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)3.75 N·mn0​n = ?

A DC motor (k = 0.75 N·m/A, Ra = 2 Ω) runs from 240 V and drives a load torque of 3.75 N·m. Find its speed n in rpm.

V = 240 Vk = 0.75 N·m/ARa = 2 ΩT = 3.75 N·m
n = rpm
2.DC motorsHard
+−220 VRa​ = 0.5 ΩM+−En = ?T = 60 N·m

A separately excited DC motor (k = 1.5 V·s/rad at full field, Ra = 0.5 Ω, V = 220 V) drives a constant load torque of 60 N·m. The field flux is reduced to 70 % of its full value. Find the new speed n.

V = 220 Vk = 1.5 V·s/radRa = 0.5 ΩT = 60 N·mΦ = 70 %
n = rpm
3.Gears & gear trainsMedium
N1​ = 20, m = 5 mm

A 5.5 kW motor running at 1450 rpm drives a pinion with 20 teeth and a module of 5 mm. Find the tangential force Ft on the pinion teeth.

P = 5.5 kWn = 1450 rpmN1 = 20m = 5 mm
Ft = N
4.Gears & gear trainsHard
d1​ = 160 mmd2​ = 400 mm1500 rpmC = 1000 mmT1​ = 1000 NT2​P = ?

An open belt connects a 160 mm driving pulley at 1500 rpm to a 400 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 160 mmn1 = 1500 rpmd2 = 400 mmC = 1000 mmμ = 0.3T1 = 1000 N
P = kW
5.Data acquisition & samplingMedium
SD0​D1​D2​D3​D4​D5​D6​D7​Tb​10·Tb​38400 Bd, 8N1

A UART sends bytes at 38400 baud in the format 8N1: 8 data bits, no parity and 1 stop bit(s), plus the start bit. Find the largest number of data bytes per second it can transfer.

baud = 38400 Bd8N1
RB = B/s
6.Data acquisition & samplingHard
T20 mV/°C×20∿01N = 8, 10 V? °C/LSB

A temperature sensor gives 20 mV/°C (0 V at 0 °C). Its signal is amplified 20 times and fed to an 8-bit ADC with a range of 0 to 10 V (1 LSB = VFSR/2ᴺ). Find the temperature change ΔTLSB that corresponds to one LSB.

S = 20 mV/°CG = 20N = 8VFSR = 10 V
ΔTLSB = °C
7.Wheatstone bridge & strain gaugesMedium
+−10 V100 Ω100 Ω + ΔR100 Ω100 Ωab+−Vab​ = ?

A bridge has four 100 Ω arms and Vs = 10 V. The sensor R₂ increases by ΔR = 0.8 Ω. Using the small-change approximation, find the output Vab in mV.

Vs = 10 VR = 100 ΩΔR = 0.8 Ω
Vab = mV
8.Wheatstone bridge & strain gaugesHard
+−2.5 V350 Ω350 Ω + ΔR350 Ω350 Ωab+−0.5 mV

A quarter bridge with one active gauge (GF = 2) is excited with 2.5 V. Under load the output reads 0.5 mV. Find the strain ε in µε.

Vs = 2.5 VGF = 2Vo = 0.5 mV
ε = µε
9.Steady-state error & PID controlMedium
R(s)+K(s + 1)(s + 4)C(s)−

A unity-feedback loop has G(s) = K/((s + 1)(s + 4)). Find the gain K that gives a steady-state error of 2 % for a step input.

G(s) = K/((s + 1)(s + 4))ess = 2 %
K =
10.Steady-state error & PID controlHard
R(s)+K = 162s(s + 4)C(s)−

A proportional controller K = 16 drives the plant G(s) = 2/(s(s + 4)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 2/(s(s + 4))K = 16
ts = s
11.Step responseMedium
t (s)y(t)2ts​ = ?

A system has the transfer function G(s) = 14/(s² + 4s + 7) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 14/(s² + 4s + 7)
ts = s
12.Step responseHard
t (s)y(t)115 %

A second-order system without zeros must have a step response with 15 % overshoot. Find the damping ratio ζ.

%OS = 15 %
ζ =

3Solutions

1.DC motors

Answer: n = 2928 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.

Armature current

Ia = T/k = 3.75/0.75 = 5 A

Back-EMF and speed

E = 240 − 5·2 = 230 V
ω = E/k = 306.7 rad/s → n = 2928 rpm

Answer

n = 2928 rpm
2.DC motors

Answer: n = 1741 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.

Weakened motor constant

k₂ = 0.7·1.5 = 1.05 V·s/rad

New armature current

Ia = 60/1.05 = 57.14 A

New speed

ω = (220 − 57.14·0.5)/1.05 = 182.3 rad/s → n = 1741 rpm

Answer

n = 1741 rpm

Field weakening is how a DC drive runs above its base speed.

3.Gears & gear trains

Answer: Ft = 724.4 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·1450/60 = 151.8 rad/s
T = P/ω = 5500/151.8 = 36.22 N·m

Pitch diameter

d = m·N = 5·20 = 100 mm

Tangential force

Ft = 2T/d = 2·36.22/0.1 = 724.4 N

Answer

Ft = 724.4 N
4.Gears & gear trains

Answer: P = 7.303 kW

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Angle of wrap

θ = 2.901 rad

Slack-side tension

eμθ = e0.8703 = 2.388, T₂ = 1000/2.388 = 418.8 N

Power

P = (1000 − 418.8)·12.57 = 7.303 kW

Answer

P = 7.303 kW
5.Data acquisition & sampling

Answer: RB = 3840 B/s

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.

Bits per frame

1 + 8 + 0 + 1 = 10

Byte rate

38400/10 = 3840 B/s

Answer

RB = 3840 B/s
6.Data acquisition & sampling

Answer: ΔTLSB = 0.09766 °C

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.

Step size

1 LSB = 10000/256 = 39.06 mV

Sensitivity at the ADC

20·20 = 400 mV/°C

Temperature resolution

ΔTLSB = 39.06/400 = 0.09766 °C

Answer

ΔTLSB = 0.09766 °C

The ADC reaches full scale at 25 °C.

7.Wheatstone bridge & strain gauges

Answer: Vab = 20 mV

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.

Relative change

ΔR/R = 0.8/100 = 0.008

Linearised output

Vab ≈ 10·0.008/4 = 20 mV

Answer

Vab = 20 mV

The exact value is 19.92 mV; the approximation is good while ΔR/R is small.

8.Wheatstone bridge & strain gauges

Answer: ε = 400 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 = 4·0.0005/(2.5·2) = 0.0004

In microstrain

ε = 400 µε

Answer

ε = 400 µε
9.Steady-state error & PID control

Answer: K = 196

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.

Required error constant

1/(1 + Kp) = 0.02 → Kp = 49

Gain

K/4 = 49 → K = 196

Answer

K = 196

Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.

10.Steady-state error & PID control

Answer: ts = 2 s

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 + 4) + 16·2 = s² + 4s + 32

ωₙ and ζ

ωₙ = √32 = 5.657 rad/s
ζ = 4/(2·5.657) = 0.3536

Settling time

ζωₙ = 4/2 = 2
ts ≈ 4/2 = 2 s

Answer

ts = 2 s

The settling time does not depend on K here: raising K only makes the response oscillate faster.

11.Step response

Answer: ts = 2 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 = 7 → ωₙ = 2.646 rad/s
2ζωₙ = 4 → ζ = 4/(2·2.646) = 0.7559

Result

ζωₙ = 2 → ts ≈ 4/2 = 2 s

Answer

ts = 2 s
12.Step response

Answer: ζ = 0.5169

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).

Damping ratio

ζ = −ln(0.15)/√(π² + ln²(0.15)) = 1.897/√(9.87 + 3.599) = 0.5169

Answer

ζ = 0.5169

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