Measurement and Instrumentation: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Wheatstone bridge & strain gauges

Divider voltages: V_a = V_sR₂/(R₁ + R₂), V_b = V_sR₄/(R₃ + R₄)
Output: V_ab = V_a − V_b
Balance: R₁/R₂ = R₃/R₄
Slide-wire bridge: R/R_x = l/(100 − l)
Thévenin resistance: R_th = R₁∥R₂ + R₃∥R₄
Gauge factor: ΔR/R = GF·ε
Bridge output: quarter V_s·GF·ε/4, half V_s·GF·ε/2, full V_s·GF·ε
Small change, one arm: V_ab ≈ V_s·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 V_a and V_b 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 V_s·GF·ε/4, /2 or ×1 for a quarter, half or full bridge.

Data acquisition & sampling

Nyquist criterion: f_s > 2·f_max, f_N = f_s/2
Alias frequency: f_a = |f − k·f_s|, k = round(f/f_s)
Step size: 1 LSB = V_FSR/2ᴺ
Quantisation error (rounding): ±½ LSB
Truncating ADC code: D = ⌊V_in/LSB⌋
Bits for a resolution r: 2ᴺ ≥ span/r
Ideal SQNR, full-scale sine: SQNR = 6.02·N + 1.76 dB
Data rate: channels × f_s × N bit/s
Multiplexed ADC: f_s = 1/(n·(t_c + t_set))
UART: T_b = 1/baud; 8N1 = 10 bits per byte; bytes/s = baud/bits per frame
Baud rate divider: baud = f_clk/(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 f_s/2; a higher frequency appears at f_a = |f − k·f_s| with the nearest multiple of f_s.
For resolution questions, find 1 LSB = V_FSR/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.

Op-amp circuits

Inverting amplifier: V_out = −(R_f/R_in)·V_in
Non-inverting amplifier: V_out = (1 + R_f/R₁)·V_in
Summing amplifier: V_out = −R_f(V₁/R₁ + V₂/R₂)
Ideal op-amp: V₊ = V₋, I_in = 0

Series & parallel resistors

Resistors in series: R_eq = R₁ + R₂ + …
Resistors in parallel: 1/R_eq = 1/R₁ + 1/R₂ + …
Two in parallel: R_eq = R₁R₂ / (R₁ + R₂)
Ohm's law: V = I·R
Power: P = V·I = I²R = V²/R

Method

Start at the end of the network furthest from the terminals.
Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
Redraw the circuit after each step.
Repeat until one resistor is left between the terminals.

Thévenin & Norton

Thévenin voltage: V_th = V_{open circuit}
Thévenin resistance: R_th: all sources off, seen from a–b
Norton current: I_N = V_th/R_th
Load current: I_L = V_th/(R_th + R_L)
Maximum power transfer: R_L = R_th, P_max = V_th²/(4R_th)

Method

Remove the load and mark the two terminals a and b.
Find the open-circuit voltage between a and b: this is V_th.
Switch off all independent sources and find the resistance seen from a–b: this is R_th.
Put the load back on the equivalent circuit: V_th in series with R_th.
For the Norton equivalent, convert: I_N = V_th/R_th, with R_N = R_th in parallel.

2Practice problems

1.Wheatstone bridge & strain gaugesMedium

A 200 Ω strain gauge is one arm (R₂) of a bridge with four equal 200 Ω arms and an excitation of 3 V. Find the power P_g dissipated in the gauge, in mW. Self-heating limits the excitation.

V_s = 3 VR = 200 Ω

P_g = mW

2.Wheatstone bridge & strain gaugesHard

A 350 Ω strain gauge with gauge factor 2.05 is bonded to a steel bar. Under load its resistance rises by 0.287 Ω. Find the strain ε in microstrain (µε).

R = 350 ΩGF = 2.05ΔR = 0.287 Ω

ε = µε

3.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 0.1 kPa. Find the smallest number of bits N the ADC needs.

span = 100 kParesolution = 0.1 kPa

N =

4.Data acquisition & samplingHard

An 8 kHz square wave is sampled at 20 kHz with no anti-aliasing filter. A square wave contains the odd harmonics 3f₁, 5f₁, 7f₁, … At which frequency f_a does its harmonic 3f₁ appear in the sampled data? Give it in kHz.

f₁ = 8 kHzf_s = 20 kHzharmonic: 3f₁

f_a = kHz

5.Op-amp circuitsMedium

The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±12 V. The inverting amplifier is driven by a sine wave. Find the largest input amplitude V_in,max (peak value) for which the output does not clip.

R₁ = 2 kΩR_f = 4.7 kΩV_sat = ±12 V

V_in,max = V

6.Op-amp circuitsHard

The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the output voltage V_out.

V₁ = −0.9 VV₂ = −1.2 VR₁ = 47 kΩR₂ = 22 kΩR_f = 100 kΩ

V_out = V

7.Series & parallel resistorsMedium

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 24 VR₁ = 4 ΩR₂ = 8 ΩR₃ = 10 ΩR₄ = 15 Ω

R_eq = Ω

8.Series & parallel resistorsHard

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 5 VR₁ = 3 ΩR₂ = 2 ΩR₃ = 15 ΩR₄ = 25 ΩR₅ = 4 ΩR₆ = 15 Ω

R_eq = Ω

9.Thévenin & NortonMedium

Find the Thévenin voltage V_th seen from terminals a–b.

V₁ = 18 VR₁ = 6 ΩR₂ = 20 ΩR₃ = 2 Ω

V_th = V

10.Thévenin & NortonHard

Find the Thévenin voltage V_th seen from terminals a–b.

V₁ = 5 VI_s = 4 AR₁ = 8 ΩR₂ = 5 ΩR₃ = 10 Ω

V_th = V

3Solutions

1.Wheatstone bridge & strain gauges

Answer: P_g = 11.25 mW

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.

Voltage across the gauge

V = 3/2 = 1.5 V

Power

P_g = 1.5²/200 = 11.25 mW

Answer

P_g = 11.25 mW

2.Wheatstone bridge & strain gauges

Answer: ε = 400 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 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.

Relative resistance change

ΔR/R = 0.287/350 = 0.00082

Strain

ε = 0.00082/2.05 = 0.0004 = 400 µε

Answer

ε = 400 µε

3.Data acquisition & sampling

Answer: N = 10

ADC resolution: an N-bit converter divides its full-scale range V_FSR into 2ᴺ codes, from 0 to 2ᴺ − 1. One step is 1 LSB = V_FSR/2ᴺ: the smallest change the converter can show. A truncating ADC gives the code ⌊V_in/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

100/0.1 = 1000

Bits

log₂(1000) = 9.966

2⁹ = 512 < 1000 ≤ 2¹⁰ = 1024

N = 10

Answer

N = 10

4.Data acquisition & sampling

Answer: f_a = 4 kHz

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.

Harmonic frequency

3f₁ = 3·8 kHz = 24 kHz

Nearest multiple of the sampling rate

m = round(1.2) = 1

Alias frequency

f_a = |24 − 1·20| = 4 kHz

Answer

f_a = 4 kHz

The alias lands inside the signal band, where no digital filter can remove it. Only an analogue filter before the ADC helps.

5.Op-amp circuits

Answer: V_in,max = 5.106 V

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Gain

|A_v| = R_f/R₁ = 4.7/2 = 2.35

Largest input before clipping

V_in,max = V_sat/|A_v| = 12/2.35 = 5.106 V

Answer

V_in,max = 5.106 V

6.Op-amp circuits

Answer: V_out = 7.369 V

Input currents

V₋ = 0, so each input resistor has its full input voltage across it.

I₁ = V₁/R₁ = −0.9/47 = −0.01915 mA

I₂ = V₂/R₂ = −1.2/22 = −0.05455 mA

I_f = I₁ + I₂ = −0.07369 mA

Output voltage

V_out = −R_f·I_f = −100·(−0.07369) = 7.369 V

Answer

V_out = 7.369 V

7.Series & parallel resistors

Answer: R_eq = 10.06 Ω

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R₃ + R₄ = 10 + 15 = 25 Ω

R₂ ∥ 25 = 8·25/(8+25) = 6.061 Ω

R_eq = R₁ + 6.061 = 4 + 6.061 = 10.06 Ω

Answer

R_eq = 10.06 Ω

8.Series & parallel resistors

Answer: R_eq = 4.856 Ω

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R₅ + R₆ = 4 + 15 = 19 Ω

R₄ ∥ 19 = 10.8 Ω

R₃ + 10.8 = 25.8 Ω

R₂ ∥ 25.8 = 1.856 Ω

R_eq = R₁ + 1.856 = 3 + 1.856 = 4.856 Ω

Answer

R_eq = 4.856 Ω

9.Thévenin & Norton

Answer: V_th = 13.85 V

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source V_th in series with a resistance R_th. Norton’s theorem gives the same network as a current source I_N = V_th/R_th in parallel with R_th.

Find R_th with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R₁ ∥ R₂ = 6·20/(6+20) = 4.615 Ω

R_th = (R₁ ∥ R₂) + R₃ = 4.615 + 2 = 6.615 Ω

Find V_th, the open-circuit voltage

With a–b open, no current flows through R₃, so there is no drop across it and V_b = V_A. R₁ and R₂ form a voltage divider.

V_th = V₁·R₂/(R₁+R₂) = 18·20/(6+20) = 13.85 V

10.Thévenin & Norton

Answer: V_th = 54.23 V

Find R_th with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R₁ ∥ R₂ = 8·5/(8+5) = 3.077 Ω

R_th = (R₁ ∥ R₂) + R₃ = 3.077 + 10 = 13.08 Ω

Find V_th, the open-circuit voltage

With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.

V_A = V₁·R₂/(R₁+R₂) + I_s·(R₁∥R₂) = 1.923 + 4·3.077 = 14.23 V

V_th = V_B = V_A + I_s·R₃ = 14.23 + 4·10 = 54.23 V

ExamProblems · examproblems.com/courses/measurement-and-instrumentation

Measurement and Instrumentation: cram pack

1Formulas and methods

Wheatstone bridge & strain gauges

Method

Data acquisition & sampling

Method

Op-amp circuits

Series & parallel resistors

Method

Thévenin & Norton

Method

2Practice problems

3Solutions

Voltage across the gauge

Power

Answer

Relative resistance change

Strain

Answer

Steps needed

Bits

Answer

Harmonic frequency

Nearest multiple of the sampling rate

Alias frequency

Answer

Gain

Largest input before clipping

Answer

Input currents

Output voltage

Answer

Combine resistors from the far end

Answer

Combine resistors from the far end

Answer

Find Rth with every source switched off

Find Vth, the open-circuit voltage

Find Rth with every source switched off

Find Vth, the open-circuit voltage

Find R_th with every source switched off

Find V_th, the open-circuit voltage

Find R_th with every source switched off

Find V_th, the open-circuit voltage