Unit 1: Display technologies and 2D primitives
Computer Graphics notes · PTU syllabus (PGCA1919)
On this page
- Unit summary
- Applications of computer graphics
- Input devices
- CRT and video basics
- Raster scan and random scan displays
- Storage tube displays
- LCD and LED displays
- Colour models: RGB and CMY
- Scan conversion of a point
- Line drawing: DDA algorithm
- Bresenham's line algorithm
- Circle drawing: Bresenham's and midpoint algorithms
- Ellipse drawing: midpoint algorithm
- Key terms
- Quick revision
- Important questions
Unit summary
Computer graphics turns data and models into pictures on screens. This unit covers applications, input and output devices, storage tube, raster scan, random scan, LCD and LED displays, CRT and video basics, the RGB and CMY colour models, and scan conversion of points, lines (DDA and Bresenham), circles (Bresenham and midpoint) and ellipses.
After this unit you can
- Describe applications and graphics input and output devices
- Explain CRT, raster, random scan, storage tube, LCD and LED displays
- Explain the RGB and CMY colour models
- Scan-convert points, lines, circles and ellipses
PTU syllabus topics
- Computer graphics applications
- I/O devices
- storage tube/raster scan/random scan/LCD/LED displays
- CRT and video basics
- color models (RGB, CMY)
- scan conversion of points
- lines (DDA, Bresenham's algorithm)
- circles (Bresenham's and midpoint algorithms) and ellipses
Draws
Every pixel row by row
Lines directly
Memory
Frame buffer
Display list
Good for
Realistic shaded scenes
Line drawings
Refresh
Fixed, e.g. 60 Hz
Depends on picture complexity
Topic 1
Applications of computer graphics
Computer-aided design
Buildings, cars, circuits
Presentation graphics
Charts and graphs for reports
Entertainment
Films, animation, video games
Education and training
Simulators, scientific visualisation
Image processing
Medical scans, satellite images
Graphical user interfaces
Windows, icons, menus
Virtual and augmented reality
Immersive environments
Topic 2
Input devices
- Light pen
- Pen-shaped device detecting light from a CRT screen to select or draw
- Graphics tablet (digitiser)
- Flat surface and stylus giving precise coordinates — design and signatures
- Joystick
- Lever controlling direction and speed — games, simulators
- Trackball
- Ball rotated by hand to move the cursor — space-saving
- Digitiser
- Converts drawings or maps into digital coordinates
- Scanner
- Converts printed images and text into digital images (with OCR for text)
- Others
- Mouse, touch screen, data glove, voice input
Topic 3
CRT and video basics
- 1
Heated cathode (electron gun) emits electrons
- 2
Control grid sets beam intensity
- 3
Focusing system narrows the beam
- 4
Deflection plates or coils steer the beam
- 5
Beam strikes phosphor-coated screen
- 6
Phosphor glows to form a spot (pixel)
- Persistence: how long phosphor glows after the beam moves; refresh rate: how often the picture is redrawn (60 Hz or more avoids flicker); resolution: maximum number of points displayed without overlap.
- Video basics: a raster image is a grid of pixels refreshed many times a second; resolution (e.g., 1920 × 1080), aspect ratio (16:9), refresh rate (60–144 Hz), colour depth (24 bits for 16.7 million colours) and frame buffer size define a display. Video signals moved from analog (VGA) to digital (HDMI, DisplayPort).
Topic 4
Raster scan and random scan displays
Drawing
Beam sweeps every row from top to bottom
Beam goes only where lines are drawn
Picture stored as
Intensity of every pixel in a frame buffer
Line-drawing commands in a display file
Resolution
Lower; jagged lines (aliasing)
High; smooth lines
Realism
Shaded, filled, realistic scenes
Line drawings only
Cost
Cheaper
Costlier
Example
TVs, monitors
Early CAD systems, plotters
Memory
Horizontal pixels × vertical pixels × bits per pixel ÷ 8 bytes
Example
1024 × 768 × 24 ÷ 8 = 2,359,296 bytes ≈ 2.25 MB
- Interlacing: odd lines drawn in one pass and even lines in the next, reducing flicker at low refresh rates.
Topic 5
Storage tube displays
- A DVST stores the picture as a charge distribution on a storage grid behind the screen, so it needs no refresh.
- Advantages: no refreshing, flicker-free, complex pictures at high resolution. Disadvantages: no colour, selective erasing impossible — the whole screen must be erased and redrawn; slow erasure.
Topic 6
LCD and LED displays
Plasma panel
Emissive
Gas between glass plates glows when cells are energised
LED and OLED
Emissive
Diodes (or organic layers) emit light at each pixel; OLED needs no backlight
LCD
Non-emissive
Liquid crystals twist polarised light from a backlight; active-matrix (TFT) has a transistor per pixel
- Flat panels are thinner, lighter and use less power than CRTs, and have replaced them in almost all uses.
Topic 7
Colour models: RGB and CMY
Type
Additive — light is added
Subtractive — ink absorbs light
Primaries
Red, green, blue
Cyan, magenta, yellow
Black and white
Black (0,0,0); white (1,1,1)
White (0,0,0); black (1,1,1)
Used in
Monitors, cameras, projectors
Printers (CMYK adds black ink)
RGB to CMY
C = 1 − R, M = 1 − G, Y = 1 − B
Example
Orange RGB (1, 0.5, 0) → CMY (0, 0.5, 1)
Topic 8
Scan conversion of a point
- A point (x, y) with real coordinates is displayed by turning on the nearest pixel — (round(x), round(y)) — with putpixel. All other primitives are built from point plotting.
Topic 9
Line drawing: DDA algorithm
Slope–intercept form
y = m x + c, where m = (y2 − y1) ÷ (x2 − x1)
Direct method
For each x, compute y = m x + c and round — needs floating-point multiplication
DDA increments
steps = max(abs(Δx), abs(Δy)); x increment = Δx ÷ steps; y increment = Δy ÷ steps
- 1Read endpoints (x1, y1) and (x2, y2)
- 2Compute Δx, Δy and steps
- 3Compute x and y increments
- 4Plot (round(x), round(y))
- 5Add increments and repeat steps times
Example
Line from (2, 3) to (8, 6): Δx = 6, Δy = 3, steps = 6, x increment 1, y increment 0.5. Points: (2,3), (3,3.5→4), (4,4), (5,4.5→5), (6,5), (7,5.5→6), (8,6).
- DDA: simpler than the direct method, but uses floating-point addition and rounding, so errors accumulate.
Topic 10
Bresenham's line algorithm
- Uses only integer addition and subtraction; at each step chooses between two candidate pixels using a decision parameter.
Initial decision parameter
p0 = 2Δy − Δx
If pk < 0
Next pixel (xk + 1, yk); pk+1 = pk + 2Δy
If pk ≥ 0
Next pixel (xk + 1, yk + 1); pk+1 = pk + 2Δy − 2Δx
- Derivation outline: at x = xk + 1 the true y is m(xk + 1) + c. Distances to the two candidates are d1 = y − yk and d2 = (yk + 1) − y. pk = Δx (d1 − d2) = 2Δy·xk − 2Δx·yk + constant has the same sign as d1 − d2, so its sign picks the nearer pixel; subtracting pk from pk+1 gives the update rules above.
Example
Line (20, 10) to (30, 18): Δx = 10, Δy = 8, p0 = 6. Pixels: (21,11) p = 2; (22,12) p = −2; (23,12) p = 14; (24,13) p = 10; (25,14) p = 6; (26,15) p = 2; (27,16) p = −2; (28,16) p = 14; (29,17) p = 10; (30,18).
Topic 11
Circle drawing: Bresenham's and midpoint algorithms
Start
(0, r); p0 = 1 − r (or 5/4 − r)
If pk < 0
Next (xk + 1, yk); pk+1 = pk + 2xk+1 + 1
Else
Next (xk + 1, yk − 1); pk+1 = pk + 2xk+1 + 1 − 2yk+1
Stop
When x ≥ y
- Derivation idea: f(x, y) = x² + y² − r² is negative inside the circle, zero on it and positive outside. Evaluate f at the midpoint (xk + 1, yk − ½) between the two candidate pixels; its sign tells which pixel is closer.
Example
r = 10: p0 = −9 → (1,10) p = −6 → (2,10) p = −1 → (3,10) p = 6 → (4,9) p = −3 → (5,9) p = 8 → (6,8) p = 5 → (7,7), and the octant is complete.
- Bresenham's circle: d0 = 3 − 2r; if d < 0 then d = d + 4x + 6, else d = d + 4(x − y) + 10 and y decreases.
Topic 12
Ellipse drawing: midpoint algorithm
- Midpoint (Bresenham) ellipse algorithm: uses four-way symmetry and divides the first quadrant into two regions — region 1 where the slope magnitude is less than 1 (step in x) and region 2 where it is greater than 1 (step in y).
Region 1 start
(0, ry); p1 = ry² − rx² ry + rx²/4
Region 1 update
If p1 < 0: p1 = p1 + 2ry² x + ry²; else y decreases and p1 = p1 + 2ry² x − 2rx² y + ry²
Switch to region 2
When 2ry² x ≥ 2rx² y
Region 2
Step y down; decision p2 chooses whether x increases
Key terms
- Raster scan
- Display that sweeps the beam row by row through every pixel
- Frame buffer
- Memory holding the colour of every pixel
- Additive colour model
- RGB, mixing light
- Scan conversion
- Converting primitives into pixels
- Decision parameter
- Integer test choosing the next pixel
Quick revision
- Applications; light pen, tablet, joystick, trackball, scanner.
- CRT working; resolution, refresh, colour depth; raster vs random scan; DVST; LCD, LED, OLED.
- RGB additive, CMY subtractive; C = 1 − R.
- DDA increments; Bresenham p0 = 2Δy − Δx.
- Midpoint circle p0 = 1 − r; Bresenham circle d0 = 3 − 2r; midpoint ellipse with two regions.
Important exam questions
Practice questions written to the PTU exam pattern for this unit's syllabus: short answers (Section A style) and long answers (Sections B and C style).
Short-answer questions
- Q1.Distinguish raster and random scan displays.
- Q2.How does an LCD produce an image?
- Q3.Convert RGB (0.2, 0.4, 1) to CMY.
- Q4.State one advantage of Bresenham's algorithm over DDA.
- Q5.What is eight-way symmetry?
- Q6.Why does the ellipse algorithm use two regions?
Long-answer questions
- Q1.Explain display devices: CRT, raster, random scan, storage tube, LCD and LED.
- Q2.Explain the DDA and Bresenham's line algorithms with an example.
- Q3.Explain the midpoint circle algorithm with an example.
- Q4.Explain the midpoint ellipse algorithm.
Stuck on this unit?
Message SBS on WhatsApp for help with Computer Graphics, or to ask about studying M.Sc IT at Synetic.