Mechatronic Systems: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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)6 N·mn0​n = ?

A DC motor (k = 0.6 N·m/A, Ra = 1.5 Ω) runs from 240 V and drives a load torque of 6 N·m. Find its speed n in rpm.

V = 240 Vk = 0.6 N·m/ARa = 1.5 ΩT = 6 N·m
n = rpm
2.DC motorsHard
+−110 VRs​ = ?Ra​ = 0.2 ΩM+−En = 500 rpmT = 18 N·m

A DC motor (k = 0.6 V·s/rad, Ra = 0.2 Ω) on a fixed 110 V supply drives a load of 18 N·m. Find the series resistance Rs that slows it down to 500 rpm.

V = 110 Vk = 0.6 V·s/radRa = 0.2 ΩT = 18 N·mn = 500 rpm
Rs = Ω
3.Gears & gear trainsMedium
N1​ = ?N2​

Two parallel shafts are 54 mm apart. They must be connected by a pair of spur gears with a module of 2 mm and a speed reduction of 2 : 1. How many teeth N1 must the pinion have?

a = 54 mmm = 2 mmi = 2
N1 =
4.Gears & gear trainsHard
d1​ = 125 mmd2​ = 300 mm720 rpmC = 1000 mmθ = ?

An open belt connects a 125 mm pulley and a 300 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 125 mmd2 = 300 mmC = 1000 mm
θ = °
5.Data acquisition & samplingMedium
tTs​ = 1/fs​f = 7.4 kHzfs​ = 4 kHz

A 7.4 kHz sine wave is sampled at 4 kHz without an anti-aliasing filter. At which frequency fa does it appear in the samples? Give it in Hz.

f = 7.4 kHzfs = 4 kHz
fa = Hz
6.Data acquisition & samplingHard
SD0​D1​D2​D3​D4​D5​D6​D7​1/fbaud​8 MHz / (16·D)

A microcontroller with an 8 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 = 8 MHztarget = 2400 Bd
e = %
7.Wheatstone bridge & strain gaugesMedium
+−9 VR1​ = 50 ΩR2​ = 20 ΩR3​ = 600 ΩR4​ = 300 Ωab+−Rth​ = ?

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

Vs = 9 VR1 = 50 ΩR2 = 20 ΩR3 = 600 ΩR4 = 300 Ω
Rth = Ω
8.Wheatstone bridge & strain gaugesHard
+−2.5 V120 Ω − ΔR120 Ω + ΔR120 Ω + ΔR120 Ω − ΔRab+−Vo​ = ?

A bridge with 120 Ω arms is excited with 2.5 V and has four active gauges, two in tension and two in compression (full bridge). The gauge factor is 2.1 and the strain is 500 µε. Find the bridge output Vo in mV.

Vs = 2.5 VGF = 2.1ε = 500 µε
Vo = mV
9.Steady-state error & PID controlMedium
R(s)+K = 1410s + 1C(s)−

A process G(s) = 4/(10s + 1) is controlled by a proportional gain K = 1 in a unity-feedback loop. Find the steady-state error ess for a step, as a percentage of the step size.

G(s) = 4/(10s + 1)K = 1
ess = %
10.Steady-state error & PID controlHard
R(s)+K = 410s(s + 5)C(s)−

A proportional controller K = 4 drives the plant G(s) = 10/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).

G(s) = 10/(s(s + 5))K = 4
ts = s
11.Step responseMedium
t (s)y(t)3tp​ = ?

A system has the transfer function G(s) = 51/(s² + 2s + 17) and receives a unit step input. Find the peak time tp of the step response.

G(s) = 51/(s² + 2s + 17)
tp = s
12.Step responseHard
R(s)+115s(s + 5)G(s)C(s)−

A unity negative feedback loop has G(s) = 115/(s(s + 5)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).

G(s) = 115/(s(s + 5))
ts = s

3Solutions

1.DC motors

Answer: n = 3581 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 = 6/0.6 = 10 A

Back-EMF and speed

E = 240 − 10·1.5 = 225 V
ω = E/k = 375 rad/s → n = 3581 rpm

Answer

n = 3581 rpm
2.DC motors

Answer: Rs = 2.419 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 = 18/0.6 = 30 A
E = 0.6·52.36 = 31.42 V

Series resistance

Ra + Rs = (110 − 31.42)/30 = 2.619 Ω
Rs = 2.419 Ω

Answer

Rs = 2.419 Ω

The resistor wastes 2178 W; a chopper controls the speed with far less loss.

3.Gears & gear trains

Answer: N1 = 18

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 = 2·54/2 = 54

Split by the ratio

N1 = 54/(1 + 2) = 18

Answer

N1 = 18
4.Gears & gear trains

Answer: θ = 170 °

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.

Belt angle

asin(0.0875) = 5.02°

Angle of wrap

θ = 180 − 2·5.02 = 170°

Answer

θ = 170°
5.Data acquisition & sampling

Answer: fa = 600 Hz

Sampling: an ADC measures the signal fs times per second. By the Nyquist–Shannon theorem a frequency f is captured correctly only while f < fs/2, the Nyquist (folding) frequency, so the sampling rate must exceed twice the highest signal frequency. A higher frequency is not lost: it shows up as an alias at fa = |f − k·fs|, with the whole number k that brings the result between 0 and fs/2. The samples cannot tell the two apart, so an analogue anti-aliasing filter must remove everything above fs/2 before the ADC.

Nyquist frequency

fs/2 = 2 kHz < f = 7.4 kHz

Nearest multiple of the sampling rate

k = round(7400/4000) = 2

Alias frequency

fa = |7400 − 2·4000| = 600 Hz

Answer

fa = 600 Hz

The dashed curve in the figure is the alias: it passes through exactly the same samples.

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 = 8·10⁶/(16·2400) = 208.3 → 208

Real baud rate

8·10⁶/(16·208) = 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 = 214.3 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 = 50·20/70 = 14.29 Ω
R₃∥R₄ = 600·300/900 = 200 Ω

Thévenin resistance

Rth = 214.3 Ω

Answer

Rth = 214.3 Ω
8.Wheatstone bridge & strain gauges

Answer: Vo = 2.625 mV

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.

Relative resistance change

ΔR/R = 2.1·500·10⁻⁶ = 0.00105

Bridge output

Vo = Vs·GF·ε = 2.5·0.00105 = 2.625 mV

Answer

Vo = 2.625 mV
9.Steady-state error & PID control

Answer: ess = 20 %

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.

Position constant

Kp = 1·4 = 4

Step error

ess = 100/(1 + 4) = 20 %

Answer

ess = 20 %
10.Steady-state error & PID control

Answer: ts = 1.6 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 + 5) + 4·10 = s² + 5s + 40

ωₙ and ζ

ωₙ = √40 = 6.325 rad/s
ζ = 5/(2·6.325) = 0.3953

Settling time

ζωₙ = 5/2 = 2.5
ts ≈ 4/2.5 = 1.6 s

Answer

ts = 1.6 s

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

11.Step response

Answer: tp = 0.7854 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 = 17 → ωₙ = 4.123 rad/s
2ζωₙ = 2 → ζ = 2/(2·4.123) = 0.2425

Result

ωd = 4.123·√(1 − 0.2425²) = 4 rad/s → tp = π/ωd = 0.7854 s

Answer

tp = 0.7854 s
12.Step response

Answer: ts = 1.6 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).

Closed-loop transfer function

T(s) = 115/(s² + 5s + 115): ωₙ = 10.72 rad/s, ζ = 0.2331

Settling time

ζωₙ = 5/2 = 2.5
ts ≈ 4/(ζωₙ) = 1.6 s

Answer

ts = 1.6 s

For this loop ζωₙ = a/2 whatever K is: raising K adds overshoot but does not shorten the settling time.

ExamProblems · examproblems.com/courses/mechatronic-systems