Measurements, Instrumentation, and Controls

5% of exam

Sensors and transducers, signal conditioning and data acquisition, feedback control and block diagrams, dynamic system response, and measurement uncertainty.

3 concepts

Exam areas

A. Sensors and transducers

Sensors, Signal Conditioning and Measurement Uncertainty

Strain gauges, thermocouples, RTDs and LVDTs; Wheatstone bridges, sampling and A/D conversion; calibration and Kline-McClintock uncertainty propagation.

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B. Control systems

Feedback Control Systems and Stability

Block diagrams, open- vs closed-loop transfer functions, the characteristic equation and stability, PID action, and steady-state error by system type.

Control questions on the FE Mechanical exam reward fluency with one picture and one ratio: a block diagram, and the closed-loop transfer function that the diagram encodes. Almost every problem reduces to reducing a diagram to a single transfer function

G(s)

, asking whether the closed loop is stable, or computing how far the steady output sits from the command. The FE Reference Handbook lays out exactly these tools in its Instrumentation, Measurement, and Control section — the transfer-function/pole-zero form, the negative-feedback block diagram, the characteristic equation

1+G_1G_2H=0

, the system-type error table, and the PID controller — but states them tersely. This concept turns those bare statements into a procedure you can run under exam pressure, where the points are lost to sign errors at the summing junction and to confusing the loop you are analyzing.

Transfer functions, poles, and zeros

A linear time-invariant block is described by a transfer function

G(s)=Y(s)/X(s)

, the Laplace-domain ratio of output to input with zero initial conditions. Write it as a ratio of polynomials

N(s)/D(s)

; the roots of the numerator are the zeros

z_m

and the roots of the denominator are the poles

p_n

. The poles are the system's natural modes — each pole at

s=p

contributes a term

e^{pt}

to the time response — so the entire question of stability lives in where the poles sit in the complex plane. Everything downstream (DC gain, transient shape, stability) is read off this single object.

G(s)=\dfrac{Y(s)}{X(s)}=\dfrac{N(s)}{D(s)}=K\,\dfrac{\prod_{m=1}^{M}(s-z_m)}{\prod_{n=1}^{N}(s-p_n)}

Open-loop vs closed-loop

In the standard negative-feedback loop,

G_1(s)

is the controller,

G_2(s)

is the plant, and

H(s)

is the sensor/measurement dynamics. The summing junction forms the error

E(s)=R(s)-H(s)Y(s)

, the forward path acts on it, and the output is fed back. The product around the loop,

G_1G_2H

, is the open-loop transfer function — what you would measure with the loop broken. Closing the loop solves

Y=G_1G_2(R-HY)

for

Y/R

. The closed-loop denominator gains the crucial

1+G_1G_2H

term that distinguishes feedback control from open-loop operation: it relocates the poles and is the source of both stabilization and, if misdesigned, instability.

\dfrac{Y(s)}{R(s)}=\dfrac{G_1(s)\,G_2(s)}{1+G_1(s)\,G_2(s)\,H(s)}

The characteristic equation and stability

Setting the closed-loop denominator to zero gives the characteristic equation; its roots are the closed-loop poles. The system is bounded-input bounded-output stable if and only if every closed-loop pole has a strictly negative real part (lies in the left half-plane). A necessary condition is that all coefficients of the characteristic polynomial are present and positive — but for third order and above this is not sufficient, and the Routh-Hurwitz array is the standard hand test for how many roots lie in the right half-plane. For the cubic

s^3+a_2s^2+a_1s+a_0

, the Routh condition reduces to all coefficients positive and

a_2a_1>a_0

1+G_1(s)\,G_2(s)\,H(s)=0

PID control action

A PID controller sums three actions on the error: proportional (responds to present error), integral (accumulates past error and drives steady-state error to zero), and derivative (anticipates future error from its rate, adding damping). Raising proportional gain speeds response but eventually erodes stability margin; integral action eliminates offset but adds phase lag that can destabilize; derivative action improves damping but amplifies sensor noise. The handbook's form factors out the proportional gain so the integral and derivative times

T_I

and

T_D

set the corner frequencies of each action.

G_C(s)=K\left(1+\dfrac{1}{T_I\,s}+T_D\,s\right)

Steady-state error and system type

For a unity-feedback loop, the steady-state error to a standard input is fixed by the system type

T

— the number of pure integrators

1/s

in the open-loop

G(s)

, i.e. the number of open-loop poles at the origin. Define the static error constants from the Final Value Theorem: position

K_p=\lim_{s\to0}G(s)

, velocity

K_v=\lim_{s\to0}sG(s)

, acceleration

K_a=\lim_{s\to0}s^2G(s)

. A type-0 system has finite

K_p

and a nonzero step error

1/(1+K_p)

; a type-1 system tracks a step with zero error but follows a ramp with error

1/K_v

; a type-2 system tracks both step and ramp exactly and shows error

1/K_a

to a parabolic (acceleration) input. Each added integrator buys exact tracking of one higher input class — at a stability cost.

e_{ss,\text{step}}=\dfrac{1}{1+K_p},\qquad e_{ss,\text{ramp}}=\dfrac{1}{K_v},\qquad e_{ss,\text{accel}}=\dfrac{1}{K_a}

Relative stability: gain and phase margins

Being in the left half-plane answers whether the system is stable; the gain and phase margins answer by how much. The phase margin is the extra phase lag, at the gain-crossover frequency

\omega_{0\text{dB}}

where

|G(j\omega)|=1

, that would push the loop to the verge of instability:

\mathrm{PM}=180^\circ+\angle G(j\omega_{0\text{dB}})

. The gain margin is the extra gain, at the phase-crossover frequency

\omega_{180}

where

\angle G=-180^\circ

, that would do the same. Positive margins mean a stable design; larger margins mean more robustness to modeling error, at the usual cost of slower response.

\mathrm{PM}=180^\circ+\angle G(j\omega_{0\text{dB}}),\qquad \mathrm{GM}=-20\log_{10}\big|G(j\omega_{180})\big|

Exam strategy

Reduce first, then ask the question. For any single-loop diagram write

Y/R=G_{\text{fwd}}/(1+G_{\text{loop}})

, where

G_{\text{loop}}

is the full product around the loop including

H(s)

. For stability, write the characteristic polynomial

1+G_{\text{loop}}=0

and check signs of all coefficients first; for a cubic apply

a_2a_1>a_0

, and at the stability boundary the oscillation frequency is

\omega=\sqrt{a_1}

. For steady-state error, count integrators to get the type, pick the matching error constant (

K_p

for a step,

K_v

for a ramp), and remember the

+1

appears only in the step formula. Keep the summing-junction sign explicit: negative feedback gives the

1+G_{\text{loop}}

denominator; a stray sign flip turns a stable design unstable on paper.

Key equations

Transfer function (pole-zero form)

G(s)=\dfrac{N(s)}{D(s)}=K\dfrac{\prod_{m=1}^{M}(s-z_m)}{\prod_{n=1}^{N}(s-p_n)}

LTI input-output ratio in the Laplace domain. Poles $p_n$ (roots of $D$ ) set stability and transient modes; zeros $z_m$ (roots of $N$ ) shape the response.

Closed-loop transfer function (general)

\dfrac{Y(s)}{R(s)}=\dfrac{G_1G_2}{1+G_1G_2H}

$G_1$ controller, $G_2$ plant, $H$ feedback/sensor. Forward path over one-plus-loop-product.

Unity-feedback closed loop

\dfrac{Y(s)}{R(s)}=\dfrac{G(s)}{1+G(s)}

Special case $H=1$ . The denominator $1+G$ defines the closed-loop poles.

Characteristic equation

1+G_1(s)G_2(s)H(s)=0

Roots are the closed-loop poles. Stable iff all roots have negative real parts (left half-plane).

Routh condition for a cubic

s^3+a_2s^2+a_1s+a_0:\quad a_i>0\ \text{and}\ a_2a_1>a_0

Necessary-and-sufficient stability test for third order. At the boundary $a_2a_1=a_0$ the loop oscillates at $\omega=\sqrt{a_1}\,\text{rad/s}$ .

Final Value Theorem / DC gain

y(\infty)=\lim_{s\to0}sY(s),\qquad \text{dc gain}=\lim_{s\to0}G(s)

Valid only when the closed-loop poles are all in the left half-plane. Gives the steady output and the position constant $K_p$ .

Static error constants

K_p=\lim_{s\to0}G,\quad K_v=\lim_{s\to0}sG,\quad K_a=\lim_{s\to0}s^2G

Position, velocity, acceleration constants for a unity-feedback loop; $G$ is the open-loop transfer function.

Steady-state error by input

e_{ss}^{\text{step}}=\dfrac{1}{1+K_p},\quad e_{ss}^{\text{ramp}}=\dfrac{1}{K_v},\quad e_{ss}^{\text{accel}}=\dfrac{1}{K_a}

Type 0: finite $K_p$ , and $K_v = K_a = 0$ , so the ramp and acceleration errors are infinite. Each integrator raises the type and zeros the error to one higher input class.

PID controller

G_C(s)=K\left(1+\dfrac{1}{T_I s}+T_D s\right)

$K$ proportional gain, $T_I$ integral time (s), $T_D$ derivative time (s). Integral action removes offset; derivative action adds damping.

Standard second-order denominator

s^2+2\zeta\omega_n s+\omega_n^2=0

Closed-loop characteristic form: $\omega_n$ natural frequency (rad/s), $\zeta$ damping ratio. Stable for $\zeta>0$ , $\omega_n>0$ .

Gain and phase margins

\mathrm{PM}=180^\circ+\angle G(j\omega_{0\text{dB}}),\quad \mathrm{GM}=-20\log_{10}|G(j\omega_{180})|

Relative-stability measures from the open-loop frequency response. Positive margins indicate a stable closed loop.

Worked examples

Closed-loop DC gain and step error

Problem. A unity-feedback loop has forward transfer function $G(s)=\dfrac{10}{s+2}$ . Find the closed-loop transfer function, the steady-state output to a unit step command, and the steady-state error.

Solution. Close the loop:

T(s)=\dfrac{G}{1+G}=\dfrac{10/(s+2)}{1+10/(s+2)}=\dfrac{10}{s+12}

. The single closed-loop pole is at

s=-12

(left half-plane), so the Final Value Theorem applies. DC gain

=\lim_{s\to0}T(s)=10/12=0.833

, so the steady output to a unit step is

y(\infty)=0.833

. Error

e_{ss}=1-0.833=0.167

. Check the type-0 formula:

K_p=\lim_{s\to0}G=10/2=5

, so

e_{ss}=1/(1+K_p)=1/6=0.167

— agrees. Answer:

T(s)=10/(s+12)

y(\infty)=0.833

e_{ss}=0.167

. Units check: gain and error are dimensionless ratios, consistent.

T(s)=\dfrac{10}{s+12},\qquad e_{ss}=\dfrac{1}{1+K_p}=\dfrac{1}{6}

Ramp error of a type-1 system

Problem. A unity-feedback loop has open-loop $G(s)=\dfrac{20}{s(s+4)}$ . Find the steady-state error to a unit-ramp command, and the closed-loop natural frequency and damping ratio.

Solution. One pole sits at the origin, so this is a type-1 system: zero step error, finite ramp error. Velocity constant

K_v=\lim_{s\to0}sG=\lim_{s\to0}\dfrac{20}{s+4}=20/4=5\ \text{s}^{-1}

. Ramp error

e_{ss}=1/K_v=1/5=0.200

. Closed-loop denominator:

s(s+4)+20=s^2+4s+20=0

. Matching

s^2+2\zeta\omega_n s+\omega_n^2

\omega_n=\sqrt{20}=4.47\ \text{rad/s}

and

2\zeta\omega_n=4\Rightarrow \zeta=4/(2\cdot4.47)=0.447

. Answer:

e_{ss}=0.200

\omega_n=4.47\ \text{rad/s}

\zeta=0.447

(underdamped, stable). Sanity:

\zeta<1

so the step response overshoots, and both poles are in the left half-plane.

K_v=\lim_{s\to0}\dfrac{20}{s+4}=5,\qquad e_{ss}=\dfrac{1}{K_v}=0.200

Gain range for stability (Routh)

Problem. A unity-feedback loop has forward path $G(s)=\dfrac{K}{s(s+1)(s+5)}$ with $K>0$ . Find the range of $K$ for closed-loop stability and the frequency of oscillation at the stability limit.

Solution. Characteristic equation:

s(s+1)(s+5)+K=0\Rightarrow s^3+6s^2+5s+K=0

. This is the cubic

s^3+a_2s^2+a_1s+a_0

with

a_2=6

a_1=5

a_0=K

. All coefficients positive requires

K>0

. The Routh sufficiency condition is

a_2a_1>a_0

6\times5=30>K

. So the loop is stable for

0<K<30

. At the limit

K=30

the system is marginally stable. The auxiliary equation from the

s^2

row,

6s^2+K=0

, gives

s^2=-K/6=-5

, so

\omega=\sqrt{5}=2.24\ \text{rad/s}

. Answer: stable for

0<K<30

; sustained oscillation at

K=30

occurs at

\omega=2.24\ \text{rad/s}

. Sanity: at

K=30

the closed-loop poles cross the imaginary axis at

\pm j2.24

, exactly the boundary between left- and right-half-plane.

s^3+6s^2+5s+K=0,\qquad 0<K<30,\qquad \omega=\sqrt{5}\ \text{rad/s}

Common pitfalls

•Confusing the open-loop product with the closed-loop ratio: stability is decided by $1+G_1G_2H=0$ (the open-loop product set to $-1$ ), but the response and DC gain come from the closed-loop ratio $G_1G_2/(1+G_1G_2H)$ . Use the right object for the question.
•Counting all poles as the system type. The type is only the number of open-loop poles at the origin (pure $1/s$ integrators). $G=K/[(s+1)(s+5)]$ is type 0, not type 2.
•Putting the $+1$ in the wrong error formula. Step error is $1/(1+K_p)$ ; ramp error is $1/K_v$ with no added one; acceleration error is $1/K_a$ . Match the constant to the input.
•Assuming all-positive coefficients guarantees stability. For order $\ge3$ that is necessary but not sufficient — apply Routh-Hurwitz (e.g. $a_2a_1>a_0$ for a cubic).
•Sign at the summing junction: negative feedback yields the $1+G_{\text{loop}}$ denominator; a positive-feedback sign error gives $1-G_{\text{loop}}$ and a spurious unstable pole.
•Applying the Final Value Theorem to an unstable or marginally stable loop. It is valid only when every closed-loop pole has a negative real part; otherwise the limit is meaningless.
•Mixing degrees and radians, or sign, in the margins: phase margin is referenced to $-180^\circ$ and the gain margin in dB is positive for a stable design.

References

NCEES FE Reference Handbook — Instrumentation, Measurement, and Control (Control Systems)
Nise, Control Systems Engineering — Block-diagram reduction, system type/error constants, Routh-Hurwitz, gain and phase margins.
Ogata, Modern Control Engineering — Transfer functions, closed-loop poles, and stability analysis.

C. Dynamic system response

Dynamic System Response: First- and Second-Order Systems

First-order time constant and step response, second-order natural frequency and damping, overshoot and settling time, and the Laplace transfer function that ties them together.

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