Dynamic Response of Measurement System

A measurement system with input $q_i(t)$ and output $q_o(t)$ is modelled by a linear ordinary differential equation. Its order is the highest derivative of the output.

Zero-order system

$$a_0 q_o = b_0 q_i \quad\Rightarrow\quad q_o = K q_i, \qquad K = \frac{b_0}{a_0}$$

$K$ is the static sensitivity. The output follows the input instantly, with no lag or distortion. Example: linear potentiometer, resistance strain gauge at low frequency.

First-order system

$$a_1\frac{dq_o}{dt} + a_0 q_o = b_0 q_i \quad\Rightarrow\quad \tau\frac{dq_o}{dt} + q_o = K q_i$$

with $\tau = a_1/a_0$ the time constant and $K = b_0/a_0$. It has one energy-storing element. Examples: a thermometer or thermocouple in a fluid (heat capacity and film resistance), RC low-pass filter, a pressure gauge with a long tube.

Second-order system

$$a_2\frac{d^2q_o}{dt^2} + a_1\frac{dq_o}{dt} + a_0 q_o = b_0 q_i$$ $$\frac{1}{\omega_n^2}\frac{d^2q_o}{dt^2} + \frac{2\zeta}{\omega_n}\frac{dq_o}{dt} + q_o = K q_i$$

where

$$\omega_n = \sqrt{\frac{a_0}{a_2}}, \qquad \zeta = \frac{a_1}{2\sqrt{a_0 a_2}}, \qquad K = \frac{b_0}{a_0}$$

$\omega_n$ is the undamped natural frequency and $\zeta$ the damping ratio. Two energy-storing elements exchange energy. Examples: spring-mass-damper accelerometer, bourdon-tube pressure gauge with fluid, U-tube manometer, seismic transducer, RLC circuit.

Summary

The first-order model is $\tau\,\dfrac{dq_o}{dt} + q_o = K q_i$. A step input is $q_i = 0$ for $t<0$ and $q_i = q_s$ (constant) for $t\ge 0$, with $q_o(0) = 0$.

Derivation

The complementary solution is $C e^{-t/\tau}$ and the particular solution is $K q_s$. So

$$q_o(t) = K q_s + C e^{-t/\tau}$$

With $q_o(0) = 0$, $C = -K q_s$, giving

$$q_o(t) = K q_s\left(1 - e^{-t/\tau}\right)$$

The normalized response is

$$\frac{q_o}{K q_s} = 1 - e^{-t/\tau}$$

The error from the final value is $e^{-t/\tau}$, which decays exponentially.

Time constant

The time constant $\tau$ is the time taken for the output to reach 63.2% of its final value, $1 - e^{-1}$. Equivalently, it is the time the output would need to reach the final value if it continued at its initial rate. For a thermometer, $\tau = mc/(hA)$.

 q_o/Kqs
 1.0 |            ____---------
     |        _--
 .632|-----.-
     |   ./
     |  /
     | /
     +------+------+-----> t
           tau   2tau

The settling time to 2% is about $4\tau$ and to 5% about $3\tau$. A small $\tau$ means a fast instrument.

For a first-order system with unit sensitivity, a step from $T_i = 25\ ^\circ$C to $T_\infty = 200\ ^\circ$C gives

$$T(t) = T_\infty - (T_\infty - T_i)e^{-t/\tau} = 200 - 175\,e^{-t/2}$$

(a) Time to indicate 190 $^\circ$C

$$190 = 200 - 175e^{-t/2} \Rightarrow e^{-t/2} = \frac{10}{175} \Rightarrow t = -2\ln(0.05714) = 5.72\ \text{s}$$

(b) Reading after 5 s

$$T = 200 - 175e^{-5/2} = 200 - 175(0.0821) = 185.6\ ^\circ\text{C}$$

(c) Time for 99% of the step

$$e^{-t/\tau} = 0.01 \Rightarrow t = \tau\ln 100 = 2(4.605) = 9.21\ \text{s}$$

(d) Sinusoidal bath temperature at 0.1 Hz

$\omega = 2\pi(0.1) = 0.6283$ rad/s, so $\omega\tau = 1.2566$.

$$M = \frac{1}{\sqrt{1 + (\omega\tau)^2}} = \frac{1}{\sqrt{1 + 1.579}} = 0.623$$ $$\phi = -\tan^{-1}(\omega\tau) = -\tan^{-1}(1.2566) = -51.5^\circ$$

The thermocouple shows only 62.3% of the true amplitude and lags by $51.5^\circ$ (time lag $= 51.5/360 \times 10 = 1.43$ s).

Answer: (a) 5.72 s; (b) 185.6 $^\circ$C; (c) 9.21 s; (d) amplitude ratio 0.623, phase lag $51.5^\circ$.

The second-order model in standard form is

$$\frac{1}{\omega_n^2}\ddot{q}_o + \frac{2\zeta}{\omega_n}\dot{q}_o + q_o = K q_i$$

For a step input $q_i = q_s$, $q_o(0) = \dot q_o(0) = 0$. Let $y = q_o/(Kq_s)$, so $y$ tends to 1.

Underdamped case ($0 < \zeta < 1$)

The characteristic roots are $s = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}$. With damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$, the solution is $y = 1 + e^{-\zeta\omega_n t}(A\cos\omega_d t + B\sin\omega_d t)$. The conditions $y(0) = 0$ and $\dot y(0) = 0$ give $A = -1$ and $B = -\zeta\omega_n/\omega_d$:

$$y(t) = 1 - \frac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}}\sin\left(\omega_d t + \phi\right), \qquad \phi = \cos^{-1}\zeta$$

Setting $\dot y = 0$ gives peak time $t_p = \pi/\omega_d$, and the peak overshoot

$$M_p = \exp\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}\right)$$

Other cases

• Critically damped ($\zeta = 1$): $y = 1 - (1 + \omega_n t)e^{-\omega_n t}$. Fastest approach to the final value without overshoot.
• Overdamped ($\zeta > 1$): $y = 1 - \dfrac{e^{-\zeta\omega_n t}}{2\sqrt{\zeta^2-1}}\left[\dfrac{e^{\omega_n\sqrt{\zeta^2-1}\,t}}{\zeta - \sqrt{\zeta^2-1}} - \dfrac{e^{-\omega_n\sqrt{\zeta^2-1}\,t}}{\zeta + \sqrt{\zeta^2-1}}\right]$. No overshoot but slow.

 y
 1.4|  .-.            zeta = 0.2
 1.0|-/---`-.__.---- zeta = 0.7
    | /  _.--------- zeta = 1
    |/ .'  ........  zeta = 2 (slow)
    +----------------> w_n t

Why $\zeta \approx 0.7$

At $\zeta = 0.7$ the overshoot is only 4.6%, the 2% settling time is near its minimum, and the amplitude response is flat within 1% up to about $0.4\,\omega_n$ with a phase lag nearly proportional to frequency (little distortion). Hence most instruments use $\zeta = 0.6$ to $0.7$.

Given: $\omega_n = 1000$ rad/s, $\zeta = 0.7$.

(a) Frequency response at 400 rad/s

Frequency ratio $r = \omega/\omega_n = 0.4$.

$$M = \frac{1}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}} = \frac{1}{\sqrt{(0.84)^2 + (0.56)^2}} = \frac{1}{\sqrt{0.7056 + 0.3136}} = 0.9905$$ $$\phi = -\tan^{-1}\left(\frac{2\zeta r}{1 - r^2}\right) = -\tan^{-1}\left(\frac{0.56}{0.84}\right) = -33.7^\circ$$

The amplitude error is only 0.95%.

(b) Overshoot

$$M_p = \exp\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}\right) = \exp\left(\frac{-\pi(0.7)}{0.7141}\right) = \exp(-3.079) = 0.0460 = 4.60\%$$

(c) Time-domain parameters

Damped frequency:

$$\omega_d = \omega_n\sqrt{1-\zeta^2} = 1000(0.7141) = 714.1\ \text{rad/s}$$

Peak time: $t_p = \pi/\omega_d = 4.40$ ms.

Rise time (0 to 100%): $t_r = \dfrac{\pi - \cos^{-1}\zeta}{\omega_d} = \dfrac{3.1416 - 0.7954}{714.1} = 3.29$ ms.

2% settling time: $t_s \approx \dfrac{4}{\zeta\omega_n} = \dfrac{4}{700} = 5.71$ ms.

Answer: $M = 0.9905$, lag $= 33.7^\circ$, overshoot $= 4.6\%$, $\omega_d = 714$ rad/s, $t_p = 4.40$ ms, $t_r = 3.29$ ms, $t_s = 5.71$ ms.

Dynamic characteristics describe how a measurement system responds to inputs that change with time.

• Amplitude response (magnitude ratio): the ratio of output amplitude to input amplitude, $M(\omega) = |q_o|/(K|q_i|)$, for a sinusoidal input of frequency $\omega$. An ideal instrument has $M = 1$ over its working range.
• Frequency response: the variation of $M$ and phase with frequency. The range where $M$ stays within a stated tolerance (e.g. $\pm 5\%$) is the bandwidth or useful frequency range.
• Phase response: the phase angle $\phi(\omega)$ between output and input. For distortion-free measurement of complex signals, $\phi$ should be proportional to frequency (constant time delay).
• Delay time ($t_d$): the time for the output to reach 50% of its final value after a step input.
• Rise time ($t_r$): the time for the step response to rise from 10% to 90% (sometimes 0 to 100%) of the final value. For a first-order system $t_r = 2.2\tau$.
• Settling time ($t_s$): the time after which the output stays within a band (2% or 5%) of the final value.
• Dynamic error: the difference between the indicated value and the true value of a time-varying input, $e_d = (M - 1)$ times input amplitude.

 y
 1.0 |           _.-~~~~~~~~~
 0.9 |--------.-'
 0.5 |------.'        <- delay time t_d
 0.1 |---.-'
     |  |<--- t_r --->|
     +--------------------> t

Given: $m = 0.05$ kg, $k = 2000$ N/m, $c = 4$ N s/m.

(a) Natural frequency and damping ratio

$$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{2000}{0.05}} = 200\ \text{rad/s}$$ $$\zeta = \frac{c}{2\sqrt{km}} = \frac{4}{2\sqrt{2000 \times 0.05}} = \frac{4}{2(10)} = 0.2$$

The system is underdamped.

(b) Damped frequency

$$\omega_d = \omega_n\sqrt{1-\zeta^2} = 200\sqrt{0.96} = 195.96\ \text{rad/s}, \qquad f_d = \frac{195.96}{2\pi} = 31.19\ \text{Hz}$$

(c) Step response

$$M_p = \exp\left(\frac{-\pi(0.2)}{\sqrt{0.96}}\right) = \exp(-0.6413) = 0.527 = 52.7\%$$ $$t_p = \frac{\pi}{\omega_d} = \frac{\pi}{195.96} = 16.0\ \text{ms}$$ $$t_s \approx \frac{4}{\zeta\omega_n} = \frac{4}{0.2 \times 200} = 0.100\ \text{s}$$

(d) Damping for $\zeta = 0.7$

$$c = 2\zeta\sqrt{km} = 2(0.7)(10) = 14\ \text{N s/m}$$

Answer: $\omega_n = 200$ rad/s, $\zeta = 0.2$; $f_d = 31.2$ Hz; overshoot $52.7\%$, $t_p = 16.0$ ms, $t_s = 0.1$ s; required $c = 14$ N s/m.

A zero-order system has the model $a_0 q_o = b_0 q_i$, so $q_o = K q_i$ with static sensitivity $K = b_0/a_0$.

Characteristics

• The output is proportional to the input at every instant. There is no lag, delay, phase shift or distortion.
• Amplitude ratio $M = 1$ (relative to $K$) and phase angle $\phi = 0$ at all frequencies.
• The only parameter is $K$, found by static calibration.
• No energy storage (no mass, spring or capacitance) in the model.

Response

For any input, a step, ramp or sine, the output has the same shape scaled by $K$.

 q_i |  ____        q_o |  ____   (scaled by K)
     | |    |           | |    |
     +-+----+--> t      +-+----+--> t

Potentiometer as example

A linear potentiometer of length $L$ and supply $E$ gives $e_o = (E/L)\,x$, so $K = E/L$ V/m. The output follows the wiper position with no dynamic lag.

Limits

A zero-order model is an idealization valid only when the input changes slowly compared with the instrument's natural frequency. In a real potentiometer, wiper mass and inertia, wire-wound resolution, friction, stray capacitance and inductance, and the loading error of a finite load give time-dependent behaviour at high frequency, so the true response is of higher order.