Measurement and Instrumentation: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

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

  1. Treat the bridge as two voltage dividers across the same supply.
  2. For balance, set R₁/R₂ = R₃/R₄ and solve for the unknown resistor.
  3. For an unbalanced bridge, find Va and Vb from the dividers and subtract; use the Thévenin equivalent when a meter loads the output.
  4. 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.

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

  1. Write down what is given: sampling rate, signal frequencies, number of bits, input range, baud rate.
  2. For sampling questions, compare each frequency with fs/2; a higher frequency appears at fa = |f − k·fs| with the nearest multiple of fs.
  3. For resolution questions, find 1 LSB = VFSR/2ᴺ, or the smallest N with 2ᴺ ≥ span/resolution.
  4. 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
Vout = −(Rf/Rin)·Vin
Non-inverting amplifier
Vout = (1 + Rf/R1)·Vin
Summing amplifier
Vout = −Rf(V1/R1 + V2/R2)
Ideal op-amp
V+ = V−, Iin = 0

Series & parallel resistors

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

Method

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

Thévenin & Norton

Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)

Method

  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.

2Practice problems

1.Wheatstone bridge & strain gaugesMedium
+−6 V400 Ω400 Ω + ΔR400 Ω400 Ωab+−Vab​ = ?

A bridge has four 400 Ω arms and Vs = 6 V. The sensor R₂ increases by ΔR = 2 Ω. Using the small-change approximation, find the output Vab in mV.

Vs = 6 VR = 400 ΩΔR = 2 Ω
Vab = mV
2.Wheatstone bridge & strain gaugesHard
+−10 V350 Ω − ΔR350 Ω + ΔR350 Ω + ΔR350 Ω − ΔRab+−Vo​ = ?

A load cell (a full bridge of strain gauges) has a rated output of 3 mV/V at its rated load of 50 kg. It is excited with 10 V. Find its output Vo under a load of 12.5 kg.

3 mV/VFrated = 50 kgVs = 10 VF = 12.5 kg
Vo = mV
3.Data acquisition & samplingMedium
Vin​N = 125 V2ᴺ − 101 LSB3.851 V?

A 12-bit ADC with an input range of 0 to 5 V truncates: its output code is ⌊Vin/LSB⌋ with 1 LSB = VFSR/2ᴺ. Find the decimal output code D for an input of 3.851 V.

N = 12VFSR = 5 VVin = 3.851 V
D =
4.Data acquisition & samplingHard
Vin​N = ?VFSR​2ᴺ − 101 LSB

An audio measurement needs an ideal signal-to-quantisation-noise ratio of at least 65 dB for a full-scale sine wave. Find the smallest number of ADC bits N.

SQNR ≥ 65 dB
N =
5.Op-amp circuitsMedium
−+Vout​R2​ = 47 kΩV1​ = 2.4 VR1​ = 10 kΩV2​ = 1.1 VR3​ = 10 kΩR4​ = 47 kΩ

The difference amplifier uses an ideal op-amp with ±15 V supplies, with R3 = R1 and R4 = R2. Find the output voltage Vout.

V1 = 2.4 VV2 = 1.1 VR1 = R3 = 10 kΩR2 = R4 = 47 kΩ
Vout = V
6.Op-amp circuitsHard
−+Vout​Rf​ = 100 kΩV1​ = 0.3 VR1​ = 10 kΩV2​ = 1.3 VR2​ = 47 kΩIf​

The summing amplifier uses an ideal op-amp with ±15 V supplies. Find the current If through the feedback resistor, flowing towards the output.

V1 = 0.3 VV2 = 1.3 VR1 = 10 kΩR2 = 47 kΩRf = 100 kΩ
If = mA
7.Series & parallel resistorsMedium
R1​ = 15 ΩR3​ = 2 Ω+−V1​12 VR2​8 ΩIR2​AR4​6 ΩB

Find the current IR2 through R2.

V1 = 12 VR1 = 15 ΩR2 = 8 ΩR3 = 2 ΩR4 = 6 Ω
IR2 = A
8.Series & parallel resistorsHard
R1​ = 20 ΩR3​ = 3 ΩR5​ = 12 Ω+−V1​18 VR2​4 ΩIR2​AR4​20 ΩBR6​5 Ω

Find the current IR2 through R2.

V1 = 18 VR1 = 20 ΩR2 = 4 ΩR3 = 3 ΩR4 = 20 ΩR5 = 12 ΩR6 = 5 Ω
IR2 = A
9.Thévenin & NortonMedium
R1​ = 5 ΩR3​ = 2 Ω+−V1​12 VR2​20 ΩAab

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

V1 = 12 VR1 = 5 ΩR2 = 20 ΩR3 = 2 Ω
Vth = V
10.Thévenin & NortonHard
R1​ = 5 ΩR3​ = 3 Ω+−V1​24 VR2​4 ΩAIs​4 ABRL​10 ΩIL​

A load RL = 10 Ω is connected across a–b. Replace the rest of the circuit with its Thévenin equivalent and find the load current IL.

V1 = 24 VIs = 4 AR1 = 5 ΩR2 = 4 ΩR3 = 3 ΩRL = 10 Ω
IL = A

3Solutions

1.Wheatstone bridge & strain gauges

Answer: Vab = 7.5 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.

Relative change

ΔR/R = 2/400 = 0.005

Linearised output

Vab ≈ 6·0.005/4 = 7.5 mV

Answer

Vab = 7.5 mV

The exact value is 7.481 mV; the approximation is good while ΔR/R is small.

2.Wheatstone bridge & strain gauges

Answer: Vo = 7.5 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.

Full-scale output

3·10 = 30 mV

Output at this load

Vo = 30·12.5/50 = 7.5 mV

Answer

Vo = 7.5 mV
3.Data acquisition & sampling

Answer: D = 3154

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.

Step size

1 LSB = 5/212 = 1.221 mV

Code

Vin/LSB = 3851/1.221 = 3155
code = ⌊3155⌋ = 3154

Answer

D = 3154

In binary that is 110001010010.

4.Data acquisition & sampling

Answer: N = 11

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 ≥ (65 − 1.76)/6.02 = 10.5

Round up

N = 11
check: 6.02·11 + 1.76 = 67.98 dB ≥ 65 dB

Answer

N = 11
5.Op-amp circuits

Answer: Vout = −6.11 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.

Voltage at the inputs

R₃ and R₄ carry the same current, so they form a voltage divider; feedback makes V₋ = V₊.

V₊ = V2·R4/(R3 + R4) = 1.1·47/(10 + 47) = 0.907 V
V₋ = V₊

KCL at the inverting input

(2.4 − 0.907)/10 = (0.907 − Vout)/47
Vout = (R2/R1)·(V2 − V1) = (47/10)·(1.1 − 2.4) = −6.11 V

Answer

Vout = −6.11 V
6.Op-amp circuits

Answer: If = 0.05766 mA

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.

Input currents

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

I1 = V1/R1 = 0.3/10 = 0.03 mA
I2 = V2/R2 = 1.3/47 = 0.02766 mA
If = I1 + I2 = 0.05766 mA

Answer

If = 0.05766 mA
7.Series & parallel resistors

Answer: IR2 = 0.3158 A

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.

R1​ = 15 ΩR3​ = 2 Ω+−V1​12 VR2​8 ΩAR4​6 ΩB

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 = 2 + 6 = 8 Ω
R2 ∥ 8 = 8·8/(8+8) = 4 Ω
Req = R1 + 4 = 15 + 4 = 19 Ω

Current from the source

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

IR1 = V1/Req = 12/19 = 0.6316 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

VA = V1 − IR1·R1 = 12 − 0.6316·15 = 2.526 V
IR2 = VA/R2 = 2.526/8 = 0.3158 A

Answer

IR2 = 0.3158 A
8.Series & parallel resistors

Answer: IR2 = 0.5889 A

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.

R1​ = 20 ΩR3​ = 3 ΩR5​ = 12 Ω+−V1​18 VR2​4 ΩAR4​20 ΩBR6​5 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 = 12 + 5 = 17 Ω
R4 ∥ 17 = 9.189 Ω
R3 + 9.189 = 12.19 Ω
R2 ∥ 12.19 = 3.012 Ω
Req = R1 + 3.012 = 20 + 3.012 = 23.01 Ω

Current from the source

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

IR1 = V1/Req = 18/23.01 = 0.7822 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

VA = V1 − IR1·R1 = 18 − 0.7822·20 = 2.356 V
IR2 = VA/R2 = 2.356/4 = 0.5889 A

Answer

IR2 = 0.5889 A
9.Thévenin & Norton

Answer: Vth = 9.6 V

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

R1​ = 5 ΩR3​ = 2 ΩSHORTR2​20 ΩAabRth​

Find Rth 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 = 5·20/(5+20) = 4 Ω
Rth = (R1 ∥ R2) + R3 = 4 + 2 = 6 Ω
R1​ = 5 ΩR3​ = 2 Ω+−V1​12 VR2​20 ΩAab+−Vth​

Find Vth, 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.

Vth = V1·R2/(R1+R2) = 12·20/(5+20) = 9.6 V
10.Thévenin & Norton

Answer: IL = 2.073 A

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

R1​ = 5 ΩR3​ = 3 ΩSHORTR2​4 ΩAOPENBabRth​

Find Rth 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 = 5·4/(5+4) = 2.222 Ω
Rth = (R1 ∥ R2) + R3 = 2.222 + 3 = 5.222 Ω
R1​ = 5 ΩR3​ = 3 Ω+−V1​24 VR2​4 ΩAIs​4 ABab+−Vth​

Find Vth, 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.

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 10.67 + 4·2.222 = 19.56 V
Vth = VB = VA + Is·R3 = 19.56 + 4·3 = 31.56 V
Rth​ = 5.222 Ω+−Vth​31.56 VRL​10 ΩIL​

Connect the load to the Thévenin equivalent

The Thévenin source and R_th in series behave exactly like the original circuit at a–b, so the load sees a simple series loop.

IL = Vth/(Rth + RL) = 31.56/(5.222 + 10) = 2.073 A

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