2.04 Systems Described by Differential and Difference Equati

Related Concepts: 2.01 Systems Classification, LTI Properties & Stability, 2.02 The Continuous Convolution Integral, 2.03 Properties of the Convolution Integral, 2.05 Block-Diagram Representations & System Interconnections

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2.04 Systems Described by Differential & Difference Equations

1. The Classical Modeling Paradigm

Alright — today we are diving into the mathematical engine room of LTI systems. Up until now, we’ve treated systems as “black boxes” represented by their impulse response $h(t)$ or $h[n]$. But how do we model the physical reality of circuits containing inductors, capacitors, or discrete digital feedback loops?

We use Linear Constant-Coefficient Differential Equations (LCCDEs) for continuous-time systems and Linear Constant-Coefficient Difference Equations for discrete-time systems. These models are the mathematical bedrock of physical engineering.

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2. Time-Domain Solutions of Continuous LCCDEs

A general $N$-th order continuous-time system is modeled by the differential equation: ${\sum }_{k=0}^N{a}_k\frac{d^{k}y(t)}{d{t}^k}={\sum }_{m=0}^M{b}_k\frac{d^{m}x(t)}{d{t}^m}$

To find the output response $y(t)$ for a given input $x(t)$, classical time-domain analysis splits the solution into two distinct parts: $y(t)=y_h(t)+y_p(t)$

A. The Homogeneous (Complementary) Solution, $y_h(t)$

The homogeneous solution $y_h(t)$ {the natural response of the system representing its behavior without any external excitation} is found by setting the input $x(t)=0$: ${\sum }_{k=0}^N{a}_k\frac{d^k{y}_h(t)}{d{t}^k}=0$

To solve this, we assume a trial solution of the form $y_h(t)=C{e}^{\lambda t}$. Substituting this trial solution yields the characteristic equation: ${\sum }_{k=0}^N{a}_k{\lambda }^k=0⟹a_N{\lambda }^N+a_{N-1}{\lambda }^{N-1}+\cdots +a_1\lambda +a_0=0$

Solving this polynomial gives the characteristic roots (or system eigenvalues) ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$. The form of $y_h(t)$ depends entirely on the nature of these roots:

1. Distinct Real Roots (${\lambda }_1\ne {\lambda }_2\ne \dots$): $y_h(t)=C_1{e}^{{\lambda }_{1}t}+C_2{e}^{{\lambda }_{2}t}+\cdots +C_N{e}^{{\lambda }_{N}t}$
2. Repeated Real Roots (${\lambda }_1={\lambda }_2=\lambda$ with multiplicity $r$): $y_h(t)=(C_1+C_{2}t+C_3{t}^2+\cdots +C_r{t}^{r-1})e^{\lambda t}$
3. Complex Conjugate Pairs ($\lambda =\alpha \pm j\beta$): $y_h(t)=e^{\alpha t}[A\cos (\beta t)+B\sin (\beta t)]$

Physical Meaning of Roots

The real part $\alpha$ of the characteristic roots dictates the damping of the natural response. If $\alpha <0$, the natural response decays exponentially, representing a stable physical system. The imaginary part $\beta$ dictates the frequency of natural oscillations.

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B. The Particular Solution, $y_p(t)$

The particular solution $y_p(t)$ {the forced response of the system representing its behavior driven entirely by the input signal} satisfies the differential equation for the given input $x(t)$. We find $y_p(t)$ using the Method of Undetermined Coefficients by matching the mathematical form of the input $x(t)$:

Input Shape $x(t)$	Assumed Particular Solution $y_p(t)$
Constant: $A$	$K$ (Constant)
Exponential: $A{e}^{st}$	$K{e}^{st}$ (provided $s$ is not a characteristic root)
Polynomial: $A{t}^n$	$K_n{t}^n+K_{n-1}{t}^{n-1}+\cdots +K_0$
Sinusoidal: $A\cos (\omega t+\theta )$	$K_1\cos (\omega t)+K_2\sin (\omega t)$

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3. Step-by-Step Continuous Numerical Solution

Let’s apply this classical pipeline to solve a standard exam-scoring continuous-time system.

Continuous-Time Step and Impulse Responses

Question: Consider the continuous LTI system described by the first-order differential equation: $\frac{dy(t)}{dt}+y(t)=x(t)$ Determine the Step Response and the Impulse Response of this system, assuming the system is initially relaxed ($y(0)=0$ at $t=0$).

Part I: Solving for the Step Response, $s(t)$

For the step response, the input is $x(t)=u(t)$. We analyze the system for $t>0$, where $x(t)=1$.

Step 1: Find the Homogeneous Solution, $y_h(t)$

Set the right-hand side of the differential equation to zero: $\frac{d{y}_h(t)}{dt}+y_h(t)=0$ Assume $y_h(t)=C{e}^{\lambda t}$. The characteristic equation is: $\lambda +1=0⟹\lambda =-1$ Thus, the homogeneous solution is: $y_h(t)=C{e}^{-t}$

Step 2: Find the Particular Solution, $y_p(t)$

Since the input for $t>0$ is a constant ($x(t)=1$), we assume a constant particular solution: $y_p(t)=K$ Substitute $y_p(t)=K$ into our governing differential equation: $\frac{d(K)}{dt}+K=1⟹0+K=1⟹K=1$ Thus, our particular solution is: $y_p(t)=1$

Step 3: Combine and Apply Initial Conditions

The total response is the sum of the homogeneous and particular responses: $y(t)=y_h(t)+y_p(t)=C{e}^{-t}+1\text{for }t>0$

Now, we apply the initial condition $y(0)=0$ to find the undetermined constant $C$: $y(0)=C{e}^0+1=0⟹C+1=0⟹C=-1$

Substituting $C=-1$ back into our total response yields our step response: $s(t)=(1-e^{-t})u(t)$

[GRAPH: Step response curve y(t) rising exponentially from 0 at t=0 and asymptotically flattening toward 1 as t approaches infinity. — governing equation: y(t) = (1 - e^-t)u(t) — source: Rabiul Sir Notes, Lec 6]

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Part II: Solving for the Impulse Response, $h(t)$

The impulse response $h(t)$ is the system output when the input is a unit impulse $x(t)=\delta (t)$. Since a unit impulse $\delta (t)$ is the derivative of a unit step $u(t)$, and the system is linear and time-invariant, the impulse response is the time derivative of the step response: $h(t)=\frac{d}{dt}[s(t)]=\frac{d}{dt}\left[(1-e^{-t})u(t)\right]$

Applying the product rule of calculus: $h(t)=\frac{d}{dt}(1-e^{-t})\cdot u(t)+(1-e^{-t})\cdot \frac{d}{dt}[u(t)]$ $h(t)=e^{-t}u(t)+(1-e^{-t})\delta (t)$

Since the term $(1-e^{-t})$ evaluated at the impulse location $t=0$ is exactly zero, the second term vanishes completely: $h(t)=e^{-t}u(t)$

[GRAPH: Impulse response curve h(t) starting abruptly at a peak of 1 at t=0 and decaying exponentially toward 0 as t approaches infinity. — governing equation: h(t) = e^-t * u(t) — source: Rabiul Sir Notes, Lec 6]

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4. Time-Domain Solutions of Discrete Difference Equations

A discrete-time LTI system of order $N$ is represented by the difference equation: ${\sum }_{k=0}^N{a}_{k}y[n-k]={\sum }_{m=0}^M{b}_{k}x[n-m]$

Just like its continuous counterpart, we solve this classical recurrence relation by decomposing it: $y[n]=y_h[n]+y_p[n]$

A. The Discrete Homogeneous Solution, $y_h[n]$

We set $x[n]=0$ and assume a trial solution of the form $y_h[n]=C{\lambda }^n$. Substituting this trial form gives the discrete characteristic equation: ${\sum }_{k=0}^N{a}_k{\lambda }^{N-k}=0⟹a_0{\lambda }^N+a_1{\lambda }^{N-1}+\cdots +a_{N-1}\lambda +a_N=0$

Solving for the roots ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$ determines the shape of $y_h[n]$:

1. Distinct Roots: $y_h[n]=C_1{\lambda }_1^n+C_2{\lambda }_2^n+\dots$
2. Repeated Roots: $y_h[n]=(C_1+C_{2}n+C_3{n}^2){\lambda }^n$

Discrete Stability Condition

For a discrete system to be stable, the natural response must decay over time as $n\to \infty$. This requires the magnitude of all characteristic roots to be strictly less than unity ($|{\lambda }_i|<1$). On the complex z-plane, this means all poles must lie inside the unit circle.

B. The Discrete Particular Solution, $y_p[n]$

The trial particular solution $y_p[n]$ matches the discrete shape of the input $x[n]$:

Input Shape $x[n]$	Assumed Particular Solution $y_p[n]$
Constant: $A$	$K$ (Constant)
Exponential: $A{\beta }^n$	$K{\beta }^n$ (provided $\beta$ is not a characteristic root)
Polynomial: $A{n}^k$	$K_k{n}^k+K_{k-1}{n}^{k-1}+\cdots +K_0$

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5. Step-by-Step Discrete Numerical Solution

Let’s tackle a highly-tested discrete recurrence problem featuring initial conditions.

Discrete Second-Order Difference Equation

Question: Solve the second-order recursive discrete system described by: $y[n]-\frac{1}{9}y[n-2]=x[n-1]$ with step input $x[n]=u[n]$, and initial auxiliary conditions {conditions specifying the starting state of the system’s memory delay buffers} given as $y[-1]=1$ and $y[-2]=0$.

Step 1: Establish the Homogeneous Solution, $y_h[n]$

Set the right-hand side of our difference equation to zero: $y_h[n]-\frac{1}{9}{y}_h[n-2]=0$ Assume $y_h[n]=C{\lambda }^n$. The characteristic polynomial equation is: ${\lambda }^2-\frac{1}{9}=0⟹\lambda =\pm \frac{1}{3}$ Thus, our homogeneous solution has the form: $y_h[n]={\alpha }_1{\left(\frac{1}{3}\right)}^n+{\alpha }_2{\left(-\frac{1}{3}\right)}^n$

Step 2: Establish the Particular Solution, $y_p[n]$

Our input is the step function $x[n]=u[n]$, meaning $x[n-1]=u[n-1]$. For $n\ge 1$, the input is a constant ($x[n-1]=1$), so we assume a constant trial particular solution: $y_p[n]=K$ Substitute $y_p[n]=K$ into our difference equation for $n\ge 1$: $K-\frac{1}{9}K=1⟹\frac{8}{9}K=1⟹K=\frac{9}{8}$ Thus, our particular solution is: $y_p[n]=\frac{9}{8}\text{for }n\ge 1$

Step 3: Compute Discrete Boundary States via Recursion

To apply our initial values to find ${\alpha }_1$ and ${\alpha }_2$, we need the system output states at $n=0$ and $n=1$. We calculate these directly using the system difference equation and the given initial conditions $y[-1]=1$, $y[-2]=0$.

• For $n=0$: $y[0]-\frac{1}{9}y[-2]=x[-1]⟹y[0]-\frac{1}{9}(0)=u[-1]$ Since step $u[-1]=0$ in negative time: $y[0]=0$

• For $n=1$: $y[1]-\frac{1}{9}y[-1]=x[0]⟹y[1]-\frac{1}{9}(1)=u[0]$ Since step $u[0]=1$: $y[1]-\frac{1}{9}=1⟹y[1]=1+\frac{1}{9}=\frac{10}{9}$

Step 4: Combine and Solve for Coefficients

Our total response is modeled as: $y[n]={\alpha }_1{\left(\frac{1}{3}\right)}^n+{\alpha }_2{\left(-\frac{1}{3}\right)}^n+\frac{9}{8}u[n-1]$

Now we apply our calculated boundary states $y[0]$ and $y[1]$ to set up a system of linear equations:

1. At $n=0$: $y[0]={\alpha }_1{\left(\frac{1}{3}\right)}^0+{\alpha }_2{\left(-\frac{1}{3}\right)}^0+\frac{9}{8}u[-1]=0$ ${\alpha }_1+{\alpha }_2+0=0⟹{\alpha }_2=-{\alpha }_1$

2. At $n=1$: $y[1]={\alpha }_1{\left(\frac{1}{3}\right)}^1+{\alpha }_2{\left(-\frac{1}{3}\right)}^1+\frac{9}{8}u[0]=\frac{10}{9}$ $\frac{1}{3}{\alpha }_1-\frac{1}{3}{\alpha }_2+\frac{9}{8}=\frac{10}{9}$

Substitute ${\alpha }_2=-{\alpha }_1$ into the equation: $\frac{1}{3}{\alpha }_1-\frac{1}{3}(-{\alpha }_1)=\frac{10}{9}-\frac{9}{8}$ $\frac{2}{3}{\alpha }_1=\frac{80-81}{72}=-\frac{1}{72}$ ${\alpha }_1=-\frac{1}{72}\times \frac{3}{2}=-\frac{1}{48}$

Since ${\alpha }_2=-{\alpha }_1$, we find: ${\alpha }_2=\frac{1}{48}$

Step 5: Final Total Solution

Substituting our solved constants back into our equation yields our final complete response: $y[n]=-\frac{1}{48}{\left(\frac{1}{3}\right)}^n+\frac{1}{48}{\left(-\frac{1}{3}\right)}^n+\frac{9}{8}u[n-1]$

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6. Natural vs. Forced Response (An Critical Distinction)

In exams, students often confuse the classical homogeneous/particular split with the physical Natural/Forced split. They are not the same!

graph TD
    TR[Total Solution Response] --> HS_PS[Method 1: Homogeneous + Particular]
    TR --> NR_FR[Method 2: Natural + Forced]
    
    HS_PS --> HS[Homogeneous Response]
    HS_PS --> PS[Particular Response]
    
    NR_FR --> NR[Natural/Zero-Input Response]
    NR_FR --> FR[Forced/Zero-State Response]
    
    style TR fill:#f9f,stroke:#333,stroke-width:2px
    style HS_PS fill:#bbf,stroke:#333
    style NR_FR fill:#bfb,stroke:#333

1. Natural Response (Zero-Input Response, $y_{\text{zi}}(t)$): The system response due only to initial conditions, setting the input $x(t)=0$.
2. Forced Response (Zero-State Response, $y_{\text{zs}}(t)$): The system response due only to the input $x(t)$, setting all initial conditions to zero.

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7. ECE 2108 Laboratory Connection

In Experiment 03 of your lab, you will model these continuous differential equations and discrete systems using MATLAB:

% Continuous system simulation
num = [1]; den = [1, 1];          % Represents dy/dt + y = x
sys = tf(num, den);
step(sys);                       % Plots the step response
 
% Discrete recurrence simulation using filter
b = [0, 1]; a = [1, 0, -1/9];     % Represents y[n] - 1/9 y[n-2] = x[n-1]
x = ones(1, 20);                  % Step input array
y = filter(b, a, x);              % Computes recursive response
stem(y);                         % Plots discrete stems

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Common Mistakes That Cost Marks

Exam Pitfall Checklist

1. Applying Initial Conditions Too Early: This is the #1 source of lost marks in difference/differential equation questions! Students frequently apply initial conditions to the homogeneous solution $y_h$ before adding the particular solution $y_p$. Always apply initial conditions to the total combined response $y(t)=y_h(t)+y_p(t)$.
2. Ignoring the Duplicate Root Trap: If the input signal matches one of your characteristic roots, your standard particular trial solution will fail. For example, if your characteristic root is $\lambda =-1$ and your input is $x(t)=e^{-t}$, your particular trial solution cannot be $K{e}^{-t}$ (as it is already absorbed by the homogeneous solution). You must multiply the trial solution by the independent variable: $Kt{e}^{-t}$.
3. Using the Wrong Auxiliary Limits: For continuous-time systems, remember that physical variables (like current through an inductor or voltage across a capacitor) cannot change instantaneously. Thus, boundary states at $t=0^{-}$ are equated to states at $t=0^{+}$. For discrete systems, compute $y[0],y[1]$ recursively from the delay terms.

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PYQ Bank — Verbatim Questions & Answer Plans

Q1: Continuous System Representation [10 Marks] [PYQ 2024]

Represent the following differential equation in a state model: $\frac{d^{3}y}{d{t}^3}+\frac{d^{2}y}{d{t}^2}+6\frac{dy}{dt}+7y=2u(t)$

• Answer Plan:
  1. Define state variables based on the phase-variable method: $x_1=y$, $x_2=\dot{y}$, $x_3=\ddot{y}$.
  2. Write the derivatives of the state variables: ${\dot{x}}_1=x_2$, ${\dot{x}}_2=x_3$, and ${\dot{x}}_3=-7x_1-6x_2-x_3+2u(t)$.
  3. Express the equations in matrix form $\dot{\mathbf{X}}=\mathbf{AX}+\mathbf{BU}$ and $\mathbf{Y}=\mathbf{CX}+\mathbf{DU}$.

Q2: System Realization and Real-World Traps [15 Marks] [PYQ 2018]

Determine the output of the system described by: $y[n]-\frac{1}{9}y[n-2]=x[n-1]$ where $x[n]=u[n-1]$, and initial conditions are given as $y[-1]=1,y[-2]=0$.

• Answer Plan:
  1. Find the homogeneous solution $y_h[n]={\alpha }_1(1/3{)}^n+{\alpha }_2(-1/3{)}^n$.
  2. Note that the input is $x[n]=u[n-1]$, meaning $x[n-1]=u[n-2]$. For $n\ge 2$, the input is 1, so particular solution $y_p[n]=9/8$.
  3. Compute boundary outputs $y[0]$ and $y[1]$ using recursive equations from the difference relation.
  4. Combine homogeneous and particular solutions, and solve for constants ${\alpha }_1$ and ${\alpha }_2$ using the boundary states.

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Self-Check Before Moving On

• Can you explain the physical difference between the homogeneous response and the particular response?
• Do you know how to identify and avoid the “duplicate root particular solution trap”?
• Can you solve a first-order continuous differential equation step-by-step for a unit step input?
• Can you recursively calculate discrete output states $y[0]$ and $y[1]$ given arbitrary past initial conditions $y[-1]$ and $y[-2]$?

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Source: (K. Deergha Rao) Signals and Systems, Chapter 2 & 6; Rabiul Sir class notes, Lecture 7 & 8.