Signal and Noise

Learn It!

Goal

Approximately how many "raw" format 12-megapixel (\(12 \cdot 10^6\) pixels/image) images can be stored in 9 GB (\( \approx 9 \cdot 10^9\) Bytes) of storage, if 24 bits (three colors, 8 bits per color) are stored for every pixel? Hint: 1 Byte = 8 bits

+

!

What is a "raw" image?

A "raw" format image just means it is not compressed in any way. The image format stores
every piece of information as though none of them are related. You will learn a little about compression in this course .

Part 1

Arrange quantities

+

1A

Dimensional Analysis

When we don't have any "canned" equations that will solve a problem for us,
small problems like this are often solved by dimensional analysis. You can usually
get the solution by just multiplying things together in the right way without thinking much.

Having many physical quantities that you don't have equations for and can't be added is usually a signal that dimensional analysis is a good choice of technique. If there are many quantities with units "this per that", it is often worth a try.

Having many physical quantities that you don't have equations for and can't be added is usually a signal that dimensional analysis is a good choice of technique. If there are many quantities with units "this per that", it is often worth a try.

Drive: \( \frac{9 \cdot 10^9 Bytes}{Drive} \)

Images: \( \frac{12 \cdot 10^6 pixels}{image} \)

Pixels: \( 24 \frac{bits}{pixel}\)

Bytes: \( 8 \frac{bits}{byte}\)

We Want: \( \frac{images}{Drive}\)

Images: \( \frac{12 \cdot 10^6 pixels}{image} \)

Pixels: \( 24 \frac{bits}{pixel}\)

Bytes: \( 8 \frac{bits}{byte}\)

We Want: \( \frac{images}{Drive}\)

First step, we need to take inventory of what we know, making careful
note of what units each quantity has.

\( \frac{ 9 \cdot 10^9 Bytes}{Drive} \cdot \frac{image}{12 \cdot 10^6 pixels}\)

The final answer must have \(Drives\) in the denominator and \(images \) in the numerator, so I can start with the two quantities
that have those in their units. Images needs to be in the numerator, so we divide by the number of pixels per image.

\( \frac{ 9 \cdot 10^9 Bytes}{Drive} \cdot \frac{image}{12 \cdot 10^6\color{blue}{\cancel{ pixels} }}
\cdot \frac{\color{blue}{\cancel{pixel}}}{24 bits} \)

To eliminate pixels, we divide by the number of bits per pixel. Pixels cancel.

\( \frac{9 \cdot 10^9 {{Bytes}} }{ drive } \cdot \frac{image}{12 \cdot 10^6 \color{blue}{\cancel{pixels}}}
\cdot \frac{\color{blue}{\cancel{pixel}}}{24 \color{black}{\cancel{bits}}} \cdot \frac{8 \color{black}{\cancel{bits}}}{{byte}}\)

To eliminate Bits, we convert Bits to Bytes. Cancel Bits

\( \frac{9 \cdot 10^9 \color{red}{\cancel{Bytes}} }{ drive } \cdot \frac{image}{12 \cdot 10^6 \color{blue}{\cancel{pixels}}}
\cdot \frac{\color{blue}{\cancel{pixel}}}{24 \color{black}{\cancel{bits}}} \cdot \frac{8 \color{black}{\cancel{bits}}}{\color{red}{\cancel{byte}}}\)

Nicely enough, Bytes also cancel without any more work.

\( \frac{9 \cdot 10^9 \color{red}{\cancel{Bytes}} }{ drive } \cdot \frac{image}{12 \cdot 10^6 \color{blue}{\cancel{pixels}}}
\cdot \frac{\color{blue}{\cancel{pixel}}}{24 \color{black}{\cancel{bits}}} \cdot \frac{8 \color{black}{\cancel{bits}}}{\color{red}{\cancel{byte}}}=\frac{images}{drive} \)

Our final answer should have units of images per drive, so any conversion factors should be placed to get all the other units to cancel out.

+

?

Does this expression make sense?

A bigger hard drive should fit more images, so the size of the drive should be in the numerator. Good!

The bigger the images are, the less should fit, so the size of the images should be in the denominator. We're right!

The bigger the images are, the less should fit, so the size of the images should be in the denominator. We're right!

★

\( \frac{9 \cdot 10^9 \cdot 8}{12 \cdot 10^6 \cdot 24}=250 \)

Multiply everything together, respecting units, to get the answer. note that our answer is bigger than 1.