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A7.7 Analogue, digital and graphics

Data representation · A level · AQA 7517 4.5.6.1, Eduqas A500QS 2.3 · about 30 min

BugBotLab

What this lesson is about

Analogue and digital signals, ADCs and DACs, bitmapped and vector graphics and when to use each.

Questions 6 marks in all

  1. [1 mark]How many bytes does an 800 x 600 bitmap with a colour depth of 24 bits take, ignoring metadata?

    Answer: 1440000. 800 x 600 x 24 = 11,520,000 bits, and dividing by 8 gives 1,440,000 bytes.

  2. [1 mark]How many different levels can a 10-bit ADC output?

    Answer: 1024. An n-bit ADC has 2 to the power n levels, and 2 to the power 10 is 1,024.

  3. [1 mark]A company logo must look sharp on a business card and on a large banner. Which is better, and why?

    1. AA vector graphic, because the shapes are recalculated at any size with no loss
    2. BA bitmap, because it stores every pixel
    3. CA bitmap, because it has a colour depth
    4. DA vector graphic, because it is always a smaller file
    Answer: A. Scaling a vector graphic changes its object properties, so edges stay sharp; an enlarged bitmap goes blocky. Vector files are not always smaller.

  4. [1 mark]Which of these would a vector graphic store for a circle?

    Tick every answer that is true.

    1. AThe centre coordinates
    2. BThe radius
    3. CThe fill colour
    4. DThe colour of every pixel inside it
    Answer: A, B, C. A vector graphic stores each object's properties, not pixels. It is turned into pixels only when drawn.

  5. [1 mark]Where would a DAC be used?

    1. ABetween a computer's digital audio and a loudspeaker
    2. BBetween a microphone and a computer
    3. CBetween a light sensor and a microcontroller
    4. DInside a hard disk to store bits
    Answer: A. A DAC turns digital codes into an analogue voltage. The microphone and light sensor need ADCs.

  6. [1 mark]The byte 01000001 is stored in memory. What does it represent?

    1. AIt depends on how the program interprets it
    2. BThe letter A
    3. CThe number 65
    4. DA machine code instruction
    Answer: A. A bit pattern has no meaning on its own: it could be 65, the letter A, a pixel or an instruction.

The task: vector to bitmap

shapes is a vector drawing of two objects. A rectangle has x and y (its top-left corner) and w and h; a circle has a centre cx, cy and a radius r. All are in pixels. Rasterise it onto a bitmap WIDTH 32 pixels wide and HEIGHT 16 pixels tall, 1 bit per pixel. 1. The pixel in column col (0 to 31) and row row (0 to 15) has its centre at (col + 0.5, row + 0.5). It is on when that centre is inside any shape: for a rectangle, x <= px < x + w and y <= py < y + h; for a circle, the squared distance from the centre is no more than r squared. 2. Print the 16 rows, top row first, each as 32 characters: # for on and . for off. 3. Then print, calculating each number from WIDTH and HEIGHT: 1-bit bitmap: <n> bits, 24-bit bitmap: <n> bits, and 1-bit bitmap at 4 times the size: <n> bits (4 times the width and 4 times the height, still 1 bit per pixel).

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

WIDTH = 32
HEIGHT = 16
shapes = [
    {"type": "rect", "x": 2, "y": 2, "w": 10, "h": 6},
    {"type": "circle", "cx": 22, "cy": 8, "r": 5},
]

def inside(shape, px, py):
    return False

The hint students can ask for: For every pixel, take the point in the middle of it and test it against each shape: inside a rectangle means between its left and right edges and between its top and bottom; inside a circle means no further from the centre than the radius. A bitmap's size is width times height times colour depth, and scaling up multiplies both width and height.

A solution

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

WIDTH = 32
HEIGHT = 16
shapes = [
    {"type": "rect", "x": 2, "y": 2, "w": 10, "h": 6},
    {"type": "circle", "cx": 22, "cy": 8, "r": 5},
]

def inside(shape, px, py):
    if shape["type"] == "rect":
        return shape["x"] <= px < shape["x"] + shape["w"] and shape["y"] <= py < shape["y"] + shape["h"]
    return (px - shape["cx"]) ** 2 + (py - shape["cy"]) ** 2 <= shape["r"] ** 2

for row in range(HEIGHT):
    text = ""
    for col in range(WIDTH):
        hit = False
        for shape in shapes:
            if inside(shape, col + 0.5, row + 0.5):
                hit = True
        text = text + ("#" if hit else ".")
    print(text)
print("1-bit bitmap:", WIDTH * HEIGHT * 1, "bits")
print("24-bit bitmap:", WIDTH * HEIGHT * 24, "bits")
print("1-bit bitmap at 4 times the size:", (WIDTH * 4) * (HEIGHT * 4), "bits")

Any program that meets the task's checks is marked correct in the simulator; this is one way, not the only way.