5.2: Dynamic Response from Linear ODEs

4. General Measurement Model (pivot to Diff EQ-lite)
   1. Form of the most general linear differential equation: \[ a_{n} \frac{d^{n}y}{dt^{n}} + \ldots + a_{2} \frac{d^{2}y}{dt^{2}}+a_{1} \frac{dy}{dt} + a_{0}y(t) = F(t) \]
   2. LHS describes physics of inertia, dampers, springs, resistors, capacitors, etc.
   3. RHS is arbitrary forcing function (i.e. impulse, step, ramps)
      1. \(F(t) = b_{m}\frac{d^{m}x}{dt^{m}} + \ldots b_{1}\frac{dx}{dt} + b_{0}x(t)\)
      2. Need to have \(m\leq n\)
      3. Solution forms for set \(n=\{0,1,2\}\) will be given for variety of \(F(t)\)
      4. Each case of \(n\) builds to the next level
      5. Given the solution to D.E. \(\Rightarrow\) just place terms in correct parameters (language and vocabulary becomes critical!)
5. Zero-order system
   1. Direct, instantaneous response to input
   2. General case for \(n=0\) \[ a_{0} y(t) = F(t) \]
   3. Solution (simple algebraic step)\[ y(t) = kF(t) \] where
      1. Static sensitivity \(k=\frac{1}{a_{0}}\)
      2. Appropriate for static calibration but scaling of input applies to higher orders of \(n\)
      3. \(k\) is calibration slope \(k = \frac{dy}{dx}\)

   4. Exercise \(\PageIndex{1}\)

      A pressure sensor responds with sensitivity of 25.0 mV/kPa; can absolute atmospheric pressure be measured on a multimeter with 0--5 V range?

      What is lowest range of atmospheric pressures that can be measured when smallest resolution is 2 mV?

      Answer

      Static sensitivity

      \[k = 25.0\] mV/kPa \[y_{max} = 5\] V \[F_{1}(t) = 101.325\] kPa \[y = kF(t) = 25.0 (101.325) = 2.53\] V

      Yes, it can measure 1~atm

      \[F(t)_{max} = \frac{y_{max}}{k} = \frac{5}{25.0\times 10^{-3}} = 1.97\] atm \[F(t)_{min} = \frac{y_{min}}{k} = \frac{2}{25.0} = 80\] Pa

      Increasing multimeter range \(y_{max}\) increases pressure range

      Decreasing \(k\) also increases pressure range if \(y_{max} =5\) V.

6. First Order System (cases when \(n=1\) in general form)
   1. Storage elements (potential energy) eliminate instantaneous system response
   2. General form of equation: \[ a_{1} \frac{dy}{dt} + a_{0}y(t) = F(t) \] \[ a_{1}\dot{y}(t) + a_{0}y(t) = F(t) \] \[ a_{1} y'(x) + a_{0}y(x) = F(x) \]
   3. Rewritten in parameterized form: \[ \tau \dot{y}(t) + y(t) = k F(t) \] where
      1. \(\tau = \frac{a_{1}}{a_{0}}\) is the time constant; multiplies the rate of response \(\dot{y}(t)\)
      2. \(k = \frac{1}{a_{0}}\) is still static sensitivity as zero-order
   4. Homogeneous solution will always have the exponential decay \(y_{h} \approx \mathbf{e}^{-t/\tau}\)
   5. Particular solution depends on input form; choice from F.6 or F.7
   6. Step function input
      1. Discontinuous functional form: \[ F(t) = \left\{\begin{array}{lr}
         0 & t < 0\\
         AU(t) & t\geq 0
         \end{array}\right. \]
         1. \(A\) is amplitude; size of step that contains units
         2. \(U(t)\) is unit step; value of one; no units
      2. Makes the DE have the form \[ \tau \dot{y}(t) + y(t) = kAU(t) \] with \(y_{0} = y(0)\) for the initial condition
      3. System response for \(t\geq 0\): \[ y(t) = kA + (y_{0}-kA)\mathbf{e}^{\frac{-t}{\tau}} \] where first portion is steady state response with addition of exponential part as transient function with magnitude
         1. Evaluate initial condition: \(y(t=0) =\)??
         2. Evaluate end condition: \(y(t\rightarrow \infty)=\)??
         3. Sketch of general form of the response
      4. Speed of transient controlled by time constant \(\tau\)
         • show examples in different colors of larger and smaller \(\tau\)
      5. Exponential represents dynamic error between current point \(y(t)\) and end condition \(y_{\infty}\) approaching \(kA\): \[\delta_{f}(t) = \mathbf{e}^{\frac{-t}{\tau}} = \frac{y(t)-y_{\infty}}{y_{0}-y_{\infty}}\]
      6. Whereas percent of response from starting point is magnitude ratio: \[M(t) = 1-\delta_{f}(t) = 1-\mathbf{e}^{\frac{-t}{\tau}} = \frac{y(t)-y_{0}}{y_{\infty}-y_{0}} \]
      7. Special cases of \(\delta_{f}\) and \(M\):
         1. One time constant \(t=\tau\) \[\delta_{f}(\tau) = \mathbf{e}^{-\tau/\tau}= \mathbf{e}^{-1} = 0.368 \] \[ M(\tau) = 1-\mathbf{e}^{-\tau/\tau} = 0.632 \]
         2. Rise time: when at 90% response (or 10% error) \[ 0.1 = \delta_{f}(t_{rise}) = \mathbf{e}^{-t/\tau} \] \[0.9 = M(t_{rise}) = 1-\mathbf{e}^{-t/\tau} \] which can be solved for \(t_{rise} = 2.30\tau\)
   7. Sine function input [an alternative, periodic forcing function \(F(t)\)]
      1. Input has form \(F(t) = A\sin(\omega t)\) where
         1. \(A\) is amplitude of input (where units are)
         2. \(\omega\) is circular frequency (rad/s)
         3. \(f = \frac{\omega}{2\pi}\) when cyclic frequency is given in Hertz
         4. No vertical offset is included (would be part of step function above)
         5. No horizontal phase shift of start: \(F(t=0) = A\sin(\omega \times 0) = 0\)
      2. D.E. in parameterized form \[\tau \dot{y}(t) + y(t) = kA\sin(\omega t) \]
      3. Solution in full form: \[ y(t) = \frac{kA}{\sqrt{\omega^{2}\tau^{2} + 1}}\sin\left(\omega t - \tan^{-1}\left[\omega \tau\right]\right) + \left(y_{0} + \frac{\omega \tau k A}{\omega^{2}\tau^{2} +1}\right) \mathbf{e}^{\frac{-t}{\tau}} \] where the first part is oscillating steady-state response while the second portion is a transient response that decays with exponential.
      4. Using the simplified form: \[ y(t) = k A M(\omega)\sin\left(\omega t + \phi\left(\omega\right)\right) + C\mathbf{e}^{\frac{-t}{\tau}} \] where
         1. Magnitude ratio is frequency dependent \(M(\omega) = \frac{1}{\sqrt{\omega^{2}\tau^{2}+1}}\)
         2. \(M(\omega)\) represents ratio of output amplitude to input amplitude
         3. Uses same form for dynamic error that is now frequency dependent \(\delta(\omega) = 1- M(\omega)\)
         4. Phase lag causes shift of \(\phi(\omega) = -\tan^{-1}(\omega \tau)\) in radians
         5. Convert to time lag of signal response thru input frequency \(\beta = \frac{\phi}{\omega} = \frac{-\tan^{-1}(\omega \tau)}{\omega}\) in seconds
         6. Decay coefficient \(C\) depends on full combination of
            • initial conditions \(y_{0}\)
            • physics of the model: sensitivity \(k\) and time constant \(\tau\)
            • input parameters: amplitude \(A\) and frequency \(\omega\)
      5. \(M(\omega)\) sometimes expressed in logarithmic sense
         1. Consider ratios of power output to input: \[bel = \log_{10}\left(\frac{P_{out}}{P_{in}}\right) \]
         2. Scale using prefix (i.e. like 1 decimeter = 10 meter) \[ deci bel = 10 \log_{10}\left(\frac{P_{out}}{P_{in}}\right) \]
         3. Power is squared quantity of measurands \[ dB = 10 \log_{10}\left(\frac{\mathrm{Amplitude}_{out}}{\mathrm{Amplitude}_{in}}\right)^{2} \] \[ dB = 20 \log_{10}\left(M(\omega)\right) \]
   8. Summary of 1st order systems
      1. Total solution still follows form \[y(t) = y_{homogeneous}(t) + y_{particular}(t) \] where \(y_{h}\) is transient response with exponential terms and \(y_{p}\) will depend on the type of input used (step of section F.6 or sinusoid of section F.7)
      2. Given the general form of solutions, fit the provided data into appropriate coefficients.

      3. Exercise \(\PageIndex{1}\)

         During a step function calibration, a first-order vane anemometer with 10 fan blades is exposed to a step change from 12 mph initially to 88 mph. The transducer counts the number of fan blades that pass a magnetic sensor, producing a static sensitivity of \(k = 22\) clicks/mph. After 143 milliseconds, the instrument output indicates 777 clicks; after 477 milliseconds the output is 1,444 clicks. Estimate the time constant of the first-order sensor.

         What time after the step change will the dynamic error be less than 2.6%?

         Answer

         The system is first-order with a step change output making \[y(t) = kA + (y_{0}-kA)\mathrm{e}^{-t/\tau}\]

         The amplitude at the end of the step is 88 mph which would become 1936 clicks (\(kA\)) as time increases; the value at the start \(y_{0} = 264\) clicks. One could solve for the exponential function or sometimes called the error fraction: \[\delta_{f}(t) = \frac{y(t)-y_{\infty}}{y_{0}-y_{\infty}} = \mathrm{e}^{-t/\tau} \] \[
         \delta_{f}(0.143) = \frac{777 - 1936}{264- 1936} = \mathrm{e}^{-0.143/\tau} \] which can algebraically be solved for \(\tau\) to determine that \(\tau = 390\) milliseconds.

         If the error function is used, the associated time of 2.6% error can be calculated: \[0.026 = \mathrm{e}^{-t/0.390} \] \[t = -0.390\ln(0.026) \] where \(t= 1.42\) sec.

         Place two EXAMPLES
      4. At these locations
7. Second order systems (cases when \(n=2\) )
   1. Systems with inertia in the physical description that affect acceleration and not rate of response
   2. General form of differential equation \[ a_{2}\ddot{y} + a_{1}\dot{y} + a_{0}y(t) = F(t) \]
   3. Rewritten in more common form \[ \frac{1}{\omega_{n}^{2}}\ddot{y} + \frac{ 2\zeta}{\omega_{n}} \dot{y} + y(t) = k F(t) \] where
      1. Natural frequency from coefficients \(\omega_{n} = \sqrt{\frac{a_{0}}{a_{2}}}\)
      2. Dimensionless damping ratio \(\zeta = \frac{a_{1}}{2\sqrt{a_{2}a_{0}}}\)
   4. Homogeneous solution form is dependent on \(\zeta\)
      1. Under damped scenario \(0\leq \zeta < 1\) -- oscillation is present
      2. Critically damped \(\zeta = 1\) -- extremely rare scenario
      3. Overdamped \(\zeta > 1\) -- no oscillation occurs
   5. Particular solution for step input
      1. Overdamped is complicated hyperbolic functions \[ y(t) = kA\left\{1-\mathbf{e}^{-\zeta\omega_{n}t}\left[\cosh\left(\omega_{n}t\sqrt{\zeta^{2}-1}\right) + \frac{\zeta}{\sqrt{\zeta^{2}-1}}\sinh\left(\omega_{n}t\sqrt{\zeta^{2}-1}\right)\right]\right\} \]
         -- will be avoided from MEE-standpoint
      2. Critically damped appears more like 1st order system with exponential decay \[ y(t) = kA\left\{1-\mathbf{e}^{-\omega_{n}t}\left(1+\omega_{n}t\right)\right\} \]
      3. Underdamped has solution form \[ y(t) = kA\left\{1-\mathbf{e}^{-\zeta \omega_{n} t}\left[\frac{1}{\sqrt{1-\zeta^{2}}}\sin\left(\omega_{n}t\sqrt{1-\zeta^{2}} + \sin^{-1}\left[\sqrt{1-\zeta^{2}}\right]\right)\right]\right\} \]
      4. Consider figure XX from for sketch of the response plots
      5. Since no input frequency to follow, a ringing (or sometimes called damped) frequency is observed
         1. Scaled version of natural frequency \(\omega_{d} = \omega_{n}\sqrt{1-\zeta^{2}}\)
         2. Period of oscillation \(T_{d} = \frac{1}{f_{d}} = \frac{2\pi}{\omega_{d}}\)
         3. Rise time is threshold of first achieving 90% response (e.g., like when \(\delta(t_{rise})<0.1\))
         4. Settling time is when oscillations are less than \(\pm 10\) around end condition
   6. Solution for sinusoidal input of \(F(t)\) \[ y(t) = y_{h} + \frac{KA}{\sqrt{\left[1-\left(\frac{\omega}{\omega_{n}}\right)^{2}\right]^{2} + \left[2\frac{\zeta\omega}{\omega_{n}}\right]^{2}}}\sin\left(\omega t + \phi(\omega)\right) \]
      1. Phase shift is frequency dependent \[\phi(\omega) = \tan^{-1}\left(\frac{-2\frac{\zeta\omega}{\omega_{n}}}{1-\left(\frac{\omega}{\omega_{n}}\right)^{2}}\right) \] where sign of numerator and denominator necessary for 4-quadrant inverse
      2. Amplitude of steady-state condition still has magnitude ratio \[M(\omega) = \mathrm{\frac{Amplitude_{output}}{Amplitude_{input}}} =\frac{\cancel{kA}}{\sqrt{\left[1-\left(\frac{\omega}{\omega_{n}}\right)^{2}\right]^{2} + \left[2\frac{\zeta\omega}{\omega_{n}}\right]^{2}}}\frac{1}{\cancel{kA}} \] \[ M(\omega) = \frac{1}{\sqrt{\left[1-\left(\frac{\omega}{\omega_{n}}\right)^{2}\right]^{2} + \left[2\frac{\zeta\omega}{\omega_{n}}\right]^{2}}} \] produces bode plots for cases around \(\omega = \omega_{n}\) (Figs XX from Supplementary Section)
      3. Other characteristics of periodically forced 2nd order systems shown in figure
         1. Resonance \(\Rightarrow\) very small \(\zeta\) but \(\frac{\omega}{\omega_{n}} \rightarrow 1\)
            • Output becomes unstable form
         2. Small frequency ratio \(\frac{\omega}{\omega_{n}}\) makes \(M(\omega)\) nearly unity with small phase lag
            • Best input to output transmission
         3. Large frequency ratio \(\frac{\omega}{\omega_{n}}\) makes \(M(\omega)\) small with delayed phase
            • Potential to eliminate unwanted frequency data

    

    

    

    

    

    

    

   7. Summary of 2nd order systems
      1. Types of homogeneous solutions become dependent on scale of constant coefficients
      2. Particular solutions consider the long-term outcome of the response rather than instant change
      3. Sorting by ODE order and type of forcing aligns with terms defined in each of the 4 categories

      4. Exercise \(\PageIndex{1}\)

         A dynamic system is evaluated by multiple input types. When zero initial displacement \(s(0)=0\) and velocity \(\dot{s}(0)=0\) are then actuated with a step input \(U(t)\) with the following differential form: \[3.6\dot{s} + 15s + 0.6\ddot{s} = 52.5 U(t) \] Determine the following:

         • full equation of the response \(s(t)\)
         • damping ratio
         • ringing (damped) frequency
         • natural frequency

         If the same system were excited by a periodic input, determine the range of frequencies where the output amplitude would exceed the input amplitude. At the maximum frequency on that range, calculate the phase lag and time delay of the response.

         Answer

         The coefficients of \(a_{2} = 0.6\), \(a_{1} = 3.6\) and \(a_{0} = 15\) determine the terms as \[\omega_{n} = \sqrt{a_{0}/a_{2}} = \sqrt{15/0.6} = 5 \ \mathrm{rad/s} \] \[\zeta = a_{1} /2\sqrt{a_{0}a_{2}} = 3.6/2\sqrt{15(0.6)} = 0.6 \ \mathrm{(underdamped)} \] \[\omega_{d} = \omega_{n}\sqrt{1-\zeta^{2}} = 4 \ \mathrm{rad/s} \]

         
         With the amplitude of the input being 52.5, the product \(KA = 3.5\) therefore the response is \[y(t) = KA\left\{1-\mathbf{e}^{-\zeta \omega_{n} t}\left[\frac{\sin\left(\omega_{d}t+\phi\right)}{\sqrt{1-\zeta^{2}}}\right]\right\} \] \[y(t) = 3.5\left\{1-1.25\mathrm{e}^{-3t}\sin\left(4t + 0.927\right)\right\} \]
         where the phase was found by \(\phi = \sin^{-1}(\sqrt{1-.6^{2}})\).

      5. Example \(\PageIndex{1}\)

         Add example text here.

         Solution

         Add example text here.

         But Example #2 follows