Mechatronic Systems: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

DC motors

Armature circuit
V = E + IaRa
Back-EMF and torque
E = k·ω, T = k·Ia
Speed
ω = 2π·n/60, ω = (V − IaRa)/k
No-load speed and stall torque
ω0 = V/k, Tstall = k·V/Ra
Powers
Pin = V·Ia, Pdev = E·Ia = T·ω, Pout = Pdev − Prot
Starting
Istart = V/(Ra + Rst)
Speed regulation
SR = (n0 − nFL)/nFL
Maximum developed power
Pmax = V²/(4Ra)
Chopper drive
Va = D·Vdc

Method

  1. Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
  2. Use V = E + Ia·Ra to link voltage, current and back-EMF.
  3. Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
  4. For power and efficiency, start from V·Ia, subtract the copper loss Ia²·Ra and then the rotational losses.

Gears & gear trains

Meshing gears
n1·N1 = n2·N2
Torque
T2 = T1·(N2/N1)·η
Power
P = T·ω, ω = 2π·n/60
Compound train
total ratio = product of the stage ratios

Data acquisition & sampling

Nyquist criterion
fs > 2·fmax, fN = fs/2
Alias frequency
fa = |f − k·fs|, k = round(f/fs)
Step size
1 LSB = VFSR/2ᴺ
Quantisation error (rounding)
±½ LSB
Truncating ADC code
D = ⌊Vin/LSB⌋
Bits for a resolution r
2ᴺ ≥ span/r
Ideal SQNR, full-scale sine
SQNR = 6.02·N + 1.76 dB
Data rate
channels × fs × N bit/s
Multiplexed ADC
fs = 1/(n·(tc + tset))
UART
Tb = 1/baud; 8N1 = 10 bits per byte; bytes/s = baud/bits per frame
Baud rate divider
baud = fclk/(16·D)

Method

  1. Write down what is given: sampling rate, signal frequencies, number of bits, input range, baud rate.
  2. For sampling questions, compare each frequency with fs/2; a higher frequency appears at fa = |f − k·fs| with the nearest multiple of fs.
  3. For resolution questions, find 1 LSB = VFSR/2ᴺ, or the smallest N with 2ᴺ ≥ span/resolution.
  4. For data rates and UART links, count everything that is sent: channels, samples per second, bits or whole bytes per sample, and the start, parity and stop bits of every frame.

Wheatstone bridge & strain gauges

Divider voltages
Va = VsR2/(R1 + R2), Vb = VsR4/(R3 + R4)
Output
Vab = Va − Vb
Balance
R1/R2 = R3/R4
Slide-wire bridge
R/Rx = l/(100 − l)
Thévenin resistance
Rth = R1∥R2 + R3∥R4
Gauge factor
ΔR/R = GF·ε
Bridge output
quarter Vs·GF·ε/4, half Vs·GF·ε/2, full Vs·GF·ε
Small change, one arm
Vab ≈ Vs·ΔR/(4R)
Stress
σ = E·ε

Method

  1. Treat the bridge as two voltage dividers across the same supply.
  2. For balance, set R₁/R₂ = R₃/R₄ and solve for the unknown resistor.
  3. For an unbalanced bridge, find Va and Vb from the dividers and subtract; use the Thévenin equivalent when a meter loads the output.
  4. For strain gauges, convert strain to ΔR/R = GF·ε and use Vs·GF·ε/4, /2 or ×1 for a quarter, half or full bridge.

Steady-state error & PID control

System type
number of poles of G(s) at s = 0
Error constants
Kp = lim G(s), Kv = lim s·G(s), Ka = lim s²·G(s) (s → 0)
Step A
ess = A/(1 + Kp)
Ramp A·t
ess = A/Kv
Parabola A·t²/2
ess = A/Ka
PID controller
C(s) = Kp + Ki/s + Kd·s
Second-order target
s² + 2ζωns + ωn²
Overshoot and timing
%OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd, ts ≈ 4/(ζωn)

Method

  1. Count the poles of the open loop C(s)·G(s) at s = 0: that is the system type.
  2. Work out the error constant that matches the input: Kp for a step, Kv for a ramp, Ka for a parabola.
  3. Apply ess = A/(1 + Kp), A/Kv or A/Ka, or solve it for the gain you need.
  4. For a damping or speed specification, write the closed-loop characteristic polynomial and match it with s² + 2ζω_n·s + ω_n².

2Practice problems

1.DC motorsMedium
T (N·m)n (rpm)n0​ = ?

A DC motor with k = 0.12 V·s/rad runs from 24 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.

V = 24 Vk = 0.12 V·s/rad
n0 = rpm
2.DC motorsHard
+−24 VRs​ = ?Ra​ = 0.25 ΩM+−En = 800 rpmT = 0.72 N·m

A DC motor (k = 0.12 V·s/rad, Ra = 0.25 Ω) on a fixed 24 V supply drives a load of 0.72 N·m. Find the series resistance Rs that slows it down to 800 rpm.

V = 24 Vk = 0.12 V·s/radRa = 0.25 ΩT = 0.72 N·mn = 800 rpm
Rs = Ω
3.Gears & gear trainsMedium
N1​ = 24, m = 2.5 mm

A 1.5 kW motor running at 1440 rpm drives a pinion with 24 teeth and a module of 2.5 mm. Find the tangential force Ft on the pinion teeth.

P = 1.5 kWn = 1440 rpmN1 = 24m = 2.5 mm
Ft = N
4.Gears & gear trainsHard
Ns​ = 20ring fixedNp​ = 24nc​ = ?

In a planetary gear set the sun gear has 20 teeth and each planet gear has 24 teeth. The ring gear is held fixed and the sun is driven at 3000 rpm. Find the speed nc of the planet carrier.

Ns = 20Np = 24ns = 3000 rpm
nc = rpm
5.Data acquisition & samplingMedium
Vin​N = ?10 V2ᴺ − 10≤ 10 mV

An ADC with an input range of 0 to 10 V must resolve steps of 10 mV or smaller. Find the smallest number of bits N.

VFSR = 10 Vresolution = 10 mV
N =
6.Data acquisition & samplingHard
SD0​D1​D2​D3​D4​D5​D6​D7​1/fbaud​2 MHz / (16·D)

A microcontroller with a 2 MHz clock sets its UART baud rate as fclk/(16·D), with D a whole number rounded to the nearest value. The target is 2400 baud. Find the percentage error of the real baud rate (positive when it is too fast).

fclk = 2 MHztarget = 2400 Bd
e = %
7.Wheatstone bridge & strain gaugesMedium
+−6 VR1​ = 200 ΩR2​ = 300 ΩR3​ = 200 ΩR4​ = 600 Ωab+−Rth​ = ?

Find the Thévenin resistance Rth of the bridge shown, seen from the output terminals a and b.

Vs = 6 VR1 = 200 ΩR2 = 300 ΩR3 = 200 ΩR4 = 600 Ω
Rth = Ω
8.Wheatstone bridge & strain gaugesHard
+−10 V350 Ω350 Ω + ΔR350 Ω350 Ωab+−7.5 mV

A quarter bridge with one active gauge (GF = 2) is excited with 10 V. Under load the output reads 7.5 mV. Find the strain ε in µε.

Vs = 10 VGF = 2Vo = 7.5 mV
ε = µε
9.Steady-state error & PID controlMedium
r(t) = 2 + 3t+5(s + 3)s(s + 4)(s + 5)C(s)−

The input r(t) = 2 + 3t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 5(s + 3)/(s(s + 4)(s + 5)). Find the steady-state error ess.

G(s) = 5(s + 3)/(s(s + 4)(s + 5))r(t) = 2 + 3t
ess =
10.Steady-state error & PID controlHard
R(s)+K10s(s + 6)C(s)−

A proportional controller K drives the plant G(s) = 10/(s(s + 6)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.4.

G(s) = 10/(s(s + 6))ζ = 0.4
K =
11.Step responseMedium
t (s)y(t)3tp​ = ?

A system has the transfer function G(s) = 225/(s² + 3s + 75) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 225/(s² + 3s + 75)
tp = s
12.Step responseHard
k = 500 N/mc = 19 N·s/mm = 0.5 kgF = 20 Nx(t)

A mass m = 0.5 kg hangs on a spring k = 500 N/m with a viscous damper c = 19 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 20 N is suddenly applied. Find the damping ratio ζ.

m = 0.5 kgk = 500 N/mc = 19 N·s/mF = 20 N
ζ =

3Solutions

1.DC motors

Answer: n0 = 1910 rpm

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

No-load speed

ω₀ = V/k = 24/0.12 = 200 rad/s
n0 = 60·200/(2π) = 1910 rpm

Answer

n0 = 1910 rpm
2.DC motors

Answer: Rs = 2.074 Ω

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Current and back-EMF

Ia = 0.72/0.12 = 6 A
E = 0.12·83.78 = 10.05 V

Series resistance

Ra + Rs = (24 − 10.05)/6 = 2.324 Ω
Rs = 2.074 Ω

Answer

Rs = 2.074 Ω

The resistor wastes 74.68 W; a chopper controls the speed with far less loss.

3.Gears & gear trains

Answer: Ft = 331.6 N

Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.

Torque on the pinion

ω = 2π·1440/60 = 150.8 rad/s
T = P/ω = 1500/150.8 = 9.947 N·m

Pitch diameter

d = m·N = 2.5·24 = 60 mm

Tangential force

Ft = 2T/d = 2·9.947/0.06 = 331.6 N

Answer

Ft = 331.6 N
4.Gears & gear trains

Answer: nc = 681.8 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

Nr = Ns + 2·Np = 20 + 2·24 = 68

Willis equation with the ring fixed

Set nr = 0 and solve for the carrier speed.

(3000 − nc)/(0 − nc) = −68/20
nc = ns·Ns/(Ns + Nr) = 3000·20/88 = 681.8 rpm

Answer

nc = 681.8 rpm

The carrier turns in the same direction as the sun.

5.Data acquisition & sampling

Answer: N = 10

ADC resolution: an N-bit converter divides its full-scale range VFSR into 2ᴺ codes, from 0 to 2ᴺ − 1. One step is 1 LSB = VFSR/2ᴺ: the smallest change the converter can show. A truncating ADC gives the code ⌊Vin/LSB⌋; a rounding ADC is never more than ½ LSB away from the true value, its quantisation error. Every extra bit halves the LSB and adds about 6 dB to the ideal signal-to-quantisation-noise ratio SQNR = 6.02·N + 1.76 dB for a full-scale sine. To reach a resolution r over a span, choose the smallest N with 2ᴺ ≥ span/r.

Steps needed

10000 mV/10 mV = 1000

Bits

log₂(1000) = 9.966
29 = 512 < 1000 ≤ 210 = 1024
N = 10

Answer

N = 10
6.Data acquisition & sampling

Answer: e = 0.1603 %

UART: an asynchronous serial link sends each byte as a frame: one start bit (low), the data bits, an optional parity bit and one or two stop bits (high). The baud rate is the number of bits per second, so one bit lasts Tb = 1/baud and a frame of n bits lasts n/baud. The common 8N1 format has 10 bits per byte. The baud generator divides the clock: baud = fclk/(16·D) with a whole number D, so the real baud rate differs slightly from the target; an error below about 2 % is usually acceptable.

Divider

D = 2·10⁶/(16·2400) = 52.08 → 52

Real baud rate

2·10⁶/(16·52) = 2404 Bd

Error

(2404 − 2400)/2400·100 = 0.1603 %

Answer

e = 0.1603 %

An error below about 2 % is usually acceptable.

7.Wheatstone bridge & strain gauges

Answer: Rth = 270 Ω

Wheatstone bridge: two voltage dividers share one supply Vs. The left divider R1–R2 gives Va = Vs·R2/(R1 + R2) and the right divider R3–R4 gives Vb = Vs·R4/(R3 + R4); the output is Vab = Va − Vb. The bridge is balanced (Vab = 0, no current in the meter) when R1/R2 = R3/R4, that is R1·R4 = R2·R3. Seen from the output, the bridge is a Thévenin source with Rth = R1∥R2 + R3∥R4.

Supply shorted

R₁∥R₂ = 200·300/500 = 120 Ω
R₃∥R₄ = 200·600/800 = 150 Ω

Thévenin resistance

Rth = 270 Ω

Answer

Rth = 270 Ω
8.Wheatstone bridge & strain gauges

Answer: ε = 1500 µε

Strain gauges: a gauge changes its resistance with strain: ΔR/R = GF·ε, where the gauge factor GF is about 2 for metal foil. In a bridge with all arms equal to R, one active gauge (quarter bridge) gives Vo ≈ Vs·GF·ε/4. Two active gauges with opposite strains in adjacent arms (half bridge) give Vs·GF·ε/2, and four (full bridge) give Vs·GF·ε. Stress follows from Hooke’s law, σ = E·ε. Load cells state their sensitivity in mV/V: the output at full load per volt of excitation.

Solve the quarter-bridge equation

ε = 4·0.0075/(10·2) = 0.0015

In microstrain

ε = 1500 µε

Answer

ε = 1500 µε
9.Steady-state error & PID control

Answer: ess = 4

Steady-state error: in a unity-feedback loop with open-loop transfer function G(s) the error is E(s) = R(s)/(1 + G(s)), and by the final value theorem ess = lims→0 s·E(s). The system type is the number of poles of G(s) at s = 0. The error constants are Kp = lim G(s), Kv = lim s·G(s) and Ka = lim s²·G(s), all for s → 0. A step of size A gives ess = A/(1 + Kp), a ramp A·t gives A/Kv and a parabola A·t²/2 gives A/Ka. A type-0 loop follows a step with a finite error, a type-1 loop follows it exactly and a ramp with a finite error.

Velocity constant

Kv = lim s·G(s) = 0.75

Step part

Kp = ∞ → estep = 0

Ramp error

ess = 3/0.75 = 4

Answer

ess = 4
10.Steady-state error & PID control

Answer: K = 5.625

P, PI and PID control: the controller C(s) = Kp + Ki/s + Kd·s acts on the error. A proportional gain makes the loop faster and reduces the error of a type-0 plant, but never removes it. The integral term adds a pole at s = 0: the system type rises by one, and a step is followed without error. The derivative term adds damping. To choose gains, write the closed-loop characteristic polynomial and compare it with s² + 2ζωns + ωn². Then %OS = 100·e−ζπ/√(1−ζ²), tp = π/ωd with ωd = ωn√(1 − ζ²), and ts ≈ 4/(ζωn).

Characteristic polynomial

s(s + 6) + 10K = s² + 6s + 10K

Natural frequency

2·0.4·ωₙ = 6 → ωₙ = 7.5 rad/s

Gain

10K = ωₙ² = 56.25 → K = 5.625

Answer

K = 5.625
11.Step response

Answer: tp = 0.3683 s

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Compare with the standard form

The denominator must match s² + 2ζωₙs + ωₙ².

ωₙ² = 75 → ωₙ = 8.66 rad/s
2ζωₙ = 3 → ζ = 3/(2·8.66) = 0.1732

Result

ωd = 8.66·√(1 − 0.1732²) = 8.529 rad/s → tp = π/ωd = 0.3683 s

Answer

tp = 0.3683 s
12.Step response

Answer: ζ = 0.6008

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jωd with ωd = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e−ζπ/√(1−ζ²), peak time tp = π/ωd, and the 2 % settling time ts ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Natural frequency and damping ratio

ωₙ = √(k/m) = √(500/0.5) = 31.62 rad/s
ζ = c/(2√(km)) = 19/(2·√(500·0.5)) = 0.6008

Result

ζ = 0.6008

Answer

ζ = 0.6008

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