- Why physical modelling produces linear constant-coefficient differential and difference equations, and why such an equation is an implicit description of a system rather than an explicit one.
- The characteristic equation, its roots, and how each root — real, repeated or complex — contributes a term to the natural response.
- How the forced response is found for the exponential family of inputs, and how auxiliary conditions fix the arbitrary constants of the complete solution.
- The initial rest condition: the single assumption that makes an LCCDE describe a system that is linear, time-invariant and causal, and what goes wrong without it.
- The zero-input / zero-state decomposition, and why only the zero-state part is the convolution of Chapter 7.
- Solving difference equations recursively, extracting the impulse response, and the divide between FIR (non-recursive) and IIR (recursive) systems.
- Direct Form I and II realisations — how an equation becomes a diagram of adders, gains and delays, and why Form II uses half the memory.
Where the Equations Come From
Chapters 6 to 8 built a complete theory around one object. Give me \(h\), and I can compute any output by convolution, decide whether the system is causal or stable, and combine it with other systems. The theory is closed and satisfying, and it has one practical defect: nobody ever hands you \(h\).
What a physical problem hands you is a relation among a signal, its rate of change, and the excitation. Kirchhoff's voltage law applied to a series \(RC\) circuit says the source voltage equals the resistor drop plus the capacitor voltage. The resistor drop is \(Ri(t)\), and the current through a capacitor is \(C\,dv_C/dt\). Substituting one into the other produces an equation that relates the output to its own derivative.
The same thing happens everywhere. Newton's second law applied to a mass on a spring with damping gives \(m\ddot{y}+b\dot{y}+ky=f(t)\). A thermal mass losing heat to its surroundings gives a first-order equation. A savings account with monthly interest and monthly deposits gives \(y[n] = (1+r)y[n-1] + d[n]\), a difference equation. In every case, the model relates the unknown signal to shifted or differentiated versions of itself.
This is why the present chapter exists. It is the bridge between the physics, which speaks in equations, and the LTI theory, which speaks in impulse responses. We shall solve the equations, extract \(h\) from them, and — most importantly — discover that an equation on its own does not define a system at all.
The General Linear Constant-Coefficient Equation
Collect all such models into one form. A linear constant-coefficient differential equation, or LCCDE, relates a weighted sum of derivatives of the output to a weighted sum of derivatives of the input, with coefficients that do not vary with time.
Three words in the name carry all the meaning. Linear: the signals and their derivatives appear to the first power only, never squared, never inside a sine. Constant-coefficient: the \(a_k\) and \(b_k\) are numbers, not functions of \(t\) — a resistance that drifted with temperature would break this. Order \(N\): the highest derivative of the output, which counts the independent energy-storage elements in the system and, as we shall see, the number of terms in the natural response.
Now the central difficulty. Consider the first-order equation \(\dot{y}+3y = x\) with \(x(t)=0\) for all time. One solution is \(y(t)=0\). But \(y(t)=Ke^{-3t}\) is also a solution for every value of \(K\), since \(-3Ke^{-3t}+3Ke^{-3t}=0\). The equation is satisfied by an entire one-parameter family of outputs for a single input. It therefore does not, by itself, define a mapping from input to output — and a system is exactly such a mapping.
An \(N\)th-order equation admits an \(N\)-parameter family of solutions. Exactly \(N\) additional pieces of information — the auxiliary or initial conditions — are needed before the output is determined by the input.
Which \(N\) conditions you choose is not a technicality; different choices give genuinely different systems from the same equation, with different impulse responses and even different causality. Worked Example 6 exhibits two systems built from one difference equation, one causal and unstable, the other stable and non-causal. Section 9-5 identifies the choice that recovers the LTI theory of the previous chapters.
The Natural Response and the Characteristic Roots
The standard route to the solution splits it in two. The homogeneous or natural response \(y_h(t)\) solves the equation with the input set to zero; it describes what the system does when left to itself, running on stored energy. The particular or forced response \(y_p(t)\) is any one signal satisfying the full equation. Their sum satisfies the full equation too, by linearity, and contains the \(N\) free constants we need.
To find \(y_h\), guess \(y(t)=e^{st}\). The reason this guess works is the reason exponentials dominate this whole subject: differentiating an exponential returns the same exponential multiplied by \(s\), so every term in the sum becomes a power of \(s\) times a common factor.
Since \(e^{st}\) is never zero, the bracket must vanish. That polynomial is the characteristic equation, and its \(N\) roots \(s_1,\dots,s_N\) are the characteristic roots — the natural frequencies of the system. Each root supplies one term of the natural response, and because the equation is linear, any combination of those terms is again a solution.
The \(C_i\) are the free constants; they are fixed later by the auxiliary conditions, never by the equation itself.
The repeated-root form deserves a word. If \(s\) is a double root, \(e^{st}\) alone supplies only one solution where two are needed, and \(te^{st}\) supplies the missing one — a fact you can verify by direct substitution into \((d/dt - s)^2 y = 0\). Physically, a repeated root is the critically damped case, the boundary between oscillation and pure decay.
Complex roots always occur in conjugate pairs when the coefficients are real, and the pair \(s = \sigma \pm j\omega_d\) combines into a real signal. Using Euler's relation, \(C e^{(\sigma+j\omega_d)t} + C^{*}e^{(\sigma-j\omega_d)t} = 2|C|e^{\sigma t}\cos(\omega_d t + \angle C)\): a damped sinusoid of exactly the kind Chapter 3 introduced. The real part \(\sigma\) sets the envelope and the imaginary part \(\omega_d\) sets the ringing frequency.
The figure is worth memorising now, because it is the picture the rest of the book keeps returning to. A system whose characteristic roots all lie strictly in the left half-plane has a natural response that dies away; combined with initial rest, that is exactly the condition for the impulse response to be absolutely integrable and the system to be BIBO stable in the sense of Chapter 8. Roots on the imaginary axis give oscillation that never decays; roots in the right half-plane give runaway growth. Chapter 19 will re-derive all of this by calling the characteristic roots the poles of the Laplace transform, but the content will be identical.
The Forced Response and the Complete Solution
The particular solution is any single signal that satisfies the full equation with the actual input present. For the inputs that matter in this course — constants, exponentials, sinusoids, polynomials — there is a reliable recipe: assume that \(y_p\) has the same functional form as the input, with undetermined coefficients, substitute, and match.
The reason the recipe works is again the closure of the exponential family under differentiation. If the input is \(e^{-t}\), then every derivative of a trial solution \(Ke^{-t}\) is a multiple of \(e^{-t}\), so substitution produces an algebraic equation in \(K\) rather than a differential one.
| Input \(x(t)\) | Trial form for \(y_p(t)\) | Note |
|---|---|---|
| Constant \(A\) | \(K\) | Gives the DC gain \(b_0/a_0\) |
| \(Ae^{\alpha t}\) | \(Ke^{\alpha t}\) | Fails if \(\alpha\) is a characteristic root |
| \(A\cos\omega t\) | \(K_1\cos\omega t + K_2\sin\omega t\) | Both terms are needed; the phase is not known in advance |
| \(At^{m}\) | \(K_m t^{m}+\cdots+K_1t+K_0\) | All lower powers must be included |
| \(Ae^{\alpha t}\), \(\alpha\) a root of multiplicity \(r\) | \(Kt^{r}e^{\alpha t}\) | Multiply by \(t^{r}\) to escape the natural response |
The last row is the only trap. If the input happens to be an exponential at one of the system's own natural frequencies, the trial solution is already a homogeneous solution and substituting it gives \(0=A\), a contradiction. Multiplying by \(t\) — or by \(t^{r}\) for a root repeated \(r\) times — produces an independent trial that works. This is resonance: driving a system at its own natural frequency produces a response that grows linearly in time.
With both pieces in hand, the complete solution is their sum, and only now are the auxiliary conditions applied.
The order of operations matters and is the most common source of lost marks. The constants \(C_i\) multiply the natural terms, but they are determined by the conditions on the total response, not on the natural part alone. Find \(y_p\) first, add it to \(y_h\), and only then impose \(y(0)\), \(\dot{y}(0)\) and so on. Worked Example 2 follows that sequence deliberately.
For a stable system the natural terms decay and the forced term survives, which is the origin of the engineer's distinction between transient and steady-state response. Those two words are not synonyms for natural and forced, though they often coincide: "natural and forced" is a statement about where a term came from, "transient and steady-state" about whether it eventually vanishes.
Initial Rest: the Condition That Restores LTI
Return to the difficulty of Section 9-2. The equation needs \(N\) auxiliary conditions, and the choice is ours. Suppose we make the natural choice of a physicist and specify the state at \(t=0\): say \(y(0)=y_0\) with \(y_0\ne0\), fixed once and for all. Does the resulting system satisfy the definitions of Chapter 5?
It does not. Test linearity. Scale the input by 2. The forced response doubles, but the natural response is pinned by the condition \(y(0)=y_0\), which has not changed, so the total output is not twice the original. A system that carries fixed non-zero initial energy is not linear, because zero input does not produce zero output. Test time invariance similarly: the conditions are anchored to \(t=0\), so delaying the input does not delay the output. Both properties fail.
Under initial rest, the system described by an LCCDE is linear, time-invariant and causal — and only then does the convolution theory of Chapters 6 to 8 apply to it.
Read the condition carefully: it is not "\(y(0)=0\)". It says that the output is zero over whatever stretch of time precedes the input, wherever that stretch happens to be. That floating reference point is what rescues time invariance — delay the input and the rest interval moves with it. And because the output cannot begin before the input does, causality is built in by construction.
Linearity is restored for the same reason: with no stored energy anywhere, zero input gives zero output, and scaling the input scales everything. Initial rest is the assumption that the system starts from nothing every time, so that all of its behaviour is attributable to the input.
In practice, imposing initial rest for a system of order \(N\) driven by an input switched on at \(t=0\) means setting \(y(0^-)=\dot{y}(0^-)=\cdots=y^{(N-1)}(0^-)=0\), and then finding the conditions just after the switching instant. When the input contains no impulses and \(M \lt N\), the output and its first \(N-1\) derivatives are continuous at the origin, so the values at \(0^+\) equal those at \(0^-\) and are all zero. When the input does contain an impulse — as it does when we compute the impulse response — some derivative jumps, and that jump is what launches the response.
Zero-Input and Zero-State Responses
Real equipment does have stored energy — a capacitor that was left charged, a flywheel already spinning — and we still need to analyse it. The device is to keep linearity by splitting the problem rather than the solution. Because the equation is linear, the response to "initial energy plus input" is the sum of the response to the initial energy with no input and the response to the input with no initial energy.
Only the zero-state response is a convolution. Everything the previous three chapters proved applies to \(y_{\text{zs}}\) and to nothing else.
This decomposition is more useful than the natural/forced one, because its two halves are separately meaningful. The zero-input response is what the system would do if you disconnected the source and walked away: it is built entirely from characteristic-root terms, with constants set by the stored energy. The zero-state response is the LTI system's answer to the input, computable by convolution, and it generally contains characteristic-root terms of its own alongside the forced term.
That last point resolves a persistent confusion. Natural-response terms appear in the zero-state response too — they must, in order that the total satisfy the initial-rest conditions — so "natural response" and "zero-input response" are not the same object. The natural/forced split classifies terms by their mathematical form; the zero-input/zero-state split classifies them by their cause. Worked Example 3 computes both splits for one system so the difference is visible.
| Decomposition | First part | Second part | Split according to |
|---|---|---|---|
| Natural / forced | Terms \(e^{s_i t}\) from the characteristic roots | Term with the form of the input | Mathematical form of each term |
| Zero-input / zero-state | Response to stored energy alone | Response to the input alone, \(x*h\) | Physical cause of each term |
| Transient / steady-state | Everything that decays to zero | Everything that persists | Long-term behaviour |
Difference Equations and Recursive Solution
The discrete-time story runs parallel, with one enormous practical advantage: a difference equation can be rearranged into an explicit recipe for computing the next output, and that recipe is a computer program. Take the general equation, isolate the \(k=0\) term of the left side, and divide by \(a_0\).
Every quantity on the right is either a present or past input, or a previously computed output. Starting from initial rest — all stored \(y\) values zero before the input begins — the formula generates \(y[0]\), then \(y[1]\) using \(y[0]\), then \(y[2]\), and so on indefinitely. No integration, no trial solutions, no characteristic polynomial: just arithmetic, repeated. This is why digital filters are implemented exactly as written.
The recursion also makes the role of the auxiliary conditions vivid. The formula needs \(N\) past output values to get started, which is the discrete counterpart of needing \(N\) initial conditions, and initial rest supplies them as zeros. The characteristic-root machinery still exists in discrete time and is used when a closed form is wanted rather than a table of numbers. Substituting the trial solution \(y[n]=z^{n}\) into the homogeneous equation gives
with natural response \(y_h[n]=\sum_i C_i z_i^{\,n}\) for distinct roots, and \(t^{m}\) replaced by \(n^{m}\) for repeated ones. The stability boundary changes shape, though: a term \(z_i^{\,n}\) decays when \(|z_i| \lt 1\), so the stable region in discrete time is the inside of the unit circle, not a half-plane. That is the picture Chapter 25 will make central.
Impulse Response, FIR and IIR
The bridge back to Chapter 7 is now short: set \(x[n]=\delta[n]\), impose initial rest, run the recursion, and whatever comes out is \(h[n]\).
The character of the answer depends entirely on whether the equation is recursive. Suppose \(N=0\), so no past outputs appear. Dividing by \(a_0\) leaves the output as a weighted sum of a finite number of input samples, and feeding it an impulse produces exactly the list of coefficients.
The impulse response has finite length \(M+1\): the system is FIR, finite impulse response, and this equation is exactly the convolution sum written out. Such a system is always BIBO stable, since a finite sum of finite numbers cannot diverge, and it has no characteristic roots to worry about.
If \(N \ge 1\), a computed output is fed back into the next computation, and the effect of a single input sample never entirely disappears. The impulse response is a combination of terms \(z_i^{\,n}\) which continue for ever: the system is IIR, infinite impulse response. Stability is now a real question, decided by whether every characteristic root satisfies \(|z_i| \lt 1\).
| FIR (non-recursive, \(N=0\)) | IIR (recursive, \(N \ge 1\)) | |
|---|---|---|
| Impulse response | Finite length \(M+1\) | Infinite; built from \(z_i^{\,n}\) |
| Feedback | None | Past outputs are reused |
| Stability | Always BIBO stable | Stable iff every \(|z_i| \lt 1\) |
| Memory span | \(M+1\) samples, then forgets | Never forgets completely |
| Typical cost | Many coefficients for a sharp response | Sharp response from few coefficients |
The trade-off in the last row is the reason both families survive. A recursive filter achieves with three or four coefficients what a non-recursive one needs fifty to approach, because feedback lets a small structure remember a long history. The price is the possibility of instability and, in fixed-point arithmetic, the accumulation of rounding error through the feedback path. Continuous-time systems have no FIR analogue worth the name: any circuit with a capacitor is inescapably an infinite-memory system.
Block-Diagram Realisations
A difference equation can be drawn as well as written, using three elements: an adder, a constant-gain multiplier, and a unit delay \(z^{-1}\). The drawing is called a realisation, and it is what a hardware engineer builds or a programmer allocates memory for. Take the first-order equation \(y[n] + a_1 y[n-1] = b_0 x[n] + b_1 x[n-1]\), rearranged as \(y[n] = b_0x[n] + b_1x[n-1] - a_1y[n-1]\).
The literal transcription builds the input side first, forming \(w[n]=b_0x[n]+b_1x[n-1]\), and then applies the feedback. This is Direct Form I, and it needs one delay for the input history and one for the output history.
Direct Form II comes from a single observation. Direct Form I is a cascade of two systems — an all-input section followed by an all-feedback section — and by the commutativity of Chapter 8 the two may be exchanged without altering the overall response. Put the feedback section first, and both delay chains are now storing the same intermediate signal \(w[n]\); one of them is redundant and can be deleted.
Direct Form II is canonical — no realisation of the same equation uses fewer delay elements. The saving is memory, and it is bought with an argument from Chapter 8, not with any new theory.
In continuous time the same diagrams are drawn with integrators in place of delays, which is a deliberate choice: differentiators amplify high-frequency noise, whereas integrators smooth it, so a practical analogue computer or op-amp realisation always integrates. Chapter 27 will read the signals stored in those integrators as the state of the system, and the block diagram will become a state-space model.
Worked Examples
Problem. The circuit of Section 9-1 obeys \(RC\,\dfrac{dy}{dt}+y(t)=x(t)\). Writing \(\tau=RC\) and assuming initial rest, find the step response and hence the impulse response.
Solution. Take \(x(t)=u(t)\). For \(t \gt 0\) the equation is \(\tau\dot{y}+y=1\). The characteristic equation \(\tau s + 1 = 0\) has the single root \(s=-1/\tau\), so \(y_h = Ce^{-t/\tau}\). The input is constant, so try \(y_p = K\); substituting gives \(0 + K = 1\), hence \(K=1\). The complete solution is
Initial rest gives \(y(0^-)=0\). The input contains no impulse and the equation is first order with \(M=0\), so \(y\) is continuous at the origin and \(y(0^+)=0\) as well. Then \(1+C=0\), so \(C=-1\) and
Section 8-9 now supplies the impulse response for free: differentiate. The product rule gives \(h(t) = \tfrac{1}{\tau}e^{-t/\tau}u(t) + \big(1-e^{-t/\tau}\big)\delta(t)\), and the impulse term vanishes because the bracket is zero at \(t=0\). So
Two checks. The area is \(\int_0^\infty \tfrac1\tau e^{-t/\tau}dt = 1\), matching the DC gain of unity that the equation predicts at steady state. And the area is finite, so by Chapter 8 the system is BIBO stable — as it must be, since its only characteristic root, \(-1/\tau\), lies in the left half-plane.
Problem. Solve \(\dfrac{d^{2}y}{dt^{2}}+5\dfrac{dy}{dt}+6y(t)=x(t)\) for \(x(t)=u(t)\) under initial rest.
Solution. The characteristic equation is \(s^{2}+5s+6=0\), which factors as \((s+2)(s+3)=0\), giving roots \(s=-2\) and \(s=-3\). Both lie in the left half-plane, so the natural response decays:
For \(t \gt 0\) the input is the constant 1, so try \(y_p=K\). Both derivatives vanish and the equation reduces to \(6K=1\), giving \(y_p = \tfrac16\) — the DC gain, as expected from \(b_0/a_0 = 1/6\). The complete solution for \(t \gt 0\) is \(y(t) = \tfrac16 + Ae^{-2t}+Be^{-3t}\).
Apply initial rest. The input has no impulses and the order is 2 with \(M=0\), so \(y\) and \(\dot{y}\) are continuous at the origin, and both are zero there. Differentiating the solution gives \(\dot{y}(t) = -2Ae^{-2t}-3Be^{-3t}\), so the two conditions are
The second gives \(A=-\tfrac32 B\). Substituting into the first, \(\tfrac16 - \tfrac32 B + B = 0\), so \(\tfrac16 = \tfrac12 B\) and \(B = \tfrac13\); then \(A = -\tfrac12\). The complete response is
Verify at \(t=0\): \(\tfrac16-\tfrac12+\tfrac13 = \tfrac{1-3+2}{6}=0\) ✓, and \(\dot{y}(0) = 1 - 1 = 0\) ✓. As \(t\to\infty\) the exponentials die and the output settles at \(\tfrac16\), the steady-state value. The two exponential terms are the transient; the constant \(\tfrac16\) is the steady state.
Problem. A system obeys \(\dfrac{dy}{dt}+3y(t)=x(t)\) with the capacitor already charged so that \(y(0^-)=2\). The input applied at \(t=0\) is \(x(t)=e^{-t}u(t)\). Find the zero-input, zero-state and total responses, and identify the natural and forced parts of the total.
Solution — zero-input. Set \(x=0\). The characteristic root is \(s=-3\), so \(y_{\text{zi}}(t)=Ce^{-3t}\), and the stored energy fixes \(C\): with \(y(0)=2\), \(C=2\) and \(y_{\text{zi}}(t)=2e^{-3t}\) for \(t\ge0\).
Solution — zero-state. With the system at rest, Example 1's method (or the recipe of Section 9-4) gives the impulse response \(h(t)=e^{-3t}u(t)\), and \(y_{\text{zs}}=x*h\). Both signals are causal, so the convolution runs from 0 to \(t\):
Total. Add the two:
Check by substitution: \(\dot{y} = -\tfrac12 e^{-t}-\tfrac92 e^{-3t}\), and \(\dot{y}+3y = \left(-\tfrac12+\tfrac32\right)e^{-t} + \left(-\tfrac92+\tfrac92\right)e^{-3t} = e^{-t}\) ✓, with \(y(0)=\tfrac12+\tfrac32=2\) ✓.
Now read the two decompositions off the same answer. Zero-input / zero-state: \(2e^{-3t}\) and \(\tfrac12\left(e^{-t}-e^{-3t}\right)\). Natural / forced: the natural terms are those in \(e^{-3t}\), totalling \(\tfrac32 e^{-3t}\), and the forced term is \(\tfrac12 e^{-t}\), which carries the input's own exponential. The two splits are different: the zero-state response contains a natural term \(-\tfrac12 e^{-3t}\), which is there solely to satisfy the rest condition, and no rearrangement makes the halves line up.
Problem. For \(y[n]-\tfrac12 y[n-1]=x[n]\) under initial rest, find \(h[n]\) by recursion, and find the step response in closed form.
Solution — impulse response. Rearranged, \(y[n]=x[n]+\tfrac12 y[n-1]\). Put \(x[n]=\delta[n]\) with \(y[-1]=0\) and iterate: \(h[0]=1+0=1\); \(h[1]=0+\tfrac12(1)=\tfrac12\); \(h[2]=0+\tfrac12\left(\tfrac12\right)=\tfrac14\). The pattern is a geometric sequence, and it is confirmed by the characteristic equation \(z-\tfrac12=0\):
It never reaches zero, so the system is IIR; and \(\sum_n |h[n]| = 1/(1-\tfrac12) = 2\) is finite, so it is stable — consistent with the root \(z=\tfrac12\) lying inside the unit circle.
Solution — step response. Two routes agree. Recursively with \(x[n]=u[n]\): \(y[0]=1\), \(y[1]=1+\tfrac12=\tfrac32\), \(y[2]=1+\tfrac34=\tfrac74\), \(y[3]=\tfrac{15}{8}\), climbing towards 2. Analytically, the step response is the running sum of \(h\), which is a finite geometric series:
which gives \(1, \tfrac32, \tfrac74, \tfrac{15}{8}\) for \(n=0,1,2,3\), matching the recursion exactly, and tends to the DC gain \(s[\infty]=2\).
Problem. Find the impulse response of \(y[n]-\tfrac34 y[n-1]+\tfrac18 y[n-2]=x[n]\) under initial rest, and test the system for stability.
Solution. The characteristic equation is \(z^{2}-\tfrac34 z+\tfrac18=0\). Its discriminant is \(\tfrac{9}{16}-\tfrac12=\tfrac{1}{16}\), whose square root is \(\tfrac14\), so
so \(h[n] = A\left(\tfrac12\right)^{n}+B\left(\tfrac14\right)^{n}\) for \(n\ge0\). To fix \(A\) and \(B\) we need two values of \(h\), and the recursion supplies them. With \(y[-1]=y[-2]=0\) and \(x[n]=\delta[n]\): \(h[0]=1\), and \(h[1]=\tfrac34 h[0]-\tfrac18 h[-1]=\tfrac34\). So
Multiplying the second by 4 gives \(2A+B=3\); subtracting the first gives \(A=2\), hence \(B=-1\):
Check against the recursion at \(n=2\). The formula gives \(2\left(\tfrac14\right)-\tfrac{1}{16}=\tfrac12-\tfrac1{16}=\tfrac{7}{16}\); the recursion gives \(h[2]=\tfrac34\left(\tfrac34\right)-\tfrac18(1)=\tfrac{9}{16}-\tfrac{2}{16}=\tfrac{7}{16}\) ✓.
Both roots satisfy \(|z| \lt 1\), so the system is stable. Confirming directly, \(\sum_n|h[n]| \le 2\sum_n\left(\tfrac12\right)^{n} + \sum_n\left(\tfrac14\right)^{n} = 4 + \tfrac43\), comfortably finite.
Problem. The difference equation \(y[n]-2y[n-1]=x[n]\) is solved (a) forwards in \(n\) with initial rest, and (b) backwards in \(n\), with the output taken to be zero for large positive \(n\). Find both impulse responses and compare.
Solution (a) — the causal system. Rearranged, \(y[n]=x[n]+2y[n-1]\). With \(x[n]=\delta[n]\) and \(y[-1]=0\): \(h[0]=1\), \(h[1]=2\), \(h[2]=4\), and in general
This is causal, but \(\sum_n 2^{n}\) diverges, so the system is unstable — as the characteristic root \(z=2\) lying outside the unit circle already announced.
Solution (b) — the anti-causal system. Solve the same equation for the earlier sample instead: \(y[n-1]=\tfrac12\big(y[n]-x[n]\big)\), and step backwards from large \(n\), where the output is zero. For \(n \ge 1\) nothing has happened yet in this direction, so \(h_b[n]=0\) for \(n\ge0\); at \(n=0\) the recursion gives \(h_b[-1]=\tfrac12(h_b[0]-1)=-\tfrac12\), then \(h_b[-2]=\tfrac12 h_b[-1]=-\tfrac14\), and so on:
Verify that it satisfies the equation. For \(n \le -1\): \(h_b[n]-2h_b[n-1] = -2^{n}+2\cdot2^{n-1} = 0\) ✓. For \(n=0\): \(h_b[0]-2h_b[-1] = 0 - 2\left(-\tfrac12\right) = 1 = \delta[0]\) ✓. For \(n\ge1\) both terms are zero ✓.
Now compare. Summing the magnitudes, \(\sum_{n\le-1}2^{n} = \tfrac12+\tfrac14+\cdots = 1\), so system (b) is stable — but it is non-causal, responding entirely before the impulse arrives. One difference equation, two auxiliary conditions, and two systems with opposite properties. This is the strongest possible illustration of Section 9-2's warning, and it is the phenomenon that Chapter 23 will formalise as the region of convergence of the \(z\)-transform.
Chapter Summary
An \(N\)th-order LCCDE has an \(N\)-parameter family of solutions. Without \(N\) auxiliary conditions it is not a system at all.
Substituting \(e^{st}\) or \(z^{n}\) collapses the equation to a polynomial. Its roots give the natural response, one term each.
\(y = y_h + y_p\). Find the particular solution first, add the natural terms, and only then impose the conditions.
Zero output before the input begins. This one assumption makes the equation describe a linear, time-invariant, causal system.
Zero-input plus zero-state splits by cause; natural plus forced splits by form. Only \(y_{\text{zs}}\) equals \(x*h\).
\(N=0\) gives a finite \(h\) and unconditional stability; \(N\ge1\) gives feedback, an infinite \(h\), and stability only if every \(|z_i| \lt 1\).
Practice Problems
Problems 1 to 3 are continuous-time solutions; 4 to 6 are discrete; 7 and 8 test whether you have absorbed the point that an equation is only half a system. Always state the characteristic roots before writing any constants.
- Solve \(\dfrac{dy}{dt}+4y(t)=x(t)\) with \(x(t)=u(t)\) under initial rest. Identify the time constant, the DC gain, and the time at which the output reaches 98% of its final value.
- For \(\dfrac{d^{2}y}{dt^{2}}+4\dfrac{dy}{dt}+4y(t)=x(t)\), find the characteristic roots and write the general form of the natural response. Why does this system need a \(te^{-2t}\) term, and what is this damping condition called?
- A system obeys \(\dfrac{dy}{dt}+2y(t)=x(t)\) with \(y(0^-)=-1\) and \(x(t)=3u(t)\). Find the zero-input and zero-state responses separately, add them, and confirm the total satisfies both the equation and the initial condition.
- Find the impulse response of \(y[n]-\tfrac13 y[n-1]=x[n]+x[n-1]\) under initial rest, by running the recursion for \(n=0,1,2,3\) and then writing a closed form. Is the system FIR or IIR?
- The moving average \(y[n]=\tfrac14\big(x[n]+x[n-1]+x[n-2]+x[n-3]\big)\) is given. Write its impulse response, state its memory span, and explain in one sentence why no stability calculation is needed.
- For \(y[n]-y[n-1]+\tfrac14 y[n-2]=x[n]\), find the characteristic roots, note that they are repeated, and write the impulse response in the form \((A+Bn)\left(\tfrac12\right)^{n}u[n]\) by matching \(h[0]\) and \(h[1]\) from the recursion.
- Show that the system defined by \(\dot{y}+5y=x\) together with the fixed condition \(y(0)=4\) is neither linear nor time-invariant, by producing an explicit counterexample for each property.
- Draw Direct Form I and Direct Form II realisations of \(y[n]+\tfrac12 y[n-1]-\tfrac18 y[n-2] = x[n]+3x[n-1]\), count the delay elements in each, and say which form you would implement on a processor with limited memory.