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 −53 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −538 bit
U =
2.Number systemsHard
An uncompressed image is 512 × 512 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px16 bit/px1 KiB = 1024 B
M = KiB
3.Digital logic: gates, K-maps & flip-flopsMedium
Simplify F = B′C + A′BC′ + ABC to its minimal sum-of-products form. How many literals L does the simplified expression contain?
F = B′C + A′BC′ + ABC
L =
4.Digital logic: gates, K-maps & flip-flopsHard
A 5-bit ripple (asynchronous) counter is built from flip-flops with a propagation delay of 10 ns each. All outputs must have settled before the next clock edge. Find the maximum clock frequency fmax.
n = 5tpd = 10 ns
fmax = MHz
5.Data acquisition & samplingMedium
A 5 kHz sine wave is sampled at 4 kHz without an anti-aliasing filter. At which frequency fa does it appear in the samples? Give it in Hz.
f = 5 kHzfs = 4 kHz
fa = Hz
6.Data acquisition & samplingHard
A 12 kHz square wave is sampled at 50 kHz with no anti-aliasing filter. A square wave contains the odd harmonics 3f₁, 5f₁, 7f₁, … At which frequency fa does its harmonic 7f₁ appear in the sampled data? Give it in kHz.
f₁ = 12 kHzfs = 50 kHzharmonic: 7f₁
fa = kHz
3Solutions
1.Number systems
Answer: U = 203
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
53 = 0011 0101
inverted: 1100 1010
+1 → 1100 1011
Read it as unsigned
1100 1011₂ = 256 − 53 = 203
Answer
U = 203
2.Number systems
Answer: M = 512 KiB
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.
Bits and bytes
512·512·16 = 4194304 bit = 524288 B
Convert to KiB
524288/1024 = 512 KiB
Answer
M = 512 KiB
3.Digital logic: gates, K-maps & flip-flops
Answer: L = 7
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
B′C → m(1, 5)
A′BC′ → m(2)
ABC → m(7)
Group and read off
F = B′C + AC + A′BC′
Count the literals
L = 7
Answer
L = 7
4.Digital logic: gates, K-maps & flip-flops
Answer: fmax = 20 MHz
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.
Worst-case settling time
T = 5·10 ns = 50 ns
Maximum frequency
fmax = 1/50 ns = 20 MHz
Answer
fmax = 20 MHz
5.Data acquisition & sampling
Answer: fa = 1000 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.
Nyquist frequency
fs/2 = 2 kHz < f = 5 kHz
Nearest multiple of the sampling rate
k = round(5000/4000) = 1
Alias frequency
fa = |5000 − 1·4000| = 1000 Hz
Answer
fa = 1000 Hz
The dashed curve in the figure is the alias: it passes through exactly the same samples.
6.Data acquisition & sampling
Answer: fa = 16 kHz
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
7f₁ = 7·12 kHz = 84 kHz
Nearest multiple of the sampling rate
m = round(1.68) = 2
Alias frequency
fa = |84 − 2·50| = 16 kHz
Answer
fa = 16 kHz
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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