CS510 Midterm Solved Paper Spring 2012

Q1. Applications of Computer Vision (15 Points)

We devoted two lectures to student presentations of commercial applications of computer vision. Please succinctly summarize three distinct applications, one paragraph for each.

Application 1:

There is no single correct answer. Indeed, any reasonably substantive discourse on any three of the topics covered by the term papers would constitute a good answer. For the record, those topics are:

Film and Video Production Security: Monitoring and Surveillance 3D Object Modeling Medical Imaging People Tracking for Sales and Marketing Driver Assistance Biometrics Gesture Recognition Autonomous Vehicles Security: Threat Detection Sports Video -­‐ Referee Sports Video -­‐ Broadcast

Application 2:

Application 3:

Q2. Limitations of Discrete Sampling (15 Points)

The two related concepts of Nyquist Rate and aliasing play a critical role in understanding one of the most troublesome risks associated with discrete sampling of an analogue signal or even down sampling a discrete signal.

Below is a picture of an analogue signal, a pure cosine wave with a periodicity of one cycle per second. Use this signal to illustrate aliasing and precisely define the Nyquist rate under the presumption that reliable encoding of signals up to one cycle per second is desired.

Here is what happens of samples are spaced at ¾ of a cycle (second) intervals.

This is sampling below the Nyquist rate of 2 samples per second (no more than 0.5 seconds between samples). Note on the curve, over 6 seconds there are only 9 samples as opposed to the required 12. Further, 12 is a special case because it can end up sampling straight down the zero crossings, missing the sine wave entirely. Thus, one says sampling must be carried out above (not equal to) the Nyquist rate. -1.0 -0.5

1. 0
2. 5
3. 0

Seconds Signal

-1.0 -0.5

1. 0
2. 5
3. 0

Seconds Signal

Q3. Comparing Vectors / Images (20 Points)

There is a circumstance fully discussed in lecture in which maximizing correlation is mathematically equivalent to minimizing L2 distance.

First, please precisely define L2 distance between two vectors denoted as:

𝑣= 𝑣! 𝑣! 𝑣! 𝑤= 𝑤! 𝑤! 𝑤!

𝐿! =

Now precisely define Pearson’s Correlation

𝐶=

Finally, state the circumstances under which maximizing correlation is equivalent to minimizing L2 distance and use the formal statement of these circumstances to derive (prove) the equivalence.

First condition, subtract off means

𝑣= 𝑤=

Second condition, make each vector unit length

𝑣= 𝑣 𝑤 𝑤

Now, start with L2 and manipulate

𝐿! = 𝑣! ! + 𝑤! ! −2 1 + 1 −2

The sum of the squares is just one because of step two and correlation is the dot product because of both steps, so maximizing L2 above is equivalent to minimizing the dot product/correlation.