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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 delivers 5.5 kW of mechanical power at 750 rpm. Find the shaft torque T.
P = 5.5 kWn = 750 rpm
T = N·m
2.DC motorsHard
A DC motor (k = 0.12 V·s/rad, Ra = 0.25 Ω) on a fixed 48 V supply drives a load of 0.72 N·m. Find the series resistance Rs that slows it down to 1200 rpm.
A 0.75 kW motor running at 1440 rpm drives a pinion with 25 teeth, which meshes with a gear with 36 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 0.75 kWn1 = 1440 rpmN1 = 25N2 = 36η = 97%
T2 = N·m
4.Gears & gear trainsHard
In a compound gear train, gear A (20 teeth) on the input shaft turning at 3000 rpm drives gear B (72 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (25 teeth) on the output shaft. Find the output speed nout.
nin = 3000 rpmA = 20B = 72C = 18D = 25
nout = rpm
5.Data acquisition & samplingMedium
Find the ideal signal-to-quantisation-noise ratio SQNR of a 10-bit ADC for a full-scale sine wave.
N = 10
SQNR = dB
6.Data acquisition & samplingHard
An audio measurement needs an ideal signal-to-quantisation-noise ratio of at least 40 dB for a full-scale sine wave. Find the smallest number of ADC bits N.
SQNR ≥ 40 dB
N =
7.Wheatstone bridge & strain gaugesMedium
A 250 Ω strain gauge is one arm (R₂) of a bridge with four equal 250 Ω arms and an excitation of 10 V. Find the power Pg dissipated in the gauge, in mW. Self-heating limits the excitation.
Vs = 10 VR = 250 Ω
Pg = mW
8.Wheatstone bridge & strain gaugesHard
A bridge with 1000 Ω arms is excited with 3 V and has two active gauges in adjacent arms, one in tension and one in compression (half bridge). The gauge factor is 2.12 and the strain is 1000 µε. Find the bridge output Vo in mV.
Vs = 3 VGF = 2.12ε = 1000 µε
Vo = mV
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 1)(s + 2)). Find the gain K that gives a steady-state error of 10 % for a step input.
G(s) = K/((s + 1)(s + 2))ess = 10 %
K =
10.Steady-state error & PID controlHard
A proportional controller K drives the plant G(s) = 2/(s(s + 8)) in a unity-feedback loop. Find the gain K that gives the closed loop a damping ratio ζ = 0.6.
G(s) = 2/(s(s + 8))ζ = 0.6
K =
11.Step responseMedium
A system has the transfer function G(s) = 77/(s² + 5s + 77) and receives a unit step input. Find the natural frequency ωn in rad/s.
G(s) = 77/(s² + 5s + 77)
ωn = rad/s
12.Step responseHard
A unity negative feedback loop has G(s) = 30/(s(s + 4)). Find the 2 % settling time ts of the closed-loop step response, using ts ≈ 4/(ζωₙ).
G(s) = 30/(s(s + 4))
ts = s
3Solutions
1.DC motors
Answer: T = 70.03 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.
Speed in rad/s
ω = 2π·750/60 = 78.54 rad/s
Torque
T = P/ω = 5500/78.54 = 70.03 N·m
Answer
T = 70.03 N·m
2.DC motors
Answer: Rs = 5.237 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 = 0.72/0.12 = 6 A
E = 0.12·125.7 = 15.08 V
Series resistance
Ra + Rs = (48 − 15.08)/6 = 5.487 Ω
Rs = 5.237 Ω
Answer
Rs = 5.237 Ω
The resistor wastes 188.5 W; a chopper controls the speed with far less loss.
3.Gears & gear trains
Answer: T2 = 6.947 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 = 750/150.8 = 4.974 N·m
Output torque
T2 = 4.974·36/25·0.97 = 6.947 N·m
Answer
T2 = 6.947 N·m
4.Gears & gear trains
Answer: nout = 600 rpm
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.
First stage
nB = 3000·20/72 = 833.3 rpm = nC
Second stage
nout = 833.3·18/25 = 600 rpm
Answer
nout = 600 rpm
total ratio = 5 : 1
5.Data acquisition & sampling
Answer: SQNR = 61.96 dB
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.
Ideal SQNR
SQNR = 6.02·10 + 1.76 = 61.96 dB
Answer
SQNR = 61.96 dB
6.Data acquisition & sampling
Answer: N = 7
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 ≥ (40 − 1.76)/6.02 = 6.352
Round up
N = 7
check: 6.02·7 + 1.76 = 43.9 dB ≥ 40 dB
Answer
N = 7
7.Wheatstone bridge & strain gauges
Answer: Pg = 100 mW
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.
Voltage across the gauge
V = 10/2 = 5 V
Power
Pg = 5²/250 = 100 mW
Answer
Pg = 100 mW
8.Wheatstone bridge & strain gauges
Answer: Vo = 3.18 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 = 2.12·1000·10⁻⁶ = 0.00212
Bridge output
Vo = Vs·GF·ε/2 = 3·0.00212/2 = 3.18 mV
Answer
Vo = 3.18 mV
9.Steady-state error & PID control
Answer: K = 18
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 error constant
1/(1 + Kp) = 0.1 → Kp = 9
Gain
K/2 = 9 → K = 18
Answer
K = 18
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
10.Steady-state error & PID control
Answer: K = 22.22
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 + 8) + 2K = s² + 8s + 2K
Natural frequency
2·0.6·ωₙ = 8 → ωₙ = 6.667 rad/s
Gain
2K = ωₙ² = 44.44 → K = 22.22
Answer
K = 22.22
11.Step response
Answer: ωn = 8.775 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 = 77 → ωₙ = 8.775 rad/s
2ζωₙ = 5 → ζ = 5/(2·8.775) = 0.2849
Result
ωₙ = 8.775 rad/s
Answer
ωn = 8.775 rad/s
12.Step response
Answer: ts = 2 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).