Digital Electronics: cram pack

Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Number systems

Place values: binary 2^k, octal 8^k, hex 16^k
To decimal: Σ d_k·b^k
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 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; f_out = f_clk/N
Ripple counter: f_max = 1/(n·t_pd)
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: f_s > 2·f_max, f_N = f_s/2
Alias frequency: f_a = |f − k·f_s|, k = round(f/f_s)
Step size: 1 LSB = V_FSR/2ᴺ
Quantisation error (rounding): ±½ LSB
Truncating ADC code: D = ⌊V_in/LSB⌋
Bits for a resolution r: 2ᴺ ≥ span/r
Ideal SQNR, full-scale sine: SQNR = 6.02·N + 1.76 dB
Data rate: channels × f_s × N bit/s
Multiplexed ADC: f_s = 1/(n·(t_c + t_set))
UART: T_b = 1/baud; 8N1 = 10 bits per byte; bytes/s = baud/bits per frame
Baud rate divider: baud = f_clk/(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 f_s/2; a higher frequency appears at f_a = |f − k·f_s| with the nearest multiple of f_s.
For resolution questions, find 1 LSB = V_FSR/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 −113 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −1138 bit

U =

2.Number systemsHard

What is the largest unsigned integer N that fits in 10 bits?

10 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) = Σm(0, 3, 4, 5, 6). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C) = Σm(0, 3, 4, 5, 6)

L =

4.Digital logic: gates, K-maps & flip-flopsHard

An 8 kHz clock drives a mod-5 counter. Find the frequency f_out of its most significant output, which completes one cycle per counting cycle.

f_clk = 8 kHzN = 5

f_out = kHz

5.Data acquisition & samplingMedium

An 8-bit ADC with an input range of 0 to 2.5 V truncates: its output code is ⌊V_in/LSB⌋ with 1 LSB = V_FSR/2ᴺ. Find the decimal output code D for an input of 2.35 V.

N = 8V_FSR = 2.5 VV_in = 2.35 V

D =

6.Data acquisition & samplingHard

A microcontroller with a 32 MHz clock sets its UART baud rate as f_clk/(16·D), with D a whole number rounded to the nearest value. The target is 9600 baud. Find the percentage error of the real baud rate (positive when it is too fast).

f_clk = 32 MHztarget = 9600 Bd

e = %

3Solutions

1.Number systems

Answer: U = 143

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

113 = 0111 0001

inverted: 1000 1110

+1 → 1000 1111

Read it as unsigned

1000 1111₂ = 256 − 113 = 143

Answer

U = 143

2.Number systems

Answer: N = 1023

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

2¹⁰ − 1 = 1023

Answer

N = 1023

3.Digital logic: gates, K-maps & flip-flops

Answer: L = 9

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′

AC′

AB′

A′BC

Minimal SOP form

F = B′C′ + AC′ + AB′ + A′BC

L = 9

Answer

L = 9

4.Digital logic: gates, K-maps & flip-flops

Answer: f_out = 1.6 kHz

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 f_clk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

Frequency division

f_out = 8 kHz / 5 = 1.6 kHz

Answer

f_out = 1.6 kHz

5.Data acquisition & sampling

Answer: D = 240

ADC resolution: an N-bit converter divides its full-scale range V_FSR into 2ᴺ codes, from 0 to 2ᴺ − 1. One step is 1 LSB = V_FSR/2ᴺ: the smallest change the converter can show. A truncating ADC gives the code ⌊V_in/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 = 2.5/2⁸ = 9.766 mV

Code

V_in/LSB = 2350/9.766 = 240.6

code = ⌊240.6⌋ = 240

Answer

D = 240

In binary that is 11110000.

6.Data acquisition & sampling

Answer: e = 0.1603 %

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 T_b = 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 = f_clk/(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.

Divider

D = 32·10⁶/(16·9600) = 208.3 → 208

Real baud rate

32·10⁶/(16·208) = 9615 Bd

Error

(9615 − 9600)/9600·100 = 0.1603 %

Answer

e = 0.1603 %

An error below about 2 % is usually acceptable.

ExamProblems · examproblems.com/courses/digital-electronics