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 −69 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −698 bit
U =
2.Number systemsHard
Set the bits to the 8-bit result of 0101 1000 XOR 0001 1111. What is the result's value R in decimal?
0101 1000XOR0001 1111
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) = Σm(0, 1, 2, 3, 6). How many product terms N does it have?
F(A, B, C) = Σm(0, 1, 2, 3, 6)
N =
4.Digital logic: gates, K-maps & flip-flopsHard
A 4-bit binary down counter starts at 3 (decimal, Q = 0011). Which count Q (in decimal) does it show after 31 clock pulses?
Q0 = 331n = 4
Q =
5.Data acquisition & samplingMedium
A recorder samples 2 channels at 20 kHz each 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 = 2fs = 20 kHzN = 10t = 10 min
M = MB
6.Data acquisition & samplingHard
A temperature sensor gives 5 mV/°C (0 V at 0 °C). Its signal is amplified 10 times and fed to a 12-bit ADC with a range of 0 to 3.3 V (1 LSB = VFSR/2ᴺ). Find the temperature change ΔTLSB that corresponds to one LSB.
S = 5 mV/°CG = 10N = 12VFSR = 3.3 V
ΔTLSB = °C
3Solutions
1.Number systems
Answer: U = 187
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
69 = 0100 0101
inverted: 1011 1010
+1 → 1011 1011
Read it as unsigned
1011 1011₂ = 256 − 69 = 187
Answer
U = 187
2.Number systems
Answer: R = 71
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
XOR gives 1 where the two bits differ.
0101 1000
XOR 0001 1111
= 0100 0111
Value of the result
0100 0111₂ = 71
Answer
R = 71
3.Digital logic: gates, K-maps & flip-flops
Answer: N = 2
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.
BC′
A′
Minimal SOP form
F = BC′ + A′
N = 2
Answer
N = 2
4.Digital logic: gates, K-maps & flip-flops
Answer: Q = 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.
Full cycles
Every 16 pulses bring the counter back to the same state.
31 = 1·16 + 15
Count after the remaining pulses
Q = (3 − 15) mod 16 = 4
Q = 0100₂
Answer
Q = 4
5.Data acquisition & sampling
Answer: M = 48 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
⌈10/8⌉ = 2 B
Byte rate
2·20000·2 = 80000 B/s
Memory
M = 80000·600 = 4.8 × 10⁷ B = 48 MB
Answer
M = 48 MB
6.Data acquisition & sampling
Answer: ΔTLSB = 0.01611 °C
ADC resolution: an N-bit converter divides its full-scale range VFSR into 2ᴺ codes, from 0 to 2ᴺ − 1. One step is 1 LSB = VFSR/2ᴺ: the smallest change the converter can show. A truncating ADC gives the code ⌊Vin/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 = 3300/4096 = 0.8057 mV
Sensitivity at the ADC
5·10 = 50 mV/°C
Temperature resolution
ΔTLSB = 0.8057/50 = 0.01611 °C
Answer
ΔTLSB = 0.01611 °C
The ADC reaches full scale at 66 °C.
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