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 draws 50 A from a 240 V supply. Its armature resistance is 0.25 Ω and its rotational (friction, windage and iron) losses are 683 W. Find its efficiency η.
V = 240 VIa = 50 ARa = 0.25 ΩProt = 683 W
η = %
2.DC motorsHard
A DC chopper feeds a motor (k = 1.2 V·s/rad, Ra = 2 Ω) from a 220 V DC bus. The average armature voltage is D·220 V. Find the duty cycle D that runs the motor at 1000 rpm with a load of 14.4 N·m.
A 2.2 kW motor running at 1440 rpm drives a pinion with 12 teeth, which meshes with a gear with 20 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 2.2 kWn1 = 1440 rpmN1 = 12N2 = 20η = 95%
T2 = N·m
4.Gears & gear trainsHard
A 1.5 kW motor at 2900 rpm drives gear A (20 teeth), which meshes with gear B (32 teeth). Gear C (18 teeth) on the same shaft as B drives gear D (72 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.
A 10-bit ADC with an input range of 0 to 5 V truncates: its output code is ⌊Vin/LSB⌋ with 1 LSB = VFSR/2ᴺ. Find the decimal output code D for an input of 3.849 V.
N = 10VFSR = 5 VVin = 3.849 V
D =
6.Data acquisition & samplingHard
A temperature sensor gives 10 mV/°C (0 V at 0 °C). Its signal is amplified 10 times and fed to a 10-bit ADC with a range of 0 to 10 V (1 LSB = VFSR/2ᴺ). Find the temperature change ΔTLSB that corresponds to one LSB.
S = 10 mV/°CG = 10N = 10VFSR = 10 V
ΔTLSB = °C
7.Wheatstone bridge & strain gaugesMedium
A galvanometer with a resistance of 100 Ω is connected between a and b of the bridge shown (Vs = 3 V). Find the galvanometer current Ig from a to b, in µA.
A load cell (a full bridge of strain gauges) has a rated output of 1 mV/V at its rated load of 200 kg. It is excited with 5 V. Find its output Vo under a load of 50 kg.
1 mV/VFrated = 200 kgVs = 5 VF = 50 kg
Vo = mV
9.Steady-state error & PID controlMedium
A ramp r(t) = 2t is applied to a unity-feedback loop with G(s) = 50/(s(s + 1)(s + 2)). Find the steady-state error ess.
G(s) = 50/(s(s + 1)(s + 2))r(t) = 2t
ess =
10.Steady-state error & PID controlHard
A PD controller C(s) = 10 + Kd·s drives the plant G(s) = 5/(s(s + 2)) in a unity-feedback loop. Find the derivative gain Kd that gives the closed loop a damping ratio ζ = 0.9.
G(s) = 5/(s(s + 2))Kp = 10ζ = 0.9
Kd = s
11.Step responseMedium
A system has the transfer function G(s) = 62/(s² + 6s + 62) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 62/(s² + 6s + 62)
ts = s
12.Step responseHard
A mass m = 20 kg hangs on a spring k = 200 N/m with a viscous damper c = 51 N·s/m (m·x″ + c·x′ + k·x = F). At rest in equilibrium, a constant force F = 50 N is suddenly applied. Find the damping ratio ζ.
m = 20 kgk = 200 N/mc = 51 N·s/mF = 50 N
ζ =
3Solutions
1.DC motors
Answer: η = 89.1 %
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 = 240·50 = 12000 W
Pdev = 227.5·50 = 11375 W
Pout = 11375 − 683 = 10692 W
Efficiency
η = 10692/12000·100 = 89.1 %
Answer
η = 89.1 %
2.DC motors
Answer: D = 68.03 %
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.
Required armature voltage
Va = 1.2·104.7 + 12·2 = 149.7 V
Duty cycle
D = 149.7/220 = 68.03 %
Answer
D = 68.03 %
3.Gears & gear trains
Answer: T2 = 23.1 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Motor torque
ω₁ = 2π·1440/60 = 150.8 rad/s
T1 = P/ω₁ = 2200/150.8 = 14.59 N·m
Output torque
T2 = 14.59·20/12·0.95 = 23.1 N·m
Answer
T2 = 23.1 N·m
4.Gears & gear trains
Answer: Tout = 29.74 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Output speed
nout = 2900·(20/32)·(18/72) = 453.1 rpm
ωout = 2π·453.1/60 = 47.45 rad/s
Output power and torque
Pout = 1500·0.97² = 1411 W
Tout = Pout/ωout = 1411/47.45 = 29.74 N·m
Answer
Tout = 29.74 N·m
5.Data acquisition & sampling
Answer: D = 788
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 = 5/210 = 4.883 mV
Code
Vin/LSB = 3849/4.883 = 788.3
code = ⌊788.3⌋ = 788
Answer
D = 788
In binary that is 1100010100.
6.Data acquisition & sampling
Answer: ΔTLSB = 0.09766 °C
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 = 10000/1024 = 9.766 mV
Sensitivity at the ADC
10·10 = 100 mV/°C
Temperature resolution
ΔTLSB = 9.766/100 = 0.09766 °C
Answer
ΔTLSB = 0.09766 °C
The ADC reaches full scale at 100 °C.
7.Wheatstone bridge & strain gauges
Answer: Ig = −2727 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 = 3·(60/260 − 1000/1500) = −1.308 V
Thévenin resistance
Rth = R₁∥R₂ + R₃∥R₄ = 379.5 Ω
Galvanometer current
Ig = −1.308/(379.5 + 100) = −2727 µA
Answer
Ig = −2727 µA
8.Wheatstone bridge & strain gauges
Answer: Vo = 1.25 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.
Full-scale output
1·5 = 5 mV
Output at this load
Vo = 5·50/200 = 1.25 mV
Answer
Vo = 1.25 mV
9.Steady-state error & PID control
Answer: ess = 0.08
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) = 25
Ramp error
ess = 2/25 = 0.08
Answer
ess = 0.08
10.Steady-state error & PID control
Answer: Kd = 2.146 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 + 2) + 5(10 + Kds) = s² + (2 + 5Kd)s + 50
Natural frequency
ωₙ = √50 = 7.071 rad/s
Derivative gain
2 + 5Kd = 2·0.9·7.071 = 12.73
Kd = 2.146 s
Answer
Kd = 2.146 s
The derivative term adds damping without changing ωₙ.
11.Step response
Answer: ts = 1.333 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 = 62 → ωₙ = 7.874 rad/s
2ζωₙ = 6 → ζ = 6/(2·7.874) = 0.381
Result
ζωₙ = 3 → ts ≈ 4/3 = 1.333 s
Answer
ts = 1.333 s
12.Step response
Answer: ζ = 0.4032
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).