Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- 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 Electronics: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
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; fout = fclk/N
- Ripple counter
- fmax = 1/(n·tpd)
- 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
- 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
- Write down what is given: sampling rate, signal frequencies, number of bits, input range, baud rate.
- For sampling questions, compare each frequency with fs/2; a higher frequency appears at fa = |f − k·fs| with the nearest multiple of fs.
- For resolution questions, find 1 LSB = VFSR/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
Write −119 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1198 bit
U =
2.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px24 bit/px1 KiB = 1024 B
M = KiB
3.Digital logic: gates, K-maps & flip-flopsMedium
Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 6, 7, 8, 10, 11, 14, 15). How many sum terms (factors) N does it have?
F(A, B, C, D) = Σm(0, 6, 7, 8, 10, 11, 14, 15)
N =
4.Digital logic: gates, K-maps & flip-flopsHard
A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.
Q₀ = 0JK
- 1clock edge 1
- 2clock edge 2
- 3clock edge 3
- 4clock edge 4
- 5clock edge 5
- 6clock edge 6
- 7clock edge 7
- 8clock edge 8
5.Data acquisition & samplingMedium
A recorder samples 4 channels at 50 kHz each with a 24-bit ADC. Each sample is stored in whole bytes. Find the memory M needed for a 5 min recording, in MB (1 MB = 10⁶ bytes).
n = 4fs = 50 kHzN = 24t = 5 min
M = MB
6.Data acquisition & samplingHard
A 3 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 fa does its harmonic 5f₁ appear in the sampled data? Give it in kHz.
f₁ = 3 kHzfs = 20 kHzharmonic: 5f₁
fa = kHz
3Solutions
1.Number systems
Answer: U = 137
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.
Invert and add one
119 = 0111 0111
inverted: 1000 1000
+1 → 1000 1001
Read it as unsigned
1000 1001₂ = 256 − 119 = 137
Answer
U = 137
2.Number systems
Answer: M = 2304 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
1024·768·24 = 18874368 bit = 2359296 B
Convert to KiB
2359296/1024 = 2304 KiB
Answer
M = 2304 KiB
3.Digital logic: gates, K-maps & flip-flops
Answer: N = 3
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 0s
Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.
(C + D′)
(B′ + C)
(A + B + C′)
Minimal POS form
F = (C + D′)(B′ + C)(A + B + C′)
N = 3
Answer
N = 3
4.Digital logic: gates, K-maps & flip-flops
Answer: clock edge 4
Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.
Q after every edge
A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.
1: J = 1, K = 0 → Q = 1
2: J = 1, K = 0 → Q = 1
3: J = 0, K = 0 → Q = 1
4: J = 1, K = 1 → Q = 0
5: J = 0, K = 0 → Q = 0
6: J = 1, K = 0 → Q = 1
7: J = 0, K = 1 → Q = 0
8: J = 1, K = 0 → Q = 1
First change from 1 to 0
clock edge 4
Answer
clock edge 4
5.Data acquisition & sampling
Answer: M = 180 MB
Data rates: every channel delivers fs samples per second of N bits, so the raw rate is channels × fs × 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 × fs × 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
⌈24/8⌉ = 3 B
Byte rate
4·50000·3 = 600000 B/s
Memory
M = 600000·300 = 1.8 × 10⁸ B = 180 MB
Answer
M = 180 MB
6.Data acquisition & sampling
Answer: fa = 5 kHz
Sampling: an ADC measures the signal fs times per second. By the Nyquist–Shannon theorem a frequency f is captured correctly only while f < fs/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 fa = |f − k·fs|, with the whole number k that brings the result between 0 and fs/2. The samples cannot tell the two apart, so an analogue anti-aliasing filter must remove everything above fs/2 before the ADC.
Harmonic frequency
5f₁ = 5·3 kHz = 15 kHz
Nearest multiple of the sampling rate
m = round(0.75) = 1
Alias frequency
fa = |15 − 1·20| = 5 kHz
Answer
fa = 5 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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