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

The 8-bit pattern 1100 0101 stores a signed integer in two's complement. What is its value N in decimal?

1100 01018 bit

N =

2.Number systemsHard

Set the bits to the 8-bit result of 0110 1011 >> 2. What is the result's value R in decimal?

0110 1011>> 2

R =

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

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 1, 2, 7, 8, 13, 14). How many product terms N does it have?

F(A, B, C, D) = Σm(0, 1, 2, 7, 8, 13, 14)

N =

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

Find the minimal SOP form of F(A, B, C, D) = Σm(3, 4, 9, 13) + d(0, 5, 10, 11), where d marks the don't-care conditions. How many literals L does it contain?

F(A, B, C, D) = Σm(3, 4, 9, 13) + d(0, 5, 10, 11)

L =

5.Data acquisition & samplingMedium

A 10-bit ADC with an input range of 0 to 3.3 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 2.026 V.

N = 10V_FSR = 3.3 VV_in = 2.026 V

D =

6.Data acquisition & samplingHard

A 2 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 7f₁ appear in the sampled data? Give it in kHz.

f₁ = 2 kHzf_s = 20 kHzharmonic: 7f₁

f_a = kHz

3Solutions

1.Number systems

Answer: N = −59

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.

Add the place values

The other bits keep their normal positive values.

−128 + 64 + 4 + 1 = −59

Answer

N = −59

2.Number systems

Answer: R = 26

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.

Bit by bit

Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.

0110 1011 >> 2

= 0001 1010

Value of the result

0001 1010₂ = 26

Answer

R = 26

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

Answer: N = 6

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.

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

B′C′D′

A′B′D′

A′B′C′

A′BCD

ABC′D

ABCD′

Minimal SOP form

F = B′C′D′ + A′B′D′ + A′B′C′ + A′BCD + ABC′D + ABCD′

N = 6

Answer

N = 6

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

Answer: L = 9

Prime implicants

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

B′CD, BC′D, A′C′D′, AC′D, AB′D, A′BC′

Minimal cover

F = B′CD + A′C′D′ + AC′D

L = 9

Answer

L = 9

Another grouping of the same size exists; it has the same number of terms and literals.

5.Data acquisition & sampling

Answer: D = 628

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 = 3.3/2¹⁰ = 3.223 mV

Code

V_in/LSB = 2026/3.223 = 628.7

code = ⌊628.7⌋ = 628

Answer

D = 628

In binary that is 1001110100.

6.Data acquisition & sampling

Answer: f_a = 6 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

7f₁ = 7·2 kHz = 14 kHz

Nearest multiple of the sampling rate

m = round(0.7) = 1

Alias frequency

f_a = |14 − 1·20| = 6 kHz

Answer

f_a = 6 kHz

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

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