Signals and Systems

Systems Characterized by Linear Constant-Coefficient Differential Equations (LCCDEs)

Imron Rosyadi

Systems Characterized by Linear Constant-Coefficient Differential Equations

Introduction to LCCDEs in LTI Systems

• Linear Constant-Coefficient Differential Equations (LCCDEs) are fundamental to describing many continuous-time LTI systems.
• They model physical systems such as electrical circuits, mechanical systems, and more.
• Understanding their behavior is crucial for analyzing and designing ECE systems.

LCCDEs are a cornerstone for understanding continuous-time LTI systems. They provide a mathematical framework that directly relates the input and output of a system through derivatives. This structure is powerful because it allows us to model a vast array of real-world phenomena, from simple RC circuits to complex control systems. Our goal today is to see how Fourier analysis simplifies working with these equations.

General Form of an LCCDE

• An LTI system’s input \(x(t)\) and output \(y(t)\) can be related by the following general form:

\[ \sum_{k=0}^{N} a_{k} \frac{d^{k} y(t)}{d t^{k}}=\sum_{k=0}^{M} b_{k} \frac{d^{k} x(t)}{d t^{k}} \quad (4.72) \]

• Here, \(a_k\) and \(b_k\) are constant coefficients.
• \(N\) represents the order of the differential equation, corresponding to the highest derivative of the output.

Equation 4.72 is the canonical form for an LCCDE. The left side involves derivatives of the output \(y(t)\), and the right side involves derivatives of the input \(x(t)\). The coefficients \(a_k\) and \(b_k\) are constants, which is why it’s called “constant-coefficient.” The order \(N\) is particularly important as it often dictates the complexity and characteristics of the system, such as the number of energy storage elements in a circuit.

Frequency Response and the Fourier Transform

• For an LTI system, the output’s Fourier Transform \(Y(j\omega)\) relates to the input’s Fourier Transform \(X(j\omega)\) through the frequency response \(H(j\omega)\).

\[ Y(j \omega)=H(j \omega) X(j \omega) \]

• This implies that the frequency response can be expressed as:

\[ H(j \omega)=\frac{Y(j \omega)}{X(j \omega)} \quad (4.73) \]

Recall the convolution property: convolution in the time domain becomes multiplication in the frequency domain. This is incredibly powerful for LTI systems. If we know the Fourier transform of the input \(X(j\omega)\) and the system’s frequency response \(H(j\omega)\), we can immediately find the Fourier transform of the output \(Y(j\omega)\) by simple multiplication. Conversely, if we know \(X(j\omega)\) and \(Y(j\omega)\), we can determine \(H(j\omega)\).

Deriving \(H(j\omega)\) from LCCDEs: Step 1

• To find \(H(j\omega)\), we apply the Fourier Transform to both sides of the LCCDE (4.72):

\[ \mathcal{F}\left\{\sum_{k=0}^{N} a_{k} \frac{d^{k} y(t)}{d t^{k}}\right\}=\mathfrak{F}\left\{\sum_{k=0}^{M} b_{k} \frac{d^{k} x(t)}{d t^{k}}\right\} \quad (4.74) \]

• Using the linearity property of the Fourier Transform, we can move the sum and constants outside:

\[ \sum_{k=0}^{N} a_{k} \mathcal{F}\left\{\frac{d^{k} y(t)}{d t^{k}}\right\}=\sum_{k=0}^{M} b_{k} \mathcal{F}\left\{\frac{d^{k} x(t)}{d t^{k}}\right\} \quad (4.75) \]

The linearity property of the Fourier Transform is key here. It allows us to transform each term in the sum independently and then combine them. This simplifies the process significantly, as we don’t need to transform the entire sum at once.

Deriving \(H(j\omega)\) from LCCDEs: Step 2

• Now, apply the differentiation property of the Fourier Transform, which states \(\mathcal{F}\left\{\frac{d^k f(t)}{d t^k}\right\} = (j\omega)^k F(j\omega)\):

\[ \sum_{k=0}^{N} a_{k}(j \omega)^{k} Y(j \omega)=\sum_{k=0}^{M} b_{k}(j \omega)^{k} X(j \omega) \]

• Factor out \(Y(j\omega)\) and \(X(j\omega)\):

\[ Y(j \omega)\left[\sum_{k=0}^{N} a_{k}(j \omega)^{k}\right]=X(j \omega)\left[\sum_{k=0}^{M} b_{k}(j \omega)^{k}\right] \]

• Finally, solve for \(H(j\omega) = Y(j\omega)/X(j\omega)\):

\[ H(j \omega)=\frac{\sum_{k=0}^{M} b_{k}(j \omega)^{k}}{\sum_{k=0}^{N} a_{k}(j \omega)^{k}} \quad (4.76) \]

Tip

The frequency response \(H(j\omega)\) for an LCCDE can be written directly by inspection from the differential equation coefficients!

This is the most crucial step. The differentiation property transforms time-domain derivatives into simple multiplications by powers of \(j\omega\) in the frequency domain. This converts a differential equation into an algebraic equation. The resulting \(H(j\omega)\) is a rational function, meaning a ratio of polynomials in \(j\omega\). The coefficients of these polynomials are precisely the coefficients \(b_k\) and \(a_k\) from the original differential equation. This direct relationship is incredibly powerful for analysis.

LCCDE to Frequency Response: Flowchart

• Visualizing the process of converting an LCCDE into its frequency response.

graph LR
    A["Linear Constant-Coefficient Differential Equation (LCCDE)"] --> B{Apply Fourier Transform to both sides};
    B --> C{Use Linearity Property};
    C --> D{Use Differentiation Property};
    D --> E{Rearrange Algebraically};
    E --> F["Frequency Response H(jω) = Y(jω)/X(jω)"];

This flowchart summarizes the elegant path from a time-domain differential equation to its frequency-domain representation. Each step relies on fundamental properties of the Fourier Transform, transforming a complex calculus problem into a straightforward algebraic one. This is why Fourier analysis is so powerful in Signals and Systems.

Example 4.24: First-Order System

• Consider a stable LTI system described by:

\[ \frac{d y(t)}{d t}+a y(t)=x(t) \quad (4.77) \]

• Using equation (4.76), we can directly find the frequency response.
• Comparing with the general form, \(N=1, M=0\).
• Coefficients: \(a_1=1, a_0=a\), and \(b_0=1\).
• Substitute into (4.76):

\[ H(j \omega)=\frac{b_0 (j\omega)^0}{a_1 (j\omega)^1 + a_0 (j\omega)^0} = \frac{1}{j \omega+a} \quad (4.78) \]

• The impulse response \(h(t)\) is the inverse Fourier Transform of \(H(j\omega)\).
• Recognizing this form (from Example 4.1), the impulse response is:

\[ h(t)=e^{-a t} u(t) \]

• For stability, we require \(a>0\).

This example demonstrates the direct application of the formula. For a first-order system, we have derivatives up to order 1 for the output and order 0 for the input. By simply plugging in the coefficients, we instantly get the frequency response. This particular form of \(H(j\omega)\) is very common and its inverse Fourier transform, \(e^{-at}u(t)\), is a fundamental signal in ECE. The condition \(a>0\) ensures that the impulse response decays over time, leading to a stable system.

Interactive Plot: First-Order Impulse Response

• Explore how the parameter ‘a’ affects the impulse response \(h(t) = e^{-at}u(t)\).

viewof a_slider = Inputs.range([0.1, 5.0], {label: "Parameter 'a'", step: 0.1, value: 1.0})

This interactive plot helps visualize the impact of the parameter ‘a’ on the system’s impulse response. As you increase ‘a’, the exponential decay becomes faster, meaning the system responds more quickly and settles sooner. This corresponds to a wider bandwidth in the frequency domain. Conversely, a smaller ‘a’ leads to a slower, more spread-out response in the time domain, indicating a narrower bandwidth. This directly relates to the system’s time constant, which is \(1/a\).

Example 4.25: Second-Order System

• Consider a stable LTI system characterized by:

\[ \frac{d^{2} y(t)}{d t^{2}}+4 \frac{d y(t)}{d t}+3 y(t)=\frac{d x(t)}{d t}+2 x(t) \]

• From eq. (4.76), the frequency response is:

\[ H(j \omega)=\frac{(j \omega)+2}{(j \omega)^{2}+4(j \omega)+3} \quad (4.79) \]

• To find the impulse response \(h(t)\), we use partial-fraction expansion.
• First, factor the denominator:

\[ H(j \omega)=\frac{j \omega+2}{(j \omega+1)(j \omega+3)} \quad (4.80) \]

Here we have a second-order system because the highest derivative of \(y(t)\) is two. The process to find \(H(j\omega)\) remains the same: identify coefficients and plug them into the formula. However, finding the inverse Fourier Transform of this \(H(j\omega)\) is not as direct as the first-order case. This is where partial-fraction expansion becomes an indispensable tool. It allows us to break down complex rational functions into simpler terms whose inverse transforms are known. Factoring the denominator is the first critical step in this process.

Partial-Fraction Expansion Process

• A visual representation of how partial-fraction expansion helps find \(h(t)\) from \(H(j\omega)\).

graph TD
    A["H(jω) as Ratio of Polynomials"] --> B{Factor Denominator Polynomial};
    B --> C{Apply Partial-Fraction Expansion};
    C --> D["Sum of Simpler Terms (e.g., 1/(jω+a))"];
    D --> E{Find Inverse Fourier Transform for Each Term};
    E --> F["Impulse Response h(t)"];

This flowchart outlines the general procedure for finding the impulse response when \(H(j\omega)\) is a rational function. Partial-fraction expansion is a technique taught in calculus and is heavily used in signals and systems to simplify complex expressions into forms that are easily invertible using a table of Fourier Transform pairs.

Impulse Response for Example 4.25

• Applying partial-fraction expansion to \(H(j\omega)\):

\[ H(j \omega)=\frac{\frac{1}{2}}{j \omega+1}+\frac{\frac{1}{2}}{j \omega+3} . \]

• Recognizing the inverse transforms of each term:

\[ h(t)=\frac{1}{2} e^{-t} u(t)+\frac{1}{2} e^{-3 t} u(t) \]

Important

Partial-fraction expansion is a critical technique for transforming frequency-domain expressions into their time-domain equivalents!

After performing the partial-fraction expansion, we get two simpler terms, each resembling the frequency response from Example 4.24. We can then easily find the inverse Fourier transform of each term and sum them up to get the total impulse response \(h(t)\). This illustrates the power of breaking down a complex problem into simpler, manageable parts.

Example 4.26: System Response to an Input

• Consider the system from Example 4.25 with the input:

\[ x(t)=e^{-t} u(t) \]

• Its Fourier Transform is \(X(j\omega) = \frac{1}{j\omega+1}\).
• The output’s Fourier Transform \(Y(j\omega) = H(j\omega)X(j\omega)\):

\[ \begin{align*} Y(j \omega) & =H(j \omega) X(j \omega)=\left[\frac{j \omega+2}{(j \omega+1)(j \omega+3)}\right]\left[\frac{1}{j \omega+1}\right] \\ & =\frac{j \omega+2}{(j \omega+1)^{2}(j \omega+3)} \quad (4.81) \end{align*} \]

Now we’re moving from finding the impulse response to finding the system’s output for a specific input. The process is similar: multiply \(H(j\omega)\) by \(X(j\omega)\) to get \(Y(j\omega)\). Notice that the denominator now has a repeated pole at \(j\omega = -1\). This requires a slightly modified partial-fraction expansion technique, as discussed in the appendix.

Partial-Fraction for \(Y(j\omega)\)

• For repeated poles, the partial-fraction expansion takes a specific form:

\[ Y(j \omega)=\frac{A_{11}}{j \omega+1}+\frac{A_{12}}{(j \omega+1)^{2}}+\frac{A_{21}}{j \omega+3} \quad (4.82) \]

• The constants are found to be:
  • \(A_{11}=\frac{1}{4}\)
  • \(A_{12}=\frac{1}{2}\)
  • \(A_{21}=-\frac{1}{4}\)
• So, \(Y(j\omega)\) becomes:

\[ Y(j \omega)=\frac{\frac{1}{4}}{j \omega+1}+\frac{\frac{1}{2}}{(j \omega+1)^{2}}-\frac{\frac{1}{4}}{j \omega+3} \quad (4.83) \]

The presence of a repeated pole \((j\omega+1)^2\) means we need two terms in the partial fraction expansion for that pole: one with \((j\omega+1)\) in the denominator and one with \((j\omega+1)^2\). The coefficients \(A_{11}, A_{12}, A_{21}\) are determined using standard partial-fraction expansion techniques, usually involving solving a system of linear equations or specific evaluation methods.

Output Signal \(y(t)\) for Example 4.26

• Taking the inverse Fourier Transform of each term in \(Y(j\omega)\) (Eq. 4.83):

\[ y(t)=\left[\frac{1}{4} e^{-t}+\frac{1}{2} t e^{-t}-\frac{1}{4} e^{-3 t}\right] u(t) \]

• The term \(\frac{1}{(j\omega+1)^2}\) corresponds to \(t e^{-t} u(t)\) in the time domain.

Each term in the partial fraction expansion of \(Y(j\omega)\) has a known inverse Fourier transform. The first and third terms are simple exponentials. The second term, \(\frac{1/2}{(j\omega+1)^2}\), is a common Fourier transform pair that can be derived using the differentiation property in the frequency domain or found in transform tables. The \(t e^{-t} u(t)\) term indicates a response that initially grows before decaying, a characteristic often seen in systems with repeated poles or specific damping conditions.

Interactive Plot: Input, Impulse Response, and Output

• Visualize \(x(t)\), \(h(t)\), and \(y(t)\) for Example 4.26.

This plot visually connects the input, the system’s impulse response, and the resulting output in the time domain. Notice how the output \(y(t)\) is a ‘smoothed’ or ‘filtered’ version of the input, influenced by the characteristics of the impulse response. The shape of \(y(t)\) shows the system’s transient behavior, demonstrating how the system responds to the sudden onset of the input. This interaction highlights the convolution operation in the time domain, which is simplified to multiplication in the frequency domain.

Summary and Key Takeaways

• Algebraic Simplification: Fourier analysis transforms LCCDEs into algebraic equations in the frequency domain.
• Direct Derivation: \(H(j\omega)\) can be directly obtained by inspection from the LCCDE coefficients.
• Partial-Fraction Expansion: A powerful technique for finding time-domain responses from rational frequency-domain expressions.
• Frequency-Domain Insights: Analyzing \(H(j\omega)\) provides deep understanding of system characteristics (e.g., stability, filtering properties).

In conclusion, the Fourier Transform provides an incredibly powerful framework for analyzing LTI systems described by LCCDEs. It converts complex differential equations into simple algebraic ones, allowing us to easily determine frequency responses, impulse responses, and system outputs. This frequency-domain perspective is invaluable for understanding system behavior and forms the basis for many design principles in ECE.