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Let First we use the roots (2) to solve equation (1). = ωd | b2 − 4mk | /2m. Then we have − b. characteristic roots: ± iωd. leading to 2m complex exponential solutions: e(−b/2m+iωd)t , e(−b/2m−iωd)t . The basic real solutions are e−bt/2m cos(ωdt) and e−bt/2m sin(ωdt).
The response is referred to as a critically damped response, and the system is called a critically damped system, with a double pole: x1 = x2 = −b 2a x 1 = x 2 = − b 2 a. if Δ <0 Δ <0, there are two complex, conjugate roots, and the response is a sinusoid with an exponential envelope.
Figure 2 shows the response of the series RLC circuit with L=47mH, C=47nF and for three different values of R corresponding to the under damped, critically damped and over damped case. We will construct this circuit in the laboratory and examine its behavior in more detail. (a) Under Damped. R=500Ω (b) Critically Damped. R=2000 Ω (c) Over Damped.
Learning Objectives. Learn to analyze a general second order system and to obtain the general solution. Identify the over-damped, under-damped, and critically damped solutions. Convert complex solution to real solution. Suspended “mass-spring-damper” equivalent system.
An example of a critically damped system is the shock absorbers in a car. It is advantageous to have the oscillations decay as fast as possible. Here, the system does not oscillate, but asymptotically approaches the equilibrium condition as quickly as possible.
The response of a system to an impulse looks identical to its response to an initial velocity. The impulse acts over such a short period of time that it essentially serves to give the system an initial velocity. Fig. 3 shows the impulse response of three systems: under-damped, critically damped, and over-damped.
This document discusses the response of a second-order system, such as the mass-spring-dashpot shown in Fig. 1, to a step function. The modeling of a step response in MATLAB and SIMULINK will also be discussed. Fig. 1. Single-degree-of-freedom mass-spring-dashpot system.
