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General Solution To Second Order Homogeneous LTI System

General Solution to Second-Order Homogeneous LTI System

We consider the zero-input response of a standard second-order linear time-invariant (LTI) system. The homogeneous second-order differential equation in standard form is given by the following relation:

General Solution to Second-Order Homogeneous LTI System

To find the general homogeneous solution, assume a trial solution of exponential form.

xh(t) = eλt

Differentiating the assumed solution with respect to time gives the first and second derivatives required to substitute into the differential equation.

General Solution to Second-Order Homogeneous LTI System

Substituting these derivatives into the homogeneous differential equation yields an algebraic equation in λ (because eλt is never zero for all t, it can be divided out).

General Solution to Second-Order Homogeneous LTI System
General Solution to Second-Order Homogeneous LTI System

The resulting algebraic equation is the characteristic equation. Its roots determine the behaviour of the zero-input response.

The roots of the characteristic equation, obtained from the quadratic formula, are

General Solution to Second-Order Homogeneous LTI System

The nature of these roots depends on the damping ratio ζ. The sign of ζ2 - 1 leads to three distinct cases: (1) 0 ≤ ζ < 1 (underdamped), (2) ζ = 1 (critically damped), and (3) ζ > 1 (overdamped). Each case is treated below.

Case 1: 0 ≤ ζ < 1 - Underdamping

When 0 ≤ ζ < 1 the system is underdamped and ζ2 - 1 < 0, so the characteristic roots are complex conjugates.

Case 1: 0 ≤ ζ < 1 - Underdamping
Case 1: 0 ≤ ζ < 1 - Underdamping

The complex conjugate roots can be written in the form -ζωn ± jωd, where ωd is the damped natural frequency.

Case 1: 0 ≤ ζ < 1 - Underdamping

The general form of the real zero-input response corresponding to the complex roots is

xh(t) = e-ζωnt(c1 cos ωdt + c2 sin ωdt)

The damped natural frequency ωd is given by

Case 1: 0 ≤ ζ < 1 - Underdamping

To determine the constants c1 and c2, use the initial conditions x(0) = x0 and ẋ(0) = ẋ0.

Apply x(0) = x0 to the general solution.

xh(0) = x0 = c1

Differentiate the general solution to obtain ẋh(t).

Case 1: 0 ≤ ζ < 1 - Underdamping

Apply the initial velocity ẋ(0) = ẋ0.

Case 1: 0 ≤ ζ < 1 - Underdamping
Case 1: 0 ≤ ζ < 1 - Underdamping

Substituting c1 = x0 into the expression for ẋ(0) gives an equation for c2.

Case 1: 0 ≤ ζ < 1 - Underdamping

Solving for c2 yields

Case 1: 0 ≤ ζ < 1 - Underdamping

Thus the zero-input response for the underdamped case can be written as

Case 1: 0 ≤ ζ < 1 - Underdamping

The response is an exponentially decaying sinusoid with envelope determined by e-ζωnt. A schematic illustrating typical underdamped responses for different ζ in 0 ≤ ζ < 1 is shown below.

Case 1: 0 ≤ ζ < 1 - Underdamping

Figure 1-2 Schematic of the zero input response of an underdamped second-order linear time-invariant system.

Case 2: ζ = 1 - Critical Damping

When ζ = 1 the system is critically damped. In this case ζ2 - 1 = 0, so the two roots coincide and are real and equal.

Case 2: ζ = 1 - Critical Damping

The repeated root is

λ1 = λ2 = -ζωn = -ωn

For a repeated real root, the general homogeneous solution is

xh(t) = ent(c1 + c2 t)

Use the initial conditions x(0) = x0 and ẋ(0) = ẋ0 to find c1 and c2.

Apply x(0) = x0 to the general solution.

xh(0) = x0 = c1

Differentiate xh(t) to get ẋh(t).

Case 2: ζ = 1 - Critical Damping

Apply ẋ(0) = ẋ0.

Case 2: ζ = 1 - Critical Damping
Case 2: ζ = 1 - Critical Damping

Substituting c1 = x0 into the derivative equation produces a linear equation for c2.

Case 2: ζ = 1 - Critical Damping

Solving that equation for c2 gives

Case 2: ζ = 1 - Critical Damping

Therefore the critically damped zero-input response becomes

Case 2: ζ = 1 - Critical Damping

A schematic of the critically damped response is shown below.

Case 2: ζ = 1 - Critical Damping

Figure 1-3 Schematic of the zero input response of a critically damped second-order linear time-invariant system.

Case 3: ζ > 1 - Overdamping

When ζ > 1 the system is overdamped. In this case ζ2 - 1 > 0 and the characteristic equation has two distinct real roots.

Case 3: ζ > 1 - Overdamping
Case 3: ζ > 1 - Overdamping

Let these distinct real roots be λ1 and λ2. The general homogeneous solution is

xh(t) = c1 eλ1 t + c2 eλ2 t

Determine c1 and c2 from the initial conditions x(0) = x0 and ẋ(0) = ẋ0.

Apply x(0) = x0 to the general solution.

xh(0) = x0 = c1 + c2

Differentiate to obtain ẋh(t).

Case 3: ζ > 1 - Overdamping

Apply the initial velocity condition ẋ(0) = ẋ0.

Case 3: ζ > 1 - Overdamping

Solve the two linear simultaneous equations for c1 and c2.

Case 3: ζ > 1 - Overdamping

Thus the overdamped zero-input response can be expressed explicitly as

Case 3: ζ > 1 - Overdamping

A schematic showing overdamped responses for various ζ > 1 is provided below.

Case 3: ζ > 1 - Overdamping

Figure 1-4 Schematic of the zero input response of an overdamped second-order linear time-invariant system.

Remarks and applications

The three canonical forms - underdamped, critically damped, and overdamped - exhaust the possible zero-input responses of a second-order homogeneous LTI system. The damping ratio ζ and natural frequency ωn are key design parameters in mechanical and aerospace applications: they determine transient settling time, overshoot, and whether oscillations occur. Typical uses include vibration analysis of single-degree-of-freedom systems, suspension design, and transient analysis of control systems.

The document General Solution To Second Order Homogeneous LTI System is a part of the Mechanical Engineering Course Theory of Machines (TOM).
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FAQs on General Solution To Second Order Homogeneous LTI System

1. What is a second order homogeneous LTI system?
Ans. A second order homogeneous LTI system is a linear time-invariant system whose input and output are related by a second order linear differential equation with constant coefficients.
2. How do you find the general solution to a second order homogeneous LTI system?
Ans. To find the general solution, we first solve the characteristic equation associated with the differential equation. The roots of the characteristic equation determine the form of the solution, which is a linear combination of exponential functions involving the roots.
3. What is a characteristic equation of a second order homogeneous LTI system?
Ans. The characteristic equation of a second order homogeneous LTI system is obtained by replacing the derivatives in the differential equation with the corresponding powers of the complex variable s. It is a polynomial equation whose roots determine the behavior of the system.
4. Can a second order homogeneous LTI system have complex roots in its characteristic equation?
Ans. Yes, a second order homogeneous LTI system can have complex roots in its characteristic equation. Complex roots result in oscillatory behavior in the system's response, leading to sinusoidal or exponential solutions.
5. Is the general solution to a second order homogeneous LTI system unique?
Ans. Yes, the general solution to a second order homogeneous LTI system is unique. It can be expressed in terms of arbitrary constants, but the form and structure of the solution remain the same for all valid choices of these constants.
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