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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 with k = 0.5 V·s/rad runs from 110 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
V = 110 Vk = 0.5 V·s/rad
n0 = rpm
2.DC motorsHard
A separately excited DC motor on 220 V (Ra = 1.2 Ω) runs at 1500 rpm with an armature current of 6 A. The load changes and the current becomes 3.6 A, with the field unchanged. Find the new speed n2.
V = 220 VRa = 1.2 Ωn1 = 1500 rpmIa1 = 6 AIa2 = 3.6 A
n2 = rpm
3.Gears & gear trainsMedium
A 7.5 kW motor running at 1450 rpm drives a pinion with 20 teeth and a module of 4 mm. Find the tangential force Ft on the pinion teeth.
P = 7.5 kWn = 1450 rpmN1 = 20m = 4 mm
Ft = N
4.Gears & gear trainsHard
An open belt connects a 120 mm driving pulley at 1500 rpm to a 400 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 800 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
A 12-bit ADC with an input range of 0 to 3.3 V truncates: its output code is ⌊Vin/LSB⌋ with 1 LSB = VFSR/2ᴺ. Find the decimal output code D for an input of 0.675 V.
N = 12VFSR = 3.3 VVin = 0.675 V
D =
6.Data acquisition & samplingHard
A microcontroller with a 10 MHz clock sets its UART baud rate as fclk/(16·D), with D a whole number rounded to the nearest value. The target is 38400 baud. Find the percentage error of the real baud rate (positive when it is too fast).
fclk = 10 MHztarget = 38400 Bd
e = %
7.Wheatstone bridge & strain gaugesMedium
A galvanometer with a resistance of 500 Ω is connected between a and b of the bridge shown (Vs = 9 V). Find the galvanometer current Ig from a to b, in µA.
A quarter bridge with one active gauge (GF = 2.12) is excited with 3 V. Under load the output reads 0.636 mV. Find the strain ε in µε.
Vs = 3 VGF = 2.12Vo = 0.636 mV
ε = µε
9.Steady-state error & PID controlMedium
The input r(t) = 1 + 1t (a step plus a ramp) is applied to a unity-feedback loop with G(s) = 50(s + 4)/(s(s + 2)). Find the steady-state error ess.
G(s) = 50(s + 4)/(s(s + 2))r(t) = 1 + 1t
ess =
10.Steady-state error & PID controlHard
A PI controller C(s) = 2 + 4/s drives the plant G(s) = 20/((s + 1)(s + 2)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 5t.
C(s) = 2 + 4/sG(s) = 20/((s + 1)(s + 2))r(t) = 5t
ess =
11.Step responseMedium
A system has the transfer function G(s) = 128/(s² + 4s + 64) and receives a unit step input. Find the peak time tp of the step response.
G(s) = 128/(s² + 4s + 64)
tp = s
12.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 20)). Find the gain K that gives a step response with 10 % overshoot.
G(s) = K/(s(s + 20))%OS = 10 %
K =
3Solutions
1.DC motors
Answer: n0 = 2101 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.
No-load speed
ω₀ = V/k = 110/0.5 = 220 rad/s
n0 = 60·220/(2π) = 2101 rpm
Answer
n0 = 2101 rpm
2.DC motors
Answer: n2 = 1520 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 − 6·1.2 = 212.8 V
E2 = 220 − 3.6·1.2 = 215.7 V
New speed
n2 = 1500·215.7/212.8 = 1520 rpm
Answer
n2 = 1520 rpm
3.Gears & gear trains
Answer: Ft = 1235 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 = 7500/151.8 = 49.39 N·m
Pitch diameter
d = m·N = 4·20 = 80 mm
Tangential force
Ft = 2T/d = 2·49.39/0.08 = 1235 N
Answer
Ft = 1235 N
4.Gears & gear trains
Answer: P = 3.852 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 = 2.861 rad
Slack-side tension
eμθ = e0.7152 = 2.045, T₂ = 800/2.045 = 391.3 N
Power
P = (800 − 391.3)·9.425 = 3.852 kW
Answer
P = 3.852 kW
5.Data acquisition & sampling
Answer: D = 837
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 = 3.3/212 = 0.8057 mV
Code
Vin/LSB = 675/0.8057 = 837.8
code = ⌊837.8⌋ = 837
Answer
D = 837
In binary that is 001101000101.
6.Data acquisition & sampling
Answer: e = 1.725 %
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 = 10·10⁶/(16·38400) = 16.28 → 16
Real baud rate
10·10⁶/(16·16) = 39063 Bd
Error
(39063 − 38400)/38400·100 = 1.725 %
Answer
e = 1.725 %
An error below about 2 % is usually acceptable.
7.Wheatstone bridge & strain gauges
Answer: Ig = 1006 µA
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.
Thévenin voltage
Vth = 9·(150/350 − 400/1200) = 0.8571 V
Thévenin resistance
Rth = R₁∥R₂ + R₃∥R₄ = 352.4 Ω
Galvanometer current
Ig = 0.8571/(352.4 + 500) = 1006 µA
Answer
Ig = 1006 µA
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 = 4·0.000636/(3·2.12) = 0.0004
In microstrain
ε = 400 µε
Answer
ε = 400 µε
9.Steady-state error & PID control
Answer: ess = 0.01
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) = 100
Step part
Kp = ∞ → estep = 0
Ramp error
ess = 1/100 = 0.01
Answer
ess = 0.01
10.Steady-state error & PID control
Answer: ess = 0.125
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).
Velocity constant
s·C(s) = 2s + 4 → 4 as s → 0.
Kv = 4·20/2 = 40
Ramp error
ess = 5/40 = 0.125
Answer
ess = 0.125
The PI controller removes the step error completely; a ramp is followed with a constant lag.
11.Step response
Answer: tp = 0.4056 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).
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).