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 −114 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1148 bit
U =
2.Number systemsHard
What is the largest unsigned integer N that fits in 12 bits?
12 bit
N =
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, D) = Σm(0, 1, 5, 7, 9, 10, 14, 15). How many literals L does it contain in total (a primed variable counts as one literal)?
F(A, B, C, D) = Σm(0, 1, 5, 7, 9, 10, 14, 15)
L =
4.Digital logic: gates, K-maps & flip-flopsHard
Find the minimal SOP form of F(A, B, C, D) = Σm(6, 10, 11, 12, 13) + d(0, 2, 3, 4, 8, 9, 15), where d marks the don't-care conditions. How many literals L does it contain?
F(A, B, C, D) = Σm(6, 10, 11, 12, 13) + d(0, 2, 3, 4, 8, 9, 15)
L =
5.Data acquisition & samplingMedium
A UART sends bytes at 115200 baud in the format 8N2: 8 data bits, no parity and 2 stop bit(s), plus the start bit. Find the largest number of data bytes per second it can transfer.
baud = 115200 Bd8N2
RB = B/s
6.Data acquisition & samplingHard
A temperature sensor gives 1 mV/°C (0 V at 0 °C). Its signal is amplified 5 times and fed to a 10-bit ADC with a range of 0 to 5 V (1 LSB = VFSR/2ᴺ). Find the temperature change ΔTLSB that corresponds to one LSB.
S = 1 mV/°CG = 5N = 10VFSR = 5 V
ΔTLSB = °C
3Solutions
1.Number systems
Answer: U = 142
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
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: N = 4095
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.
Largest value
212 − 1 = 4095
Answer
N = 4095
3.Digital logic: gates, K-maps & flip-flops
Answer: L = 15
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.
B′C′D
BCD
A′C′D
ACD′
A′B′C′
Minimal SOP form
F = B′C′D + BCD + A′C′D + ACD′ + A′B′C′
L = 15
Answer
L = 15
Another grouping of the same size exists; it has the same number of terms and literals.
4.Digital logic: gates, K-maps & flip-flops
Answer: L = 6
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′, B′D′, B′C, A′D′, AD, AC′, AB′
Minimal cover
F = B′C + A′D′ + AC′
L = 6
Answer
L = 6
Another grouping of the same size exists; it has the same number of terms and literals.
5.Data acquisition & sampling
Answer: RB = 10473 B/s
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 frame
1 + 8 + 0 + 2 = 11
Byte rate
115200/11 = 10473 B/s
Answer
RB = 10473 B/s
6.Data acquisition & sampling
Answer: ΔTLSB = 0.9766 °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 = 5000/1024 = 4.883 mV
Sensitivity at the ADC
1·5 = 5 mV/°C
Temperature resolution
ΔTLSB = 4.883/5 = 0.9766 °C
Answer
ΔTLSB = 0.9766 °C
The ADC reaches full scale at 1000 °C.
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