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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 (k = 0.3 N·m/A, Ra = 0.4 Ω) runs from 110 V and drives a load torque of 7.5 N·m. Find its speed n in rpm.
V = 110 Vk = 0.3 N·m/ARa = 0.4 ΩT = 7.5 N·m
n = rpm
2.DC motorsHard
A separately excited DC motor on 110 V (Ra = 0.2 Ω) runs at 1000 rpm with an armature current of 40 A. The load changes and the current becomes 20 A, with the field unchanged. Find the new speed n2.
V = 110 VRa = 0.2 Ωn1 = 1000 rpmIa1 = 40 AIa2 = 20 A
n2 = rpm
3.Gears & gear trainsMedium
A 0.75 kW motor running at 1450 rpm drives a pinion with 20 teeth, which meshes with a gear with 48 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 0.75 kWn1 = 1450 rpmN1 = 20N2 = 48η = 98%
T1 = N·m
4.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turning at 1450 rpm drives gear B (48 teeth). Gear C (30 teeth) is fixed to the same shaft as B and drives gear D (80 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 12B = 48C = 30D = 80
nout = rpm
5.Data acquisition & samplingMedium
Find the ideal signal-to-quantisation-noise ratio SQNR of an 8-bit ADC for a full-scale sine wave.
N = 8
SQNR = dB
6.Data acquisition & samplingHard
An audio measurement needs an ideal signal-to-quantisation-noise ratio of at least 55 dB for a full-scale sine wave. Find the smallest number of ADC bits N.
SQNR ≥ 55 dB
N =
7.Wheatstone bridge & strain gaugesMedium
A bridge has four 200 Ω arms and Vs = 10 V. The sensor R₂ increases by ΔR = 40 Ω. Find the exact output Vab in mV.
Vs = 10 VR = 200 ΩΔR = 40 Ω
Vab = mV
8.Wheatstone bridge & strain gaugesHard
A quarter bridge with one active gauge (GF = 2) is excited with 2.5 V. Under load the output reads 1.875 mV. Find the strain ε in µε.
Vs = 2.5 VGF = 2Vo = 1.875 mV
ε = µε
9.Steady-state error & PID controlMedium
A unity-feedback loop has G(s) = K/((s + 4)(s + 5)). Find the gain K that gives a steady-state error of 20 % for a step input.
G(s) = K/((s + 4)(s + 5))ess = 20 %
K =
10.Steady-state error & PID controlHard
A PI controller C(s) = 2 + 0.5/s drives the plant G(s) = 4/((s + 4)(s + 10)) in a unity-feedback loop. Find the steady-state error ess for the ramp input r(t) = 2t.
A system has the transfer function G(s) = 108/(s² + 5s + 36) and receives a unit step input. Find the 2 % settling time ts of the step response, using ts ≈ 4/(ζωₙ).
G(s) = 108/(s² + 5s + 36)
ts = s
12.Step responseHard
A unity negative feedback loop has G(s) = K/(s(s + 4)). Find the gain K that gives a step response with 25 % overshoot.
G(s) = K/(s(s + 4))%OS = 25 %
K =
3Solutions
1.DC motors
Answer: n = 3183 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.
Armature current
Ia = T/k = 7.5/0.3 = 25 A
Back-EMF and speed
E = 110 − 25·0.4 = 100 V
ω = E/k = 333.3 rad/s → n = 3183 rpm
Answer
n = 3183 rpm
2.DC motors
Answer: n2 = 1039 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 − 40·0.2 = 102 V
E2 = 110 − 20·0.2 = 106 V
New speed
n2 = 1000·106/102 = 1039 rpm
Answer
n2 = 1039 rpm
3.Gears & gear trains
Answer: T1 = 4.939 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·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 750/151.8 = 4.939 N·m
Answer
T1 = 4.939 N·m
4.Gears & gear trains
Answer: nout = 135.9 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 = 1450·12/48 = 362.5 rpm = nC
Second stage
nout = 362.5·30/80 = 135.9 rpm
Answer
nout = 135.9 rpm
total ratio = 10.67 : 1
5.Data acquisition & sampling
Answer: SQNR = 49.92 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 = 6.02·8 + 1.76 = 49.92 dB
Answer
SQNR = 49.92 dB
6.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.
Solve for N
N ≥ (55 − 1.76)/6.02 = 8.844
Round up
N = 9
check: 6.02·9 + 1.76 = 55.94 dB ≥ 55 dB
Answer
N = 9
7.Wheatstone bridge & strain gauges
Answer: Vab = 454.5 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 = 10·240/440 = 5.455 V
Vb = 5 V
Output
Vab = 454.5 mV
Answer
Vab = 454.5 mV
The linear estimate Vs·ΔR/(4R) = 500 mV is 10 % too high: a quarter bridge is not quite linear.
8.Wheatstone bridge & strain gauges
Answer: ε = 1500 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 = 4·0.001875/(2.5·2) = 0.0015
In microstrain
ε = 1500 µε
Answer
ε = 1500 µε
9.Steady-state error & PID control
Answer: K = 80
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) = 0.2 → Kp = 4
Gain
K/20 = 4 → K = 80
Answer
K = 80
Check the closed-loop stability as well: for this second-order loop any K > 0 is stable.
10.Steady-state error & PID control
Answer: ess = 40
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 + 0.5 → 0.5 as s → 0.
Kv = 0.5·4/40 = 0.05
Ramp error
ess = 2/0.05 = 40
Answer
ess = 40
The PI controller removes the step error completely; a ramp is followed with a constant lag.
11.Step response
Answer: ts = 1.6 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 = 36 → ωₙ = 6 rad/s
2ζωₙ = 5 → ζ = 5/(2·6) = 0.4167
Result
ζωₙ = 2.5 → ts ≈ 4/2.5 = 1.6 s
Answer
ts = 1.6 s
12.Step response
Answer: K = 24.54
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).