Digital Electronics: cram pack

Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.

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1Formulas and methods

Number systems

Place values: binary 2^k, octal 8^k, hex 16^k
To decimal: Σ d_k·b^k
From decimal: divide by the base, read the remainders bottom-up
Hex ↔ binary: one hex digit = 4 bits
Two’s complement: invert the bits and add 1

Digital logic: gates, K-maps & flip-flops

Basic gates: A·B, A + B, A′, (A·B)′, (A + B)′, A ⊕ B
Boolean rules: X + X′ = 1, X·X′ = 0, X + XY = X, XY + XY′ = X
De Morgan: (X·Y)′ = X′ + Y′, (X + Y)′ = X′·Y′
Karnaugh map order: 00, 01, 11, 10 (Gray code)
Group of 2ᵏ cells: loses k variables
4-to-1 multiplexer: F = A′B′I₀ + A′BI₁ + AB′I₂ + ABI₃
Flip-flops: D: Q⁺ = D, T: Q⁺ = T ⊕ Q, JK: Q⁺ = JQ′ + K′Q
Counters: n-bit: 2ⁿ states; mod-N: smallest n with 2ⁿ ≥ N; f_out = f_clk/N
Ripple counter: f_max = 1/(n·t_pd)
Ring and Johnson counters: n and 2n states

Method

For a circuit, work from the inputs to the output one gate at a time and write the value on every wire, or write its expression and fill in the truth table.
For simplification, put the 1s (and the don’t-cares as X) into a Karnaugh map with Gray-code rows and columns: 00, 01, 11, 10.
Cover the 1s with as few and as large groups of 1, 2, 4, 8 or 16 cells as possible, wrapping round the edges; each group is one product term made of the variables that stay constant. Group the 0s for the POS form.
For flip-flops and counters, read the inputs at each active clock edge, write the state after every edge, and remember that an n-bit counter wraps round after 2ⁿ states.

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.

2Practice problems

1.Number systemsMedium

Set the bits so they store −81 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −818 bit

U =

2.Number systemsHard

An uncompressed image is 800 × 600 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

800 × 600 px32 bit/px1 KiB = 1024 B

M = KiB

3.Digital logic: gates, K-maps & flip-flopsMedium

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(2, 3, 4, 5, 9, 10)

1AB = 00, CD = 00
2AB = 00, CD = 01
3AB = 00, CD = 11
4AB = 00, CD = 10
5AB = 01, CD = 00
6AB = 01, CD = 01
7AB = 01, CD = 11
8AB = 01, CD = 10
9AB = 11, CD = 00
10AB = 11, CD = 01
11AB = 11, CD = 11
12AB = 11, CD = 10
13AB = 10, CD = 00
14AB = 10, CD = 01
15AB = 10, CD = 11
16AB = 10, CD = 10

4.Digital logic: gates, K-maps & flip-flopsHard

For F(A, B, C, D) = Σm(0, 2, 4, 5, 8) + d(6, 11), where d marks the don't-care conditions, find the number N of essential prime implicants.

F(A, B, C, D) = Σm(0, 2, 4, 5, 8) + d(6, 11)

N =

5.Data acquisition & samplingMedium

A recorder samples one channel at 5 kHz with a 10-bit ADC. Each sample is stored in whole bytes. Find the memory M needed for a 10 min recording, in MB (1 MB = 10⁶ bytes).

n = 1f_s = 5 kHzN = 10t = 10 min

M = MB

6.Data acquisition & samplingHard

A temperature sensor gives 20 mV/°C (0 V at 0 °C). Its signal is amplified 10 times and fed to a 10-bit ADC with a range of 0 to 10 V (1 LSB = V_FSR/2ᴺ). Find the temperature change ΔT_LSB that corresponds to one LSB.

S = 20 mV/°CG = 10N = 10V_FSR = 10 V

ΔT_LSB = °C

3Solutions

1.Number systems

Answer: U = 175

Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.

Invert and add one

81 = 0101 0001

inverted: 1010 1110

+1 → 1010 1111

Read it as unsigned

1010 1111₂ = 256 − 81 = 175

Answer

U = 175

2.Number systems

Answer: M = 1875 KiB

Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.

Bits and bytes

800·600·32 = 15360000 bit = 1920000 B

Convert to KiB

1920000/1024 = 1875 KiB

Answer

M = 1875 KiB

3.Digital logic: gates, K-maps & flip-flops

Answer: AB = 10, CD = 01

Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m9: m8, m11, m13, m1 → 0

The isolated 1

m9 = AB′C′D → AB = 10, CD = 01

Answer

AB = 10, CD = 01

4.Digital logic: gates, K-maps & flip-flops

Answer: N = 3

Prime implicants

The largest groups that can be formed from 1s and Xs.

B′C′D′, A′D′, A′BC′

Essential prime implicants

B′C′D′, A′D′, A′BC′

N = 3

Answer

N = 3

5.Data acquisition & sampling

Answer: M = 6 MB

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).

Bytes per sample

⌈10/8⌉ = 2 B

Byte rate

1·5000·2 = 10000 B/s

Memory

M = 10000·600 = 6 × 10⁶ B = 6 MB

Answer

M = 6 MB

6.Data acquisition & sampling

Answer: ΔT_LSB = 0.04883 °C

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 = 10000/1024 = 9.766 mV

Sensitivity at the ADC

20·10 = 200 mV/°C

Temperature resolution

ΔT_LSB = 9.766/200 = 0.04883 °C

Answer

ΔT_LSB = 0.04883 °C

The ADC reaches full scale at 50 °C.

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