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
Set the bits so they store −30 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −308 bit
U =
2.Number systemsHard
An uncompressed image is 800 × 600 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px16 bit/px1 KiB = 1024 B
M = KiB
3.Digital logic: gates, K-maps & flip-flopsMedium
Simplify F = ABC + ABC′ + A′B′C + AC′ to its minimal sum-of-products form. How many literals L does the simplified expression contain?
F = ABC + ABC′ + A′B′C + AC′
L =
4.Digital logic: gates, K-maps & flip-flopsHard
Find the minimal SOP form of F(A, B, C, D) = Σm(1, 2, 4, 8, 9, 11, 12) + d(5, 10, 13), where d marks the don't-care conditions. How many literals L does it contain?
F(A, B, C, D) = Σm(1, 2, 4, 8, 9, 11, 12) + d(5, 10, 13)
L =
5.Data acquisition & samplingMedium
A recorder samples 2 channels at 5 kHz each with a 14-bit ADC. Each sample is stored in whole bytes. Find the memory M needed for a 15 min recording, in MB (1 MB = 10⁶ bytes).
n = 2fs = 5 kHzN = 14t = 15 min
M = MB
6.Data acquisition & samplingHard
A 10-bit ADC streams every sample over a UART at 57600 baud, format 8E1 (start bit, 8 data bits, even parity, 1 stop bit). Each sample is sent as 2 whole byte(s) and the link is busy all the time. Find the highest signal frequency fmax that can be sampled without aliasing, in Hz.
baud = 57600 BdN = 108E1
fmax = Hz
3Solutions
1.Number systems
Answer: U = 226
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
30 = 0001 1110
inverted: 1110 0001
+1 → 1110 0010
Read it as unsigned
1110 0010₂ = 256 − 30 = 226
Answer
U = 226
2.Number systems
Answer: M = 937.5 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
800·600·16 = 7680000 bit = 960000 B
Convert to KiB
960000/1024 = 937.5 KiB
Answer
M = 937.5 KiB
3.Digital logic: gates, K-maps & flip-flops
Answer: L = 7
Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.
Minterms of the expression
ABC → m(7)
ABC′ → m(6)
A′B′C → m(1)
AC′ → m(4, 6)
Group and read off
F = AC′ + AB + A′B′C
Count the literals
L = 7
Answer
L = 7
4.Digital logic: gates, K-maps & flip-flops
Answer: L = 9
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.
Prime implicants
The largest groups that can be formed from 1s and Xs.
C′D, BC′, B′CD′, AC′, AB′
Minimal cover
F = C′D + BC′ + B′CD′ + AB′
L = 9
Answer
L = 9
5.Data acquisition & sampling
Answer: M = 18 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
⌈14/8⌉ = 2 B
Byte rate
2·5000·2 = 20000 B/s
Memory
M = 20000·900 = 1.8 × 10⁷ B = 18 MB
Answer
M = 18 MB
6.Data acquisition & sampling
Answer: fmax = 1309 Hz
UART: an asynchronous serial link sends each byte as a frame: one start bit (low), the data bits, an optional parity bit and one or two stop bits (high). The baud rate is the number of bits per second, so one bit lasts Tb = 1/baud and a frame of n bits lasts n/baud. The common 8N1 format has 10 bits per byte. The baud generator divides the clock: baud = fclk/(16·D) with a whole number D, so the real baud rate differs slightly from the target; an error below about 2 % is usually acceptable.
Bits per sample on the line
2·11 = 22 bit
Highest sampling rate
fs,max = 57600/22 = 2618 Hz
Nyquist limit
fmax = 2618/2 = 1309 Hz
Answer
fmax = 1309 Hz
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