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 −19 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −198 bit
U =
2.Number systemsHard
Set the bits to the 8-bit result of 1101 1100 AND 0010 1010. What is the result's value R in decimal?
1101 1100AND0010 1010
R =
3.Digital logic: gates, K-maps & flip-flopsMedium
Simplify F = A′BC + B′C + A′BC′ + AB to its minimal sum-of-products form. How many literals L does the simplified expression contain?
F = A′BC + B′C + A′BC′ + AB
L =
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 29 clock pulses?
Q0 = 329n = 4
Q =
5.Data acquisition & samplingMedium
A 12-bit ADC with an input range of 0 to 3.3 V truncates: its output code is ⌊Vin/LSB⌋ with 1 LSB = VFSR/2ᴺ. Find the decimal output code D for an input of 2.107 V.
N = 12VFSR = 3.3 VVin = 2.107 V
D =
6.Data acquisition & samplingHard
A 16-bit ADC streams every sample over a UART at 57600 baud, format 8N1 (start bit, 8 data bits, 1 stop bit). Each sample is sent as 2 whole byte(s) and the link is busy all the time. Find the highest signal frequency fmax that can be sampled without aliasing, in Hz.
baud = 57600 BdN = 168N1
fmax = Hz
3Solutions
1.Number systems
Answer: U = 237
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
19 = 0001 0011
inverted: 1110 1100
+1 → 1110 1101
Read it as unsigned
1110 1101₂ = 256 − 19 = 237
Answer
U = 237
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
AND gives 1 only where both bits are 1.
1101 1100
AND 0010 1010
= 0000 1000
Value of the result
0000 1000₂ = 8
Answer
R = 8
3.Digital logic: gates, K-maps & flip-flops
Answer: L = 2
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′BC → m(3)
B′C → m(1, 5)
A′BC′ → m(2)
AB → m(6, 7)
Group and read off
F = C + B
Count the literals
L = 2
Answer
L = 2
4.Digital logic: gates, K-maps & flip-flops
Answer: Q = 6
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.
29 = 1·16 + 13
Count after the remaining pulses
Q = (3 − 13) mod 16 = 6
Q = 0110₂
Answer
Q = 6
5.Data acquisition & sampling
Answer: D = 2615
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 = 3.3/212 = 0.8057 mV
Code
Vin/LSB = 2107/0.8057 = 2615
code = ⌊2615⌋ = 2615
Answer
D = 2615
In binary that is 101000110111.
6.Data acquisition & sampling
Answer: fmax = 1440 Hz
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 sample on the line
2·10 = 20 bit
Highest sampling rate
fs,max = 57600/20 = 2880 Hz
Nyquist limit
fmax = 2880/2 = 1440 Hz
Answer
fmax = 1440 Hz
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