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
+−VRa​M+−En = 1200 rpmT = ?

A DC motor delivers 3 kW of mechanical power at 1200 rpm. Find the shaft torque T.

P = 3 kWn = 1200 rpm
T = N·m
2.DC motorsHard
+−220 VRa​ = 1.2 ΩIa​: 15 → 7.5 AM+−En = ?T

A separately excited DC motor on 220 V (Ra = 1.2 Ω) runs at 1500 rpm with an armature current of 15 A. The load changes and the current becomes 7.5 A, with the field unchanged. Find the new speed n2.

V = 220 VRa = 1.2 Ωn1 = 1500 rpmIa1 = 15 AIa2 = 7.5 A
n2 = rpm
3.Gears & gear trainsMedium
N1​N2​ = ?

Two parallel shafts are 150 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 2 : 1. How many teeth N2 must the gear have?

a = 150 mmm = 5 mmi = 2
N2 =
4.Gears & gear trainsHard
d1​ = 150 mmd2​ = 400 mm1450 rpmC = 800 mmT1​ = 1000 NT2​P = ?

An open belt connects a 150 mm driving pulley at 1450 rpm to a 400 mm pulley; the centres are 800 mm apart. The coefficient of friction is 0.35 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 = 150 mmn1 = 1450 rpmd2 = 400 mmC = 800 mmμ = 0.35T1 = 1000 N
P = kW
5.Data acquisition & samplingMedium
Vin​N = 102.5 V2ᴺ − 101 LSB2.17 V?

A 10-bit ADC with an input range of 0 to 2.5 V truncates: its output code is ⌊Vin/LSB⌋ with 1 LSB = VFSR/2ᴺ. Find the decimal output code D for an input of 2.17 V.

N = 10VFSR = 2.5 VVin = 2.17 V
D =
6.Data acquisition & samplingHard
SD0​D1​D2​D3​D4​D5​D6​D7​PTb​11·Tb​57600 Bd, 8E1

A 24-bit ADC streams every sample over a UART at 57600 baud, format 8E1 (start bit, 8 data bits, even parity, 1 stop bit). Each sample is sent as 3 whole byte(s) and the link is busy all the time. Find the highest signal frequency fmax that can be sampled without aliasing, in Hz.

baud = 57600 BdN = 248E1
fmax = Hz
7.Wheatstone bridge & strain gaugesMedium
+−2 V1000 Ω1000 Ω1000 Ω1000 Ωab+−

A 1000 Ω strain gauge is one arm (R₂) of a bridge with four equal 1000 Ω arms and an excitation of 2 V. Find the power Pg dissipated in the gauge, in mW. Self-heating limits the excitation.

Vs = 2 VR = 1000 Ω
Pg = mW
8.Wheatstone bridge & strain gaugesHard
+−3 V1000 Ω1000 Ω + ΔR1000 Ω1000 Ωab+−1.59 mV

A quarter bridge with one active gauge (GF = 2.12) is excited with 3 V. Under load the output reads 1.59 mV. Find the strain ε in µε.

Vs = 3 VGF = 2.12Vo = 1.59 mV
ε = µε
9.Steady-state error & PID controlMedium
r(t) = 4t+4(s + 4)s(s + 2)(s + 5)C(s)−

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

G(s) = 4(s + 4)/(s(s + 2)(s + 5))r(t) = 4t
ess =
10.Steady-state error & PID controlHard
R(s)+Kp​ + Ki​/s4s + 2C(s)−

A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 4/(s + 2) in a unity-feedback loop. The closed loop must have ωₙ = 6 rad/s and ζ = 0.8. Find the integral gain Ki.

G(s) = 4/(s + 2)ωₙ = 6 rad/sζ = 0.8
Ki =
11.Step responseMedium
t (s)y(t)0.5tp​ = ?

A system has the transfer function G(s) = 38.5/(s² + 9s + 77) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 38.5/(s² + 9s + 77)
tp = s
12.Step responseHard
k = 500 N/mc = 30 N·s/mm = 20 kgF = 100 Nx(t)

A mass m = 20 kg hangs on a spring k = 500 N/m with a viscous damper c = 30 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 100 N is suddenly applied. Find the damping ratio ζ.

m = 20 kgk = 500 N/mc = 30 N·s/mF = 100 N
ζ =

3Solutions

1.DC motors

Answer: T = 23.87 N·m

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.

Speed in rad/s

ω = 2π·1200/60 = 125.7 rad/s

Torque

T = P/ω = 3000/125.7 = 23.87 N·m

Answer

T = 23.87 N·m
2.DC motors

Answer: n2 = 1567 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.

Back-EMF before and after

E1 = 220 − 15·1.2 = 202 V
E2 = 220 − 7.5·1.2 = 211 V

New speed

n2 = 1500·211/202 = 1567 rpm

Answer

n2 = 1567 rpm
3.Gears & gear trains

Answer: N2 = 40

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.

Total number of teeth

N1 + N2 = 2a/m = 2·150/5 = 60

Split by the ratio

N1 = 60/(1 + 2) = 20
N2 = 2·20 = 40

Answer

N2 = 40
4.Gears & gear trains

Answer: P = 7.156 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 = 2.828 rad

Slack-side tension

eμθ = e0.9897 = 2.691, T₂ = 1000/2.691 = 371.7 N

Power

P = (1000 − 371.7)·11.39 = 7.156 kW

Answer

P = 7.156 kW
5.Data acquisition & sampling

Answer: D = 888

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 = 2.5/210 = 2.441 mV

Code

Vin/LSB = 2170/2.441 = 888.8
code = ⌊888.8⌋ = 888

Answer

D = 888

In binary that is 1101111000.

6.Data acquisition & sampling

Answer: fmax = 872.7 Hz

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 sample on the line

3·11 = 33 bit

Highest sampling rate

fs,max = 57600/33 = 1745 Hz

Nyquist limit

fmax = 1745/2 = 872.7 Hz

Answer

fmax = 872.7 Hz
7.Wheatstone bridge & strain gauges

Answer: Pg = 1 mW

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.

Voltage across the gauge

V = 2/2 = 1 V

Power

Pg = 1²/1000 = 1 mW

Answer

Pg = 1 mW
8.Wheatstone bridge & strain gauges

Answer: ε = 1000 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 = 4·0.00159/(3·2.12) = 0.001

In microstrain

ε = 1000 µε

Answer

ε = 1000 µε
9.Steady-state error & PID control

Answer: ess = 2.5

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) = 1.6

Ramp error

ess = 4/1.6 = 2.5

Answer

ess = 2.5
10.Steady-state error & PID control

Answer: Ki = 9

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

1 + C(s)G(s) = 0, multiplied by s(s + 2).

s² + (2 + 4Kp)s + 4Ki

Target polynomial

s² + 2·0.8·6s + 6² = s² + 9.6s + 36

Match the coefficients

4Ki = 36 → Ki = 9

Answer

Ki = 9

The integral term makes the loop type 1, so a step is followed without steady-state error.

11.Step response

Answer: tp = 0.417 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 = 77 → ωₙ = 8.775 rad/s
2ζωₙ = 9 → ζ = 9/(2·8.775) = 0.5128

Result

ωd = 8.775·√(1 − 0.5128²) = 7.533 rad/s → tp = π/ωd = 0.417 s

Answer

tp = 0.417 s
12.Step response

Answer: ζ = 0.15

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) = √(500/20) = 5 rad/s
ζ = c/(2√(km)) = 30/(2·√(500·20)) = 0.15

Result

ζ = 0.15

Answer

ζ = 0.15

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