Data storage — OCR GCSE Computer Science
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Data storage explained
Data storage covers 13 specification points, including How to convert positive denary whole numbers… and How to add two binary integers together (up to….
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How to convert positive denary whole numbers to binary numbers (up to and including 8 bits) and vice versa
Denary is base 10, using digits 0–9. Binary is base 2, using only 0 and 1. An 8-bit binary number uses place values 128, 64, 32, 16, 8, 4, 2 and 1. To convert denary to binary, subtract the largest place value that fits, put 1 in that column, and repeat with the remainder until zero, filling unused columns with 0. For example, 157 = 128 + 16 + 8 + 4 + 1, so 157 in 8-bit binary is 10011101. To convert binary to denary, add the place values where the bit is 1. For example, 00101101 = 32 + 8 + 4 + 1 = 45. Always write exactly 8 bits when the question asks for 8-bit binary.
How to add two binary integers together (up to and including 8 bits) and explain overflow errors which may occur
Binary addition operates column by column from the least significant bit (rightmost), similar to denary addition but using only 0 and 1. The single-bit rules are: 0+0=0, 0+1=1, 1+0=1, 1+1=10₂ (write 0, carry 1), and 1+1+1=11₂ (write 1, carry 1). For example, adding 0110 1101₂ and 0011 0111₂: starting at the right, 1+1=10₂ (write 0, carry 1); next 0+1+1=10₂ (write 0, carry 1); then 1+1+1=11₂ (write 1, carry 1); continuing leftwards yields 1010 0100₂. In an 8-bit system, the maximum representable value is 255. If an addition exceeds this (e.g., 200+100=300), a carry is generated out of the most significant bit (the 128s column). Because there is no ninth bit to store this carry, it is discarded, causing an overflow error. The stored result is incorrect, specifically 256 less than the true sum (e.g., 300 becomes 44).
How to convert positive denary whole numbers into 2-digit hexadecimal numbers and vice versa
Hexadecimal is base 16, using the digits 0–9 and the letters A–F, where A = 10, B = 11, C = 12, D = 13, E = 14 and F = 15. A 2-digit hexadecimal number has place values 16¹ = 16 and 16⁰ = 1, so it represents values from 0 to 255. To convert denary to hexadecimal, divide the number by 16: the quotient is the first digit and the remainder is the second. For example, 202 ÷ 16 = 12 remainder 10, so 202 = CA₁₆. To convert hexadecimal to denary, multiply the first digit by 16, multiply the second digit by 1, then add: CA₁₆ = 12 × 16 + 10 × 1 = 192 + 10 = 202.
How to convert binary integers to their hexadecimal equivalents and vice versa
Hexadecimal is base 16, so each hex digit stands for a group of four binary bits (a nibble). To convert binary to hex, split the binary integer into nibbles from the right, pad the leftmost group with leading zeros if needed, then replace each nibble with its hex digit: 1010 becomes A, 1111 becomes F. For example 11011011 splits as 1101 1011, giving DB. To convert hex to binary, do the reverse: write each hex digit as its four-bit pattern and join them, so 2F becomes 0010 1111, or 101111 after dropping leading zeros. This works because 16 = 2⁴, so four binary place values map exactly onto one hex place value.
Binary shifts
A binary shift moves every bit in a binary pattern a stated number of places left or right. A left shift by n places multiplies the value by 2ⁿ, because each bit moves to a place value n positions higher. A right shift by n places divides the value by 2ⁿ, discarding any bits shifted past the right-hand end and losing fractional parts. For example, 00010110 (22) shifted left by 2 gives 01011000 (88), which is 22 × 2². The same pattern shifted right by 1 gives 00001011 (11), which is 22 ÷ 2. Bits shifted out of the fixed-width register are lost, and zeros fill the vacated positions. This means a left shift can cause an overflow error if significant bits are lost, changing the value unexpectedly, while a right shift causes a loss of precision.
The use of binary codes to represent characters
Computers store characters as binary codes because a byte can hold only 0s and 1s. A character set defines which code maps to which character. For example, in ASCII the letter A is decimal 65, which is 01000001 in 8-bit binary. To encode a word, each character is converted to its code and the codes are stored in order. To decode, the binary is split into equal-length codes and each code is looked up in the character set. The same binary pattern can represent different characters in different character sets, so the correct character set must be known. Unicode uses more bits, allowing many more characters from different languages and symbols.
The term ‘character set’
A character set is a defined list of characters and the unique binary code assigned to each one. It allows a computer to convert between binary and text. For example, in ASCII the character A is code 65, which is 01000001 in 8-bit binary. The character set must be known by both the sender and receiver; otherwise the same binary code may be displayed as a different character. ASCII is a small character set, while Unicode is a much larger character set that includes characters from many languages and many symbols. The size of a character set affects how many bits are needed for each code.
The relationship between the number of bits per character in a character set, and the number of characters which can be represented, e.g. ASCII Unicode
A character set maps each character to a unique binary code. If a set uses n bits per character, each character is one n-bit pattern, so the number of distinct patterns is 2ⁿ. With 7 bits, 2⁷ = 128 characters can be represented, which covers standard ASCII. With 8 bits, 2⁸ = 256 characters, covering extended ASCII. Unicode uses more bits per character, so it can represent far more characters, including symbols and scripts from many languages. To work out capacity, identify the bits per character, then evaluate 2ⁿ. To work out bits needed for a required number of characters, find the smallest n where 2ⁿ is at least that number. For example, 300 characters need 9 bits because 2⁸ = 256 is too few and 2⁹ = 512 is enough.
How an image is represented as a series of pixels, represented in binary
A bitmap image is a grid of pixels. Each pixel stores one colour value as a binary code. The number of bits per pixel is the colour depth, so n bits per pixel allow 2ⁿ different colours. For example, 1 bit per pixel gives 2 colours, 3 bits gives 8 colours, and 8 bits gives 256 colours. The image file stores the binary codes for the pixels, usually row by row. Image resolution is the number of pixels in the width and height. File size in bits can be estimated by multiplying width by height by colour depth. For example, a 10 by 10 pixel image at 3 bits per pixel needs 10 × 10 × 3 = 300 bits. More pixels or more bits per pixel increase file size and can improve quality, but use more storage.
Metadata
Metadata is data about data: it describes a file's content, format, origin, and properties so that software and users can interpret and manage it. In a bitmap image, metadata may include width and height in pixels, colour depth, resolution, file format, creation date, camera settings, and geolocation. For a sound file, it may include sample rate, bit depth, duration, and artist. Metadata does not usually form the main content, but without it the raw data may be meaningless. For example, a bitmap stores pixel values, but the image viewer needs metadata to know how many pixels wide the image is and how many bits represent each colour. Metadata also supports searching, sorting, copyright tracking, and automatic processing. In exams, you should explain what metadata is, give relevant examples, and describe how it is used, rather than simply listing file properties.
The effect of colour depth and resolution on The quality of the image The size of an image file
In a bitmap image, resolution is the number of pixels, often expressed as width × height. Colour depth is the number of bits used to represent each pixel's colour. Increasing resolution adds more pixels, allowing finer detail, but increases file size. Increasing colour depth allows more distinct colours, improving colour accuracy, but also increases file size as each pixel needs more bits. Uncompressed file size in bits is calculated as width × height × colour depth. For example, a 1000 × 800 image at 24-bit colour depth has 19,200,000 bits. Dividing by 8 gives 2,400,000 bytes, which is 2.4 MB (using base-10 conversions: 1000 bytes = 1 kB, 1000 kB = 1 MB). Higher resolution or colour depth improves quality but requires more storage.
How sound can be sampled and stored in digital form
Sound is a continuous pressure wave, so a computer cannot record every instant of it. Instead, a microphone converts the wave into a varying analogue voltage, and an analogue-to-digital converter measures that voltage at regular intervals. Each measurement is a sample, and the number of samples taken each second is the sample rate, measured in hertz (Hz). Each sample is rounded to the nearest available level, and the number of bits used per sample is the bit depth. The resulting stream of binary values is stored as a file, for example a WAV file. Playback reverses the process: a digital-to-analogue converter turns each stored value back into a voltage, and a speaker produces the sound. A higher sample rate and bit depth capture the wave more accurately but create a larger file.
The effect of sample rate, duration and bit depth on The playback quality The size of a sound file
Three factors control the size of a sound file, while two of these also determine the quality of played-back sound. Sample rate is how many samples are taken each second, measured in hertz (Hz); a higher rate captures rapid changes in the wave more accurately. Bit depth is the number of bits per sample; a greater depth gives more amplitude levels, so each sample is rounded less coarsely and quiet detail is preserved. Duration is how long the sound lasts. File size in bits is calculated as sample rate × bit depth × duration in seconds × number of channels. Doubling any of these three factors roughly doubles the file size. Higher sample rates and bit depths improve playback quality, but all three factors increase storage and processing demands.
Your focus
- Convert a positive denary whole number up to 255 into 8-bit binary.
- Convert an 8-bit binary number into denary.
- Use place values to check a conversion is correct.
Show all 39 objectives
- Add two binary integers of up to 8 bits correctly, showing carries between columns.
- Identify when a carry out of the most significant bit produces an overflow error.
- Explain the effect of a discarded carry on the value of the stored result.
- Convert a positive denary whole number in the range 0 to 255 into a 2-digit hexadecimal number.
- Convert a 2-digit hexadecimal number into its denary equivalent.
- Use the place values 16 and 1 and the digit values A–F correctly in both directions.
- Convert a binary integer of any length into its hexadecimal equivalent by grouping bits into nibbles.
- Convert a hexadecimal integer into binary by expanding each hex digit into four bits.
- Justify why four binary bits correspond to one hexadecimal digit using the relationship 16 = 2⁴.
- Perform a left or right binary shift by a given number of places on a fixed-width binary pattern.
- Calculate the denary value after a shift and relate it to multiplication or division by a power of two.
- Explain the effect of bits being shifted out of a fixed-width register, including overflow and truncation.
- State that characters are represented by binary codes.
- Convert a short word into binary codes using a given character set.
- Explain why the same binary code can represent different characters in different character sets.
- Define the term character set.
- Give examples of character sets such as ASCII and Unicode.
- Explain why the same binary code can represent different characters in different character sets.
- State the relationship between bits per character and the number of characters representable.
- Calculate the number of characters from a given number of bits per character using 2ⁿ.
- Compare ASCII and Unicode in terms of bits per character and range of characters supported.
- Describe how a bitmap image is stored as pixels with binary colour values.
- Calculate the number of colours from a given colour depth using 2ⁿ.
- Calculate image file size in bits from width, height and colour depth.
- Define metadata and distinguish it from the main content of a file.
- Identify appropriate examples of metadata for bitmap images and sound files.
- Explain how metadata enables software and users to interpret, search, and manage digital data.
- Define resolution and colour depth and describe how each affects a bitmap image.
- Explain how changing resolution or colour depth affects image quality and file size.
- Calculate the uncompressed file size of a bitmap image using width, height, and colour depth.
- Describe how a microphone and ADC convert a continuous sound wave into a sequence of binary samples.
- Define sample rate and bit depth and state the units used for each.
- Explain how stored samples are converted back into an analogue signal for playback.
- Explain how changing sample rate and bit depth affects playback quality.
- Explain how sample rate, bit depth and duration affect the size of a sound file.
- Calculate or estimate a sound file's size from sample rate, bit depth, duration and number of channels.
Data storage flashcards
Card 1 of 12
Hexadecimal
Data storage exam tips
Marking Points
- Identify the place values for 8-bit binary: 128, 64, 32, 16, 8, 4, 2, 1.
- Convert denary to binary by repeatedly subtracting the largest possible place value and recording 1s and 0s.
- Convert binary to denary by adding the place values of all bits that are 1.
- Produce an 8-bit result, including leading zeros where necessary, for example 45 as 00101101.
- Check the result by converting back to the original number or by ensuring the sum of place values equals the denary value.
- State the single-bit addition rules, explicitly showing that 1+1=10₂ and 1+1+1=11₂.
- Add two 8-bit binary numbers column by column from the least significant bit, correctly processing carries.
- Define an overflow error as occurring when a carry is generated out of the most significant bit (eighth bit) and cannot be stored.
- Explain that an overflow error results in an inaccurate answer because the discarded ninth bit has a place value of 256.
- Recall that hexadecimal is base 16 and that A–F represent the denary values 10–15.
- Convert denary to hexadecimal by dividing by 16, using the quotient as the first digit and the remainder as the second.
- Convert hexadecimal to denary by multiplying the first digit by 16 and the second digit by 1, then adding the two products.
- Convert any remainder or digit greater than 9 into its letter equivalent, for example 12 becomes C and 15 becomes F.
- Check a conversion by reversing the method, such as converting CA₁₆ back to 202 to confirm the answer.
- Splitting a binary integer into groups of four bits starting from the right-hand end, because each hex digit represents exactly four binary bits.
- Padding the leftmost nibble with leading zeros when the binary integer does not contain a multiple of four bits, for example 10111 becomes 0001 0111.
- Recalling the hex digit values 0-9 and A-F, where A = 10, B = 11, C = 12, D = 13, E = 14 and F = 15.
- Converting each nibble independently to its hex digit and joining the digits in the same left-to-right order as the nibbles.
- Reversing the method for hex to binary by expanding every hex digit into a four-bit pattern and concatenating the patterns.
- Removing leading zeros only after the full conversion, so that 0001 0111 is written as 10111 rather than with an incorrect internal gap such as 1 0111.
- Moving every bit the stated number of positions in the stated direction, keeping the remaining bits in their relative order.
- Filling vacated positions with zeros rather than copying existing bits or leaving gaps.
- Recognising that a left shift by n places multiplies the original value by 2ⁿ, for example a left shift by 3 multiplies by 8.
- Recognising that a right shift by n places divides the original value by 2ⁿ, discarding any remainder, for example a right shift by 2 divides by 4.
- Identifying bits shifted beyond the available width as lost, which can cause overflow on a left shift or loss of precision on a right shift.
- Calculating the new denary value by applying the multiplication or division, or by re-evaluating the place values of the shifted pattern.
- Characters are represented by unique binary codes, not by the shape of the character itself.
- A character set provides the mapping between each binary code and a character.
- In ASCII, each character is represented by a 7-bit or 8-bit code, for example A is 01000001.
- To encode text, each character is converted to its binary code and the codes are stored in sequence.
- To decode text, the binary data is divided into equal-length codes and each code is matched to a character using the same character set.
- Unicode uses more bits than ASCII, so it can represent a much larger range of characters, including symbols and characters from many languages.
- A character set is a defined list of characters and their binary codes.
- Each character in a character set has a unique binary code.
- A character set is used to encode text into binary and decode binary back into text.
- ASCII is an example of a character set that uses 7 or 8 bits per character.
- Unicode is an example of a larger character set that uses more bits and can represent many more characters.
- If the wrong character set is used, the decoded text may contain incorrect characters.
- States that a character set assigns a unique binary code to each character.
- Uses the relationship that n bits per character give 2ⁿ possible characters.
- Correctly evaluates powers of two, such as 2⁷ = 128 and 2⁸ = 256.
- Explains that more bits per character increase the number of representable characters.
- Distinguishes ASCII, typically 7 or 8 bits per character, from Unicode, which uses more bits and supports many more characters.
- Applies the relationship to find the minimum bits needed for a given number of characters.
- States that an image is made of pixels arranged in a grid.
- Explains that each pixel is represented by a binary code.
- Links the number of bits per pixel to the number of available colours using 2ⁿ.
- Describes how the binary codes for all pixels are stored to represent the image.
- Calculates image file size in bits using width × height × colour depth.
- Explains that greater resolution or colour depth increases file size and can improve image quality.
- Defines metadata as data about data, describing the content, format, or properties of a file rather than the main user content itself.
- Gives at least one accurate example of metadata in a bitmap image, such as image width, image height, colour depth, resolution, file format, or creation date.
- Gives at least one accurate example of metadata in a sound file, such as sample rate, bit depth, duration, or artist name.
- Explains how metadata is used, for example to allow software to interpret raw data correctly, to enable searching and sorting, or to record ownership and copyright.
- Distinguishes metadata from the actual data, for example pixel colour values in an image or amplitude samples in a sound file.
- Applies metadata to a scenario, such as a photo library using date and location metadata to group images automatically.
- States that resolution is the number of pixels in an image, often given as width × height, and that higher resolution means more pixels.
- Explains that increasing resolution improves image quality by allowing finer detail and sharper edges, but increases file size because more pixels are stored.
- States that colour depth is the number of bits used per pixel and that a greater colour depth allows more distinct colours.
- Explains that increasing colour depth can improve colour accuracy and reduce banding, but increases file size because each pixel requires more bits.
- Uses the relationship file size in bits = width × height × colour depth to calculate or compare image file sizes.
- Explains that file size also depends on compression and metadata, so the calculated value is an estimate of uncompressed size.
- Compares the relative effects of changing resolution versus changing colour depth on quality and file size.
- Sound is a continuous analogue wave, so it must be converted into discrete digital values before a computer can store it.
- A microphone converts the sound wave into a varying analogue electrical signal.
- An analogue-to-digital converter (ADC) samples the analogue signal at regular time intervals.
- Each sample is a measurement of the signal's amplitude at that instant, and it is quantised to the nearest representable binary value.
- The sample rate is the number of samples taken per second, measured in hertz (Hz).
- The bit depth is the number of bits used to store each sample, which determines how many amplitude levels are available.
- The sequence of binary sample values is stored in a sound file, such as a WAV file.
- Playback uses a digital-to-analogue converter (DAC) to rebuild an analogue signal that a speaker outputs as sound.
- A higher sample rate records the wave more often, so fast changes in the waveform are represented more accurately and playback is closer to the original.
- A higher bit depth gives more possible amplitude values per sample, reducing quantisation error and improving dynamic detail.
- Increasing duration increases the number of samples stored, so the file becomes larger, but it does not affect the playback quality.
- File size in bits can be estimated as sample rate × bit depth × duration in seconds × number of channels.
- Doubling sample rate, bit depth or duration approximately doubles the file size, assuming the other factors stay constant.
- Lower sample rate or bit depth reduces file size but can make playback sound less accurate, for example losing high frequencies or adding quantisation noise.
- The choice of values is a trade-off between acceptable playback quality and available storage or transmission capacity.
Examiner Tips
- 💡Write the place values 128, 64, 32, 16, 8, 4, 2, 1 above your working to avoid column errors.
- 💡For denary to binary, work left to right, placing 1 where the place value fits and 0 where it does not.
- 💡Check your answer by converting back; the sum of the place values with 1s must equal the original denary number.
- 💡Align the two 8-bit numbers vertically and write any carries in a clear row above or below the main columns.
- 💡If a calculation produces a ninth bit, explicitly state that an overflow error has occurred because the register is limited to 8 bits.
- 💡Write out the hexadecimal digit values 0–15 before starting, so letter conversions are quick and accurate.
- 💡For denary to hexadecimal, show the division by 16 with its quotient and remainder so your method is visible.
- 💡For hexadecimal to denary, write the multiplication by 16 and by 1 before adding, then state the final denary value.
- 💡Show the nibble grouping explicitly, for example 1101 1011, so the examiner can follow your method even if a digit is wrong.
- 💡Check your answer by converting back: if DB is correct, then D = 1101 and B = 1011 should rebuild the original binary integer.
- 💡Keep place values in mind: the rightmost hex digit has value 16⁰, the next 16¹, then 16², which confirms why four bits are needed per digit.
- 💡Write the original pattern with place values above the bits so you can track each bit as it moves.
- 💡State the arithmetic effect as well as the new pattern, for example left shift by 2 multiplies by 4, to show full understanding.
- 💡Check for overflow by comparing the number of bits needed for the new value with the fixed width of the register.
- 💡When asked to encode a word, convert each character separately and keep the codes in the same order as the characters.
- 💡When asked to decode binary, group the bits into equal-length codes before looking up characters.
- 💡Use the term character set when explaining how binary codes are mapped to characters.
- 💡Give a clear definition of character set and then give one example such as ASCII or Unicode.
- 💡When explaining why a character set is needed, refer to encoding and decoding text.
- 💡Use the term unique binary code when describing the code for each character.
- 💡Show the power-of-two calculation, for example 2⁷ = 128, rather than only stating the final number.
- 💡When asked for the minimum bits, test powers of two in order until the required number of characters is reached.
- 💡Use the terms bits per character and characters representable precisely, and link them with the 2ⁿ relationship.
- 💡Define pixel, resolution and colour depth clearly before using them in an explanation.
- 💡Show the file-size calculation with units, for example 300 bits, and convert to bytes only if asked.
- 💡Use the 2ⁿ relationship to justify how many colours a given colour depth provides.
- 💡Start with a precise definition: metadata is data about data that describes a file's content, format, or properties.
- 💡Use examples from the specification context, such as image width, height, colour depth, and resolution for bitmaps, or sample rate and bit depth for sound.
- 💡Explain the purpose of metadata in a scenario, for example how a media player uses sample rate metadata to play a sound file correctly.
- 💡Avoid simply listing file extensions; explain what the metadata describes and why it is needed.
- 💡Write the formula clearly: file size in bits = width × height × colour depth, then convert to bytes or larger units if required.
- 💡When comparing, state which factor changes and how it affects both quality and file size, for example doubling colour depth doubles the uncompressed file size.
- 💡Use correct units and show your working, especially when converting bits to bytes, kB, or MB using base-10 (1000).
- 💡Link quality to the specific factor: resolution affects detail and sharpness, while colour depth affects the range and accuracy of colours.
- 💡Use the full sequence microphone → ADC → binary storage → DAC → speaker when explaining the process.
- 💡Define sample rate in Hz and bit depth in bits, and keep the two terms clearly separate in your answer.
- 💡When asked to compare, refer to the shape of the reconstructed wave and to file size rather than saying simply 'better quality'.
- 💡Show the file-size relationship as sample rate × bit depth × duration × channels, and substitute values carefully with units.
- 💡When explaining quality, refer to how closely the reconstructed wave matches the original, not just to 'better sound'.
- 💡For trade-off questions, give a context such as streaming versus archiving and justify a suitable sample rate and bit depth.
Common Mistakes
- Writing fewer than 8 bits when 8-bit binary is requested. Correction: pad with leading zeros on the left until there are exactly 8 bits.
- Using the wrong place values, such as starting at 1 instead of 128. Correction: write the place values 128 to 1 above the columns before converting.
- Adding all place values instead of only those with a 1. Correction: for binary to denary, sum only the columns containing 1.
- Adding from left to right: this causes carries to move in the wrong direction. Correction: always start at the least significant bit (rightmost) and move left.
- Writing '2' in a binary column: using a denary digit instead of binary. Correction: write 0 and carry 1 to the next column, as 1+1=10₂.
- Ignoring the final carry out of the 8th bit: dropping the carry without explanation. Correction: explicitly state that the carry is discarded, resulting in an overflow error.
- Using base 10 place values: the error is treating the digits as tens and units. Correction: use 16 and 1 as the place values.
- Writing 10, 11 or 12 as two digits: the error is leaving a denary value in a single hexadecimal position. Correction: write A, B or C instead.
- Swapping quotient and remainder: the error is putting the remainder first. Correction: the quotient is the first digit and the remainder is the second.
- Grouping bits from the left instead of the right, which misaligns every nibble; correct this by always starting the grouping at the least significant bit.
- Writing a hex digit as a decimal number, such as writing 12 instead of C; correct this by learning the A-F values and using single hex characters.
- Dropping leading zeros inside a nibble during hex-to-binary conversion, such as writing 111 for 7 instead of 0111; correct this by always writing four bits per hex digit before removing only the very first leading zeros.
- Shifting in the wrong direction, which changes multiplication into division; correct this by checking whether the value should increase (left) or decrease (right).
- Forgetting to fill empty positions with zeros, which produces a pattern with missing bits; correct this by writing the full-width result with zeros in every vacated position.
- Assuming a right shift rounds to the nearest integer, when it truncates the fractional part; correct this by discarding the remainder, so 00000101 (5) shifted right by 1 gives 00000010 (2).
- Writing that the letter A is stored as the shape of the letter A. Correction: the letter A is stored as a binary code such as 01000001.
- Assuming that a binary code always represents the same character in every character set. Correction: the same code can represent different characters in different character sets, so the character set must be known.
- Confusing the number of bits used with the number of characters available. Correction: more bits allow more possible codes, but the actual number of characters depends on the character set.
- Defining a character set as just a list of letters. Correction: a character set includes letters, digits, punctuation and other symbols, each with a unique binary code.
- Thinking that a character set is the same as a font. Correction: a character set defines binary codes for characters, while a font defines their visual appearance.
- Believing that ASCII and Unicode are the same. Correction: ASCII is a smaller character set, while Unicode is a larger character set that can represent many more characters.
- Multiplying the number of bits by 2 instead of using 2ⁿ; correct by treating each bit as doubling the number of patterns.
- Confusing bits per character with total bits in a message; correct by separating the code length for one character from the total for a string.
- Assuming Unicode uses a fixed small number of bits like ASCII; correct by noting Unicode uses more bits per character and therefore supports many more characters.
- Treating a pixel as a single bit rather than a code of one or more bits; correct by saying each pixel uses a fixed number of bits called the colour depth.
- Confusing resolution with colour depth; correct by defining resolution as the number of pixels and colour depth as bits per pixel.
- Forgetting to multiply by colour depth when estimating file size; correct by using width × height × bits per pixel.
- Confusing metadata with the main content of a file. Correction: metadata describes the content, while the content is the actual pixel values, audio samples, or text.
- Assuming metadata is always visible to the user. Correction: metadata is often hidden and used by software, though some applications display it in file properties.
- Giving only one example and no explanation. Correction: provide a definition, at least two examples, and a clear use, such as enabling a viewer to display an image correctly.
- Confusing resolution with colour depth. Correction: resolution is the number of pixels, while colour depth is the number of bits per pixel.
- Thinking that increasing colour depth always makes an image look sharper. Correction: colour depth affects the number of colours and colour accuracy, while resolution affects detail and sharpness.
- Forgetting to convert between bits, bytes, and larger units correctly. Correction: divide by 8 to convert bits to bytes, then by 1000 for kB, and 1000 again for MB, as OCR uses base-10.
- Assuming file size is exactly width × height × colour depth in all cases. Correction: this gives uncompressed size; compression and metadata can change the actual file size.
- Thinking the whole sound wave is stored continuously: the wave is sampled at intervals, so only discrete values are kept. Correction: describe sampling as taking regular measurements of amplitude.
- Confusing sample rate with bit depth: sample rate is how often samples are taken per second, while bit depth is how many bits represent each sample. Correction: link rate to time and depth to amplitude precision.
- Believing a higher sample rate alone gives perfect reproduction: quantisation to a limited number of levels still loses detail. Correction: explain that both sample rate and bit depth affect accuracy.
- Saying a higher sample rate makes the file smaller: it increases the number of samples per second, so the file grows. Correction: state that higher sample rate increases both quality and size.
- Mixing up the units of sample rate and bit depth: sample rate is in hertz (Hz) and bit depth is in bits. Correction: quote the correct unit with each term.
- Stating that duration affects sound quality: a longer file just plays for more time. Correction: only link duration to the file size, while linking sample rate and bit depth to both size and quality.