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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 = 0.15 V·s/rad runs from 48 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
V = 48 Vk = 0.15 V·s/rad
n0 = rpm
2.DC motorsHard
A separately excited DC motor on 110 V (Ra = 0.5 Ω) runs at 1200 rpm with an armature current of 30 A. The load changes and the current becomes 12 A, with the field unchanged. Find the new speed n2.
V = 110 VRa = 0.5 Ωn1 = 1200 rpmIa1 = 30 AIa2 = 12 A
n2 = rpm
3.Gears & gear trainsMedium
A 150 mm pulley at 2880 rpm drives a 450 mm pulley by a belt with 2.5 % slip. Find the speed n2 of the driven pulley.
n1 = 2880 rpmd1 = 150 mmd2 = 450 mms = 2.5 %
n2 = rpm
4.Gears & gear trainsHard
In a planetary gear set the sun gear has 20 teeth and each planet gear has 15 teeth. The planet carrier is held fixed and the sun is driven at 1440 rpm. Find the speed nr of the ring gear (magnitude).
Ns = 20Np = 15ns = 1440 rpm
nr = rpm
5.Data acquisition & samplingMedium
A sensor measures pressure over a span of 100 kPa, and its output fills the whole input range of an ADC. The reading must resolve 0.2 kPa. Find the smallest number of bits N the ADC needs.
span = 100 kParesolution = 0.2 kPa
N =
6.Data acquisition & samplingHard
A 15 kHz square wave is sampled at 50 kHz with no anti-aliasing filter. A square wave contains the odd harmonics 3f₁, 5f₁, 7f₁, … At which frequency fa does its harmonic 7f₁ appear in the sampled data? Give it in kHz.
f₁ = 15 kHzfs = 50 kHzharmonic: 7f₁
fa = kHz
7.Wheatstone bridge & strain gaugesMedium
Find the Thévenin resistance Rth of the bridge shown, seen from the output terminals a and b.
Vs = 2 VR1 = 200 ΩR2 = 300 ΩR3 = 400 ΩR4 = 300 Ω
Rth = Ω
8.Wheatstone bridge & strain gaugesHard
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 2.05 and the strain is 1000 µε. Find the bridge output Vo in mV.
Vs = 2.5 VGF = 2.05ε = 1000 µε
Vo = mV
9.Steady-state error & PID controlMedium
The parabolic input r(t) = 0.5·t²/2 is applied to a unity-feedback loop with G(s) = 10(s + 1)/(s²(s + 5)). Find the steady-state error ess.
G(s) = 10(s + 1)/(s²(s + 5))r(t) = 0.5·t²/2
ess =
10.Steady-state error & PID controlHard
A proportional controller K = 20 drives the plant G(s) = 1/(s(s + 5)) in a unity-feedback loop. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 1/(s(s + 5))K = 20
ts = s
11.Step responseMedium
A system has the transfer function G(s) = 84/(s² + 15s + 84) and receives a unit step input. Find the percent overshoot %OS of the step response.
G(s) = 84/(s² + 15s + 84)
%OS = %
12.Step responseHard
The step response of a second-order system without zeros shows 40 % overshoot, and the first peak occurs at t = 0.5 s. Find the natural frequency ωn.
%OS = 40 %tp = 0.5 s
ωn = rad/s
3Solutions
1.DC motors
Answer: n0 = 3056 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 = 48/0.15 = 320 rad/s
n0 = 60·320/(2π) = 3056 rpm
Answer
n0 = 3056 rpm
2.DC motors
Answer: n2 = 1314 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 = 110 − 30·0.5 = 95 V
E2 = 110 − 12·0.5 = 104 V
New speed
n2 = 1200·104/95 = 1314 rpm
Answer
n2 = 1314 rpm
3.Gears & gear trains
Answer: n2 = 936 rpm
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.
Speed without slip
2880·150/450 = 960 rpm
With slip
n2 = 960·0.975 = 936 rpm
Answer
n2 = 936 rpm
4.Gears & gear trains
Answer: nr = 576 rpm
Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.
Ring gear teeth
Nr = Ns + 2·Np = 20 + 2·15 = 50
Willis equation with the carrier fixed
With nc = 0 the sun drives the ring through the planets, which act as idlers.
nr = −ns·Ns/Nr = −1440·20/50 = −576 rpm
|nr| = 576 rpm
Answer
nr = 576 rpm
The ring turns opposite to the sun.
5.Data acquisition & sampling
Answer: N = 9
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.
Steps needed
100/0.2 = 500
Bits
log₂(500) = 8.966
28 = 256 < 500 ≤ 29 = 512
N = 9
Answer
N = 9
6.Data acquisition & sampling
Answer: fa = 5 kHz
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.
Harmonic frequency
7f₁ = 7·15 kHz = 105 kHz
Nearest multiple of the sampling rate
m = round(2.1) = 2
Alias frequency
fa = |105 − 2·50| = 5 kHz
Answer
fa = 5 kHz
The alias lands inside the signal band, where no digital filter can remove it. Only an analogue filter before the ADC helps.
7.Wheatstone bridge & strain gauges
Answer: Rth = 291.4 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 = 200·300/500 = 120 Ω
R₃∥R₄ = 400·300/700 = 171.4 Ω
Thévenin resistance
Rth = 291.4 Ω
Answer
Rth = 291.4 Ω
8.Wheatstone bridge & strain gauges
Answer: Vo = 5.125 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 = 2.05·1000·10⁻⁶ = 0.00205
Bridge output
Vo = Vs·GF·ε = 2.5·0.00205 = 5.125 mV
Answer
Vo = 5.125 mV
9.Steady-state error & PID control
Answer: ess = 0.25
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.
Acceleration constant
Ka = 10·1/5 = 2
Parabola error
ess = 0.5/2 = 0.25
Answer
ess = 0.25
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) + 20·1 = s² + 5s + 20
ωₙ and ζ
ωₙ = √20 = 4.472 rad/s
ζ = 5/(2·4.472) = 0.559
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: %OS = 1.142 %
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 = 84 → ωₙ = 9.165 rad/s
2ζωₙ = 15 → ζ = 15/(2·9.165) = 0.8183
Result
%OS = 100·e−ζπ/√(1−ζ²) = 100·e−4.473 = 1.142 %
Answer
%OS = 1.142 %
12.Step response
Answer: ωn = 6.545 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).