IoT & Robotics crash course

Assessment

60 questions with answers, two sample papers and an answer key, mapped to the syllabus’s assessment split. Every numeric answer is computed by the course code when the page is built; short answers are model answers.

Sample papers (questions) (PDF)172 KBSample papers (questions) (Word)17 KBAnswer key (PDF)169 KBAnswer key (Word)14 KBQuestion bank with answers (PDF)185 KBQuestion bank with answers (Word)20 KB

How the syllabus weights assessment

ComponentWeight
Internal / continuous assessment30%
Laboratory20%
Mid-semester examination15%
End-semester examination25%
Capstone / project10%

The two sample papers cover the written parts (mid-semester 15%, end-semester 25%). The laboratory part is the lab track, and the capstone is Unit 12.

The sample papers

These are samples written by the course author for study. They are not past papers of any university, and the marks are a suggestion.

Question bank

Think first, then click a question to reveal the answer.

Unit 1

In the four-layer architecture of an IoT system, a soil-moisture node belongs to the: (1 mark, easy)
Answer. A. It senses the world; the network carries its data, the edge and cloud process it.
A node sends a 20-byte payload with 10 bytes of overhead every 60 s. How many bytes does it send in a day? (2 marks, medium)
Answer. (20 + 10) x 86400 / 60 = 43,200 bytes per day.
A battery of 2000 mAh (80% usable) feeds a load of 100 mA average. How long does it last? (2 marks, medium)
Answer. 2000 x 0.8 / 100 = 16 h = 0.67 days.
A bang-bang thermostat with a hysteresis band swings wider than the band when its sensor reading is delayed. Why? (3 marks, medium-hard)
Answer. After the temperature crosses the switching level the heater stays in its old state until the delayed reading arrives, so the room keeps moving during the delay. The extra swing is about the rate of change times the delay.
A room obeys dT/dt = -(T - T_out)/tau + k u with tau = 60 min, k = 0.5 degC/min, T_out = 5 degC. With the heater on, how long does it take to warm from 15 to 21 degC? (4 marks, hard)
Answer. The heater-on equilibrium is T_out + k tau = 35 degC. T(t) = 35 - (35 - 15) e^(-t/60) = 21 gives t = 60 ln(20/14) = 21.4 min.

Unit 2

A 3.3 V supply drives an LED (forward drop 2.0 V) at 10 mA. What series resistor? (1 mark, easy)
Answer. R = (3.3 - 2.0) / 0.010 = 130 ohm.
A divider of 10 kohm (top) and 20 kohm (bottom) is fed from 5 V. What is the output unloaded, and with a 20 kohm load across the bottom resistor? (2 marks, medium)
Answer. Unloaded: 5 x 20/30 = 3.333 V. Loaded: the bottom becomes 10 kohm, so 5 x 10/20 = 2.500 V.
A 12-bit ADC has a 3.3 V reference. Give the size of one step (LSB) and the RMS quantisation error of an ideal converter. (2 marks, medium)
Answer. LSB = 3.3 / 4096 = 0.806 mV; RMS error = LSB / sqrt(12) = 0.233 mV.
A node wakes for 0.2 s at 50 mA, connects for 1.3 s at 100 mA, publishes for 0.3 s at 140 mA, shuts down for 0.2 s at 50 mA and sleeps 58 s at 0.01 mA. Average current and life of a 2000 mAh battery (80% usable)? (3 marks, medium-hard)
Answer. Charge per 60 s cycle = 10 + 130 + 42 + 10 + 0.58 = 192.58 mA s; average 3.210 mA; life = 1600 / 3.210 / 24 = 20.8 days. The 2 s awake dominate.
A motor needs an average of 6 V. The pack is 7.4 V. What PWM duty is needed if the bridge drops 0.4 V, and what if the bridge drops 2 V? (4 marks, hard)
Answer. Duty = 6 / (7.4 - 0.4) = 0.857. With a 2 V drop: 6 / (7.4 - 2.0) = 1.11 > 1, so even full duty cannot give 6 V: that bridge is unsuitable.

Unit 3

A sensor reads 25.3, 25.2, 25.3, 25.2 when the true value is 20.0. It is: (1 mark, easy)
Answer. B. The readings agree with each other (precise) but are far from the truth (biased).
A soil sensor reads raw 1200 in air (0%) and 3000 in water (100%). Using a two-point linear calibration, what is the reading for raw 2100? (2 marks, medium)
Answer. slope = 100/(3000 - 1200); percent = (2100 - 1200) x slope = 50%.
A 70 Hz signal is sampled at 100 Hz. At what frequency does it appear? (2 marks, medium)
Answer. Aliased to |70 - 100| = 30 Hz.
An exponential filter is designed for a time constant of 5 s at a 1 s sample period. Give its coefficient and the factor by which it reduces white-noise variance. (3 marks, medium-hard)
Answer. alpha = 1 - e^(-1/5) = 0.1813; noise variance gain = alpha/(2 - alpha) = 0.316.
A 1320 counts-per-revolution encoder is on a wheel of radius 33 mm. What distance does one count represent, and what is the speed resolution if speed is the count change over 100 ms? (4 marks, hard)
Answer. Distance per count = 2 pi x 0.033 / 1320 = 0.1571 mm; speed step = 0.1571 mm / 0.1 s = 1.57 mm/s.

Unit 4

A mechanical button is debounced to: (1 mark, easy)
Answer. B.
How many bits does one UART byte occupy on the wire in the 8N1 format, and how long does it take at 115200 baud? (2 marks, medium)
Answer. 10 bits (start + 8 data + stop); 10/115200 = 86.8 microseconds.
Using the course's I2C model, how long does a 2-byte read take at a 100 kHz clock? (2 marks, medium)
Answer. 480 microseconds (a register read: 3 start/restart/stop clocks, 27 clocks for address+W, register and address+R, and 9 clocks per data byte: 48 clocks at 10 microseconds).
An encoder produces 7643 transitions per second and the interrupt handler takes 5 microseconds. What fraction of the CPU does it use? (3 marks, medium-hard)
Answer. 7643 x 5e-6 = 3.82% of the CPU.
A node retries a failed publish up to 3 times and each attempt fails with probability 0.3, independently. What fraction of cycles delivers? (4 marks, hard)
Answer. 1 - 0.3^3 = 0.973 (97.3%). Each extra attempt costs energy at the waiting current.

Unit 5

Which MQTT quality-of-service level guarantees delivery at least once? (1 mark, easy)
Answer. B. QoS 0 is at most once; QoS 1 at least once (duplicates possible); QoS 2 exactly once.
How many bytes is an MQTT PUBLISH at QoS 1 with the 10-byte topic 'gh/p1/soil' and a 7-byte payload (fixed header, length fields and packet id included)? (2 marks, medium)
Answer. 23 bytes (computed with the course's MQTT packet encoder).
With independent loss p = 0.1 per packet, how many packets (both directions) does one QoS 1 message cost on average, with unlimited retries? (2 marks, medium)
Answer. 1/(1-p)^2 + 1/(1-p) = 2.346 packets.
Free-space path loss at 100 m and 2.4 GHz? (3 marks, medium-hard)
Answer. 20 log10(4 pi d f / c) = 80.1 dB.
Does the subscription 'greenhouse/+/soil' receive a message published on 'greenhouse/zone1/soil/raw'? What would match both? (4 marks, hard)
Answer. No: '+' matches exactly one level, so the topic has an extra level 'raw'. 'greenhouse/#' (or 'greenhouse/+/soil/#') matches both; '#' must be the last character and also matches the parent level.

Unit 6

Compared with JSON, CBOR is mainly: (1 mark, easy)
Answer. C.
For the reading {zone: 1, soil_raw: 1832, temp_c: 21.5, batt_v: 3.71}, how many bytes does it take as compact JSON, as CBOR and as the packed binary of the course? (2 marks, medium)
Answer. JSON 54 bytes, CBOR 51 bytes, packed 7 bytes.
100 nodes each send 20 bytes every 60 s. How many bytes per day reach the cloud, and how many if each node sends one 20-byte aggregate every 600 s? (2 marks, medium)
Answer. Every 60 s: 2,880,000 bytes per day; every 600 s: 288,000 bytes per day (10 times less).
A z-score alarm at k = 3 on Gaussian noise is checked once a minute for one zone. How many false alarms per day should you expect? (3 marks, medium-hard)
Answer. P(|z| > 3) = 0.00270; times 1440 samples = 3.89 false alarms per day.
Which detector finds a small persistent shift in the mean sooner, an EWMA or a CUSUM, for the same false-alarm rate, and why? (4 marks, hard)
Answer. Both accumulate evidence; the CUSUM, designed for a shift of a chosen size, adds up every small deviation beyond a reference value and is the optimal sequential test for that shift, so it usually reacts sooner to a small, steady change. A threshold on single samples is much slower.

Unit 7

The Mirai botnet spread mainly by: (1 mark, easy)
Answer. D. It tried a short list of default username and password pairs on telnet.
How many bits of entropy has a random password of 8 lowercase letters? (2 marks, medium)
Answer. 8 x log2(26) = 37.6 bits.
At 10^9 guesses per second, how long on average to guess it? (2 marks, medium)
Answer. Expected guesses = 2^(bits-1); time = 104 s (1.7 minutes).
A command is authenticated by an HMAC truncated to 8 bytes, or to 4 bytes. What is the probability that one random guess of the tag is accepted? (3 marks, medium-hard)
Answer. 8 bytes: 2^-64 = 5.42e-20; 4 bytes: 2^-32 = 2.33e-10.
Why does the course put TLS on the gateway rather than on the battery node, and how does the node still authenticate commands? (4 marks, hard)
Answer. A TLS handshake costs the node about 1.5 kB and an extra round trip on every connection, which dominates its energy. The node instead verifies a short HMAC envelope (counter, command, 8-byte tag) with a pre-shared per-device key: 12 bytes and no extra round trips; the gateway terminates TLS toward the cloud.

Unit 8

How many degrees of freedom has a planar 2-link arm (two revolute joints)? (1 mark, easy)
Answer. A.
A 2-link arm with links 0.30 m and 0.20 m has joint angles 30 and 45 degrees. Where is the tip? (2 marks, medium)
Answer. x = 0.3 cos 30 + 0.2 cos 75 = 0.3116 m; y = 0.3 sin 30 + 0.2 sin 75 = 0.3432 m.
A differential drive with track 0.16 m has left wheel speed 0.4 m/s and right 0.6 m/s. Give the body speed and turn rate. (2 marks, medium)
Answer. v = (0.4 + 0.6)/2 = 0.50 m/s; omega = (0.6 - 0.4)/0.16 = 1.25 rad/s.
Inverse kinematics (elbow up) for the target (0.35, 0.20) m with links 0.30 and 0.20 m: give both joint angles in degrees. (3 marks, medium-hard)
Answer. q1 = 1.22 degrees, q2 = 74.29 degrees.
The right wheel of a rover is 0.5% larger than assumed (left exact), track 0.16 m, and the odometry believes it drove 20 m straight. How far is the true end point from where the odometry says it is? (4 marks, hard)
Answer. True end pose (arc): (18.77, 6.06) m; the odometry says (20, 0): the error is 6.19 m (31% of the distance).

Unit 9

A proportional-only controller on a motor speed loop leaves: (1 mark, easy)
Answer. A. The command is Kp times the error, so a nonzero command needs a nonzero error.
A plant has DC gain K = 3.491 rad/s per volt. With P control Kp = 1, what fraction of the setpoint is the steady-state error? (2 marks, medium)
Answer. 1/(1 + K Kp) = 0.2227 (22.3%).
Design a PI by pole-zero cancellation for tau = 112.5 ms, K = 3.491 and a closed-loop time constant of 30 ms. Give Kp and Ti. (2 marks, medium)
Answer. Ti = tau = 112.5 ms; Kp = tau/(K tau_c) = 1.074 V per rad/s.
For the wheel loop (10 ms sample, one sample of delay) the course finds the ultimate gain from the loop's characteristic polynomial. Give Ku (V per rad/s) and the oscillation period. (3 marks, medium-hard)
Answer. Ku = 3.368; Tu = 57.3 ms.
The rover brakes at the traction limit with mu = 0.6 from 0.5 m/s. If the reaction takes 150 ms, how far does it travel from detection to rest? (4 marks, hard)
Answer. v t + v^2/(2 mu g) = 0.5 x 0.15 + 0.25/(2 x 0.6 x 9.81) = 9.62 cm.

Unit 10

For Gaussian noise, the fraction of readings within two standard deviations of the mean is about: (1 mark, easy)
Answer. D. 68%, 95% and 99.7% for one, two and three standard deviations.
Dry soil has a 5% prior. A sensor flags 'dry' in 90% of dry cases and 8% of wet ones. What is P(dry) after one flag? (2 marks, medium)
Answer. 0.9 x 0.05 / (0.9 x 0.05 + 0.08 x 0.95) = 0.3719.
Fuse an ultrasonic range of 2.00 m (std 1 cm) with an infrared range of 2.04 m (std 3 cm). Give the fused value and its standard deviation. (2 marks, medium)
Answer. Variance = 1/(1/1e-4 + 1/9e-4) = 9.00e-05; std = 0.95 cm; mean = 2.004 m.
A scalar Kalman filter has process variance q = 1e-4 per step and measurement variance r = 0.04. What is its steady-state gain, and the equivalent EMA coefficient? (3 marks, medium-hard)
Answer. P- = (q + sqrt(q^2 + 4 q r))/2; K = P-/(P- + r) = 0.0488; an EMA with alpha = 0.049.
Why was the course's EKF overconfident about its position along the wall, though accurate across it? (4 marks, hard)
Answer. Wall ranges constrain only the distance perpendicular to the wall and the heading, so the along-wall position is corrected by nothing but odometry. The filter's model contained only random slip, not the systematic wheel-radius error, so its reported along-wall uncertainty (1.3 cm) was far below the actual error (5.3 cm): the model was incomplete.

Unit 11

Which search is guaranteed to find the path with the fewest steps on an unweighted grid? (1 mark, easy)
Answer. B.
On an 8-connected grid the goal is 7 cells right and 2 up. Give the octile heuristic. (2 marks, medium)
Answer. (7 + 2) + (sqrt 2 - 2) x 2 = 7.828 cells.
In the rover frame the pure-pursuit target is at (0.4, -0.2) m. What curvature does the rover steer at? (2 marks, medium)
Answer. kappa = 2 y / L^2 = 2(-0.2)/(0.4^2 + 0.2^2) = -2.00 per metre (turn right); omega = v kappa.
The rover brakes at mu = 0.6 from 0.5 m/s with a 15 ms edge loop. What is the longest ultrasonic ping period that still stops it within 5 cm in the worst case? (3 marks, medium-hard)
Answer. T <= (0.05 - v^2/(2 mu g))/v - 0.015 = 42.5 ms (23.5 Hz).
A watchdog stops the rover when no command arrives for 0.2 s. At 0.5 m/s (15 ms edge loop, mu = 0.6), how far does it travel after the last good command? (4 marks, hard)
Answer. v (0.2 + 0.015) + v^2/(2 mu g) = 12.9 cm.

Unit 12

In the capstone, 'coverage' means the fraction of: (1 mark, easy)
Answer. A.
MQTT QoS 1 tries a PUBLISH up to 5 times; each attempt is lost with probability 0.3. What fraction of messages is lost for good? (2 marks, medium)
Answer. 0.3^5 = 0.00243 = 0.243%.
Using the course's node power budget, how long does the 2000 mAh battery (80% usable) last if the node reads once every 300 s instead of 60 s? (2 marks, medium)
Answer. Average current = (192 + 0.01 x 298)/300 = 0.650 mA; life = 102.6 days.
Over 40 days the success rate has a mean of 99.6% and a 95% interval of +-0.8 percentage points (1.96 s/sqrt(n)). What is the standard deviation of the daily success rate? (3 marks, medium-hard)
Answer. s = 0.8 x sqrt(40)/1.96 = 2.6 percentage points.
The first end-to-end design served only 78% of dry events within 30 minutes although every component passed its tests. What was wrong and how was it fixed? (4 marks, hard)
Answer. The detector fired only when the filtered moisture crossed the dry level itself, so the delays of the 10-minute sampling, the EMA filter and the trip together used up the 30-minute deadline. Raising the detector threshold to 0.33 gives an earlier warning (a lead of about 0.15 tau); it raised the served fraction to 99.6% at the price of more water and trips. The flaw is in the composition, which no unit test sees.