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 −113 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1138 bit
U =
2.Number systemsHard
Set the bits to the 8-bit result of 0011 1011 >> 3. What is the result's value R in decimal?
0011 1011>> 3
R =
3.Digital logic: gates, K-maps & flip-flopsMedium
Simplify F = AB′C + AC + A′C + A′BC to its minimal sum-of-products form. How many literals L does the simplified expression contain?
F = AB′C + AC + A′C + A′BC
L =
4.Digital logic: gates, K-maps & flip-flopsHard
Find the minimal SOP form of F(A, B, C, D) = Σm(1, 2, 3, 5, 8, 13) + d(6, 9, 14), where d marks the don't-care conditions. How many literals L does it contain?
F(A, B, C, D) = Σm(1, 2, 3, 5, 8, 13) + d(6, 9, 14)
L =
5.Data acquisition & samplingMedium
A recorder samples one channel at 10 kHz with a 12-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 = 1fs = 10 kHzN = 12t = 15 min
M = MB
6.Data acquisition & samplingHard
A 2.4 kHz square wave is sampled at 8 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 Hz.
f₁ = 2.4 kHzfs = 8 kHzharmonic: 5f₁
fa = Hz
3Solutions
1.Number systems
Answer: U = 143
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
113 = 0111 0001
inverted: 1000 1110
+1 → 1000 1111
Read it as unsigned
1000 1111₂ = 256 − 113 = 143
Answer
U = 143
2.Number systems
Answer: R = 7
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 3 moves every bit 3 places to the right; bits that fall off the right end are lost and zeros come in on the left.
0011 1011 >> 3
= 0000 0111
Value of the result
0000 0111₂ = 7
Answer
R = 7
3.Digital logic: gates, K-maps & flip-flops
Answer: L = 1
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
AB′C → m(5)
AC → m(5, 7)
A′C → m(1, 3)
A′BC → m(3)
Group and read off
F = C
Count the literals
L = 1
Answer
L = 1
4.Digital logic: gates, K-maps & flip-flops
Answer: L = 8
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, A′CD′, A′B′D, A′B′C, AB′C′
Minimal cover
F = C′D + A′B′C + AB′C′
L = 8
Answer
L = 8
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
⌈12/8⌉ = 2 B
Byte rate
1·10000·2 = 20000 B/s
Memory
M = 20000·900 = 1.8 × 10⁷ B = 18 MB
Answer
M = 18 MB
6.Data acquisition & sampling
Answer: fa = 4000 Hz
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 = 5·2.4 kHz = 12 kHz
Nearest multiple of the sampling rate
m = round(1.5) = 2
Alias frequency
fa = |12000 − 2·8000| = 4000 Hz
Answer
fa = 4000 Hz
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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