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 galvanometer with a resistance of 50 Ω is connected between a and b of the bridge shown (V_s = 2 V). Find the galvanometer current I_g from a to b, in µA.

V_s = 2 VR₁ = 500 ΩR₂ = 20 ΩR₃ = 150 ΩR₄ = 40 ΩR_g = 50 Ω

I_g = µA

2.Wheatstone bridge & strain gaugesHard

A quarter bridge (one active gauge, GF = 2.12) is excited with 2.5 V. The largest strain to be measured is 1000 µε, and it must give 5 V at the input of an analogue-to-digital converter. Find the amplifier gain G.

V_s = 2.5 VGF = 2.12ε_max = 1000 µεV_FS = 5 V

G =

3.Data acquisition & samplingMedium

An 8-bit ADC with an input range of 0 to 2.5 V truncates: its output code is ⌊V_in/LSB⌋ with 1 LSB = V_FSR/2ᴺ. Find the decimal output code D for an input of 1.844 V.

N = 8V_FSR = 2.5 VV_in = 1.844 V

D =

4.Data acquisition & samplingHard

A 16-channel multiplexer feeds one ADC. Each conversion takes 5 µs, and after every channel switch the input needs 5 µs to settle. The channels are scanned in turn without pause. Find the sampling rate f_s of each channel, in kHz.

n = 16t_c = 5 µst_set = 5 µs

f_s = kHz

5.Op-amp circuitsMedium

The op-amp is ideal and runs from ±15 V supplies; its output saturates at ±14 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 = 4.7 kΩR_f = 10 kΩV_sat = ±14 V

V_in,max = V

6.Op-amp circuitsHard

The op-amp is ideal and runs from ±15 V supplies. The resistor ratios are not matched. Find the output voltage V_out.

V₁ = −0.5 VV₂ = −0.7 VR₁ = 1 kΩR₂ = 100 kΩR₃ = 47 kΩR₄ = 100 kΩ

V_out = V

7.Series & parallel resistorsMedium

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 12 VR₁ = 3 ΩR₂ = 8 ΩR₃ = 12 ΩR₄ = 4 Ω

R_eq = Ω

8.Series & parallel resistorsHard

Find the current I_R1 delivered by the source, which flows through R₁.

V₁ = 9 VR₁ = 15 ΩR₂ = 30 ΩR₃ = 25 ΩR₄ = 10 ΩR₅ = 4 ΩR₆ = 12 Ω

I_R1 = A

9.Thévenin & NortonMedium

Find the Thévenin resistance R_th seen from terminals a–b.

V₁ = 6 VR₁ = 10 ΩR₂ = 3 ΩR₃ = 4 Ω

R_th = Ω

10.Thévenin & NortonHard

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

V₁ = 36 VI_s = 4 AR₁ = 6 ΩR₂ = 20 ΩR₃ = 3 Ω

V_th = V

3Solutions

1.Wheatstone bridge & strain gauges

Answer: I_g = −3414 µA

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.

Thévenin voltage

V_th = 2·(20/520 − 40/190) = −0.3441 V

Thévenin resistance

R_th = R₁∥R₂ + R₃∥R₄ = 50.81 Ω

Galvanometer current

I_g = −0.3441/(50.81 + 50) = −3414 µA

Answer

I_g = −3414 µA

2.Wheatstone bridge & strain gauges

Answer: G = 3774

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.

Bridge output at full scale

V_o = 1.325 mV

Gain

G = 5/0.001325 = 3774

Answer

G = 3774

3.Data acquisition & sampling

Answer: D = 188

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.

Step size

1 LSB = 2.5/2⁸ = 9.766 mV

Code

V_in/LSB = 1844/9.766 = 188.8

code = ⌊188.8⌋ = 188

Answer

D = 188

In binary that is 10111100.

4.Data acquisition & sampling

Answer: f_s = 6.25 kHz

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).

Time per channel

5 + 5 = 10 µs

Time per scan

16·10 = 160 µs

Sampling rate per channel

f_s = 1/160 µs = 6.25 kHz

Answer

f_s = 6.25 kHz

5.Op-amp circuits

Answer: V_in,max = 6.58 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 = 10/4.7 = 2.128

Largest input before clipping

V_in,max = V_sat/|A_v| = 14/2.128 = 6.58 V

Answer

V_in,max = 6.58 V

6.Op-amp circuits

Answer: V_out = 1.905 V

Voltage at the + input

V₊ = V₂·R₄/(R₃ + R₄) = −0.7·100/(47 + 100) = −0.4762 V

Superpose the two inputs

With V₂ = 0, V₁ sees an inverting amplifier. With V₁ = 0, V₊ is amplified by the non-inverting gain 1 + R₂/R₁.

V_out = −(100/1)·(−0.5) + (1 + 100/1)·(−0.4762) = 50 + (−48.1) = 1.905 V

Answer

V_out = 1.905 V

7.Series & parallel resistors

Answer: R_eq = 8.333 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 = 12 + 4 = 16 Ω

R₂ ∥ 16 = 8·16/(8+16) = 5.333 Ω

R_eq = R₁ + 5.333 = 3 + 5.333 = 8.333 Ω

Answer

R_eq = 8.333 Ω

8.Series & parallel resistors

Answer: I_R1 = 0.2972 A

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 + 12 = 16 Ω

R₄ ∥ 16 = 6.154 Ω

R₃ + 6.154 = 31.15 Ω

R₂ ∥ 31.15 = 15.28 Ω

R_eq = R₁ + 15.28 = 15 + 15.28 = 30.28 Ω

Current from the source

The source sees only R_eq, so Ohm’s law gives the current it delivers. That current flows through R₁.

I_R1 = V₁/R_eq = 9/30.28 = 0.2972 A

Answer

I_R1 = 0.2972 A

9.Thévenin & Norton

Answer: R_th = 6.308 Ω

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 = 10·3/(10+3) = 2.308 Ω

R_th = (R₁ ∥ R₂) + R₃ = 2.308 + 4 = 6.308 Ω

10.Thévenin & Norton

Answer: V_th = 58.15 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 = 6·20/(6+20) = 4.615 Ω

R_th = (R₁ ∥ R₂) + R₃ = 4.615 + 3 = 7.615 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) = 27.69 + 4·4.615 = 46.15 V

V_th = V_B = V_A + I_s·R₃ = 46.15 + 4·3 = 58.15 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

Thévenin voltage

Thévenin resistance

Galvanometer current

Answer

Bridge output at full scale

Gain

Answer

Step size

Code

Answer

Time per channel

Time per scan

Sampling rate per channel

Answer

Gain

Largest input before clipping

Answer

Voltage at the + input

Superpose the two inputs

Answer

Combine resistors from the far end

Answer

Combine resistors from the far end

Current from the source

Answer

Find Rth with every source switched off

Find Rth with every source switched off

Find Vth, the open-circuit voltage

Find R_th with every source switched off

Find R_th with every source switched off

Find V_th, the open-circuit voltage