Properties of Systems

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 $x_1(t)⟼y_1(t)$ and $x_2(t)⟼y_2(t)$, then $$x(t)=\alpha x_1(t)+\beta x_2(t)⟼y(t)=\alpha y_1(t)+\beta y_2(t)$$ where, with no name specified, $⟼$ 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 $$a_0\frac{d^{N}y(t)}{d{t}^N}+a_1\frac{d^{N-1}y(t)}{d{t}^{N-1}}+\cdots +a_{N}y(t)=b_{N-M}\frac{d^{M}x(t)}{d{t}^M}+\cdots +b_{N-1}\frac{dx(t)}{dt}+b_{N}x(t)$$ 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.

# TODO

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: $y(t)=tx(t)$
• System B: $y(t)=10x(t)$

---

System A: Start with a delay of the input $x_d(t)=x(t-\delta )$ by some time $\delta$ $$\begin{array}{rl}& y(t)=tx(t)\\ & y_1(t)=t{x}_d(t)=tx(t-\delta )\end{array}$$

Now delay the output by $\delta$ $$\begin{array}{rl}& y(t)=tx(t)\\ & y_2(t)=y_d(t)=y(t-\delta )=(t-\delta )x(t-\delta )\end{array}$$

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

System B: Start with a delay of the input $x_d(t)=x(t-\delta )$ $$\begin{array}{rl}& y(t)=10x(t)\\ & y_1(t)=10x_d(t)=10x(t-\delta )\end{array}$$

Now delay the output by $\delta$ $$\begin{array}{rl}& y(t)=10x(t)\\ & y_2(t)=y_d(t)=y(t-\delta )=10x(t+\delta )\end{array}$$

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

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 $$y(t)=\underset{T\to 0}{lim}\frac{x(t)-x(t-T)}{T}$$

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

3. 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 $t\le t_0$.

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. $y(t)=tx(t)$, and

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

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

2. 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 $$y_2(t)=y_d(t)=y(t-\delta )=x(-(t-\delta ))=x(-t+\delta )$$ Since $y_{(}t)\ne y_2(t)$, 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.

4. The system described by $y(t)=tx(t-2)$ :

• Time-varying: Start with a delayed input $x_d(t)=x(t-\delta )$, the output is $$y_1(t)=t{x}_d(t-2)=tx(t-2-\delta ).$$ Now we delay the output by $\delta$, getting $$y_2(t)=y_d(t)=y(t-\delta )=(t-\delta )x(t-\delta -2)$$ Since $y_{(}t)\ne y_2(t)$, 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 $y(0)=0x(0-2)=0$ and $y(t)$ for any $t\ne 0$ 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.

5. 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 $$y_2(t)=y_d(t)=y(t-\delta )={\int }_{-5}^{5}x(\tau )d\tau$$ Since $y_{(}t)\ne y_2(t)$, 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 $|y(t)|=|{\int }_{-5}^{5}x(\tau )d\tau |<10K$, which is bounded.