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4.2: Response of a First Order System to a Suddenly Applied Cosine

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    7643
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    First, we derive a complete solution in the conventional manner for the original standard 1st order ODE \(\dot{x}-a x=b u(t)\) [Equation 1.2.1], with IC \(x(0)=x_{0}\), and with the suddenly applied (at \(t\) = 0) cosine input \(u(t)=U \cos \omega t\), \(t\) > 0, where \(U\) is a constant amplitude [Problem 2.12.2].

    \[\mathrm{ODE}+\mathrm{IC}: \dot{x}-a x=b U \cos \omega t=b U \frac{e^{j \omega t}+e^{-j \omega t}}{2}, x(0)=x_{0}, \text { find } x(t) \text { for } t>0\label{eqn:4.1} \]

    Laplace transformation of ODE + IC: \(s X(s)-x_{0}-a X(s)=\frac{b U}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)\)

    Solve for \(X(s)\):

    \[X(s)=\frac{x_{0}}{s-a}+\frac{b U}{2} \frac{1}{(s-a)}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right) \nonumber \]

    Completed partial-fraction expansion (Problem 2.8):

    \[X(s)=\frac{x_{0}}{s-a}+\frac{b U}{2} \frac{1}{a^{2}+\omega^{2}}\left(\frac{2 a}{s-a}+\frac{-a-j \omega}{s-j \omega}+\frac{-a+j \omega}{s+j \omega}\right) \nonumber \]

    Inverse transform:

    \[ \begin{align} x(t) &=x_{0} e^{a t}+\frac{b U}{a^{2}+\omega^{2}}\left[a e^{a t}+\frac{1}{2}\left((-a-j \omega) e^{j \omega t}+(-a+j \omega) e^{-j \omega t}\right)\right] \\[4pt] \quad x(t)&=x_{0} e^{a t}+\frac{b U}{a^{2}+\omega^{2}}\left[a e^{a t}+\frac{1}{2}\left((-a)\left(e^{j \omega t}+e^{-j \omega t}\right)+\omega \frac{e^{j \omega t}-e^{-j \omega t}}{j}\right)\right] \\[4pt] x(t)&=x_{0} e^{a t}+\frac{b U}{a^{2}+\omega^{2}}\left(a e^{a t}-a \cos \omega t+\omega \sin \omega t\right), t \geq 0 \\[4pt] x(t) &= x_{0} e^{a t}+\left(\frac{b}{-a}\right) \frac{U}{\left[1+\left(\frac{\omega}{-a}\right)^{2}\right]}\left[-e^{a t}+\cos \omega t+\left(\frac{\omega}{-a}\right) \sin \omega t\right], t \geq 0\label{eqn:4.2} \end{align} \]

    Next, we adapt solution Equation \(\ref{eqn:4.2}\) of general 1st order problem Equation \(\ref{eqn:4.1}\) to the damper-spring system (ideal shock strut) of Figure 3.7.1 with suddenly applied cosine force input \(f_{x}(t)=F \cos \omega t\), for which the comparable ODE + IC problem is

    \[c \dot{x}+k x=F \cos \omega t, x(0)=x_{0}, \text { find position } x(t) \text { for } t>0\label{eqn:4.3} \]

    Comparing Equation \(\ref{eqn:4.3}\) with Equation \(\ref{eqn:4.1}\), we define \(U\) \(\equiv\) \(F\), then the other constants of the standard equation become

    \[a=-\frac{k}{c} \equiv-\frac{1}{\tau_{1}} \Rightarrow \text { time constant } \tau_{1} \equiv \frac{1}{-a}=\frac{c}{k} \quad \text { and } \quad b=\frac{1}{c} \text { and } \frac{b}{-a}=\frac{1}{k} \nonumber \]

    \[\Rightarrow \quad x(t)=x_{0} e^{-t / \tau_{1}}+\frac{F}{k}\left(\frac{1}{1+\left(\omega \tau_{1}\right)^{2}}\right)\left(-e^{-t / \tau_{1}}+\cos \omega t+\omega \tau_{1} \sin \omega t\right), t \geq 0\label{eqn:4.4} \]

    It is instructive to study the physical nature of response Equation \(\ref{eqn:4.4}\) in the context of a specific numerical example. Consider an ideal shock strut with the initial condition \(x_0\) = -2 m, and with the system parameters \(c=1 / \pi\) N/(m/s) = 0.3183 N-s/m and \(k\) = 1 N/m, so that time constant \(\tau_{1}=1 / \pi\) s = 0.3183 s. Let the magnitude of the cosine force be \(F\) = 1.5 N, and let the period of the cosine be \(T_p\) = 1 s/cycle. Therefore, the cyclic frequency is \(f\) = 1 Hz (cycle/s), and the circular frequency is \(\omega = 2\pi\) radians/s. (Period, frequency, and phase of periodic signals are discussed more generally in Section 4.4.) The numerical solution of Equation \(\ref{eqn:4.4}\) with these parameters for time 0 \(\leq\) \(t\) \(\leq\) 3 s is calculated and graphed in the following MATLAB operations.

    %MATLABdemo41.m
    
    %Damper-spring ideal shock strut response to IC + cosine forcing
    
    c=1/pi;k=1; %system viscous damping & stiffness constants, SI units
    
    F=1.5;Tp=1; %cosine forcing: amplitude (N), period (sec)
    
    xo=-2; %initial displacement (m)
    
    w=2*pi/Tp; %circular frequency of cosine forcing (rad/sec)
    
    t1=c/k;denom=1+(w*t1)^2;
    
    t=0:0.01:3*Tp; %time instants for forced response
    
    fx=F*cos(w*t);
    
    x=(xo-(F/k)/denom)*exp(-t/t1)+(F/k)/denom*(cos(w*t)+w*t1*sin(w*t));
    
    plot(t,fx,'k',t,x,'k.'),grid,xlabel('Time t (sec)'),... ylabel('Force input f_x(t) (N) and displacement output x(t) (m)'),... title('Time response of an ideal shock strut to IC + cosine input')
    Hello world!

    MATLAB M-file script:

    %MATLABdemo41.m

    %Damper-spring ideal shock strut response to IC + cosine forcing

    c=1/pi;k=1; %system viscous damping & stiffness constants, SI units

    F=1.5;Tp=1; %cosine forcing: amplitude (N), period (sec)

    xo=-2; %initial displacement (m)

    w=2*pi/Tp; %circular frequency of cosine forcing (rad/sec)

    t1=c/k;denom=1+(w*t1)^2;

    t=0:0.01:3*Tp; %time instants for forced response

    fx=F*cos(w*t);

    x=(xo-(F/k)/denom)*exp(-t/t1)+(F/k)/denom*(cos(w*t)+w*t1*sin(w*t));

    plot(t,fx,'k',t,x,'k.'),grid,xlabel('Time t (sec)'),... ylabel('Force input f_x(t) (N) and displacement output x(t) (m)'),... title('Time response of an ideal shock strut to IC + cosine input')

    MATLAB command/responses:

    >> MATLABdemo41

    The graph of input \(f_x(t)\) and response \(x(t)\) is below (after figure editing in MATLAB,Version 6 or later, to flatten the aspect ratio and to distinguish the two curves).

    clipboard_ec2fb8492d2b6f1f9b06808bcfc2257c7.png
    Figure \(\PageIndex{1}\)

    Let us observe from the graph some important features of the response:

    • As calculated above, the time constant of the exponential decay terms in Equation \(\ref{eqn:4.4}\) [the two terms involving \(\exp \left(-t / \tau_{1}\right)\)] is \(\tau_{1}\) = 0.3183 s, so the settling time for those terms is \(4 \tau_{1}\) = 1.273 s. In other words, the exponential decay terms in \(x(t)\) have essentially vanished after about 1.3 s of response, and this is clear from the graph. Because the exponential decay terms are relatively short-lived, we often refer to them as the “transient” part of the total solution.
    • After the exponential decay terms have vanished, only the \(\cos \omega t\) and \(\sin \omega t\) terms of Equation \(\ref{eqn:4.4}\) remain. It appears from the graph that those two terms combine to make a single sinusoid at frequency \(\omega\), and that the remaining steady-state \(x(t)\) sinusoid is displaced in time by a constant time lag from the \(F_x(t)\) sinusoid. This steady-state sinusoidal response is what we call the frequency response, and we will derive equations that describe it explicitly in the remainder of the chapter.

    This page titled 4.2: Response of a First Order System to a Suddenly Applied Cosine is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by William L. Hallauer Jr. (Virginia Tech Libraries' Open Education Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.