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
Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
Use V = E + Ia·Ra to link voltage, current and back-EMF.
Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
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
Write down what is given: sampling rate, signal frequencies, number of bits, input range, baud rate.
For sampling questions, compare each frequency with fs/2; a higher frequency appears at fa = |f − k·fs| with the nearest multiple of fs.
For resolution questions, find 1 LSB = VFSR/2ᴺ, or the smallest N with 2ᴺ ≥ span/resolution.
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
Treat the bridge as two voltage dividers across the same supply.
For balance, set R₁/R₂ = R₃/R₄ and solve for the unknown resistor.
For an unbalanced bridge, find Va and Vb from the dividers and subtract; use the Thévenin equivalent when a meter loads the output.
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
Count the poles of the open loop C(s)·G(s) at s = 0: that is the system type.
Work out the error constant that matches the input: Kp for a step, Kv for a ramp, Ka for a parabola.
Apply ess = A/(1 + Kp), A/Kv or A/Ka, or solve it for the gain you need.
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
A DC motor with k = 1.5 N·m/A and Ra = 0.4 Ω is connected to 120 V. Find its stall torque Tstall, the torque at zero speed.
V = 120 Vk = 1.5 N·m/ARa = 0.4 Ω
Tstall = N·m
2.DC motorsHard
A DC motor (k = 0.3 V·s/rad, Ra = 0.2 Ω) on a fixed 24 V supply drives a load of 1.2 N·m. Find the series resistance Rs that slows it down to 400 rpm.
An 11 kW motor running at 1450 rpm drives a pinion with 16 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.
P = 11 kWn = 1450 rpmN1 = 16m = 3 mm
Ft = N
4.Gears & gear trainsHard
An open belt connects an 80 mm pulley and a 315 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 80 mmd2 = 315 mmC = 1000 mm
θ = °
5.Data acquisition & samplingMedium
A recorder samples 4 channels at 1 kHz each with a 12-bit ADC. Each sample is stored in whole bytes. Find the memory M needed for a 60 min recording, in MB (1 MB = 10⁶ bytes).
n = 4fs = 1 kHzN = 12t = 60 min
M = MB
6.Data acquisition & samplingHard
An audio measurement needs an ideal signal-to-quantisation-noise ratio of at least 96 dB for a full-scale sine wave. Find the smallest number of ADC bits N.
SQNR ≥ 96 dB
N =
7.Wheatstone bridge & strain gaugesMedium
Find the Thévenin resistance Rth of the bridge shown, seen from the output terminals a and b.
Vs = 10 VR1 = 250 ΩR2 = 50 ΩR3 = 120 ΩR4 = 400 Ω
Rth = Ω
8.Wheatstone bridge & strain gaugesHard
A strain gauge (GF = 2) on a member with E = 210 GPa forms a quarter bridge excited with 10 V. The output reads 1 mV. Find the axial stress σ in MPa.
Vs = 10 VGF = 2Vo = 1 mVE = 210 GPa
σ = MPa
9.Steady-state error & PID controlMedium
A unity-feedback loop with G(s) = K/(s(s + 10)) must follow the ramp r(t) = 2t with a steady-state error of 0.01. Find the gain K.
G(s) = K/(s(s + 10))r(t) = 2tess = 0.01
K =
10.Steady-state error & PID controlHard
A PI controller C(s) = Kp + Ki/s drives the plant G(s) = 4/(s + 5) in a unity-feedback loop. The closed loop must have ωₙ = 8 rad/s and ζ = 0.8. Find the integral gain Ki.
G(s) = 4/(s + 5)ωₙ = 8 rad/sζ = 0.8
Ki =
11.Step responseMedium
A system has the transfer function G(s) = 12/(s² + 5s + 12) and receives a unit step input. Find the natural frequency ωn in rad/s.
G(s) = 12/(s² + 5s + 12)
ωn = rad/s
12.Step responseHard
A second-order system without zeros must have a step response with 30 % overshoot. Find the damping ratio ζ.
%OS = 30 %
ζ =
3Solutions
1.DC motors
Answer: Tstall = 450 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.
Stall current
Ia = V/Ra = 120/0.4 = 300 A
Stall torque
Tstall = 1.5·300 = 450 N·m
Answer
Tstall = 450 N·m
2.DC motors
Answer: Rs = 2.658 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 = 1.2/0.3 = 4 A
E = 0.3·41.89 = 12.57 V
Series resistance
Ra + Rs = (24 − 12.57)/4 = 2.858 Ω
Rs = 2.658 Ω
Answer
Rs = 2.658 Ω
The resistor wastes 42.53 W; a chopper controls the speed with far less loss.
3.Gears & gear trains
Answer: Ft = 3018 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 = 11000/151.8 = 72.44 N·m
Pitch diameter
d = m·N = 3·16 = 48 mm
Tangential force
Ft = 2T/d = 2·72.44/0.048 = 3018 N
Answer
Ft = 3018 N
4.Gears & gear trains
Answer: θ = 166.5 °
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(0.1175) = 6.748°
Angle of wrap
θ = 180 − 2·6.748 = 166.5°
Answer
θ = 166.5°
5.Data acquisition & sampling
Answer: M = 28.8 MB
Data rates: every channel delivers fs samples per second of N bits, so the raw rate is channels × fs × N bit/s. In memory a sample takes whole bytes (a 12-bit sample is stored in 2 bytes), so the memory for a recording is channels × fs × bytes per sample × time. A multiplexed ADC converts the channels one after another: each conversion (with settling) takes t, so one round of n channels takes n·t and each channel is sampled at 1/(n·t).
Bytes per sample
⌈12/8⌉ = 2 B
Byte rate
4·1000·2 = 8000 B/s
Memory
M = 8000·3600 = 2.88 × 10⁷ B = 28.8 MB
Answer
M = 28.8 MB
6.Data acquisition & sampling
Answer: N = 16
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.
Solve for N
N ≥ (96 − 1.76)/6.02 = 15.65
Round up
N = 16
check: 6.02·16 + 1.76 = 98.08 dB ≥ 96 dB
Answer
N = 16
7.Wheatstone bridge & strain gauges
Answer: Rth = 134 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 = 250·50/300 = 41.67 Ω
R₃∥R₄ = 120·400/520 = 92.31 Ω
Thévenin resistance
Rth = 134 Ω
Answer
Rth = 134 Ω
8.Wheatstone bridge & strain gauges
Answer: σ = 42 MPa
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.
Strain
ε = 4·0.001/(10·2) = 0.0002
Stress
σ = 210·10³ MPa·0.0002 = 42 MPa
Answer
σ = 42 MPa
9.Steady-state error & PID control
Answer: K = 2000
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 velocity constant
Kv = A/ess = 2/0.01 = 200
Gain
K = 10·Kv = 2000
Answer
K = 2000
A large gain makes this loop very lightly damped: ζ = 0.1118. That conflict is why compensators are used.
10.Steady-state error & PID control
Answer: Ki = 16
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 + 5).
s² + (5 + 4Kp)s + 4Ki
Target polynomial
s² + 2·0.8·8s + 8² = s² + 12.8s + 64
Match the coefficients
4Ki = 64 → Ki = 16
Answer
Ki = 16
The integral term makes the loop type 1, so a step is followed without steady-state error.
11.Step response
Answer: ωn = 3.464 rad/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 = 12 → ωₙ = 3.464 rad/s
2ζωₙ = 5 → ζ = 5/(2·3.464) = 0.7217
Result
ωₙ = 3.464 rad/s
Answer
ωn = 3.464 rad/s
12.Step response
Answer: ζ = 0.3579
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).