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 −47 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −478 bit
U =
2.Number systemsHard
A system must give every one of 2723 items its own binary code of the same length. What is the smallest number of bits b needed?
2723
b = bit
3.Digital logic: gates, K-maps & flip-flopsMedium
Simplify F = A′C′ + BC′ + BC + AB to its minimal sum-of-products form. How many literals L does the simplified expression contain?
F = A′C′ + BC′ + BC + AB
L =
4.Digital logic: gates, K-maps & flip-flopsHard
Find the minimal SOP form of F(A, B, C, D) = Σm(1, 4, 8, 14) + d(2, 10, 11, 13), where d marks the don't-care conditions. How many literals L does it contain?
F(A, B, C, D) = Σm(1, 4, 8, 14) + d(2, 10, 11, 13)
L =
5.Data acquisition & samplingMedium
A 10-bit ADC with an input range of 0 to 5 V truncates: its output code is ⌊Vin/LSB⌋ with 1 LSB = VFSR/2ᴺ. Find the decimal output code D for an input of 2.535 V.
N = 10VFSR = 5 VVin = 2.535 V
D =
6.Data acquisition & samplingHard
A temperature sensor gives 1 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 = 1 mV/°CG = 2.5N = 10VFSR = 10 V
ΔTLSB = °C
3Solutions
1.Number systems
Answer: U = 209
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
47 = 0010 1111
inverted: 1101 0000
+1 → 1101 0001
Read it as unsigned
1101 0001₂ = 256 − 47 = 209
Answer
U = 209
2.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 2723
212 = 4096 ≥ 2723
Answer
b = 12
3.Digital logic: gates, K-maps & flip-flops
Answer: L = 3
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
A′C′ → m(0, 2)
BC′ → m(2, 6)
BC → m(3, 7)
AB → m(6, 7)
Group and read off
F = B + A′C′
Count the literals
L = 3
Answer
L = 3
4.Digital logic: gates, K-maps & flip-flops
Answer: L = 14
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.
ACD′, AB′D′, A′B′C′D, A′BC′D′
Minimal cover
F = ACD′ + AB′D′ + A′B′C′D + A′BC′D′
L = 14
Answer
L = 14
5.Data acquisition & sampling
Answer: D = 519
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 = 5/210 = 4.883 mV
Code
Vin/LSB = 2535/4.883 = 519.2
code = ⌊519.2⌋ = 519
Answer
D = 519
In binary that is 1000000111.
6.Data acquisition & sampling
Answer: ΔTLSB = 3.906 °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
1·2.5 = 2.5 mV/°C
Temperature resolution
ΔTLSB = 9.766/2.5 = 3.906 °C
Answer
ΔTLSB = 3.906 °C
The ADC reaches full scale at 4000 °C.
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