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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
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 draws 10 A from a 120 V supply. Its armature resistance is 1.5 Ω and its rotational (friction, windage and iron) losses are 53 W. Find its efficiency η.
V = 120 VIa = 10 ARa = 1.5 ΩProt = 53 W
η = %
2.DC motorsHard
A small DC motor has an armature resistance of 4 Ω and runs from 24 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 24 VRa = 4 Ω
Pmax = W
3.Gears & gear trainsMedium
Two parallel shafts are 75 mm apart. They must be connected by a pair of spur gears with a module of 1.5 mm and a speed reduction of 4 : 1. How many teeth N1 must the pinion have?
a = 75 mmm = 1.5 mmi = 4
N1 =
4.Gears & gear trainsHard
An open belt connects a 200 mm driving pulley at 1440 rpm to a 500 mm pulley; the centres are 800 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
A recorder samples 8 channels at 20 kHz each with a 16-bit ADC. Each sample is stored in whole bytes. Find the memory M needed for a 30 min recording, in MB (1 MB = 10⁶ bytes).
n = 8fs = 20 kHzN = 16t = 30 min
M = MB
6.Data acquisition & samplingHard
A 16-channel multiplexer feeds one ADC. Each conversion takes 20 µs, and after every channel switch the input needs 0.5 µs to settle. The channels are scanned in turn without pause. Find the sampling rate fs of each channel, in kHz.
n = 16tc = 20 µstset = 0.5 µs
fs = kHz
7.Wheatstone bridge & strain gaugesMedium
A bridge has four 100 Ω arms and Vs = 15 V. The sensor R₂ increases by ΔR = 5 Ω. Find the exact output Vab in mV.
Vs = 15 VR = 100 ΩΔR = 5 Ω
Vab = mV
8.Wheatstone bridge & strain gaugesHard
A strain gauge (GF = 2.12) on a member with E = 70 GPa forms a quarter bridge excited with 5 V. The output reads 1.06 mV. Find the axial stress σ in MPa.
Vs = 5 VGF = 2.12Vo = 1.06 mVE = 70 GPa
σ = MPa
9.Steady-state error & PID controlMedium
A unity-feedback loop with G(s) = K/(s(s + 5)) must follow the ramp r(t) = 1t with a steady-state error of 0.05. Find the gain K.
G(s) = K/(s(s + 5))r(t) = 1tess = 0.05
K =
10.Steady-state error & PID controlHard
A PD controller C(s) = 20 + Kd·s drives the plant G(s) = 1/(s(s + 1)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.9.
G(s) = 1/(s(s + 1))Kp = 20ζ = 0.9
Kd = s
11.Step responseMedium
A system has the transfer function G(s) = 135/(s² + 6s + 45) and receives a unit step input. The poles are s = σ ± jωd. Find the real part σ.
G(s) = 135/(s² + 6s + 45)
σ =
12.Step responseHard
A mass m = 8 kg hangs on a spring k = 1000 N/m with a viscous damper c = 36 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 = 8 kgk = 1000 N/mc = 36 N·s/mF = 20 N
ζ =
3Solutions
1.DC motors
Answer: η = 83.08 %
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.
Powers
Pin = 120·10 = 1200 W
Pdev = 105·10 = 1050 W
Pout = 1050 − 53 = 997 W
Efficiency
η = 997/1200·100 = 83.08 %
Answer
η = 83.08 %
2.DC motors
Answer: Pmax = 36 W
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.
Power as a function of the back-EMF
P(E) = E·(V − E)/Ra
dP/dE = 0 → E = V/2 = 12 V
Maximum power
Pmax = (V/2)²/Ra = V²/(4Ra) = 24²/(4·4) = 36 W
Answer
Pmax = 36 W
At this point the efficiency is only 50 %, so real motors run well above half speed.
3.Gears & gear trains
Answer: N1 = 20
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 = 2·75/1.5 = 100
Split by the ratio
N1 = 100/(1 + 4) = 20
Answer
N1 = 20
4.Gears & gear trains
Answer: P = 3.762 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 = 2.764 rad
Slack-side tension
eμθ = e0.6911 = 1.996, T₂ = 500/1.996 = 250.5 N
Power
P = (500 − 250.5)·15.08 = 3.762 kW
Answer
P = 3.762 kW
5.Data acquisition & sampling
Answer: M = 576 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
⌈16/8⌉ = 2 B
Byte rate
8·20000·2 = 320000 B/s
Memory
M = 320000·1800 = 5.76 × 10⁸ B = 576 MB
Answer
M = 576 MB
6.Data acquisition & sampling
Answer: fs = 3.049 kHz
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).
Time per channel
20 + 0.5 = 20.5 µs
Time per scan
16·20.5 = 328 µs
Sampling rate per channel
fs = 1/328 µs = 3.049 kHz
Answer
fs = 3.049 kHz
7.Wheatstone bridge & strain gauges
Answer: Vab = 182.9 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.
Node voltages
Va = 15·105/205 = 7.683 V
Vb = 7.5 V
Output
Vab = 182.9 mV
Answer
Vab = 182.9 mV
The linear estimate Vs·ΔR/(4R) = 187.5 mV is 2.5 % too high: a quarter bridge is not quite linear.
8.Wheatstone bridge & strain gauges
Answer: σ = 28 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 = 4·0.00106/(5·2.12) = 0.0004
Stress
σ = 70·10³ MPa·0.0004 = 28 MPa
Answer
σ = 28 MPa
9.Steady-state error & PID control
Answer: K = 100
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 = 1/0.05 = 20
Gain
K = 5·Kv = 100
Answer
K = 100
A large gain makes this loop very lightly damped: ζ = 0.25. That conflict is why compensators are used.
10.Steady-state error & PID control
Answer: Kd = 7.05 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 + 1) + 1(20 + Kds) = s² + (1 + 1Kd)s + 20
Natural frequency
ωₙ = √20 = 4.472 rad/s
Derivative gain
1 + 1Kd = 2·0.9·4.472 = 8.05
Kd = 7.05 s
Answer
Kd = 7.05 s
The derivative term adds damping without changing ωₙ.
11.Step response
Answer: σ = −3
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 = 45 → ωₙ = 6.708 rad/s
2ζωₙ = 6 → ζ = 6/(2·6.708) = 0.4472
Result
σ = −ζωₙ = −6/2 = −3
Answer
σ = −3
12.Step response
Answer: ζ = 0.2012
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).