{t}\left|h_{i j}(t)\right|\nonumber\]

is the smallest constant \(C\) that satisfies the inequalty (24.3) when \(p = \infty\). We leave the proof of these facts to the reader.

Application to finite-dimensional State-Space Models

Now consider the application to the following causal CT LTI system in state-space form (and hence of finite order) :

\[\begin{aligned}
\dot{x} &=A x+B u \ (15.17) \\
y &=C x+D u \ (15.18)
\end{aligned}\nonumber\]

The impulse response of this system is given by

\[\mathcal{H}(t)=C e^{A t} B+D \delta(t) \text { for } t \geq 0\nonumber\]

which has Laplace transform

\[H(s)=C(s I-A)^{-1} B+D\nonumber\]

The system (15.18) is BIBO stable if and only if the poles of \(H(s)\) are in the open left half plane. (We leave the proof to you.) This is in turn guaranteed if the system is asymptotically stable, i.e. if \(A\) has all its eigenvalues in the open left half plane.

Example 15.5 BIBO Stability Doesn't Imply Asymptotic Stability

It is possible that a system be BIBO stable and not asymptotically stable. Consider the system

\[\begin{array}{l}
\dot{x}=\left(\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}\right) x+\left(\begin{array}{l}
0 \\
1
\end{array}\right) u \\
y=\left(\begin{array}{cc}
1 & -1
\end{array}\right) x
\end{array}\nonumber\]

This system is not stable since \(A\) has an eigenvalue at 1. Nevertheless, thanks to a pole-zero cancellation, the only pole that \(H(s)\) has is at \(-1\), so the system is BIBO stable. We shall have much more to say about such cancellations in the context of reachability, observability, and minimality (the example here turns out to be unobservable).

Marginal stability of an LTI system, i.e., stability in the sense of Lyapunov but without asymptotic stability, is not sufficient to guarantee BIBO stability. For instance, consider a simple integrator, whose transfer function is \(1/s\).

Time-Varying and Nonlinear Systems

Although there are results connecting Lyapunov stability with I/O stability for general time-varying and nonlinear systems, they are not as powerful as the linear time-invariant case. In particular, systems may be I/O stable with respect to one norm and not stable with respect to another. Below are some examples illustrating these facts.

Example 15.6 A Time- Varying System

Consider the time-varying DT system given by:

\[y(t)=H(u)(t)=u\tag{0}\]

\(H\) is obviously 1-stable with gain less than 1. However, it is not 2-stable.

Example 15.7 A Nonlinear System

Consider the nonlinear system given by:

\[\dot{x}=-x+e^{x} u, \quad y=x\nonumber\]

The unforced system is linear and is asymptotically stable. On the other hand the system is not I/O stable. To see this, consider the input \(u(t) = 1\). Since \(e^{x}>x, \dot{x}\) is always strictly positive, indicating that \(x\) is strictly increasing. Hence, for a bounded input, the output is not bounded.

15.3.2 p-Stability of LTI Systems (optional)

In this section we will continue our analysis of the p-stability of systems described through input-output relations. Let us start with the continuous-time case, and restrict ourselves to single-input single-output. The input \(u(t)\) is related to the output \(y(t)\) by

\[y(t)=\int h(t-\tau) u(\tau) d \tau\nonumber\]

where \(h(t)\) is the impulse response. The following theorem shows that the constant \(C\) in 24.3 is always bounded above by \(\|h\|_{1}\).

Theorem 15.3

If \(\|h\|_{1}<\infty\) and \(\|u\|_{p}<\infty\) then \(\|y\|_{p}<\infty\) and furthermore

\[\|y\|_{p} \leq\|h\|_{1}\|u\|_{p}\nonumber\]

Proof

In Theorem 15.1 we have already established this result for \(p = \infty\). In what follows \(p = 1, 2\). The output \(y(t)\) satisfies

\[|y(t)|^{p}=|(h * u)(t)|^{p}=\left|\int_{-\infty}^{\infty} h(t-\tau) u(\tau) d \tau\right|^{p} \leq\left(\int_{-\infty}^{\infty}|h(t-\tau)||u(\tau)| d \tau\right)^{p}\nonumber\]

therefore,

\[\|h * u\|_{p}^{p}=\int_{-\infty}^{\infty}|(h * u)(t)|^{p} d t \leq \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}|h(t-\tau)||u(\tau)| d \tau\right)^{p} d t\nonumber\]

Next we analyze the inner integral

\[\begin{aligned}
\int_{-\infty}^{\infty}|h(t-\tau)||u(\tau)| d \tau &=\int_{-\infty}^{\infty}|h(t-\tau)|^{1 / q}|h(t-\tau)|^{1 / p}|u(\tau)| d \tau \\
& \leq\left(\int_{-\infty}^{\infty}|h(t-\tau)| d \tau\right)^{1 / q}\left(\int_{-\infty}^{\infty}|h(t-\tau)||u(\tau)|^{p} d \tau\right)^{1 / p}
\end{aligned}\nonumber\]

where the last inequality follows from Minkowski's inequalities, and \(\frac{1}{p}+\frac{1}{q}=1\). Hence,

\[\begin{aligned}
\|h * u\|_{p}^{p} & \leq \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}|h(t-\tau)| d \tau\right)^{p / q}\left(\int_{-\infty}^{\infty}|h(t-\tau)||u(\tau)|^{p} d \tau\right) d t \\
&=\int_{-\infty}^{\infty}\left(\|h\|_{1}\right)^{p / q}\left(\int_{-\infty}^{\infty}|h(t-\tau)||u(\tau)|^{p} d \tau\right) d t \\
&=\|h\|_{1}^{p / q} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}|h(t-\tau)||u(\tau)|^{p} d \tau d t \\
&=\|h\|_{1}^{p / q} \int_{-\infty}^{\infty}|u(\tau)|^{p}\left(\int_{-\infty}^{\infty}|h(t-\tau)| d t\right) d \tau \\
&=\|h\|_{1}^{p / q+1} \int_{-\infty}^{\infty}|u(\tau)|^{p} d \tau \\
&=\|h\|_{1}^{p}\|u\|_{p}^{p}
\end{aligned}\nonumber\]

Therefore

\[\|h * u\|_{p} \leq\|h\|_{1}\|u\|_{p}\nonumber\]

Recall that when \(p=\infty,\|h\|_{1}\) was the smallest constant for which the inequality \(\|y\|_{p} \leq C\|u\|_{p}\) for all \(u\). This is not the case for \(p = 2\), and we will see later that a smaller constant can be found. We will elaborate on these issues when we discuss systems' norms later on in the course. The discrete-time case follows in exactly the same fashion.

Example 15.8

For a finite-dimensional state-space model, a system H is p-stable if and only if all the poles of of H(s) are in the LHP. This coincides with BIBO stability.