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Bounded-input bounded-output stability means that every bounded plant input produces a bounded plant output from zero initial state. For a causal continuous-time linear time-invariant system with a proper rational transfer function, BIBO stability is equivalent to every pole of the reduced transfer function having strictly negative real part. Cancel common factors first: an apparent denominator root may be absent from the reduced transfer function. For $K/(\tau s+1)$ with $\tau >0$, the only pole is $-1/\tau$, so the first-order low-pass filter is BIBO stable. ^[ref]

Internal asymptotic stability also requires every unforced internal state to decay. A canceled factor can hide a growing internal state. For example, consider the linear system of ordinary differential equationsIts zero-state transfer function can be written $(s-1)/((s+1)(s-1))=1/(s+1)$ and has BIBO stability. Yet from zero initial conditions and a unit-step plant input, $x_2(t)=e^t-1$ grows without bound while the plant output remains bounded. The reduced transfer function conceals that second state. Even a transfer function with no visible cancellations can be implemented with additional hidden states; its poles alone do not establish internal asymptotic stability of an arbitrary implementation.

The continuous-time frequency response is the frequency response of a continuous-time linear time-invariant system. From zero initial state, a BIBO stable system driven by a sustained sinusoid $u(t)=A\cos (\omega t)$ approaches the steady responseAny decaying initial or switching transient is separate. The modulus $|G(i\omega )|$ is the amplitude gain; the complex argument is the shift in oscillation phase. Thus evaluating a transfer function at $s=i\omega$ measures both how much of an oscillation passes through and how far it lags.

For the first-order low-pass filter, with $K>0$, direct complex number arithmetic givesThe phase uses the inverse tangent branch between $-\pi /2$ and $\pi /2$. Low angular frequencies have gain close to $K$ and little lag. High angular frequencies have gain approximately $K/(\omega \tau )$ and lag approaching $\pi /2$. This is why the model is a first-order low-pass filter: it suppresses fast variation compared with slow variation. Rowell relates frequency response gain and phase to poles and then introduces Bode plots in MIT 2.004, lecture 32, sections 1 and 2.

At the corner angular frequency,For $K=2$ and $\tau =3\mathrm{s}$, ${\omega }_c=1/3\mathrm{rad}{\mathrm{s}}^{-1}$, the gain is $\sqrt{2}\approx 1.4142$, and the lag is $45$ degrees. The angular frequency $\omega$ is measured in radians per second; ordinary frequency $f$ is measured in hertz (cycles per second), with $\omega =2\pi f$. Consequently the same corner is $f_c=1/(2\pi \tau )\approx 0.0531\mathrm{Hz}$, not $1/3\mathrm{Hz}$.

Controllability asks whether the available control inputs can move a dynamical system between every pair of states in its declared state space. Fixing a duration $T>0$ gives the precise condition ${\mathcal{R}}_T(x_0)=X$ for every $x_0\in X$, where $X$ is the state space and ${\mathcal{R}}_T(x_0)$ its reachable set. With freely chosen duration, every pair must instead be connectable in some permitted finite time, which may depend on the pair. Always specify admissible control inputs, deadlines and any constraints along the path. ^[ref]

The surprising part is that one control input can control several state variables by acting at different times. For the double integrator, acceleration changes velocity directly, while the accumulated velocity changes position. Starting from rest, accelerate during the first half of the journey and brake during the second: the final velocity can vanish although the final position has changed. The figure is an analytic mathematical example in dimensionless quantities, with no experimental measurements.

Controllability is useful before trying to design a controller: a requested endpoint outside the reachable set cannot be attained by clever choice of feedback control. The Kalman rank criterion makes this an exact linear algebra calculation for the stated linear time-invariant system model. It concerns whole states, so attaining one plant output is not enough to establish controllability. In quantum control, the declared state space may be an orbit constrained by conservation laws; see quantum controllability and unitary orbit of a density operator.

For the linear time-invariant system $\dot{x}=Ax+Bu$, with $A\in {\mathbb{R}}^{n\times n}$ and $B\in {\mathbb{R}}^{n\times m}$, the controllability matrix has $n$ rows and $nm$ columns. The columns of $B$ are directions influenced directly by the control inputs; $AB,A^{2}B,\dots$ account for how the dynamics carry their influence into further directions. The column space of the controllability matrix is the zero-state reachable set for any $T>0$ when inputs are unrestricted. ^[ref]

Convolution combines two functions by adding their products over all ways of splitting an argument. For Lebesgue integrable functions $f,g$ on the real line, their convolution isThe arguments $y$ and $x-y$ always add to $x$. For $f,g$ in the L1 space, the integral exists almost everywhere and defines another element of the L1 space, with $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$ by Young's convolution inequality. Particular examples below have a value at every $x$.

To see what the integral counts, take the indicator function $q=1_{[0,1]}$. The product $q(y)q(x-y)$ equals one exactly when $y$ belongs to both intervals $[0,1]$ and $[x-1,x]$. Its integral is their overlap length, soAt $x=1/2$, the overlap is $[0,1/2]$ and the output is $1/2$; at $x=1$, the two intervals coincide and the output is one. The result is the shifted triangular function $\mathrm{tri}(x-1)$. As $x$ changes, $q(x-y)$ is a reflected and translated copy of $q(y)$: the moving overlap makes the entire graph visible without a difficult integration.

The same calculation gives the probability distribution of a sum of two independent random variables. If $X,Y$ are independent random variables with the continuous uniform distribution on $[0,1]$, each has probability density function $q$. Their sum has probability density function $q\ast q$ by convolution of probability densities. The probability density function at one is larger than near zero or two because there are more combinations of the two inputs giving that sum. Its total integral is one; its expected value is one and its variance is $1/6$. These follow equally from integrating the triangular function or from linearity of expectation and variance additivity for independent random variables.

Convolution also gives a useful averaging operation. For $\epsilon >0$, let $k_{\epsilon }(y)=1_{[-\epsilon ,\epsilon ]}(y)/(2\epsilon )$. Thenthe mean of $f$ in a neighborhood of $x$. A smaller $\epsilon$ examines a smaller neighborhood. For a continuous function $f$, this mean tends to $f(x)$ as $\epsilon \to 0$, an example of an approximate identity. Integrating a sharp input against a distributed convolution kernel can produce a smoother output; the triangular example turns two discontinuous inputs into a continuous output.

In signal processing, a linear time-invariant system combines shifted copies of its impulse response by the same rule. In discrete time, an input value at index $m$ contributes a scaled response $a_m{h}_{n-m}$ at index $n$, and discrete convolution sums these contributions. This is the superposition interpretation developed in Oppenheim, MIT Signals and Systems, Lecture 4, convolution sums and integrals. The convolution theorem then gives another description: transform both inputs with a Fourier transform and multiply their transforms. Combining shifts, combining independent quantities and multiplying frequency descriptions are three faces of the same operation.

A convolution kernel is the fixed function $h$ in an operator $T_{h}f=f\ast h$. It specifies how shifted input values are weighted. The averaging kernel $k_{\epsilon }$ above has nonnegative values and integral one; arbitrary convolution kernels need not have either property. A linear time-invariant system uses its impulse response as a kernel when its output admits this representation.

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

The double integrator is the linear time-invariant system in which one control input is integrated into velocity, and velocity into position. It is a mathematical idealization of one-dimensional motion under commanded acceleration; for a constant mass $m$ and net force $F$, Newton's second law gives $u=F/m$. The worked examples here use normalized dimensionless quantities and assume no control disturbances, rather than describing a laboratory record. ^[ref]

Writing $x=(p,v{)}^T$ givesThe Kalman rank criterion therefore gives controllability in every $T>0$ for unrestricted real control inputs. More concretely, choose constants $a,b$ on the first and second halves. Starting at $x_0=(p_0,v_0{)}^T$, direct integration givesThe endpoint map has determinant $T^3/8\ne 0$, so these two amplitudes can supply any desired endpoint displacement. For a rest-to-rest move of distance $d$, take $a=4d/T^2$, $b=-4d/T^2$. At $T=2$ and $d=1$ these are the two commands in the first figure. No impulse is needed; shorter deadlines require larger finite commands.

Electrical impedance is the ratio of terminal electric potential difference and electric current phasors at one specified positive angular frequency $\omega$. For an electrical circuit modeled as a linear time-invariant system, use the sinusoidal steady state after any decaying transient has settled. With the passive sign convention, positive electric current enters the terminal assigned positive electric potential difference. Define peak complex amplitudes byThe physical $v(t)$ and $i(t)$ are real; the complex numbers encode their amplitudes and wave phases. The unit of electrical impedance is the ohm, $1\Omega =1\mathrm{V}/\mathrm{A}$. Both phasors may instead use root mean square magnitudes, with the same ratio $Z$. A nonlinear diode can generate additional frequencies; a single $Z(\omega )$ then does not describe the whole response.

Write the electrical impedance in Cartesian and polar forms asHere $R$ is the resistive real part, $X$ the signed electrical reactance, and $\varphi$ the voltage-minus-current phase difference, defined modulo a full turn. The Pythagorean theorem gives the modulus from a right triangle in the complex plane. The complex argument gives the rotation from the current phasor to the voltage phasor. Thus $V_{\mathrm{rms}}=|Z|I_{\mathrm{rms}}$, while negative $\varphi$ means electric current leads terminal electric potential difference. For $R>0$, $\varphi =\arctan (X/R)$ lies between $-\pi /2$ and $\pi /2$; outside that case, use the quadrant of the complex number rather than a bare inverse tangent. The geometric magnitude and phase difference are the two pieces of the AC relation.^[ref]

Multiplication by $Z$ acts on phasors, not on instantaneous physical signals. For example, when $i(t)=I_0\cos \omega t$,The second term is in phase quadrature with electric current; it cannot be represented by pointwise multiplication $v(t)=Zi(t)$ when $X\ne 0$. This is the useful compression: one complex multiplication replaces separate amplitude and phase difference calculations, while the real part restores the physical signal.

An ideal resistor, inductor and capacitor have $Z_R=R$, $Z_L=\mathrm{i}\omega L$ and $Z_C=1/(\mathrm{i}\omega C)$. Their electrical impedances add in a series RLC circuit, giving $Z=R+\mathrm{i}(\omega L-1/(\omega C))$. The current amplitude is $V_0/|Z|$, so cancellation of the imaginary parts explains the resonant series RLC response. The component-law derivation and opposite complex exponential convention are in reactance phase convention.

Let $K(f)$ be the complete current-to-voltage frequency response from detector input to readout, in $V/A$. For a real stable linear time-invariant system, the shot contribution to the output voltage variance isIf $K(f)=R_{F}GH(f)$ with feedback electrical resistance $R_F$, a reference voltage gain $G$, and dimensionless normalized frequency response $H$, thenThis is the effective noise bandwidth for unit reference magnitude. The transimpedance amplifier factor $R_F$ converts electric current to voltage; $G$ multiplies voltage. Neither is an additional noise source merely because it amplifies an existing fluctuation. In contrast, a real amplifier adds input voltage noise and input current noise, and its feedback resistor contributes thermal electrical noise. Those terms need their own transfers to the output. In particular, input voltage noise generally has a different transfer from an injected detector electric current. Simply multiplying every input specification by $R_{F}G$ is incorrect.

For a unit-gain exponential low-pass filter, with time constant $\tau >0$,The cutoff frequency $f_c$ is not $B_{\mathrm{n}}$. An independent time-domain derivation sums independent event contributions to giveThe same reasoning for a rectangular averaging gate has $h=1/T$ on $[0,T]$ and zero elsewhere, hence effective noise bandwidth $1/(2T)$ and variance $qI/T$, agreeing with the shot noise counting model. Equal numerical durations $T=\tau$ do not make these two different linear filters equivalent.

A linear time-invariant system has finite impulse response when its impulse response is nonzero at only finitely many indices. For a causal filter supported at indices $0,\dots ,M-1$,This is a finite discrete convolution. The coefficients $h[j]$ specify the complete operation; they need not be positive or equal. Such a filter has a simple bounded-input estimate, $|y[n]|\le \Vert x{\Vert }_{\infty }{\sum }_j|h[j]|$, so bounded inputs remain bounded. The moving average chooses $h[j]=1/M$.

A Fourier series describes a periodic function through sines and cosines whose frequencies are integer multiples of one base frequency. These Fourier harmonics give a useful alternative to describing the value at every input: each Fourier coefficient measures how much of one oscillating pattern is present. For a real, integrable function of period $2\pi$, the real form isThe cosine formula includes $n=0$; the sine formula uses $n\ge 1$. Thus $a_0/2$ is the mean value over a period, not $a_0$ itself. The symbol $\sim$ associates a function with its Fourier coefficients; an equality requires a specified kind of convergence, such as pointwise convergence. For general period $T>0$, replace $nx$ by $2\pi nt/T$ and the coefficient prefactor by $2/T$. Fourier orthogonality explains why integrating against one mode removes the others. ^[ref].

For a finite example, takeIts mean is $1/2$, its real Fourier coefficients are $a_0=1$, $a_1=1$, $b_2=1/2$, and all other $a_n,b_n$ vanish. At $x=\pi /4$, the three contributions add to $1/2+1/\sqrt{2}+1/2=1+1/\sqrt{2}\approx 1.70710678$. The function is already a trigonometric polynomial, so retaining these modes reconstructs it exactly. Here $x$ is an angular coordinate in radians; if it instead represents time, a base angular frequency must supply the corresponding units.

The change of viewpoint is useful because many linear problems act separately on these modes. The second derivative maps $\sin (nx)$ to $-n^2\sin (nx)$ and maps $\cos (nx)$ in the same way. Consequently a differential equation that is complicated for a whole profile can become a simpler calculation for each Fourier harmonic. A linear time-invariant system similarly has a separate response to each complex exponential. Superposition combines the responses when the relevant sums and operations are justified. ^[ref].

The infinite case contains a subtlety absent from the finite example. Under pointwise convergence of a piecewise smooth Fourier series, the value at a jump discontinuity is the average of the two one-sided limits. For a square-integrable function, the complex Fourier series instead converges in the $L^2$ norm: the integrated squared error tends to zero. Neither statement says that the error becomes small everywhere at once. A square wave shows the difference vividly through the Gibbs phenomenon.

A historical source for this connection between profiles and modes is Joseph Fourier's 1822 book Théorie analytique de la chaleur. Chapter III develops expansions in sines and cosines while studying the propagation of heat. The heat equation gives a continuing application of this approach. ^[ref].

The frequency response gives the change in complex amplitude of a sustained oscillation under a linear time-invariant system. If the impulse response is finite or has absolute summability, substituting $x[n]=e^{\mathrm{i}\omega n}$ into its discrete convolution yieldsThus $|H(\omega )|$ is the amplitude gain and its complex argument is the phase change. Real sinusoids follow by taking real or imaginary parts. Here $\omega$ is radians per sample; $\omega =2\pi f$ for $f$ cycles per sample. A physical sampling rate converts that normalized frequency into cycles per second.

For a trailing moving average,with removable singularities evaluated from the finite sum. In particular $H(0)=1$. For $M=5$ and $\omega =2\pi /5$, the five roots of unity sum to zero, so the oscillation is canceled. At small positive frequencies in the first pass region, the real sine ratio is positive and the phase factor gives a delay of $(M-1)/2$ samples.

Increasing $M$ moves the first zero to $\omega =2\pi /M$, narrowing the main region of large gain. It does not create an ideal low-frequency cutoff: later side lobes transmit some higher-frequency components, and suppressing a component is only helpful when that component is unwanted. The frequency plot is the same operation as the delayed-sample average; the five-vector diagram explains one exact zero geometrically.

The impulse response records the response to a unit impulse, with specified initial or boundary conditions. In continuous time the impulse is a Dirac delta function; in discrete time it is the Kronecker delta $\delta [n]$. For a linear time-invariant system with zero initial state, the response to a general input is its convolution with the impulse response. The impulse response of a causal filter is zero before the impulse.

For the five-sample moving average, a discrete impulse produces $h[n]=1/5$ at $n=0,1,2,3,4$ and zero otherwise. The response lasts five samples, so this is a finite impulse response system. Its frequency response is the finite sum of the five delayed phase contributions: one short input experiment determines how every sustained sinusoid is filtered.

For a finite-dimensional continuous-time linear time-invariant system $\dot{x}=Ax+Bu$ on ${\mathbb{R}}^n$, with constant matrices $A,B$ over the real numbers, no state constraints, and every piecewise continuous function $u:[0,T]\to {\mathbb{R}}^m$ allowed as a control input without an amplitude bound, controllability in any fixed $T>0$ is equivalent to the controllability matrix having rank of a matrix $n$. The same equivalence holds with all square-integrable functions as inputs. A stronger useful statement iswhere $e^{AT}$ is a matrix exponential and $\mathrm{im}\mathcal{C}$ is the column space. Thus a rank of a matrix $r<n$ leaves only an $r$-dimensional affine subspace reachable at that time. ^[ref]

The Kalman rank criterion allows arbitrarily large commands as $T$ shrinks. It does not assert controllability with bounded control inputs, nor test a general nonlinear dynamical system just by counting its inputs. For the double integrator, $B=(0,1{)}^T$ and $AB=(1,0{)}^T$ give two independent directions, despite there being only one control input.

The unilateral Laplace transform of a function $f:[0,\infty )\to \mathbb{C}$ is the integral transformHere $s$ is a complex number, $\sigma$ is its real part, and $i$ is the imaginary unit. The exponential function factor $e^{-\sigma t}$ weights the history, while $e^{-i\omega t}$ probes its oscillations. The integral starts at zero, so it uses only the history from that starting time onward. This section concerns ordinary functions; their value at the single endpoint does not change the integral. See MIT 18.03SC, definition and elementary examples.

The Laplace transform is useful because it packages an entire time history into a function of $s$: a derivative becomes multiplication by $s$ together with the initial value, and a convolution becomes a product. Thus a linear differential equation can become an algebraic equation that already remembers its initial value problem. The worked example below shows both descriptions of the same response.

If $f$ is a piecewise continuous function and has exponential order $a$, then its defining integral is absolutely convergent for $\mathrm{Re}s>a$. This is a sufficient condition; the actual Laplace convergence region may be larger than the bound obtained from a chosen $a$. ^[ref].

The bound has a short explanation. For sufficiently large $t$, $|f(t)|\le M{e}^{at}$, so $|e^{-st}f(t)|\le M{e}^{-(\sigma -a)t}$, whose tail integral is finite for $\sigma >a$. The finite initial interval is covered by piecewise continuity.

Three calculations make the convergence condition concrete. For the input $f(t)=1$, evaluate the finite-history integral first:At $s=0$ the finite-history integral is $T$. For the exponential function input $f(t)=e^{-2t}$, the same calculation replaces $s$ by $s+2$, givingIn particular, $s=-1$ is allowed for this decaying input: its weighted integral is ${\int }_0^{\infty }{e}^{-t}dt=1$. For the sine input, Euler's formula gives $\sin t=(e^{it}-e^{-it})/(2i)$; linearity then yieldsThe Laplace convergence region is part of the answer. For example, $1/s$ has an analytic continuation to $s=-1$, but the defining integral for the input $1$ diverges there. The sine calculation is also developed in MIT 18.03SC, further elementary transforms.

To see why the initial value appears, suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, with $f(0+)$ its right one-sided limit. At the chosen $s$, assume $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions and $e^{-sT}f(T)\to 0$. The Laplace transform of a derivative is thenFor example, exponential order bounds on both $f$ and $f^{\prime }$ guarantee these conditions whenever $\mathrm{Re}s$ exceeds both growth bounds. The initial value is the lower-end boundary contribution of integration by parts.

Consider the initial value problemWrite $Y=\mathcal{L}\{y\}$. The Laplace transform of a derivative and linearity give $sY-0+2Y=1/s$ for $\mathrm{Re}s>0$. A partial fraction decomposition therefore givesWe recognize the two elementary Laplace transforms already computed. Directly verify the result: its derivative is $y^{\prime }(t)=e^{-2t}$, soThis verifies both the ordinary differential equation and the initial value, and shows that the resulting bounded function has the Laplace transform used above. The response rises from zero toward $1/2$: at that value the input $1$ balances the term $2y$. The general method is described in MIT 18.03SC, solving initial value problems.

The same solution is a convolution. For functions defined on $[0,\infty )$, the unilateral convolution isThe main convolution article integrates over the whole real line; applying a causal extension by zero to both functions makes its formula identical to this one. Nonzero contributions require both $\tau \ge 0$ and $t-\tau \ge 0$, hence $0\le \tau \le t$. If $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ are Lebesgue integrable functions on $[0,\infty )$, the convolution theorem for Laplace transforms statesFor piecewise continuous functions of exponential order $a$ and $b$, it holds throughout $\mathrm{Re}s>\max (a,b)$. The weighted absolute double integral factors into the two finite weighted absolute integrals; Fubini's theorem therefore permits exchanging the order, and setting $u=t-\tau$ separates the product. ^[ref].

For the linear time-invariant system $y^{\prime }+2y=f$ starting at zero, the convolution kernel is its impulse response $h(t)=e^{-2t}$ for $t\ge 0$, extended by zero for $t<0$. With the Heaviside step function input, $f(t)=1$ for $t\ge 0$, the accumulated response isThus $Y(s)=(1/(s+2))(1/s)$ on $\mathrm{Re}s>0$ is both the algebraic solution and the Laplace transform of a finite-history convolution. Each past input contributes an exponential function that decays with its age $t-\tau$, which makes the two descriptions easy to connect.

Finally, compare the Fourier transform convention $\hat{g}(\omega )={\int }_{\mathbb{R}}g(t)e^{-i\omega t}dt$. If $f_0$ is the causal extension by zero of $f$, define $g_{\sigma }(t)=e^{-\sigma t}{f}_0(t)$. Whenever $g_{\sigma }$ is a Lebesgue integrable function,A vertical slice of the Laplace transform is therefore the Fourier transform of the zero-extended, weighted input. Positive $\sigma$ damps its positive-time history. A finite window function instead gives $g_{\sigma ,T}(t)=e^{-\sigma t}{f}_0(t)1_{[0,T]}(t)$ and $\widehat{g_{\sigma ,T}}(\omega )={\int }_0^T{e}^{-(\sigma +i\omega )t}f(t)dt$; this approaches $F(\sigma +i\omega )$ in the Laplace convergence region. Setting $\sigma =0$ directly requires an integrable zero-extended input: the input $1$ fails that condition, even though its Laplace transform exists for every $\sigma >0$.

The history of the Laplace transform spans several stages. Schiff's historical summary notes that Laplace's 1812 treatise acknowledged earlier work on integral transforms and describes later development of the modern Laplace transform method. ^[ref].

A linear filter obeys the superposition principle: filtering $\alpha x+\beta z$ gives $\alpha T[x]+\beta T[z]$. Many familiar examples are also linear time-invariant systems, represented by an impulse response and convolution. Linearity alone does not imply time invariance: multiplying $x[n]$ by the index $n$ is linear but changes its behavior when the same input is shifted.

The moving average is a finite impulse response example implemented by arithmetic; a band-pass filter can instead be built from a series RLC circuit. Their frequency responses explain which oscillations pass through. Choosing a filter means choosing how much amplitude, timing and detail may change, not simply removing every visible fluctuation.

A moving average replaces each value by an average over a moving window of neighboring values. The simple trailing version with positive integer window length $M$ is the causal filterIt is a finite impulse response linear time-invariant system. A constant input remains unchanged because the weights sum to one. Weighted and centered averages are other choices; the definition of the window determines both the weights and whether future values are needed.

For an exact five-sample example, letdefined at every integer index. The first component completes one cycle in forty samples and the second in five. Averaging five consecutive samples cancels the second component exactly. The first becomesThe remaining component is slightly smaller and delayed by two samples; the filter does not recover the original component unchanged.

The example is in steady operation: its early output values use the specified negative-index samples. Starting a device at $n=0$ with an unknown past requires an initialization rule. Treating earlier samples as zero, shortening the initial window, and postponing output until a full window is available produce different initial values.

For zero-mean uncorrelated random variables $\epsilon [n]$ with common variance ${\sigma }^2$, the averaged disturbance hasThe standard deviation decreases by $\sqrt{M}$, but successive filtered errors share samples and are generally correlated. This variance calculation also assumes the wanted component is not damaged by averaging. A rapidly changing feature can be suppressed along with the disturbance.

The frequency response explains the cancellation and limits of this method. A moving average preserves low frequencies comparatively well, yet its side lobes make it a poor substitute for a sharply selective frequency cutoff. Smith develops this time-domain versus frequency-domain distinction in The Scientist and Engineer's Guide to Digital Signal Processing, chapter 15: moving-average filters.

Convolution describes the response of a linear time-invariant system to an arbitrary input. If $h$ is its impulse response and $f$ its input,This is a weighted sum of responses to shifted impulses. For integrable functions the operation is commutative and associative. A causal system has $h(t)=0$ for $t<0$, restricting the integral to past inputs. With the Fourier convention $F(\omega )=\int f(t)e^{-i\omega t}dt$, the convolution theorem gives $G=HF$.

For example, an RC low-pass frequency response has $h(t)={\tau }_0^{-1}{e}^{-t/{\tau }_0}\Theta (t)$ and $H(\omega )=(1+i\omega {\tau }_0{)}^{-1}$, where $\Theta$ is the Heaviside step function. Fast input variations are smoothed. The same mathematics describes optical blurring by a point spread function and many other linear response measurements.

Deconvolution tries to recover $f$ from measured $g$ and known $h$. In noise-free data one formally sets $F=G/H$ where $H\ne 0$. Zeros of $H$ lose corresponding information; small values amplify measurement noise. If the measured spectrum is $G+N$, the direct estimate has error $N/H$, which can be very large at frequencies strongly suppressed by the instrument. An inverse is therefore not automatically unique or stable.

One regularized spectral deconvolution reconstruction isThis minimizes $|HF-G{|}^2+\lambda |F{|}^2$ at each frequency and limits noise amplification, at the cost of bias and lost resolution. Choosing the regularization requires assumptions about noise and the desired signal; deconvolution cannot restore components that the data do not determine without such additional information.

A proper rational transfer function is a rational function whose numerator polynomial has polynomial degree no greater than its denominator polynomial, after common factors are canceled. Every finite-dimensional continuous-time linear time-invariant system of the form $\dot{x}=Ax+Bu$, $y=Cx+Du$ has such a transfer function $G(s)=C(sI-A{)}^{-1}B+D$. ^[ref]