Linear time-invariant system

The behavior of a linear, continuous-time, time-invariant system with input signal x(t) and output signal y(t) is described by the convolution integral:[3]

${\displaystyle y(t)=(x*h)(t)}$	${\displaystyle \mathrel {\stackrel {\mathrm {def} }{=}} \int \limits _{-\infty }^{\infty }x(t-\tau )\cdot h(\tau )\,\mathrm {d} \tau }$
	${\displaystyle =\int \limits _{-\infty }^{\infty }x(\tau )\cdot h(t-\tau )\,\mathrm {d} \tau ,}$       (using commutativity)

where ${\textstyle h(t)}$ is the system's response to an impulse: ${\textstyle x(\tau )=\delta (\tau )}$. ${\textstyle y(t)}$ is therefore proportional to a weighted average of the input function ${\textstyle x(\tau )}$. The weighting function is ${\textstyle h(-\tau )}$, simply shifted by amount ${\textstyle t}$. As ${\textstyle t}$ changes, the weighting function emphasizes different parts of the input function. When ${\textstyle h(\tau )}$ is zero for all negative ${\textstyle \tau }$, ${\textstyle y(t)}$ depends only on values of ${\textstyle x}$ prior to time ${\textstyle t}$, and the system is said to be causal.

To understand why the convolution produces the output of an LTI system, let the notation ${\textstyle \{x(u-\tau );\ u\}}$ represent the function ${\textstyle x(u-\tau )}$ with variable ${\textstyle u}$ and constant ${\textstyle \tau }$. And let the shorter notation ${\textstyle \{x\}}$ represent ${\textstyle \{x(u);\ u\}}$. Then a continuous-time system transforms an input function, ${\textstyle \{x\},}$ into an output function, ${\textstyle \{y\}}$. And in general, every value of the output can depend on every value of the input. This concept is represented by:

$${\displaystyle y(t)\mathrel {\stackrel {\text{def}}{=}} O_{t}\{x\},}$$

where ${\textstyle O_{t}}$ is the transformation operator for time ${\textstyle t}$. In a typical system, ${\textstyle y(t)}$ depends most heavily on the values of ${\textstyle x}$ that occurred near time ${\textstyle t}$. Unless the transform itself changes with ${\textstyle t}$, the output function is just constant, and the system is uninteresting.

For a linear system, ${\textstyle O}$ must satisfy Eq.1:

${\displaystyle O_{t}\left\{\int \limits _{-\infty }^{\infty }c_{\tau }\ x_{\tau }(u)\,\mathrm {d} \tau$ ;\ u\right\}=\int \limits _{-\infty }^{\infty }c_{\tau }\ \underbrace {y_{\tau }(t)} _{O_{t}\{x_{\tau }\}}\,\mathrm {d} \tau .}

(Eq.2)

And the time-invariance requirement is:

${\displaystyle {\begin{aligned}O_{t}\{x(u-\tau );\ u\}&\mathrel {\stackrel {\quad }{=}} y(t-\tau )\\&\mathrel {\stackrel {\text{def}}{=}} O_{t-\tau }\{x\}.\,\end{aligned}}}$

(Eq.3)

In this notation, we can write the impulse response as ${\textstyle h(t)\mathrel {\stackrel {\text{def}}{=}} O_{t}\{\delta (u);\ u\}.}$

Similarly:

${\displaystyle h(t-\tau )}$	${\displaystyle \mathrel {\stackrel {\text{def}}{=}} O_{t-\tau }\{\delta (u);\ u\}}$
	${\displaystyle =O_{t}\{\delta (u-\tau );\ u\}.}$       (using Eq.3)

Substituting this result into the convolution integral:

$${\displaystyle {\begin{aligned}(x*h)(t)&=\int _{-\infty }^{\infty }x(\tau )\cdot h(t-\tau )\,\mathrm {d} \tau \\[4pt]&=\int _{-\infty }^{\infty }x(\tau )\cdot O_{t}\{\delta (u-\tau );\ u\}\,\mathrm {d} \tau ,\,\end{aligned}}}$$

which has the form of the right side of Eq.2 for the case ${\textstyle c_{\tau }=x(\tau )}$ and ${\textstyle x_{\tau }(u)=\delta (u-\tau ).}$

Eq.2 then allows this continuation:

$${\displaystyle {\begin{aligned}(x*h)(t)&=O_{t}\left\{\int _{-\infty }^{\infty }x(\tau )\cdot \delta (u-\tau )\,\mathrm {d} \tau$$ ;\ u\right\}\\[4pt]&=O_{t}\left\{x(u);\ u\right\}\\&\mathrel {\stackrel {\text{def}}{=}} y(t).\,\end{aligned}}}

In summary, the input function, ${\textstyle \{x\}}$, can be represented by a continuum of time-shifted impulse functions, combined "linearly", as shown at Eq.1. The system's linearity property allows the system's response to be represented by the corresponding continuum of impulse responses, combined in the same way. And the time-invariance property allows that combination to be represented by the convolution integral.

The mathematical operations above have a simple graphical simulation.[4]

An eigenfunction is a function for which the output of the operator is a scaled version of the same function. That is,

$${\displaystyle {\mathcal {H}}f=\lambda f,}$$

where f is the eigenfunction and ${\displaystyle \lambda }$ is the eigenvalue, a constant.

The exponential functions ${\displaystyle Ae^{st}}$, where ${\displaystyle A,s\in \mathbb {C} }$, are eigenfunctions of a linear, time-invariant operator. A simple proof illustrates this concept. Suppose the input is ${\displaystyle x(t)=Ae^{st}}$. The output of the system with impulse response ${\displaystyle h(t)}$ is then

$${\displaystyle \int _{-\infty }^{\infty }h(t-\tau )Ae^{s\tau }\,\mathrm {d} \tau }$$

which, by the commutative property of convolution, is equivalent to

$${\displaystyle {\begin{aligned}\overbrace {\int _{-\infty }^{\infty }h(\tau )\,Ae^{s(t-\tau )}\,\mathrm {d} \tau } ^{{\mathcal {H}}f}&=\int _{-\infty }^{\infty }h(\tau )\,Ae^{st}e^{-s\tau }\,\mathrm {d} \tau \\[4pt]&=Ae^{st}\int _{-\infty }^{\infty }h(\tau )\,e^{-s\tau }\,\mathrm {d} \tau \\[4pt]&=\overbrace {\underbrace {Ae^{st}} _{\text{Input}}} ^{f}\overbrace {\underbrace {H(s)} _{\text{Scalar}}} ^{\lambda },\\\end{aligned}}}$$

where the scalar

$${\displaystyle H(s)\mathrel {\stackrel {\text{def}}{=}} \int _{-\infty }^{\infty }h(t)e^{-st}\,\mathrm {d} t}$$

is dependent only on the parameter s.

So the system's response is a scaled version of the input. In particular, for any ${\displaystyle A,s\in \mathbb {C} }$, the system output is the product of the input ${\displaystyle Ae^{st}}$ and the constant ${\displaystyle H(s)}$. Hence, ${\displaystyle Ae^{st}}$ is an eigenfunction of an LTI system, and the corresponding eigenvalue is ${\displaystyle H(s)}$.

It is also possible to directly derive complex exponentials as eigenfunctions of LTI systems.

Let's set ${\displaystyle v(t)=e^{i\omega t}}$ some complex exponential and ${\displaystyle v_{a}(t)=e^{i\omega (t+a)}}$ a time-shifted version of it.

${\displaystyle H[v_{a}](t)=e^{i\omega a}H[v](t)}$ by linearity with respect to the constant ${\displaystyle e^{i\omega a}}$.

${\displaystyle H[v_{a}](t)=H[v](t+a)}$ by time invariance of ${\displaystyle H}$.

So ${\displaystyle H[v](t+a)=e^{i\omega a}H[v](t)}$. Setting ${\displaystyle t=0}$ and renaming we get:

$${\displaystyle H[v](\tau )=e^{i\omega \tau }H[v](0)}$$

i.e. that a complex exponential ${\displaystyle e^{i\omega \tau }}$ as input will give a complex exponential of same frequency as output.

The eigenfunction property of exponentials is very useful for both analysis and insight into LTI systems. The one-sided Laplace transform

$${\displaystyle H(s)\mathrel {\stackrel {\text{def}}{=}} {\mathcal {L}}\{h(t)\}\mathrel {\stackrel {\text{def}}{=}} \int _{0}^{\infty }h(t)e^{-st}\,\mathrm {d} t}$$

is exactly the way to get the eigenvalues from the impulse response. Of particular interest are pure sinusoids (i.e., exponential functions of the form ${\displaystyle e^{j\omega t}}$ where ${\displaystyle \omega \in \mathbb {R} }$ and ${\displaystyle j\mathrel {\stackrel {\text{def}}{=}} {\sqrt {-1}}}$). The Fourier transform ${\displaystyle H(j\omega )={\mathcal {F}}\{h(t)\}}$ gives the eigenvalues for pure complex sinusoids. Both of ${\displaystyle H(s)}$ and ${\displaystyle H(j\omega )}$ are called the system function, system response, or transfer function.

The Laplace transform is usually used in the context of one-sided signals, i.e. signals that are zero for all values of t less than some value. Usually, this "start time" is set to zero, for convenience and without loss of generality, with the transform integral being taken from zero to infinity (the transform shown above with lower limit of integration of negative infinity is formally known as the bilateral Laplace transform).

The Fourier transform is used for analyzing systems that process signals that are infinite in extent, such as modulated sinusoids, even though it cannot be directly applied to input and output signals that are not square integrable. The Laplace transform actually works directly for these signals if they are zero before a start time, even if they are not square integrable, for stable systems. The Fourier transform is often applied to spectra of infinite signals via the Wiener–Khinchin theorem even when Fourier transforms of the signals do not exist.

Due to the convolution property of both of these transforms, the convolution that gives the output of the system can be transformed to a multiplication in the transform domain, given signals for which the transforms exist

$${\displaystyle y(t)=(h*x)(t)\mathrel {\stackrel {\text{def}}{=}} \int _{-\infty }^{\infty }h(t-\tau )x(\tau )\,\mathrm {d} \tau \mathrel {\stackrel {\text{def}}{=}} {\mathcal {L}}^{-1}\{H(s)X(s)\}.}$$

One can use the system response directly to determine how any particular frequency component is handled by a system with that Laplace transform. If we evaluate the system response (Laplace transform of the impulse response) at complex frequency s = jω, where ω = 2πf, we obtain |H(s)| which is the system gain for frequency f. The relative phase shift between the output and input for that frequency component is likewise given by arg(H(s)).

Some of the most important properties of a system are causality and stability. Causality is a necessity for a physical system whose independent variable is time, however this restriction is not present in other cases such as image processing.

A system is causal if the output depends only on present and past, but not future inputs. A necessary and sufficient condition for causality is

$${\displaystyle h(t)=0\quad \forall t<0,}$$

where ${\displaystyle h(t)}$ is the impulse response. It is not possible in general to determine causality from the two-sided Laplace transform. However when working in the time domain one normally uses the one-sided Laplace transform which requires causality.

A system is bounded-input, bounded-output stable (BIBO stable) if, for every bounded input, the output is finite. Mathematically, if every input satisfying

$${\displaystyle \ \|x(t)\|_{\infty }<\infty }$$

leads to an output satisfying

$${\displaystyle \ \|y(t)\|_{\infty }<\infty }$$

(that is, a finite maximum absolute value of ${\displaystyle x(t)}$ implies a finite maximum absolute value of ${\displaystyle y(t)}$), then the system is stable. A necessary and sufficient condition is that ${\displaystyle h(t)}$, the impulse response, is in L¹ (has a finite L¹ norm):

$${\displaystyle \|h(t)\|_{1}=\int _{-\infty }^{\infty }|h(t)|\,\mathrm {d} t<\infty .}$$

In the frequency domain, the region of convergence must contain the imaginary axis ${\displaystyle s=j\omega }$.

As an example, the ideal low-pass filter with impulse response equal to a sinc function is not BIBO stable, because the sinc function does not have a finite L¹ norm. Thus, for some bounded input, the output of the ideal low-pass filter is unbounded. In particular, if the input is zero for ${\displaystyle t<0}$ and equal to a sinusoid at the cut-off frequency for ${\displaystyle t>0}$, then the output will be unbounded for all times other than the zero crossings.

Given is an explicit linear system of differential equations in the form:

${\displaystyle {\begin{aligned}{\frac {dx(t)}{dt}}=A\,x(t)+b\,u(t),\,x(t=0)=x_{0}\end{aligned}}}$

with the state vector ${\displaystyle x(t)\in \mathbb {R} ^{n}}$, the system matrix ${\displaystyle A\in \mathbb {R} ^{n\times n}}$, the input ${\displaystyle u(t)\in \mathbb {R} }$, the input vector ${\displaystyle b\in \mathbb {R} ^{n}}$ and the initial condition ${\displaystyle x_{0}\in \mathbb {R} ^{n}}$. The solution consists of a homogeneous and a particulate part.

The homogeneous differential equation is obtained by setting the input equal to zero.

${\displaystyle {\begin{aligned}{\frac {dx(t)}{dt}}=A\,x(t),\,x(t=0)=x_{0}\end{aligned}}}$

This solution can now be described using a Taylor series representation:

${\displaystyle {\begin{aligned}x(t)=\phi (t)x_{0}=(E+\phi _{1}t+\phi _{2}t^{2}+...+\phi _{n}t^{n}+...)x_{0}\end{aligned}}}$

where ${\displaystyle E}$ is the unit matrix. Substituting this solution into the above equation, one obtains:

${\displaystyle {\begin{aligned}{\frac {d}{dt}}(\phi (t)x_{0})&=A\,\phi (t)\,x_{0}\\(\phi _{1}+2\,\phi _{2}\,t+...+n\,\phi _{n}\,t^{n-1}+...)x_{0}&=(A+A\phi _{1}t+A\phi _{2}t^{2}+...+A\phi _{n}t^{n}+...)\,x_{0}\end{aligned}}}$

Now the unknown matrices ${\displaystyle \phi _{n}}$ can be determined by comparing coefficients:

${\displaystyle {\begin{aligned}\phi _{1}&=A\\\phi _{2}&={\frac {1}{2}}A\,\phi _{1}={\frac {1}{2!}}A^{^{2}}\\\phi _{3}&={\frac {1}{3}}A\,\phi _{2}={\frac {1}{3!}}A^{3}\\&...\\\phi _{n}&={\frac {1}{n!}}A^{n}.\end{aligned}}}$

The following notation is commonly used for the fundamental matrix ${\displaystyle \phi _{n}}$:

${\displaystyle {\begin{aligned}\phi (t)=e^{At}=E+At+{\frac {1}{2!}}A^{2}t^{2}+{\frac {1}{3!}}A^{3}t^{3}+...+{\frac {1}{n!}}A^{n}t^{n}+...\end{aligned}}}$

Assuming ${\displaystyle u(t)\neq 0}$ and ${\displaystyle x_{0}=0}$, follows:

${\displaystyle {\begin{aligned}{\frac {d}{dt}}x(t)=A\,x(t)+b\,u(t)\end{aligned}}}$

The particulate solution is obtained in the form:

${\displaystyle {\begin{aligned}x_{p}(t)=\phi (t)\xi (t)=e^{At}\xi (t),\end{aligned}}}$

where ${\displaystyle \xi (t)}$ is an unknown function vector with ${\displaystyle \xi (0)=0}$. From the above two equations follows:

${\displaystyle {\begin{aligned}{\frac {d}{dt}}x_{p}(t)&=A\,x_{p}(t)+b\,u(t)\\\xi (t){\frac {d}{dt}}\phi (t)+\phi (t){\frac {d}{dt}}\xi (t)&=A\,x_{p}(t)+b\,u(t)\\A\,\phi (t)\xi (t)+\phi (t){\frac {d}{dt}}\xi (t)&=A\,x_{p}(t)+b\,u(t)\\A\,x_{p}(t)+\phi (t){\frac {d}{dt}}\xi (t)&=A\,x_{p}(t)+b\,u(t)\\\end{aligned}}}$

Thus ${\displaystyle \xi (t)}$ can be determined:

${\displaystyle {\begin{aligned}{\frac {d}{dt}}\xi (t)=\phi ^{-1}(t)bu(t)\\\end{aligned}}}$

One obtains by integration utilizing the properties of the fundamental matrix:

${\displaystyle {\begin{aligned}\phi (t)\xi (t)&=\phi (t)\int _{0}^{t}\phi ^{-1}(\tau )bu(\tau )d\tau \\x_{p}(t)&=\int _{0}^{t}\phi (t-\tau )bu(\tau )d\tau \\x_{p}(t)&=\int _{0}^{t}e^{A(t-\tau )}bu(\tau )d\tau \\\end{aligned}}}$

Thus, we finally obtain the solution of a linear time-invariant differential equation:

${\displaystyle {\begin{aligned}x(t)=e^{At}x_{0}(t)+\int _{0}^{t}e^{A(t-\tau )}bu(\tau )d\tau \\\end{aligned}}}$

In many contexts, a discrete time (DT) system is really part of a larger continuous time (CT) system. For example, a digital recording system takes an analog sound, digitizes it, possibly processes the digital signals, and plays back an analog sound for people to listen to.

In practical systems, DT signals obtained are usually uniformly sampled versions of CT signals. If ${\displaystyle x(t)}$ is a CT signal, then the sampling circuit used before an analog-to-digital converter will transform it to a DT signal:

$${\displaystyle x_{n}\mathrel {\stackrel {\text{def}}{=}} x(nT)\qquad \forall \,n\in \mathbb {Z} ,}$$

where T is the sampling period. Before sampling, the input signal is normally run through a so-called Nyquist filter which removes frequencies above the "folding frequency" 1/(2T); this guarantees that no information in the filtered signal will be lost. Without filtering, any frequency component above the folding frequency (or Nyquist frequency) is aliased to a different frequency (thus distorting the original signal), since a DT signal can only support frequency components lower than the folding frequency.

Let ${\displaystyle \{x[m-k];\ m\}}$ represent the sequence ${\displaystyle \{x[m-k];{\text{ for all integer values of }}m\}.}$

And let the shorter notation ${\displaystyle \{x\}}$ represent ${\displaystyle \{x[m];\ m\}.}$

A discrete system transforms an input sequence, ${\displaystyle \{x\}}$ into an output sequence, ${\displaystyle \{y\}.}$ In general, every element of the output can depend on every element of the input. Representing the transformation operator by ${\displaystyle O}$, we can write:

$${\displaystyle y[n]\mathrel {\stackrel {\text{def}}{=}} O_{n}\{x\}.}$$

Note that unless the transform itself changes with n, the output sequence is just constant, and the system is uninteresting. (Thus the subscript, n.) In a typical system, y[n] depends most heavily on the elements of x whose indices are near n.

For the special case of the Kronecker delta function, ${\displaystyle x[m]=\delta [m],}$ the output sequence is the impulse response:

$${\displaystyle h[n]\mathrel {\stackrel {\text{def}}{=}} O_{n}\{\delta [m];\ m\}.}$$

For a linear system, ${\displaystyle O}$ must satisfy:

${\displaystyle O_{n}\left\{\sum _{k=-\infty }^{\infty }c_{k}\cdot x_{k}[m];\ m\right\}=\sum _{k=-\infty }^{\infty }c_{k}\cdot O_{n}\{x_{k}\}.}$

(Eq.4)

And the time-invariance requirement is:

${\displaystyle {\begin{aligned}O_{n}\{x[m-k];\ m\}&\mathrel {\stackrel {\quad }{=}} y[n-k]\\&\mathrel {\stackrel {\text{def}}{=}} O_{n-k}\{x\}.\,\end{aligned}}}$

(Eq.5)

In such a system, the impulse response, ${\displaystyle \{h\}}$, characterizes the system completely. That is, for any input sequence, the output sequence can be calculated in terms of the input and the impulse response. To see how that is done, consider the identity:

$${\displaystyle x[m]\equiv \sum _{k=-\infty }^{\infty }x[k]\cdot \delta [m-k],}$$

which expresses ${\displaystyle \{x\}}$ in terms of a sum of weighted delta functions.

Therefore:

$${\displaystyle {\begin{aligned}y[n]=O_{n}\{x\}&=O_{n}\left\{\sum _{k=-\infty }^{\infty }x[k]\cdot \delta [m-k];\ m\right\}\\&=\sum _{k=-\infty }^{\infty }x[k]\cdot O_{n}\{\delta [m-k];\ m\},\,\end{aligned}}}$$

where we have invoked Eq.4 for the case ${\displaystyle c_{k}=x[k]}$ and ${\displaystyle x_{k}[m]=\delta [m-k]}$.

And because of Eq.5, we may write:

$${\displaystyle {\begin{aligned}O_{n}\{\delta [m-k];\ m\}&\mathrel {\stackrel {\quad }{=}} O_{n-k}\{\delta [m];\ m\}\\&\mathrel {\stackrel {\text{def}}{=}} h[n-k].\end{aligned}}}$$

Therefore:

${\displaystyle y[n]}$	${\displaystyle =\sum _{k=-\infty }^{\infty }x[k]\cdot h[n-k]}$
	${\displaystyle =\sum _{k=-\infty }^{\infty }x[n-k]\cdot h[k],}$       (commutativity)

which is the familiar discrete convolution formula. The operator ${\displaystyle O_{n}}$ can therefore be interpreted as proportional to a weighted average of the function x[k]. The weighting function is h[−k], simply shifted by amount n. As n changes, the weighting function emphasizes different parts of the input function. Equivalently, the system's response to an impulse at n=0 is a "time" reversed copy of the unshifted weighting function. When h[k] is zero for all negative k, the system is said to be causal.

An eigenfunction is a function for which the output of the operator is the same function, scaled by some constant. In symbols,

$${\displaystyle {\mathcal {H}}f=\lambda f,}$$

where f is the eigenfunction and ${\displaystyle \lambda }$ is the eigenvalue, a constant.

The exponential functions ${\displaystyle z^{n}=e^{sTn}}$, where ${\displaystyle n\in \mathbb {Z} }$, are eigenfunctions of a linear, time-invariant operator. ${\displaystyle T\in \mathbb {R} }$ is the sampling interval, and ${\displaystyle z=e^{sT},\ z,s\in \mathbb {C} }$. A simple proof illustrates this concept.

Suppose the input is ${\displaystyle x[n]=z^{n}}$. The output of the system with impulse response ${\displaystyle h[n]}$ is then

$${\displaystyle \sum _{m=-\infty }^{\infty }h[n-m]\,z^{m}}$$

which is equivalent to the following by the commutative property of convolution

$${\displaystyle \sum _{m=-\infty }^{\infty }h[m]\,z^{(n-m)}=z^{n}\sum _{m=-\infty }^{\infty }h[m]\,z^{-m}=z^{n}H(z)}$$

where

$${\displaystyle H(z)\mathrel {\stackrel {\text{def}}{=}} \sum _{m=-\infty }^{\infty }h[m]z^{-m}}$$

is dependent only on the parameter z.

So ${\displaystyle z^{n}}$ is an eigenfunction of an LTI system because the system response is the same as the input times the constant ${\displaystyle H(z)}$.

The eigenfunction property of exponentials is very useful for both analysis and insight into LTI systems. The Z transform

$${\displaystyle H(z)={\mathcal {Z}}\{h[n]\}=\sum _{n=-\infty }^{\infty }h[n]z^{-n}}$$

is exactly the way to get the eigenvalues from the impulse response. Of particular interest are pure sinusoids; i.e. exponentials of the form ${\displaystyle e^{j\omega n}}$, where ${\displaystyle \omega \in \mathbb {R} }$. These can also be written as ${\displaystyle z^{n}}$ with ${\displaystyle z=e^{j\omega }}$. The discrete-time Fourier transform (DTFT) ${\displaystyle H(e^{j\omega })={\mathcal {F}}\{h[n]\}}$ gives the eigenvalues of pure sinusoids. Both of ${\displaystyle H(z)}$ and ${\displaystyle H(e^{j\omega })}$ are called the system function, system response, or transfer function.

Like the one-sided Laplace transform, the Z transform is usually used in the context of one-sided signals, i.e. signals that are zero for t<0. The discrete-time Fourier transform Fourier series may be used for analyzing periodic signals.

Due to the convolution property of both of these transforms, the convolution that gives the output of the system can be transformed to a multiplication in the transform domain. That is,

$${\displaystyle y[n]=(h*x)[n]=\sum _{m=-\infty }^{\infty }h[n-m]x[m]={\mathcal {Z}}^{-1}\{H(z)X(z)\}.}$$

Just as with the Laplace transform transfer function in continuous-time system analysis, the Z transform makes it easier to analyze systems and gain insight into their behavior.

The input-output characteristics of discrete-time LTI system are completely described by its impulse response ${\displaystyle h[n]}$. Two of the most important properties of a system are causality and stability. Non-causal (in time) systems can be defined and analyzed as above, but cannot be realized in real-time. Unstable systems can also be analyzed and built, but are only useful as part of a larger system whose overall transfer function is stable.

A discrete-time LTI system is causal if the current value of the output depends on only the current value and past values of the input.[5] A necessary and sufficient condition for causality is

$${\displaystyle h[n]=0\ \forall n<0,}$$

where ${\displaystyle h[n]}$ is the impulse response. It is not possible in general to determine causality from the Z transform, because the inverse transform is not unique. When a region of convergence is specified, then causality can be determined.

A system is bounded input, bounded output stable (BIBO stable) if, for every bounded input, the output is finite. Mathematically, if

$${\displaystyle \|x[n]\|_{\infty }<\infty }$$

implies that

$${\displaystyle \|y[n]\|_{\infty }<\infty }$$

(that is, if bounded input implies bounded output, in the sense that the maximum absolute values of ${\displaystyle x[n]}$ and ${\displaystyle y[n]}$ are finite), then the system is stable. A necessary and sufficient condition is that ${\displaystyle h[n]}$, the impulse response, satisfies

$${\displaystyle \|h[n]\|_{1}\mathrel {\stackrel {\text{def}}{=}} \sum _{n=-\infty }^{\infty }|h[n]|<\infty .}$$

In the frequency domain, the region of convergence must contain the unit circle (i.e., the locus satisfying ${\displaystyle |z|=1}$ for complex z).