Mechatronic Systems: cram pack

DC motor: the armature is a resistance R_a in series with the back-EMF E, so V = E + I_a·R_a. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·I_a, 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·I_a = T·ω; the input power is V·I_a, and the copper loss I_a²R_a and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − R_a·T/k², from the no-load speed V/k to zero at the stall torque k·V/R_a. At standstill E = 0, so the starting current V/R_a is large unless a resistor limits it.

DC motor: the armature is a resistance R_a in series with the back-EMF E, so V = E + I_a·R_a. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·I_a, 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·I_a = T·ω; the input power is V·I_a, and the copper loss I_a²R_a and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − R_a·T/k², from the no-load speed V/k to zero at the stall torque k·V/R_a. At standstill E = 0, so the starting current V/R_a is large unless a resistor limits it.

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.

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.

Data rates: every channel delivers f_s samples per second of N bits, so the raw rate is channels × f_s × N bit/s. In memory a sample takes whole bytes (a 12-bit sample is stored in 2 bytes), so the memory for a recording is channels × f_s × bytes per sample × time. A multiplexed ADC converts the channels one after another: each conversion (with settling) takes t, so one round of n channels takes n·t and each channel is sampled at 1/(n·t).

Sampling: an ADC measures the signal f_s times per second. By the Nyquist–Shannon theorem a frequency f is captured correctly only while f < f_s/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 f_a = |f − k·f_s|, with the whole number k that brings the result between 0 and f_s/2. The samples cannot tell the two apart, so an analogue anti-aliasing filter must remove everything above f_s/2 before the ADC.

Wheatstone bridge: two voltage dividers share one supply V_s. The left divider R₁–R₂ gives V_a = V_s·R₂/(R₁ + R₂) and the right divider R₃–R₄ gives V_b = V_s·R₄/(R₃ + R₄); the output is V_ab = V_a − V_b. The bridge is balanced (V_ab = 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 R_th = R₁∥R₂ + R₃∥R₄.

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 V_o ≈ V_s·GF·ε/4. Two active gauges with opposite strains in adjacent arms (half bridge) give V_s·GF·ε/2, and four (full bridge) give V_s·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.

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 e_ss = lim_s→0 s·E(s). The system type is the number of poles of G(s) at s = 0. The error constants are K_p = lim G(s), K_v = lim s·G(s) and K_a = lim s²·G(s), all for s → 0. A step of size A gives e_ss = A/(1 + K_p), a ramp A·t gives A/K_v and a parabola A·t²/2 gives A/K_a. 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.

P, PI and PID control: the controller C(s) = K_p + K_i/s + K_d·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 + 2ζω_ns + ω_n². Then %OS = 100·e^{−ζπ/√(1−ζ²)}, t_p = π/ω_d with ω_d = ω_n√(1 − ζ²), and t_s ≈ 4/(ζω_n).

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 ± jω_d with ω_d = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e^{−ζπ/√(1−ζ²)}, peak time t_p = π/ω_d, and the 2 % settling time t_s ≈ 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 ± jω_d with ω_d = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e^{−ζπ/√(1−ζ²)}, peak time t_p = π/ω_d, and the 2 % settling time t_s ≈ 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).