General Solution for a Critically Damped System

Critical Damping

2mKn: critical damping coefficient in vibration system of one degree-of-freedom consisting of a mass, a spring and a dashpot, N•s/m

From: Handbook of Powder Technology , 2001

CRITICAL DAMPING

D. Inman , in Encyclopedia of Vibration, 2001

Definition Of Critical Damping

Critical damping is defined for a single-degree-of-freedom, spring-mass-damper arrangement, as illustrated in Figure 1. The equation of motion for this system is found from Newton's law and the free-body diagram to be:

Figure 1. A single-degree-of-freedom system and free-body diagram.

(1) $m\ddot{x}(t)+c\stackrel{.}{x}(t)+kx(t)=0$

Here x(t) is the displacement in meters, $\stackrel{.}{x}(t)$ is the velocity in meters per second, $\ddot{x}(t)$ is the acceleration in meters per second per second, m is the mass in kilograms, k is thestiffness in Newtons per meter and c is the damping coefficient in Newton second per meter or kilograms per second. Eqn (1) is written in dimensionless form by dividing the expression by the mass. This yields:

(2) $\ddot{x}(t)+2\zeta {\omega }_n\stackrel{.}{x}(t)+{\omega }_n^{2}x(t)=0$

Here the undamped natural frequency is defined to be:

(3) ${\omega }_n=\sqrt{k/m}$

(in radians per second) and the damping ratio is defined to be:

(4) $\zeta =\frac{c}{2m{\omega }_n}=\frac{c}{2\sqrt{km}}$

which is dimensionless.

To solve the differential equation given in eqn (2), a solution of the form $x(t)=A{\text{e}}^{\lambda t}$ is assumed and substituted into eqn (2) to yield:

(5) ${\lambda }^2+2\zeta {\omega }_n\lambda +{\omega }_n^2=0$

which is called the characteristic equation in the unknown parameter λ. This process effectively changes the differential eqn (2) into a quadratic algebraic equation with a well-known solution. The value of λ satisfying eqn (5) is then:

(6) $\lambda =-\zeta {\omega }_n\pm {\omega }_n\sqrt{{\zeta }^2-1}$

which are the two roots of the characteristic equation. It is the radical expression that gives rise to the concept of critical damping. If the radical expression is zero, that is, if:

(7) ${\zeta }^2=1$

the system of Figure 1 is said to be critically damped. If ζ = 1, then the corresponding damping coefficient c is called the critical damping coefficient, c _cr, by:

(8) $c_{cr}=2m{\omega }_n=2\sqrt{km}$

obtained by setting ζ = 1 in eqn (4).

The mathematical significance of ζ = 1 is that the two roots given in eqn (6) collapse to a repeated real root of value $\lambda =-{\omega }_n$ . The solution to the vibration problem given in eqn (1) then becomes:

(9) $x(t)=(a_1+a_{2}t){\text{e}}^{-{\omega }_{\text{n}}t}$

where the constants of integration a ₁ and a ₂ are determined by the initial conditions. Substituting the initial displacement, x ₀, into eqn (10) and the initial velocity, v ₀, into the derivative of eqn (10) yields:

(10) $a_1=x_0,\mathrm{and}{a}_2={\upsilon }_0+{\omega }_n{x}_0$

Critically damped motion is plotted in Figure 2 for two different values of initial conditions. Notice that no oscillation occurs in this response.

Figure 2. The critically damped response for two different initial conditions.

The motion of critically damped systems may be thought of in several ways. First, a critically damped system represents a system with the smallest value of damping coefficient that yields aperiodic motion. If ζ > 1, the roots given in eqn (6) are distinct, real roots giving rise to solutions of the form:

(11) $x(t)={\text{e}}^{-\zeta {\omega }_{n}t}\left(a_1{\text{e}}^{-{\omega }_n\sqrt{{\zeta }^2-1}t}+a_2{\text{e}}^{+{\omega }_n}\sqrt{{\zeta }^2-1}t\right)$

which also represents a nonoscillatory response. Again, the constants of integration a ₁ and a ₂ are determined by the initial conditions. In this aperiodic case, the constants of integration are real valued and are given by:

(12) $a_1=\frac{-{\upsilon }_0+\left(-\zeta +\sqrt{{\zeta }^2-1}\right){\omega }_n{x}_0}{2{\omega }_n\sqrt{{\zeta }^2-1}}$

and

(13) $a_2=\frac{-{\upsilon }_0+\left(\zeta +\sqrt{{\zeta }^2-1}\right){\omega }_n{x}_0}{2{\omega }_n\sqrt{{\zeta }^2-1}}$

Such a response is called an overdamped system and does not oscillate, but rather returns to its rest position exponentially.

If the damping value is reduced below the critical value, so that ζ < 1, the discriminant of eqn (6) is negative, resulting in a complex conjugate pair of roots. These are:

(14) ${\lambda }_1=-\zeta {\omega }_n-{\omega }_n\sqrt{1-{\zeta }^{2}j}$

and:

(15) ${\lambda }_2=-\zeta {\omega }_n+{\omega }_n\sqrt{1-{\zeta }^2}j$

where $j=\sqrt{-1}$ and:

(16) $\sqrt{1-{\zeta }^2}j=\sqrt{(1-{\zeta }^2)(-1)}=\sqrt{{\zeta }^2-1}$

The solution of eqn (1) is then of the form:

(17) $x(t)={\text{e}}^{-\zeta {\omega }_{n}t}\left(a_1{\text{e}}^j\sqrt{1-{\zeta }^2}{\omega }_{n}t+a_2{\text{e}}^{-j}\sqrt{1-{\zeta }^2}{\omega }_{n}t\right)$

where a ₁ and a ₂ are arbitrary complex valued constants of integration to be determined by the initial conditions. Using the Euler relations for the sine function, this can be written as:

(18) $x(t)=A{\text{e}}^{-\zeta {\omega }_{n}t}sin({\omega }_{d}t+\phi )$

where A and ϕ are constants of integration and ω _d, called the damped natural frequency, is given by:

(19) ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$

The constants A and ϕ are evaluated using the initial conditions. This yields:

(20) $\begin{array}{c}A=\sqrt{\frac{{\left({\upsilon }_0+\zeta {\omega }_n{x}_0\right)}^2+{(x_0{\omega }_d)}^2}{{\omega }_d^2}},\\ \phi ={tan}^{-1}\left[\frac{x_0{\omega }_d}{{\upsilon }_0+\zeta {\omega }_n{x}_0}\right]\end{array}$

where x ₀ and v ₀ are the initial displacement and velocity. Note that for this case (0 < ζ < 1) the motion oscillates. This is called an underdamped system. Hence, if the damping is less then critical, the motion vibrates, and critical damping corresponds to the smallest value of damping that results in no vibration. Critical damping can also be thought of as the value of damping that separates nonoscillation from oscillation.

The concept of critical damping provides a useful value to discuss level of damping in a system, separating the physical phenomenon of oscillation and no oscillation. Often when reporting levels of damping, the percent of critical damping is provided. The percent critical damping is defined as 100ζ% or:

(21) $\frac{c}{c_{cr}}\times 100$

Normal values for ζ are very small, such as 0.001, so that percent critical damping is of the order of 0.1–1%. However, viscoelastic materials, hydraulic dampers, and active control systems are able to provide large values of percent critical damping.

Critical damping viewed as the minimum value of damping that prevents oscillation is a desirable solution to many vibration problems. Increased damping implies more energy dissipation, and more phase lag in the response of a system. Reduced damping means more oscillation, which is often undesirable. Adding phase to the system slows the response down and in some cases this may be undesirable. Hence, critical damping is a desirable tradeoff between having enough damping to prevent oscillation and not so much damping that the system uses too much energy or causes too large a phase lag.

An example of the use of critical damping is in the closing of a door. If the mechanism has too much damping, the door will move slowly and too much heat will exchange between the inside and outside. If too little damping is used the door will oscillate at closing and again too much air and heat is exchanged. At critical damping the door closes without oscillation and a minimum amount of air and heat are exchanged. The needle gauges used in tachometers and speedometers in automobiles provide additional examples of critically damped systems. The needle operates as a torsional spring with critical damping. Too much damping would make the gauge slow to reach the actual value and too little damping would cause the needle to vibrate.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B0122270851000618

Ceramic input capacitors can cause overvoltage transients

Goran Perica , in Analog Circuit Design, 2011

Optimizing Input Capacitors

Waveforms in Figure 1.3 show how input transients vary with the type of input capacitors used.

Optimizing the input capacitors requires clear understanding of what is happening during transients. Just as in an ordinary resonant RLC circuit, the circuit in Figure 1.1 may have an underdamped, critically damped or overdamped transient response.

Because of the objective to minimize the size of input filter circuit, the resulting circuit is usually an underdamped resonant tank. However, a critically damped circuit is actually required. A critically damped circuit will rise nicely to the input voltage without voltage overshoots or ringing.

To keep the input filter design small, it is desirable to use ceramic capacitors because of their high ripple current ratings and low ESR. To start the design, the minimum value of the input capacitor must first be determined. In the example, it has been determined that a 22   μF, 35V ceramic capacitor should be sufficient. The input transients generated with this capacitor are shown in the top trace of Figure 1.4. Clearly, there will be a problem if components that are rated for 30V are used.

Figure 1.4. Optimizing Input Circuit Waveforms for Reduced Peak Voltage

To obtain optimum transient characteristic, the input circuit has to be damped. The waveform R2 shows what happens when another 22   μF ceramic capacitor with a 0.5   Ω resistor in series is added. The input voltage transient is now nicely leveled off at 30V.

Critical damping can also be achieved by adding a capacitor of a type that already has high ESR (on the order of 0.5  Ω). The waveform R3 shows the transient response when a 22   μF, 35V TPS type tantalum capacitor from AVX is added across the input.

Table 1.3. Peak Voltages of Waveforms In Figure 1.4 with 22   μF Input Ceramic Capacitor and Added Snubber

TRACE	SNUBBER TYPE	Vin PEAK (V)
R1	None	40.8
R2	22   μF Ceramic   +   0.5   Ω In Series	30
R3	22   μF Tantalum AVX, TPS Series	33
R4	30V TVS, P6KE30A	35
Ch1	47   μF, 35V Aluminum Electrolytic Capacitor	25

The waveform R4 shows the input voltage transient with a 30V transient voltage suppressor for comparison.

Finally, an ideal waveform shown in Figure 1.4, bottom trace (Ch1) is achieved. It also turns out that this is the least expensive solution. The circuit uses a 47   μF, 35V aluminum electrolytic capacitor from Sanyo (35CV47AXA). This capacitor has just the right value of capacitance and ESR to provide critical damping of the 22   μF ceramic capacitor in conjunction with the 1   μH of input inductance. The 35CV47AXA has an ESR value of 0.44   Ω and an RMS current rating of 230mA. Clearly, this capacitor could not be used alone in an application with 1A to 2A of RMS ripple current without the 22   μF ceramic capacitor. An additional benefit is that this capacitor is very small, measuring just 6.3mm by 6mm.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780123851857000019

Mechanical Vibration

John P.T. Mo , ... Raj Das , in Demystifying Numerical Models, 2019

Case (ii): Critical Damping (δ=ω ₀)

For the case of critical damping, the analytical solution is given by Eq. (10.18) [1].

(10.18) $u(t)=\left(A+Bt\right)e^{-\delta t}$

where A and B are evaluated based on initial conditions. Fig. 10.27 shows the comparison between the analytical and two numerical solutions with different discretizations of 100 and 500 time steps, whereby the numerical solutions match closely with the analytical one.

Figure 10.27. Comparison between the numerical and analytical solutions for a free critically damped vibration with δ=ω ₀=0.1   s⁻¹, u ₀=0.1   m and ${{\displaystyle \dot{u}}}_0$ =0   m   s⁻¹.

The maximum error is affected by the natural frequency and the initial velocity. It should be noted that when the analytical solution is close to zero, the relative error becomes higher.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780081009758000102

Damping

Alvar M. Kabe , Brian H. Sako , in Structural Dynamics Fundamentals and Advanced Applications, 2020

4.2 Damping from half-power points of total response

In this section, we will derive the critical damping ratio, $\zeta$ , from the frequencies associated with the half-power points of the total response. We will also show that these frequencies are nearly the same as the frequencies that correspond to the peaks in the coincident component of response, and that the average of the coincident component of response frequencies corresponds to the half-power points. Fig. 4.2-1 shows the total response (solid line) as well as the quadrature (short dashes) and coincident (short/long dashes) components. The half-power point frequencies are labeled as ${\lambda }_1$ and ${\lambda }_2$ , and they correspond to the values, on either side of the natural frequency, of the total response curve that are $1/\sqrt{2}$ , or 0.707, of the peak response value.

Figure 4.2-1. Quadrature (short dash line), coincident (long/short dashed line), and total (solid line) components of the acceleration response of a single-degree-of-freedom system with $\zeta =0.02$ and ${\lambda }_i={\omega }_i/{\omega }_n$ .

We begin the derivation by seeking the fraction, ${\alpha }_i$ , of the peak total response that corresponds to the frequencies associated with each of the peaks in the coincident component of response, which we have labeled as ${\lambda }_1$ and ${\lambda }_2$ . We will start with ${\lambda }_1$ ; and note that we have introduced functional notation for the quadrature and coincident components to explicitly indicate that they are both functions of $\lambda$ , i.e.,

(4.2-1) $\begin{array}{c}{\alpha }_1\sqrt{{\left\{Q{d}_{\ddot{x}}\left({\lambda }_n\right)\right\}}^2}=\sqrt{{\left\{Q{d}_{\ddot{x}}\left({\lambda }_1\right)\right\}}^2+{\left\{C{o}_{\ddot{x}}\left({\lambda }_1\right)\right\}}^2}\\ {\alpha }_1^2{\left\{Q{d}_{\ddot{x}}\left({\lambda }_n\right)\right\}}^2={\left\{Q{d}_{\ddot{x}}\left({\lambda }_1\right)\right\}}^2+{\left\{C{o}_{\ddot{x}}\left({\lambda }_1\right)\right\}}^2\end{array}$

For lightly damped systems, the peak total response can be computed with $\lambda ={\lambda }_n=1$ , and the coincident component of response will, therefore, be zero. The left-hand side of Eq. (4.2-1), which is the peak of the total response, only involves the quadrature component evaluated at ${\lambda }_n=1$ . The right-hand side is the total response at ${\lambda }_1$ .

Substituting from Chapter 2,

(4.2-2) ${\alpha }_1^2\frac{1}{4{\zeta }^2}={\left\{{\lambda }_1^2\frac{2\zeta {\lambda }_1}{{\left(1-{\lambda }_1^2\right)}^2+{\left(2\zeta {\lambda }_1\right)}^2}\right\}}^2+{\left\{-{\lambda }_1^2\frac{\left(1-{\lambda }_1^2\right)}{{\left(1-{\lambda }_1^2\right)}^2+{\left(2\zeta {\lambda }_1\right)}^2}\right\}}^2$

Performing the indicated algebraic operations yields

(4.2-3) ${\alpha }_1^2\frac{1}{4{\zeta }^2}={\lambda }_1^4\frac{1}{{\left(1-{\lambda }_1^2\right)}^2+{\left(2\zeta {\lambda }_1\right)}^2}$

Next, we substitute for ${\lambda }_1^2$ the value from Eq. (4.1-4) that corresponds to the first peak in the coincident component of response,

(4.2-4) $\begin{array}{c}{\alpha }_1^2\frac{1}{4{\zeta }^2}={\left(\frac{1}{1+2\zeta }\right)}^2\frac{1}{{\left(1-\frac{1}{1+2\zeta }\right)}^2+\frac{4{\zeta }^2}{1+2\zeta }}\\ =\frac{1}{4{\zeta }^2+4{\zeta }^2\left(1+2\zeta \right)}\end{array}$

Solving for ${\alpha }_1$ we obtain

(4.2-5) ${\alpha }_1=\frac{1}{\sqrt{2}}\sqrt{\frac{1}{1+\zeta }}$

Following the same procedure for ${\lambda }_2$ , we get

(4.2-6) ${\alpha }_2=\frac{1}{\sqrt{2}}\sqrt{\frac{1}{1-\zeta }}$

${\alpha }_1$ and ${\alpha }_2$ are the fractions of the peak response at the frequencies where the coincident component of response peaks. For lightly damped systems, ${\alpha }_1\approx {\alpha }_2\approx \frac{1}{\sqrt{2}}$ . However, it's worth noting that ${\alpha }_2$ will be slightly greater than ${\alpha }_1$ . So what if we used the average of the two, i.e.,

(4.2-7) $\alpha =\frac{1}{2}\left({\alpha }_1+{\alpha }_2\right)=\frac{1}{\sqrt{2}}\frac{1}{2}\left\{\sqrt{\frac{1}{1+\zeta }}+\sqrt{\frac{1}{1-\zeta }}\right\}$

Taking the Taylor series expansion (see Appendix 4.1) of the term in the braces, and noting that for lightly damped system we can neglect the higher-order terms associated with $\zeta$ , we obtain

(4.2-8) $\begin{array}{c}\alpha =\frac{1}{\sqrt{2}}\frac{1}{2}2\left\{1+\frac{3}{16}{\zeta }^2+\frac{35}{128}{\zeta }^4+f\left({\zeta }^6\right)\right\}\\ \approx \frac{1}{\sqrt{2}}\end{array}$

Therefore, given a total response function we can take the frequencies corresponding to $1/\sqrt{2}$ times the peak response, and use these frequencies in either Eqs. (4.1-6), (4.1-7) or (4.1-8) to compute the critical damping ratio, $\zeta$ . In Volume II, we will show how we can use the above derivations to establish damping for large, complex, multi-degree-of-freedom systems, such as buildings, airplanes, launch vehicles, and satellites.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128216149000045

Transient and Frequency Response

Dennis L. Feucht , in Handbook of Analog Circuit Design, 1990

5.7 Time-Domain Response to a Unit Step Function

To find the time-domain response to a given input function, we can take the inverse Laplace transform $L^{-1}$ of X _o(s) in (5.68) (or apply the convolution integral). After X _o(s) is in the form of a known transform, it is inverse Laplace transformed to produce the time-domain response. Since M(s) is a rational function, partial-fraction expansion is the usual method of expressing X _o in terms that can be inverse transformed.

The Laplace-transformed impulse function, when multiplied by M(s), yields the s-domain transient response. Because it is such a difficult function to generate and observe, the step function is the dominant alternative. It is approximated in practice by a square wave with a period much longer than the duration of significant transient response (and thereby is effectively aperiodic). Various characteristics of circuit response to the step are of interest and all are time related. This approach to circuit characterization is time-domain analysis.

Transfer functions represent the complex dynamic behavior of circuits but are an abstraction of actual circuit behavior. The response of a circuit under controlled conditions produces features that characterize the circuit. We now investigate the characterization of circuits by their time-domain response to a unit step input, u(t). The time-domain response can be determined by multiplying the transfer function by the Laplace transform of u(t), or 1/s, and inverse transforming the result.

The RC integrator response is calculated as

(5.97) ${\upsilon }_{\text{o}}(t)={ℒ}^{-1}\left\{\frac{1}{sRC+1}·\frac{1}{s}\right\}$

The s-domain expression is partial-fraction expanded to

(5.98) $\frac{A}{s}+\frac{B}{sRC+1}=\frac{1}{s}-\frac{RC}{sRC+1}$

This inverse transforms, using (5.71) and (5.72), to

(5.99) ${\upsilon }_{\text{o}}(t)=u(t)-e^{t/RC}=1-e^{-t/RC},t>0$

which is plotted against t/RC in Fig. 5.9, curve a.

FIG. 5.9. First-order responses of (a) RC integrator and (b) RC differentiator. These curves apply to any first-order circuit and allow rapid determination of fractional decay versus time constant.

For the RC differentiator,

(5.100) ${\upsilon }_{\text{o}}(t)={ℒ}^{-1}\left\{\frac{RCs}{sRC+1}·\frac{1}{s}\right\}={ℒ}^{-1}\left\{\frac{1}{s+(1/RC)}\right\}=e^{-t/RC}$

which is shown in 5.9, curve b, with time scaled in time constants.

The response of a circuit with complex poles is demonstrated by the RLC circuit of Fig. 5.10. Its transfer function can be written by treating it as a voltage divider. Then

FIG. 5.10. RLC circuit with quadratic pole factor.

(5.101) $\begin{array}{l}\frac{V_{\text{o}}(s)}{V_{\text{i}}(s)}=\frac{(1/sC)\Vert R}{(1/sC)\Vert R+sL}=\frac{1}{LC}·\frac{1}{s^2+s(1/RC)+(1/LC)}\\ =\frac{1}{s^2+LC+s(L/R)+1}\end{array}$

where

(5.102) $\begin{array}{ccc}K=1,& {\omega }_n=\frac{1}{\sqrt{LC}},& \alpha =\frac{1}{2RC}\end{array}$

For V _i(s) = 1/s, the step response of the RLC circuit is

(5.103) ${\upsilon }_{\text{step}}(t)={\mathcal{L}}^{-1}\left\{\frac{1}{s}·\frac{1}{s^{2}LC+s(L/R)+1}\right\}$

The quadratic factor is of the form

(5.104) $\frac{N(s)}{s^2+2\alpha s+\omega \begin{array}{c}2\\ n\end{array}}=\frac{1}{\omega \begin{array}{c}2\\ n\end{array}}·\frac{N(s)}{(s^2/\omega \begin{array}{c}2\\ n\end{array})+(2\alpha /\omega \begin{array}{c}2\\ n\end{array})s+1}$

The denominator can be factored into

(5.105) $\begin{array}{l}\frac{N(s)}{(s+\alpha +j{\omega }_{\text{d}})(s+\alpha -j{\omega }_{\text{d}})}\\ =\frac{1}{\omega \begin{array}{c}2\\ n\end{array}}·\frac{N(s)}{[(s/{\omega }_{\text{n}})+(\alpha /{\omega }_{\text{n}})s+j({\omega }_d/{\omega }_n)][(s/{\omega }_n)+(\alpha /{\omega }_{\text{n}})s-j({\omega }_{\text{d}}/{\omega }_n)]}\end{array}$

This can be expressed as a partial fraction expansion:

(5.106) $\frac{N(s)}{(s-p)(s-p*)}=\frac{A*}{(s-p)}+\frac{A}{(s-p*)}$

where X* is the complex conjugate of X and

(5.107) $p=-\alpha +j{\omega }_{\text{d}}$

This form can be shown to be valid by letting

$\{\begin{array}{c}A=a+jb\\ N(s)=cs+d\end{array}$

This is the most general form N can take, with its degree one less than the denominator. [If (5.106) is a transfer function of a circuit with zero magnitude at infinite frequency, the fraction must be less than 1, or m < n of (5.61).] Then the partial-fraction expansion coefficients are

(5.108) $a=\frac{c}{2},b=\frac{-\alpha c+d}{2{\omega }_{\text{d}}}$

The time-domain response of the right side of (5.106) is found by making use of the polar form of A. Substituting A = ‖A‖ e ^jv and A* = ‖A‖e ^−jv , we obtain

(5.109a) ${\mathcal{L}}^{-1}\left\{\frac{A*}{s-p}+\frac{A}{s-p*}\right\}=A*e^{pt}+A{e}^{p*t}$

(5.109b) $\begin{array}{l}=\Vert A\Vert e^{-\alpha t}(e^{-j\vartheta }{e}^{j{\omega }_{d^t}}+e^{j\vartheta }{e}^{-j{\omega }_{d^t}})\\ =\Vert A\Vert e^{-\alpha t}(e^{j({\omega }_{{\text{d}}^{t-\vartheta )}}}+e^{-j({\omega }_{{\text{d}}^{t-\vartheta )}}})\\ =2\Vert A\Vert e^{-\alpha t}\cos ({\omega }_{\text{d}}t-\vartheta )\end{array}$

Therefore, the general transform involving complex pole pairs is

(5.110) $\frac{\Vert A\Vert e^{j\vartheta }}{s+\alpha +j{\omega }_{\text{d}}}+\frac{\Vert A\Vert e^{-j\vartheta }}{s+\alpha -j{\omega }_{\text{d}}}\stackrel{L^{-1}}{\Rightarrow }2\Vert A\Vert e^{-\alpha t}\cos ({\omega }_{\text{d}}t-\vartheta )$

Returning to (5.103), its partial fraction expansion is

(5.111) $\frac{1}{s}·\frac{1}{s^{2}LC+s(L/R)+1}=\frac{1}{LC}\left\{|\frac{A}{s}+\frac{B*}{s-p}+\frac{B}{s-p*}\right\}$

where p = −α +jω _d. Solving for the numerators of (5.111) gives

(5.112) $A=\frac{1}{{{\displaystyle \omega }}_{\text{n}}^2},B=-\frac{1}{2\omega \begin{array}{c}2\\ n\end{array}}\left[1+j\left(\frac{\alpha }{{\omega }_d}\right)\right]=-\frac{1}{2\omega \begin{array}{c}2\\ n\end{array}}·\frac{1}{\cos \gamma }·e^{j\gamma },\gamma ={\tan }^{-1}\left\{\frac{\alpha }{{\omega }_{\text{d}}}\right\}$

Inverse transforming (5.111), using (5.71) and (5.110), gives

(5.113) $\begin{array}{cc}{\upsilon }_{step}(t)=1-\frac{1}{\sin \gamma }·e^{-\alpha t}\cos ({\omega }_{d}t-\gamma ),& \gamma ={\tan }^{-1}\left\{\frac{\alpha }{{\omega }_{\text{d}}}\right\}\end{array}$

With the trigonometric relation

$\tan \vartheta =\frac{1}{\tan ({90}^{°}-\vartheta )}$

(5.113) becomes

(5.114) $\begin{array}{cc}{\upsilon }_{\text{step}}(t)=1-\frac{1}{\sin \varphi }·e^{-\alpha t}\sin ({\omega }_{d}t+\varphi ),& \varphi ={\tan }^{-1}\left\{\frac{{\omega }_d}{\alpha }\right\}\end{array}$

The factor 1/sin ϕ can be expressed as

(5.115) $\frac{1}{\sin \varphi }+\sqrt{1+{\left(\frac{\alpha }{{\omega }_d}\right)}^2={[1-{\zeta }^2]}^{-1/2}}$

In circuit element values, for the RLC circuit of Fig. 5.10,

(5.116) $\zeta =\frac{Z_n}{2R}=\sqrt{\frac{L/C}{2R}}$

An alternative approach to the inverse Laplace transformation of (5.104) is to complete the square for the quadratic denominator:

(5.117) $s^2+s\alpha s+\omega \begin{array}{c}2\\ \text{n}\end{array}=(s+\alpha ){}^2-({\alpha }^2-\omega \begin{array}{c}2\\ \text{n}\end{array})=(s+\alpha ){}^2+\omega \begin{array}{c}2\\ \text{d}\end{array}$

Then, for N(s) = cs + d, (5.104) becomes

(5.118) $\begin{array}{l}\frac{N(s)}{s^2+2\alpha s+\omega \begin{array}{c}2\\ \text{n}\end{array}}=\frac{cs}{{(s+\alpha )}^2+{\omega }_{\text{d}}^2}+\frac{d}{{(s+\alpha )}^2+\omega \begin{array}{c}2\\ \text{d}\end{array}}\\ =c·\frac{(s+\alpha )}{{(s+\alpha )}^2+{{\displaystyle \omega }}_{\text{d}}^2}-\left(\frac{\alpha c}{{\omega }_{\text{d}}}\right)\frac{{\omega }_{\text{d}}}{{(s+\alpha )}^2+\omega \begin{array}{c}2\\ \text{d}\end{array}}+\left(\frac{d}{{\omega }_{\text{d}}}\right)\frac{{\omega }_d}{{(s+\alpha )}^2+{{\displaystyle \omega }}_{\text{d}}^2}\end{array}$

(5.119) $\frac{N(s)}{s^2+2\alpha s+{{\displaystyle \omega }}_{\text{n}}^2}=c·\frac{(s+\alpha )}{{(s+\alpha )}^2+{{\displaystyle \omega }}_{\text{d}}^2}+\left(\frac{d-\alpha c}{{\omega }_{\text{d}}}\right)\frac{{\omega }_{\text{d}}}{{(s+\alpha )}^2+{\omega }_{\text{d}}^2}$

Using (5.75) and a similar extension of (5.74), we obtain

(5.120) ${\mathcal{L}}^{-1}(5.119)=c{e}^{-\alpha t}\cos {\omega }_{\text{d}}t+\left(\frac{d-\beta c}{{\omega }_{\text{d}}}\right)e^{-\alpha t}\sin {\omega }_{d}t$

For the sum of a sine and cosine,

(5.121) $a\cos \vartheta +b\sin \vartheta =\sqrt{a^2+b^2}\sin \left[\vartheta +{\tan }^{-1}\left(\frac{1}{b}\right)\right]$

Equation (5.121) can be applied to (5.120) to express it as a single sinusoid. After some manipulation,

(5.122) $\begin{array}{cc}\frac{cs+d}{s^2+2\alpha s+\omega \begin{array}{c}2\\ \text{n}\end{array}}\stackrel{\mathcal{L}-1}{\Rightarrow }\frac{c}{\sin \varphi }·e^{-\alpha t}\sin ({\omega }_{\text{d}}t-\varphi ),& \varphi ={\tan }^{-1}\left(\frac{c{\omega }_d}{\alpha c-d}\right)\end{array}$

If we apply this method to (5.103), the partial-fraction expansion is

(5.123) $\frac{1}{s}-\frac{s+2\alpha }{s^2+2\alpha s+{{\displaystyle \omega }}_{\text{n}}^2}$

From this expression, c = 1 and d = 2α. Substituting into (5.122) for the quadratic term yields

(5.124) $\begin{array}{cc}{\mathcal{L}}^{-1}(5.123)=1-\frac{1}{\sin \varphi }·e^{-\alpha t}\sin ({\omega }_{d}t+\varphi ),& \varphi ={\tan }^{-1}\left\{\frac{{\omega }_{\text{d}}}{\alpha }\right\}\end{array}$

This is the same result as (5.114).

For the case of repeated real poles (critical damping),

(5.125) ${\upsilon }_{step}(t)=1-(1+\alpha t)e^{-\alpha t}$

and for distinct real poles, since ζ > 1, ω_d is imaginary and

(5.126) $p_{1,2}=-\alpha \mp {\omega }_{\text{d}}=-\alpha \mp {\omega }_{\text{n}}\sqrt{{\zeta }^2-1}$

These are real roots. The step response for them is

(5.127) ${\upsilon }_{\text{step}}(t)=1-\left(\frac{p_1}{p_1-p_2}{e}^{p_{2^t}}-\frac{p_2}{p_1-p_2}{e}^{p_{1^t}}\right)$

Example 5.6

Magnetic Deflection Yoke Coil Circuit

Figure E5.6 shows a simplified CRT deflection circuit. The deflection yoke consists or horizontal and vertical deflection coils that magnetically deflect the CRT electron beam. A yoke coil has significant series resistance and intra winding capacitance, modeled as shown. If i _i(t) is a ramp function (producing a horizontal or vertical sweep needed for raster scanning of the CRT screen by the electron beam), then it can be expressed as

$i_i(t)=\left(\frac{1}{T}\right)t=mt$

where I is the peak ramp current and T the ramp duration (or the period of an ideal sawtooth function). The output current i _o(t) is the current that flows through L, creating the deflection field. Our goal is to find a general expression in s for I _o(s) and also to find the time-domain response.

FIG. E5.6.

The current divider formula is used here and yields

$\frac{I_{\text{o}}(s)}{I_i(s)}=\frac{1}{s^2\mathcal{L}C+sRC+1}$

For I _i(s) = m/s ²,

$I_{\text{o}}(s)=\frac{I}{s^2\mathcal{L}C+sRC+1}·\frac{m}{s^2}$

and ω _n ² = 1/LC, α = ζω _n = 1/(2L/R). I _o can be written by completing the square of the quadratic pole and expressing I _o as a partial-fraction expansion. Because of multiple roots at s = 0, it is necessary to take the derivative of the partial-fraction equation to find the coefficient k for the k/s term. Then,

$I_{\text{o}}(s)=\frac{(2\zeta /{\omega }_{\text{n}})·s+m(4{\zeta }^{2-1})}{{(s+\alpha )}^2+\omega \begin{array}{c}2\\ \text{d}\end{array}}+\frac{m}{s^2}-\frac{2\zeta m/{\omega }_{\text{n}}}{s}$

Using (5.122) to perform ∠⁻¹ on the first term, we find that

$i_{\text{o}}(t)=\left(\frac{2\zeta }{{\omega }_{\text{n}}\sin \varphi }\right)e^{ot}\sin ({\omega }_{\text{d}}t-\varphi )+m\left(\frac{2\zeta }{{\omega }_{\text{n}}}\right)$

i _o(t) is a ramp delayed by 2ζ/ω_n. Superimposed on this ramp is a decaying sinusoid, the first term. When the response of the horizontal deflector is too underdamped, the resulting ringing causes the picture on the left side of the CRT screen to show an alternating compaction and expansion until the sinusoid dies out.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780122542404500102

Single-degree-of-freedom systems

Alvar M. Kabe , Brian H. Sako , in Structural Dynamics Fundamentals and Advanced Applications, 2020

2.3.2 Nonoscillatory damped vibration

In the previous section, we covered the case where the critical damping ratio, $\zeta$ , was less than 1.0, and as such the system oscillated when set in motion. In this section, we will first cover the case where $\zeta$ is equal to 1.0, and then the case where $\zeta$ is greater than 1.0. For the case where $\zeta =1.0$ , which is referred to as the critically damped condition because $c=c_c$ , the assumed solution, Eqs. (2.3-2) and (2.3-4), will yield two identical roots; that is, Eq. (2.3-7) becomes $s_1,s_2=-{\omega }_n$ . This results in a solution with only one arbitrary constant, which would not be sufficient to allow for the application of two initial conditions. To resolve this, we start with the solution to the case where $\zeta <1.0$ , Eq. (2.3-13), and determine the solution as $\zeta$ approaches 1.0 in the limit,

(2.3-14) $x\left(t\right)=\underset{\zeta \to 1.0}{\lim }{e}^{-\zeta {\omega }_{n}t}\left(x\left(0\right)\cos \sqrt{1-{\zeta }^2}{\omega }_{n}t+\frac{\dot{x}\left(0\right)+\zeta {\omega }_{n}x\left(0\right)}{\sqrt{1-{\zeta }^2}{\omega }_n}\sin \sqrt{1-{\zeta }^2}{\omega }_{n}t\right)$

In Eq. (2.3-14), the second term in the parenthesis will have both its denominator and numerator approach zero as $\zeta$ approaches 1.0. There are two ways to resolve this. First, since the sine of a small angle approaches the angle, one can argue that

(2.3-15) $\underset{\zeta \to 1.0}{\lim }\sin \sqrt{1-{\zeta }^2}{\omega }_{n}t=\underset{\zeta \to 1.0}{\lim }\sqrt{1-{\zeta }^2}{\omega }_{n}t$

and Eq. (2.3-14), as $\zeta$ approaches 1.0 in the limit, reduces to

(2.3-16) $x\left(t\right)=e^{-{\omega }_{n}t}\left(x\left(0\right)+\left[\dot{x}\left(0\right)+{\omega }_{n}x\left(0\right)\right]t\right)$

The second and more rigorous approach is to use L'Hôpital's Rule (Crowell and Slesnick, 1968). Differentiating with respect to $\zeta$ the numerator and denominator of the second term in the parenthesis of Eq. (2.3-14), and then setting $\zeta$ equal to 1.0, will yield the proper quantity for this term in the limit (see Appendix 2.1). By also applying the limits to the other terms in Eq. (2.3-14) we obtain Eq. (2.3-16) again. Hence, Eq. (2.3-16) provides the response of a single-degree-of-freedom system that is critically damped.

Fig. 2.3-3 shows the response of the system from Fig. 2.3-2, except that $\zeta$ has been increased from 0.10 to 1.0. A system that has critical damping will return to its equilibrium point in the shortest amount of time without oscillating. Hence, many practical components are intentionally designed to have damping as close to critical damping as possible; examples include shock absorbers, door dampers, and spacecraft deployment mechanism dampers.

Figure 2.3-3. Response of a critically damped ( $c=c_c\Rightarrow \zeta =1.0$ ), single-degree-of-freedom system with ${\omega }_n=2\pi$ . Motion initiated by $x\left(0\right)=2.0$ —solid line; $x\left(0\right)=2.0$ and $\dot{x}\left(0\right)=10.0$ —dashed line.

For the case where $\zeta >1.0$ Eq. (2.3-7) yields two real roots, and our solution, from Eq. (2.3-8), becomes

(2.3-17) $\begin{array}{c}x\left(t\right)=e^{-\zeta {\omega }_{n}t}\left(A{e}^{\sqrt{{\zeta }^2-1}\text{undefined}{\omega }_{n}t}+B{e}^{-\sqrt{{\zeta }^2-1}\text{undefined}{\omega }_{n}t}\right)\\ =e^{-\zeta {\omega }_{n}t}\left(A{e}^{{\hat{\omega }}_{d}t}+B{e}^{-{\hat{\omega }}_{d}t}\right)\end{array}$

where ${\hat{\omega }}_d={\omega }_n\sqrt{{\zeta }^2-1}$ . Since $e^{\pm \theta }=\cosh \theta \pm \sinh \theta$ , we can write the solution as

(2.3-18) $\begin{array}{c}x\left(t\right)=e^{-\zeta {\omega }_{n}t}\left(A\cosh {\hat{\omega }}_{d}t+A\sinh {\hat{\omega }}_{d}t+B\cosh {\hat{\omega }}_{d}t-B\sinh {\hat{\omega }}_{d}t\right)\\ =e^{-\zeta {\omega }_{n}t}((A+B)\cosh {\hat{\omega }}_{d}t+(A-B)\sinh {\hat{\omega }}_{d}t)\end{array}$

Starting with $x\left(t\right)=A{e}^{s_{1}t}+B{e}^{s_{2}t}$ we can solve for $A$ and $B$ using the initial conditions,

(2.3-19) $\begin{array}{ccc}x\left(0\right)=A+B& & A=\frac{x\left(0\right)s_2-\dot{x}\left(0\right)}{s_2-s_1}\\ & \Rightarrow & \\ \dot{x}\left(0\right)=A{s}_1+B{s}_2& & B=\frac{\dot{x}\left(0\right)-x\left(0\right)s_1}{s_2-s_1}\end{array}$

Substituting $s_1=-\zeta {\omega }_n+{\hat{\omega }}_d$ and $s_2=-\zeta {\omega }_n-{\hat{\omega }}_d$ yields

(2.3-20) $\begin{array}{l}A+B=\frac{x\left(0\right)s_2-\dot{x}\left(0\right)+\dot{x}\left(0\right)-x\left(0\right)s_1}{s_2-s_1}=x\left(0\right)\\ A-B=\frac{x\left(0\right)s_2-\dot{x}\left(0\right)-\dot{x}\left(0\right)+x\left(0\right)s_1}{s_2-s_1}=\frac{x\left(0\right)\left(-2\zeta {\omega }_n\right)-2\dot{x}\left(0\right)}{-2{\hat{\omega }}_d}=\frac{x\left(0\right)\zeta {\omega }_n+\dot{x}\left(0\right)}{{\hat{\omega }}_d}\end{array}$

Finally, substituting into Eq. (2.3-18) produces the sought-after solution,

(2.3-21) $x\left(t\right)=e^{-\zeta {\omega }_{n}t}\left(x\left(0\right)\cosh {\hat{\omega }}_{d}t+\frac{x\left(0\right)\zeta {\omega }_n+\dot{x}\left(0\right)}{{\hat{\omega }}_d}\sinh {\hat{\omega }}_{d}t\right)$

Fig. 2.3-4 shows the response of a single-degree-of-freedom system with critical damping ratios of 1.0, 1.5, 2.0, and 5. The $\zeta =1.0$ case was computed with Eq. (2.3-16), whereas the others with Eq. (2.3-21). The undamped natural frequency is ${\omega }_n=2\pi$ , and the initial conditions are $x\left(0\right)=0$ and $\dot{x}\left(0\right)=30$ . The item to observe is that the system with $\zeta =1.0$ , although yielding a higher response, approaches zero in the shortest amount of time. Hence, many mechanical systems are designed with a critical damping ratio as near 1.0 as possible so that once disturbed, the system returns to zero as quickly as possible without oscillating. In the case of car shock absorbers, the value of $\zeta$ would establish how many times a car would bounce after hitting a bump.

Figure 2.3-4. Responses of a single-degree-of-freedom system with critical damping ratios as shown.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128216149000021

Single-degree-of-freedom systems: Free vibrations

John T. Katsikadelis , in Dynamic Analysis of Structures, 2020

2.3.1 Critically damped system

The value of the damping coefficient $c$ for which the discriminant vanishes is called critical damping and it is denoted by $c_{\mathit{cr}}$ . The system under critical damping is called a critically damped system. This designation is justified because, as we shall see below, this value represents the transition threshold from oscillatory to nonoscillatory motion and vice versa. Eq. (2.3.4) for $\mathrm{\Delta }=0$ determines this value as

(2.3.5) $c=c_{\mathit{cr}}=2\mathit{m\omega }$

For this value of $c$ , Eq. (2.3.3) gives the double root

(2.3.6) $\lambda =-\omega$

and the general solution of Eq. (2.3.1) is

(2.3.7) $u\left(t\right)=\left(A+\mathit{Bt}\right)e^{-\mathit{\omega t}}$

The arbitrary constants are evaluated from the initial conditions $u\left(0\right)=u_0$ and $\dot{u}\left(0\right)={\dot{u}}_0$ . Thus, we obtain

(2.3.8) $A=u_0,B=\omega u_0+{\dot{u}}_0$

and Eq. (2.3.7) becomes

(2.3.9) $u\left(t\right)=\left[u_0+\left(\omega u_0+{\dot{u}}_0\right)t\right]e^{-\mathit{\omega t}}$

Fig. 2.3.1 shows the plot of the displacement given by Eq. (2.3.9) if $u_0=0.05\mathrm{m}$ , ${\dot{u}}_0=0.2\mathrm{m}{\mathrm{s}}^{-1}$ , $\omega =8{\mathrm{s}}^{-1}$ . We see that the motion of the critically damped system is nonoscillatory, but the displacement vanishes exponentially, that is, the system returns to static equilibrium in infinite time.

Fig. 2.3.1. Response of a system with critical damping.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128186435000029

Circuit Fundamentals

Martin Plonus , in Electronics and Communications for Scientists and Engineers (Second Edition), 2020

A Slightly Underdamped Case

In the above underdamped example, we considered a large resistance R   =   56 Ω (when compared to R in critical damping) which resulted in greatly decreased losses; such light damping also resulted in a voltage v response which looks almost sinusoidal. Recall that critical damping is between exponential and oscillatory behavior of voltage v. Perhaps we can better understand the effects of varying losses by first using an R  =   13 Ω which is only slightly larger than resistance R   =   10.58 Ω used for critical damping. The damping coefficient then becomes α   =   1/2RC   =   1/2(13)(1/56)   =   2.15 and the damped frequency ${\omega }_d=\sqrt{{{\omega }_0}^2-{\alpha }^2}=1.54$ . The voltage response for this slightly underdamped case, using Eq. (1.75), is then

(1.76) $\mathrm{v}\left(\mathrm{t}\right)=109e^{-2.15t}sin1.54t$

where 168/1.54   =   109 and v(t) is sketched in Fig. 1.29. We can see that it still overshoots but only slightly and oscillations are barely recognizable on the sketch. We can now make an interesting observation: if a system with a fast response time is desired, faster than critical damping, one could allow a moderate overshoot (assuming the system can tolerate a small overshoot), that is, underdamp the system slightly. For example, in Fig. 1.29, the settling time for overdamped is 5.09   s, for critical damping it is 2.89   s and for slight underdamping it is 1.56   s. Thus, allowing some overshoot, results in the fastest response time.

Another application of underdamping is in real world systems such as bridges, airplane wings, automotive suspensions etc., which when slightly underdamped can be made lighter in weight. Consider aircraft where weight is crucial. Allowing a wing to flex (underdamped) so it can be made lighter is the norm, considering that advances in modern materials makes such construction possible. If you have ever flown in stormy or turbulent weather it is unnerving to see a wing twist and flap. Some people handle such a scary situation by drawing the shade and ordering another drink. But not to worry, the underdamped wing is built sufficiently strong to withstand turbulent weather (Nevertheless, look up Tacoma Narrows Suspension Bridge Collapse in 1940).

Another characteristic of interest is the effect that damping has on the frequency of an underdamped system. The resonant frequency, in our examples, is ${\omega }_0=\sqrt{7}=2.64$ , and is the frequency if damping is absent, in other words, no losses or equivalently R   =   ∞. The voltage response in this case is a pure, un-attenuated sinusoid (see next section). When losses enter, the frequency decreases, that is, ω _d   < ω ₀, because friction hinders motion. Therefore, for small losses or small damping, as is the case for R   =   56 Ω, frequency ω _d decreased only slightly to 2.60, as was shown in Eq. (1.72). However, increasing damping to R   =   13Ω, damped frequency ω _d decreased considerably to 1.54. If we continue increasing damping, oscillations would stop as the system becomes overdamped.

In summary we can say that the degree of damping, which strictly depends on the parallel resistance R in all three cases, causes the different voltage shapes of the over, critically, and underdamped (R   <   1/2RC, R   =   1/2RC,   R   >   1/2RC) circuit shown in Fig. 1.29. So why does the voltage overshoot and start to oscillate in one case whereas in the other, voltage rises and simply decays to zero. In the overdamped case, by the time voltage v peaks, much energy is already lost to heat while at the same time the remaining stored energy is transferred to capacitor C. All of this energy is then consumed during the exponential decay of voltage. In the underdamped case, however, losses are so small that by the time voltage peaks, most of the stored energy in L is transferred to C. Voltage then declines and overshoots. But before it overshoots, at v   =   0, energy is transferred back to L again with only small losses. After overshooting, when v is a negative max, energy is again transferred to C, then as voltage moves back to zero, energy is back in L, thus completing a period at the damped frequency ω _d , with this process repeating. The oscillatory nature of voltage means that energy (or current) flow from L to C, then from C to L continues until all energy is dissipated in heat losses. As a matter of fact, if resistance R were infinite, this process of energy exchange between L and C would continue ad-infinitum. Such pure sinusoids were already discussed in the second paragraph following Eq. (1.75) and will again be in the following section.

The exchange of energies in L and C which causes the sinusoidal variation in voltage and current in an underdamped circuit can be perplexing and mysterious to a student at first. It might be helpful to compare this to the sinusoidal or harmonic motion of a simple pendulum which is also caused by energy exchange between two energies, that is, by potential and kinetic energy. If a pendulum initially at rest is lifted to a higher position and let go, it will swing back and forth. At the lifted, high position it is motionless and possesses only potential energy. When let go it will swing across, reach maximum speed (at its rest position) at which it possesses only kinetic energy. This exchange process between potential and kinetic energy continuous until bearing friction consumes pendulum energy and the pendulum comes to rest. Similar to continuous sinusoidal motion in a circuit, the pendulum can be made to oscillate continuously by providing external energy that will negate friction, i.e., by periodic kicking of the pendulum or by some other means.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128170083000012

Transient excitation

Alvar M. Kabe , Brian H. Sako , in Structural Dynamics Fundamentals and Advanced Applications, 2020

Problem 5.19

A test is to be performed on a system whose natural frequency of vibration is 10 Hz. The system has a critical damping ratio $\zeta =0.01$ . The excitation will be sinusoidal and the frequency of excitation will start at zero  Hz. At what linear rate can the frequency increase so that the excitation will generate at least 95% of the steady-state response? If the excitation starts at 0.125   Hz and increases at an octave sweep rate, what rate must it be limited to in order to achieve the same results as for the linear sweep test.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780128216149000057

The Root Locus Method

Mark A. Haidekker , in Linear Feedback Controls, 2013

12.3 Design Example: Positioner with PI Control

We will return to the positioner example to illustrate how the root locus method can help us in the design process. Referring to Figure 12.4, we recall that the positioner consists of a first-order armature motor with inertia $J$ and friction $R_F$ . The motor constant is $k_M$ . The leadscrew has a pitch of $r$ (this includes the factor of $2\pi$ ) and acts as an integrator to obtain a position from the rotational speed of the motor. A feedback control system with a feedback path calibrated to unity and a controller with the transfer function $H(s)$ are provided.

Figure 12.4. Positioner (motor with leadscrew) and feedback control.

We now assume that the controller is a PI controller with the transfer function

(12.22) $H(s)=k_p+\frac{k_I}{s}=\frac{k_{p}s+k_I}{s}$

which leads to the closed-loop transfer function

(12.23) $\frac{X(s)}{V_{\mathit{set}}(s)}=\frac{k_{M}r(k_{p}s+k_I)}{{\mathit{Js}}^3+R_F{s}^2+k_{M}r(k_{p}s+k_I)}$

We can now define $\alpha =k_{M}r/J$ and $R_J=R_F/J$ . We divide the equation by $J$ and re-interpret $k_p$ and $k_I$ as $\alpha k_p$ and $\alpha k_I$ , respectively. In this case, the transfer function can be simplified to

(12.24) $\frac{X(s)}{V_{\mathit{set}}(s)}=\frac{k_{p}s+k_I}{s^3+R_J{s}^2+k_{p}s+k_I}$

We will now examine how $k_I$ influences the poles of the closed-loop system when $k_p$ is chosen to provide critical damping in the absence of integral control. ¹ Setting $k_I=0$ reduces the order of the system, and we obtain the poles of the second-order system as

(12.25) $-\frac{R_J}{2}\pm \sqrt{\frac{R_J^2}{4}-k_p}$

which provides us with the condition $k_p=R_J^2/4$ for critical damping. Setting the denominator polynomial of Eq. (12.24) to zero provides us with the equation for the poles, and we treat this equation as suggested in Eq. (12.8) to obtain $F(s)$ :

(12.26) $q(s)=s^3+R_J{s}^2+R_J^{2}s/4+k_I=0$

(12.27) $1+k_I·\frac{1}{s^3+R_J{s}^2+R_J^{2}s/4}=0$

Following the steps described above, we see that $F(s)$ has no zeros and three poles, namely $p_1=0$ and a double pole $p_{2/3}=-R_J/2$ . The double pole for $k_I=0$ is consistent with our initial critically damped system. With $M=0$ and $N=3$ , we have three asymptotes (see Figure 12.2), and the asymptote centroid is at ${\sigma }_A=-R_J/3$ . The entire real axis left of the origin is a root locus. Without knowing numerical values for $R_F$ and $\alpha$ , calculation of ${\sigma }_B$ is tedious, but a rough estimate would be ${\sigma }_B={\sigma }_A/2$ . We can now sketch the entire root locus plot for $k_I$ (Figure 12.5).

Figure 12.5. Root locus plot for $k_I$ . The poles $p_1$ and $p_{2/3}$ are the poles for the system when $k_I=0$ . Asymptotes are indicated by dashed lines. Pole $p_1$ emits a segment to the left, and $p_2$ emits a segment to the right. These segments meet at the branchoff point and move into the complex plane, where the segments follows the asymptotes. At high $k_I$ , the root locus follows the asymptotes into the right half-plane, and the closed-loop system becomes unstable. The third pole, $p_3$ , emits a segment to the left, which follows the third asymptote along the negative real axis.

Two branches of the root locus follow the complex conjugate asymptotes into the right half-plane, and we can see that large values of $k_I$ make the system unstable. Furthermore, the system reveals that a new double pole (critical damping) can be achieved, but with two differences to the critically damped P controlled system: (1) A third pole exists, but this is a fast pole with a very short transient component; (2) the new double pole is closer to the origin, and its transient component is slower than that of the original P controlled system. The advantage of the newly introduced integral component, however, is the suppression of steady-state tracking errors.

It is possible to use the root locus diagram to obtain the values of $\kappa$ for a desired pole location $s_d$ . We can use the magnitude criterion, Eq. (12.5), which can be rewritten as

(12.28) $\kappa =\frac{{\prod }_{n=1}^N\mid s_d-p_n\mid }{{\prod }_{m=1}^M\mid s_d-z_m\mid }$

where $\mid s_d-p_n\mid$ is the length of the vector that connects the desired pole location with the pole $p_n$ of the function $F(s)$ and $\mid s_d-z_m\mid$ is the vector that connects the desired pole location with the zero $z_m$ of the function $F(s)$ . This equation readily lends itself for a semi-graphical solution, and we will examine this approach in the next examples.

In the first example, we wish to obtain the values of $k_I$ where the system becomes unstable. From the sketch, we estimate that the asymptotes cross into the right half-plane at $s_I=\pm j{\sigma }_A·\tan 60°\approx 0.58·j·R_J$ . There are three vectors from the poles of $F(s)$ to the point $0.58·j·R_J$ on the imaginary axis. The first has a length of $0.58R_J$ , and the two vectors from the double pole to the imaginary axis have the length $\sqrt{{0.5}^2+{0.58}^2}{R}_J$ . The product of the three vector lengths is $0.34·R_J^3$ , and we know that for $k_I=0.34·R_J^3$ , a complex conjugate pole pair lies on the imaginary axis.

The special case where a system becomes unstable can much better be solved with the Routh-Hurwitz scheme, however, which provides us with $k_I<0.25R_J^3$ for stability. The deviations from the value obtained from the root locus sketch may primarily be attributed to rounding errors and our use of the asymptote rather than the actual root locus trace, which, when raised to the third power, may deviate significantly from the true value.

Another example is the graphical solution of the magnitude component of Eq. (12.9) to place closed-loop poles in a desired location. In Figure 12.6, this process is demonstrated for an underdamped pole pair with $\zeta =0.707$ , that is, a pole pair on the diagonal where $\mathfrak{R}(s)=\mathfrak{I}(s)$ . We designate the desired pole location $s_d$ . We can measure the lengths of the vectors $q_1\text{,}{q}_2$ , and $q_3$ and insert these into (12.9), For this example, the magnitude component of Eq. (12.9) becomes

Figure 12.6. We can use the magnitude criterion to graphically determine $\kappa$ (in this case, $k_I$ ), to place poles in a desired location $s_d$ . For example, if we wanted to obtain a complex conjugate pole pair with a damping factor $\zeta =0.707$ , indicated by the gray rectangles, the lengths of the three vectors that connect the "open-loop" poles (i.e., those with $k_I=0$ ) with the desired pole location can be used in the magnitude component of Eq. (12.9) to solve for $k_I$ . Note that a third closed-loop pole exists to the left of the open-loop pole pair $p_{2/3}$ .

(12.29) $1+\mid k_I\mid ·\frac{1}{\mid (s_d-p_1)\mid ·\mid (s_d-p_2)\mid ·\mid (s_d-p_3)\mid }=0$

We can measure the length of the three lines to obtain $q_1\approx 0.28R_J$ and $q_2=q_3\approx 0.36R_J$ , and find the corresponding value for $k_I=0.036R_J^3$ . In our third-order system, the choice for $k_I$ will create a third pole to the left of the open-loop pole pair $p_{2/3}$ , but this fast pole has a short transient response and therefore a small influence on the overall dynamic response.

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/B9780124058750000127

General Solution for a Critically Damped System

Source: https://www.sciencedirect.com/topics/engineering/critical-damping

Share :