Systems Described by Differential and Difference Equations

Linear constant-coefficient equations give a compact time-domain model of an LTI system. A differential equation describes a continuous-time (CT) system, whereas a difference equation describes a discrete-time (DT) system. Their Laplace- and $Z$-domain ratios characterize the zero-state input–output behavior; initial stored energy must be handled separately.

Continuous-Time LCCDE Model

The general $N$th-order linear constant-coefficient differential equation (LCCDE) is

Caution

Key Formula — General continuous-time system equation

$$\sum_{k=0}^{N}a_k\frac{\mathrm{d}^k y(t)}{\mathrm{d}t^k} =\sum_{m=0}^{M}b_m\frac{\mathrm{d}^m x(t)}{\mathrm{d}t^m}, \qquad a_N\neq 0.$$

Here $a_k$ and $b_m$ are constants, $x(t)$ is the input, and $y(t)$ is the output. With the differential operator $D=\mathrm{d}/\mathrm{d}t$, the same equation is written

$$A(D)y(t)=B(D)x(t), \quad A(D)=\sum_{k=0}^{N}a_kD^k, \quad B(D)=\sum_{m=0}^{M}b_mD^m.$$

The coefficients and the required initial values $y(0^-),y'(0^-),\ldots,y^{(N-1)}(0^-)$ determine a unique response for a specified input under the usual regularity assumptions.

Transfer function and the zero-state qualification

With all initial conditions zero, the Laplace differentiation property gives

$$A(s)Y(s)=B(s)X(s), \quad A(s)=\sum_{k=0}^{N}a_ks^k, \quad B(s)=\sum_{m=0}^{M}b_ms^m.$$

Therefore

Caution

Key Formula — Continuous-time transfer function

$$H(s)\equiv\left.\frac{Y(s)}{X(s)}\right|_{\text{zero initial state}} =\frac{B(s)}{A(s)} =\frac{\sum_{m=0}^{M}b_ms^m} {\sum_{k=0}^{N}a_ks^k}.$$

After common factors are cancelled, roots of $B(s)$ are the zeros and roots of $A(s)$ are the poles. Pole locations set the natural modes and transient decay; zeros shape how the input excites those modes and how frequencies are attenuated.

The zero-initial-condition phrase is essential. For the unilateral Laplace transform,

$$\mathcal{L}_{u}\!\left\{y^{(k)}(t)\right\} =s^kY(s)-\sum_{r=0}^{k-1}s^{k-1-r}y^{(r)}(0^-).$$

Nonzero initial values consequently appear as additional source terms. In general the transformed equation has the form

$$A(s)Y(s)=B(s)X(s)+I_y(s)-I_x(s),$$

where $I_y(s)$ contains output initial conditions and $I_x(s)$ contains any input prehistory terms introduced by derivatives of $x(t)$. Thus

$$Y(s)=H(s)X(s)+Y_{\mathrm{ic}}(s),$$

not merely $Y(s)=H(s)X(s)$. The ratio in the equation is a system property only when the initial state is zero; dividing the total $Y(s)$ by $X(s)$ when initial energy is present does not produce the transfer function.

Tip

Exam Focus — Unilateral-transform trap

The unilateral transform is convenient because it exposes initial conditions. Do not silently replace $\mathcal{L}_{u}\{y'(t)\}=sY(s)-y(0^-)$ by $sY(s)$. Conversely, when deriving $H(s)$, deliberately set the initial state to zero.

Response Components and Characteristic Modes

Two useful decompositions answer different questions:

$$\begin{aligned} y(t)&=y_{\mathrm{zi}}(t)+y_{\mathrm{zs}}(t), \\ y(t)&=y_{\mathrm{natural}}(t)+y_{\mathrm{forced}}(t). \end{aligned}$$

• Zero-input response $y_{\mathrm{zi}}$: Set $x(t)=0$ and use the actual initial conditions. It is caused by stored energy.

• Zero-state response $y_{\mathrm{zs}}$: Set every initial condition to zero and retain the input. For an LTI system, $y_{\mathrm{zs}}(t)=x(t)*h(t)$.

• Natural response: Solve the homogeneous equation $A(D)y=0$. Its form is fixed by the characteristic roots.

• Forced response: Choose a particular solution associated with the applied input.

The characteristic equation is

$$A(\lambda)=a_N\lambda^N+a_{N-1}\lambda^{N-1}+\cdots+a_0=0.$$

If $p_i$ is a pole of multiplicity $q_i$, its natural contribution is

$$\left(C_{i,0}+C_{i,1}t+\cdots+C_{i,q_i-1}t^{q_i-1}\right)e^{p_it}.$$

Constants are found from initial and continuity conditions. A forced form that duplicates a natural mode must be multiplied by a sufficient power of $t$ (the resonance rule).

The two decompositions should not be identified term by term. The zero-input response is homogeneous, but the zero-state response can also contain natural transient terms needed to satisfy its zero initial conditions. Likewise, “natural” and “transient” are not exact synonyms: a pole on the imaginary axis creates a natural mode that does not decay.

Frequency Response, Causality, and Stability in the s-Plane

If the region of convergence (ROC) of $H(s)$ includes the imaginary axis, the CT frequency response is

Caution

Key Formula — Continuous-time frequency response

$$H(\mathrm{j}\omega)=H(s)\big|_{s=\mathrm{j}\omega}.$$

For a stable LTI system driven by $x(t)=A\cos(\omega_0t+\phi)$, the steady-state output is

$$y_{\mathrm{ss}}(t) =A\left\lvert H(\mathrm{j}\omega_0)\right\rvert \cos\!\left(\omega_0t+\phi+\angle H(\mathrm{j}\omega_0)\right).$$

The frequency remains $\omega_0$; only amplitude and phase change. This statement concerns steady state, after decaying natural transients have died out.

For a rational transform, the algebraic expression alone does not uniquely specify $h(t)$: the ROC must also be known.

• Causal system: For a rational $H(s)$, a right-sided impulse response has an ROC to the right of the rightmost pole.

• BIBO-stable system: The impulse response obeys $\int_{-\infty}^{\infty}\left\lvert h(t)\right\rvert\mathrm{d}t<\infty$; equivalently, the ROC of $H(s)$ contains the entire $\mathrm{j}\omega$ axis.

• Causal and stable rational system: Every pole of the reduced $H(s)$ lies strictly in the open left half-plane.

A right-half-plane pole does not by itself rule out BIBO stability for an anti-causal realization: a left-sided impulse response may have an ROC to the left of that pole and include the imaginary axis. A two-sided response has a vertical-strip ROC. Therefore “left-half-plane poles” is the stability test only after causality has been stated. Poles on the imaginary axis are excluded from the ROC and prevent BIBO stability. A proper rational function ($M\leq N$) has an ordinary causal realization; an improper ratio contains differentiator behavior and is not BIBO stable as an ordinary input–output filter.

Note

Example — First-Order Continuous-Time System

Consider

$$\frac{\mathrm{d}y(t)}{\mathrm{d}t}+ay(t)=bx(t),\qquad a>0.$$

Transfer function and impulse response. With zero initial state,

$$(s+a)Y(s)=bX(s), \qquad \boxed{H(s)=\frac{b}{s+a}}.$$

For the causal ROC $\operatorname{Re}\{s\}>-a$,

$$\boxed{h(t)=b\mathrm{e}^{-at}u(t)}.$$

The sole pole is $s=-a$, which is in the open left half-plane. Also,

$$\int_0^\infty\left\lvert b\mathrm{e}^{-at}\right\rvert\mathrm{d}t=\frac{\left\lvert b\right\rvert}{a}<\infty,$$

so the causal system is BIBO stable.

Step response with a nonzero initial value. Let $x(t)=u(t)$ and $y(0^-)=y_0$. The unilateral transform gives

$$sY(s)-y_0+aY(s)=\frac{b}{s},$$

hence

$$\begin{aligned} Y(s) &=\underbrace{\frac{y_0}{s+a}}_{Y_{\mathrm{zi}}(s)} +\underbrace{\frac{b}{s(s+a)}}_{Y_{\mathrm{zs}}(s)},\\ y_{\mathrm{zi}}(t)&=y_0\mathrm{e}^{-at}u(t),\\ y_{\mathrm{zs}}(t)&=\frac{b}{a}\left(1-\mathrm{e}^{-at}\right)u(t). \end{aligned}$$

Therefore

$$\boxed{y(t)=\left[y_0\mathrm{e}^{-at} +\frac{b}{a}\left(1-\mathrm{e}^{-at}\right)\right]u(t)}.$$

Equivalently, for $t\geq0$,

$$y(t)=\underbrace{\left(y_0-\frac ba\right)\mathrm{e}^{-at}}_{\text{natural}} +\underbrace{\frac ba}_{\text{forced}}.$$

The time constant is $\tau=1/a$, the transient decays, and $y(\infty)=b/a$.

Discrete-Time LCC Difference Equation

A causal finite-order LTI recursion is commonly normalized as

Caution

Key Formula — General discrete-time system equation

$$y[n]+\sum_{k=1}^{N}a_ky[n-k] =\sum_{m=0}^{M}b_mx[n-m].$$

The present output is obtained from delayed outputs and present or delayed inputs. The $N$ past output samples are the recursive memory; a feedforward-only equation has $N=0$.

With zero initial state, time shifting gives

$$\left(1+\sum_{k=1}^{N}a_kz^{-k}\right)Y(z) =\left(\sum_{m=0}^{M}b_mz^{-m}\right)X(z).$$

Thus

Caution

Key Formula — Discrete-time system function

$$H(z)\equiv\left.\frac{Y(z)}{X(z)}\right|_{\text{zero initial state}} =\frac{\sum_{m=0}^{M}b_mz^{-m}} {1+\sum_{k=1}^{N}a_kz^{-k}}.$$

Multiplying numerator and denominator by $z^L$, where $L\geq\max(M,N)$, exposes the same poles and zeros as polynomials in $z$. Any common factors must be cancelled for the external BIBO test, although an exact cancellation can still conceal an unstable internal mode in a physical realization.

One-sided Z-transform and initial samples

For $Y^+(z)=\sum_{n=0}^{\infty}y[n]z^{-n}$,

$$\mathcal{Z}_{u}\{y[n-k]\} =z^{-k}\left[Y^+(z)+\sum_{r=-k}^{-1}y[r]z^{-r}\right].$$

The samples $y[-1],\ldots,y[-N]$ therefore appear explicitly. They must be retained when finding the total response and set to zero only when deriving $H(z)$ or the zero-state response. Detailed $Z$-transform properties are developed in the later dedicated section; here the transform is used only to solve the system equation.

As in CT,

$$y[n]=y_{\mathrm{zi}}[n]+y_{\mathrm{zs}}[n], \qquad y_{\mathrm{zs}}[n]=x[n]*h[n].$$

The homogeneous equation has characteristic polynomial

$$\lambda^N+a_1\lambda^{N-1}+\cdots+a_N=0.$$

For a root $p_i$ of multiplicity $q_i$, the natural sequence contains

$$\left(C_{i,0}+C_{i,1}n+\cdots+C_{i,q_i-1}n^{q_i-1}\right)p_i^n.$$

A particular sequence supplies the forced response. If the input mode is a characteristic mode, multiply the trial particular sequence by the required power of $n$.

Frequency Response, Causality, and Stability in the z-Plane

If the ROC contains the unit circle, the DT frequency response is

Caution

Key Formula — Discrete-time frequency response

$$H(\mathrm{e}^{\mathrm{j}\Omega})=H(z)\big|_{z=\mathrm{e}^{\mathrm{j}\Omega}}, \qquad -\pi\leq\Omega\leq\pi,$$

and is $2\pi$-periodic in the normalized digital frequency $\Omega$. For an input $A\cos(\Omega_0n+\phi)$, a stable system produces the steady-state sinusoid

$$A\left\lvert H(\mathrm{e}^{\mathrm{j}\Omega_0})\right\rvert \cos\!\left(\Omega_0n+\phi+\angle H(\mathrm{e}^{\mathrm{j}\Omega_0})\right).$$

The ROC qualifications mirror the $s$-plane case:

• Causal rational system: Its impulse response is right-sided, and the ROC is outside the outermost pole.

• BIBO-stable system: It obeys $\sum_{n=-\infty}^{\infty}\left\lvert h[n]\right\rvert<\infty$; equivalently, the ROC contains the unit circle.

• Causal and stable rational system: Every pole of the reduced $H(z)$ lies strictly inside the unit circle.

An anti-causal system may be stable with poles outside the unit circle if its inside-the-innermost-pole ROC contains $|z|=1$. A two-sided sequence has an annular ROC between poles. Hence a pole plot without an ROC cannot establish causality, and “poles inside the unit circle” is specifically the causal rational stability rule. A pole on the unit circle prevents the unit circle from lying in the ROC and is not BIBO stable.

Note

Example — First-Order Discrete-Time System

Consider the causal recursion

$$y[n]-ay[n-1]=bx[n],\qquad \left\lvert a\right\rvert<1.$$

Zero-state system function. Setting the initial state to zero gives

$$Y(z)-az^{-1}Y(z)=bX(z), \qquad \boxed{H(z)=\frac{b}{1-az^{-1}}}.$$

For the causal ROC $|z|>|a|$,

$$\boxed{h[n]=ba^nu[n]}.$$

Its pole is $z=a$. Since $|a|<1$, the causal ROC contains the unit circle and

$$\sum_{n=0}^{\infty}\left\lvert ba^n\right\rvert =\frac{\left\lvert b\right\rvert}{1-|a|}<\infty;$$

the system is BIBO stable.

Step response and initial state. For $x[n]=u[n]$ and $y[-1]=y_{-1}$, the one-sided transform is

$$Y^+(z)-a\left[z^{-1}Y^+(z)+y_{-1}\right] =\frac{b}{1-z^{-1}}.$$

Consequently

$$\begin{aligned} Y^+(z) &=\underbrace{\frac{ay_{-1}}{1-az^{-1}}}_{Y_{\mathrm{zi}}(z)} +\underbrace{\frac{b}{(1-z^{-1})(1-az^{-1})}}_{Y_{\mathrm{zs}}(z)},\\ y_{\mathrm{zi}}[n]&=a^{n+1}y_{-1}u[n],\\ y_{\mathrm{zs}}[n]&=b\sum_{r=0}^{n}a^r u[n] =\boxed{\frac{b(1-a^{n+1})}{1-a}u[n]}. \end{aligned}$$

The total response is therefore

$$\boxed{y[n]=\left[a^{n+1}y_{-1} +\frac{b(1-a^{n+1})}{1-a}\right]u[n]}.$$

The $a^{n+1}$ terms are natural transients, and $\lim_{n\to\infty}y[n]=b/(1-a)$.

Continuous- and Discrete-Time Correspondence

Feature	Continuous time	Discrete time
Time-domain model	$\sum a_k y^{(k)}=\sum b_m x^{(m)}$	$y[n]+\sum a_ky[n-k]=\sum b_mx[n-m]$
Transform ratio	$H(s)=B(s)/A(s)$, zero initial state	$H(z)=B(z)/A(z)$, zero initial state
Natural mode	$t^r\mathrm{e}^{pt}$ for a repeated pole $p$	$n^rp^n$ for a repeated pole $p$
Frequency response	Set $s=\mathrm{j}\omega$ if the ROC includes the imaginary axis	Set $z=\mathrm{e}^{\mathrm{j}\Omega}$ if the ROC includes the unit circle
Causal rational ROC	Right of the rightmost pole	Outside the outermost pole
Causal BIBO test	Poles strictly in the open left half-plane	Poles strictly inside the unit circle

Parallel system-equation facts in continuous and discrete time.

Tip

Exam Focus — System-equation exam traps

1. State “zero initial conditions” before forming $H(s)=Y/X$ or $H(z)=Y/X$.

2. Keep unilateral-transform initial terms when the total response is requested; $y(0^-)$ and $y[-1]$ are not automatically zero.

3. Do not confuse zero-input/zero-state with natural/forced response.

4. A frequency response exists only when the relevant frequency contour lies in the ROC.

5. Pole location alone is incomplete without causality or the ROC. The familiar left-half-plane and inside-unit-circle tests assume a causal rational system.

6. Read the sign convention from the equation. In $y[n]+\sum a_ky[n-k]=\sum b_mx[n-m]$, feedback multipliers are $-a_k$.

[!NOTE]

Key Point — Quick Review

An LCCDE or LCC difference equation determines both forced behavior and characteristic modes. The transfer/system function describes only the zero-state response. Initial energy produces a separate zero-input response, and the ROC is indispensable when deciding causality, stability, and whether the frequency response exists.