Properties of Systems

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Sources:

  1. B. P. Lathi & Roger Green. (2018). Chapter 1: Signals and Systems. Signal Processing and Linear Systems (3rd ed., pp. 97-107). Oxford University Press.

Linearity

Linear systems have the property that if \(x_1(t) \longmapsto y_1(t)\) and \(x_2(t) \longmapsto y_2(t)\), then \[ x(t)=\alpha x_1(t)+\beta x_2(t) \longmapsto y(t)=\alpha y_1(t)+\beta y_2(t) \] where, with no name specified, \(\longmapsto\) denotes the system under test. This mathematical condition must be true for any choice of the constants \(\alpha\) and \(\beta\), and for all signals \(x_1[n]\) and \(x_2[n]\).

It's also called the principle of superposition.

show that a system described by a differential equation of the form \[ a_0 \frac{d^N y(t)}{d t^N}+a_1 \frac{d^{N-1} y(t)}{d t^{N-1}}+\cdots+a_N y(t)=b_{N-M} \frac{d^M x(t)}{d t^M}+\cdots+b_{N-1} \frac{d x(t)}{d t}+b_N x(t) \] is a linear system. The coefficients \(a_i\) and \(b_i\) in this equation can be constants or functions of time. Although here we proved only zero-state linearity, it can be shown that such systems are also zero-input linear and have the decomposition property.

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Time-Invariant property

A system is said to be time-invariant if, when an input is delayed (shifted) by \(t_0\), the output is delayed by the same amount. The condition must be true for any signal \(x(t)\) and for any choice of \(t_0\).

To demonstrate how to determine if a system is time-invariant, consider the two systems:

  • System A: \(y(t)=t x(t)\)
  • System B: \(y(t)=10 x(t)\)

System A: Start with a delay of the input \(x_d(t)=x(\color{brown}{t - \delta})\) by some time \(\delta\) \[ \begin{aligned} & y(t)=t x(t) \\ & y_1(t)=t x_d(t)=t x(\color{brown}{t - \delta}) \end{aligned} \]

Now delay the output by \(\delta\) \[ \begin{aligned} & y(t)=t x(t) \\ & y_2(t)= y_d(t) = y(\color{brown}{t - \delta})=(\color{brown}{t-\delta}) x(\color{brown}{t-\delta}) \end{aligned} \]

Clearly \(y_1(t) \neq y_2(t)\), therefore the system is not time-invariant.

System B: Start with a delay of the input \(x_d(t)=x(\color{brown}{t-\delta})\) \[ \begin{aligned} & y(t)=10 x(t) \\ & y_1(t)=10 x_d(t)=10 x(\color{brown}{t-\delta}) \end{aligned} \]

Now delay the output by \(\delta\) \[ \begin{aligned} & y(t)=10 x(t) \\ & y_2(t)= y_d(t) = y(\color{brown}{t-\delta})=10 x({t+\delta}) \end{aligned} \]

Clearly \(y_1(t)=y_2(t)\), therefore the system is time-invariant.

Memoryless property

A system is said to be instantaneous (or memoryless) if its output at any instant \(t\) depends, at most, on the strength of its input(s) at the same instant \(t\), and not on any past or future values of the input(s). Otherwise, the system is said to be dynamic (or a system with memory).

A system whose response at \(t\) is completely determined by the input signals over the past \(T\) seconds (interval from \((t-T)\) to \(t\)) is a finite-memory system with a memory of \(T\) seconds.

Examples

Determine whether the following systems are memoryless:

  1. \(y(t-1)=2 x(t-1)\),

  2. \(y(t)=\frac{d}{d} x(t)\), and

  3. \(y(t)=(t-1) x(t)\).

  1. In this case, the output at time \(t-1\) is just twice the input at the same time \(t-1\). Since the output at a particular time depends only on the strength of the input at the same time, the system is memoryless.

  2. Although it appears that the output \(y(t)\) at time \(t\) depends on the input \(x(t)\) at the same time \(t\), we know that the slope (derivative) of \(x(t)\) cannot be determined solely from a single point. There must be some memory, even if infinitesimally small, involved. This is confirmed by using the fundamental theorem of calculus to express the system as \[ y(t)=\lim _{T \rightarrow 0} \frac{x(t)-x(t-T)}{T} \]

Since the output at a particular time depends on more than just the input at the same time, the system is not memoryless.

  1. The output \(y(t)\) at time \(t\) is just the input \(x(t)\) at the same time \(t\) multiplied by the (time-dependent) coefficient \(t-1\). Since the output at a particular time depends only on the strength of the input at the same time, the system is memoryless.

Causality

A causal system is one for which the output at any instant \(t_0\) depends only on the value of the input \(x(t)\) for \(t \leq t_0\).

In other words, the value of the output at the present instant depends only on the past and present values of the input \(x(t)\), not on its future values. To put it simply, in a causal system the output cannot start before the input is

Examples

The system \(y(t)=x(-t)\) is

  1. dynamic: For example, the output at time 2, \(y(2)\), is not determined by the input at time 2, \(x(2)\).
  2. noncausal: For example, the output at time -2, \(y(-2)\), is \(x(2)\), i.e., it depends on a future input value.

Invertiblity

If we can obtain the input \(x(t)\) back from the corresponding output \(y(t)\) by some operation, the system \(\mathcal{S}\) is said to be invertible.

Examples

Determine whether the following systems are invertible:

  1. \(y(t)=x(-t)\),

  2. \(y(t)=t x(t)\), and

  3. \(y(t)=\frac{d}{d t} x(t)\).

  4. It's invertible. We can always recover the input \(x(t)\) by \(x(t) = y(-t)\).

  5. It's not invertible. Since we can't recover \(x(t)\) by \(x(t)=\frac{1}{t} y(t)\) at \(t=0\).

  6. It's not invertible. Since differentiation eliminates any constant component. For example, the inputs \(x_1(t)=1\) and \(x_2(t)=2\) both produce the same output \(y(t)=0\). Given only \(y(t)=0\), it is impossible to know if the original input was \(x_1(t)=1, x_2(t)=2\), or something else entirely.

Examples

  1. The system described by \(y(t)=x(-t)\) is:

  • Time-varying: Start with a delayed input \(x_d(t) = x(t-\delta)\), the output is \[ y_1(t) = x_d(-t) = x(-t-\delta) / \] {.} Now we delay the output by \(\delta\), getting \[ y_2(t) = y_d(t) = y(\color{brown}{t-\delta}) = x(-(\color{brown}{t-\delta})) = x(-t + \delta) \] Since \(y_(t) \neq y_2(t)\), the system is time-varying.

  • Dynamic: For example, the output at time \(2, y(2)\), is not determined by the input at time 2 .

  • Noncausal: For example, the output at time -2, y(-2), is x(2), i.e., it depends on a future input value.

  • Invertible: Since \(y(t)=x(-t)\), we can recover \(x(t)\) from \(y(t)\) using \(x(t)=y(-t)\).

  • BIBO-stable: If \(x(t)\) is bounded, i.e., there is \(K\) such that \(|x(t)|<K\) for all \(t\), then \(|y(t)|=|x(-t)|<K\) for all \(t\), i.e., \(y(t)\) is bounded.

  1. The system described by \(y(t)=t x(t-2)\) :

  • Time-varying: Start with a delayed input \(x_d(t) = x(t-\delta)\), the output is \[ y_1(t) = tx_d(t-2) = t x(t-2 - \delta) . \] Now we delay the output by \(\delta\), getting \[ y_2(t) = y_d(t) = y(\color{brown}{t-\delta}) = (\color{brown}{t-\delta})x(\color{brown}{t-\delta} -2) \] Since \(y_(t) \neq y_2(t)\), the system is time-varying.

  • Dynamic: For example, the output at time \(3, y(3)\), is not determined by the input at time 3.

  • Causal: The output at time \(t, y(t)\), does not depend on any future input value.

  • Noninvertible: The input at time \(2, x(-2)\), cannot be determined from the output \(y(t)\) since \(y(0)=0 x(0-2)=0\) and \(y(t)\) for any \(t \neq 0\) does not depend on \(x(-2)\). Hence, the input at time -2 cannot be recovered from the output \(y(t)\).

  • BIBO-unstable: For example, if \(x(t)=1\) for all \(t\), which is bounded, \(y(t)=t\), which is unbounded.

  1. The system described by \(y(t)=\int_{-5}^5 x(\tau) d \tau\) :

  • Time-varying:

    Start with a delayed input \(x_d(t) = x(t-\delta)\), the output is \[ y_1(t) = \int_{-5}^5 x_d(\tau) d \tau = \int_{-5}^5 x(\tau - \delta) d \tau = \int_{-5-\delta}^{5-\delta} x(\tau) d \tau \] Now we delay the output by \(\delta\), getting \[ y_2(t) = y_d(t) = y(\color{brown}{t-\delta}) = \int_{-5}^5 x(\tau) d \tau \] Since \(y_(t) \neq y_2(t)\), the system is time-varying.

  • Dynamic: The output at time 0 depends on the input at time 3.

  • Noncausal: The output at time 0 depends on the input at time 3 , which is a future input value.

  • Noninvertible: Different inputs give the same outputs if their integrals from -5 to 5 are the same.

  • BIBO-stable: If \(x(t)\) is bounded, i.e., there is \(K\) such that \(|x(t)|<K\) for all \(t\), then \(|y(t)|=\left|\int_{-5}^5 x(\tau) d \tau\right|<10 K\), which is bounded.