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 −114 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1148 bit
U =
2.Number systemsHard
Set the bits to the 8-bit result of 0110 0001 << 3. What is the result's value R in decimal?
0110 0001<< 3
R =
3.Digital logic: gates, K-maps & flip-flopsMedium
In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.
F = Σm(0, 3, 5, 6, 7, 12, 14)
- 1AB = 00, CD = 00
- 2AB = 00, CD = 01
- 3AB = 00, CD = 11
- 4AB = 00, CD = 10
- 5AB = 01, CD = 00
- 6AB = 01, CD = 01
- 7AB = 01, CD = 11
- 8AB = 01, CD = 10
- 9AB = 11, CD = 00
- 10AB = 11, CD = 01
- 11AB = 11, CD = 11
- 12AB = 11, CD = 10
- 13AB = 10, CD = 00
- 14AB = 10, CD = 01
- 15AB = 10, CD = 11
- 16AB = 10, CD = 10
4.Digital logic: gates, K-maps & flip-flopsHard
Find the minimal SOP form of F(A, B, C, D) = Σm(4, 6, 7, 10, 11, 14) + d(0, 9, 12), where d marks the don't-care conditions. How many product terms N does it have?
F(A, B, C, D) = Σm(4, 6, 7, 10, 11, 14) + d(0, 9, 12)
N =
5.Data acquisition & samplingMedium
Find the ideal signal-to-quantisation-noise ratio SQNR of an 18-bit ADC for a full-scale sine wave.
N = 18
SQNR = dB
6.Data acquisition & samplingHard
A temperature sensor gives 10 mV/°C (0 V at 0 °C). Its signal is amplified 2.5 times and fed to a 10-bit ADC with a range of 0 to 10 V (1 LSB = VFSR/2ᴺ). Find the temperature change ΔTLSB that corresponds to one LSB.
S = 10 mV/°CG = 2.5N = 10VFSR = 10 V
ΔTLSB = °C
3Solutions
1.Number systems
Answer: U = 142
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
114 = 0111 0010
inverted: 1000 1101
+1 → 1000 1110
Read it as unsigned
1000 1110₂ = 256 − 114 = 142
Answer
U = 142
2.Number systems
Answer: R = 8
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 left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0110 0001 << 3
= 0000 1000
Value of the result
0000 1000₂ = 8
Answer
R = 8
3.Digital logic: gates, K-maps & flip-flops
Answer: AB = 00, CD = 00
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.
Neighbours of each 1
A 1 with a neighbouring 1 can join a group of two or more.
m0: m1, m2, m4, m8 → 0
The isolated 1
m0 = A′B′C′D′ → AB = 00, CD = 00
Answer
AB = 00, CD = 00
4.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.
Prime implicants
The largest groups that can be formed from 1s and Xs.
BD′, A′C′D′, ACD′, AB′D, A′BC, AB′C
Minimal cover
F = BD′ + A′BC + AB′C
N = 3
Answer
N = 3
5.Data acquisition & sampling
Answer: SQNR = 110.1 dB
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.
Ideal SQNR
SQNR = 6.02·18 + 1.76 = 110.1 dB
Answer
SQNR = 110.1 dB
6.Data acquisition & sampling
Answer: ΔTLSB = 0.3906 °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 = 10000/1024 = 9.766 mV
Sensitivity at the ADC
10·2.5 = 25 mV/°C
Temperature resolution
ΔTLSB = 9.766/25 = 0.3906 °C
Answer
ΔTLSB = 0.3906 °C
The ADC reaches full scale at 400 °C.
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