Ict 303 : Data Communication
1) Differentiate between guided media and unguided media.
Answer:
Guided transmission media are more commonly known as the wired
communication or bounded transmission media.
The unguided media is also called wireless communication. It does not
require any physical medium to transmit electromagnetic signals.
2) Differentiate between an analog and a digital electromagnetic
signal.
Signals are used to carry information from one device to another. Analog signal
is a continuous wave that keeps on changing over a time period. Digital signal is
discrete in nature. The fundamental difference between analog and digital signal
is that analog signal is represented by the sine waves whereas, the digital signal
is represented by square waves.
3) What are three important characteristics of a periodic signal?
Three important characteristics of a periodic signal are: amplitude the instantaneous
value of a signal at any time measures in volts. Frequency the number of repetitions of the
period per second or the inverse of the period; it is expressed in cycles per second or Hertz
(Hz).
4) How many radians are there in a complete circle of 360
degrees?
The size of a radian is determined by the requirement that there are 2 radians in a circle.
Thus 2 radians equals 360 degrees. This means that 1 radian = 180/ degrees, and
1 degree = /180 radians.
5) What is the relationship between the wavelength and
frequency of a sine wave?
Assuming a sinusoidal wave moving at a fixed wave speed, wavelength is inversely
proportional to frequency of the wave: waves with higher frequencies have
shorter wavelengths, and lower frequencies have longer wavelengths.
6) Define fundamental frequency.
the lowest frequency which is produced by the oscillation of the whole of an object, as
distinct from the harmonics of higher frequency.
7) What is the relationship between a signal's spectrum and its bandwidth?
The spectrum of a signal is the range of frequencies contained in the signal.
The bandwidth is the difference between the lowest and highest frequency in
the spectrum.
8) What is attenuation?
Attenuation is the loss of signal strength in networking cables or connections. This typically is
measured in decibels (dB) or voltage and can occur due to a variety of factors.
9) Define channel capacity
Channel capacity, in electrical engineering, computer science, and information theory, is the
tight upper bound on the rate at which information can be reliably transmitted over a
communication channe
10)
The key factor affect channel capacity are
1) data rate - bits per second
2) cycles per second
3) error rate
Problem Set 22.9
Commonly, medical digital radiology ultrasound studies consist of
about 25 images extracted from a full-motion ultrasound examination.
Each image consists of 512 by 512 pixels, each with 8 bits of intensity
information.
a.How many bits are there in the 25 images?
Answer:
No of bits of intensity information in each image = 8 Hence,
the number of bits in one image = 512 * 512 * 8 = 2,097,152 bits per
image No of images = 25 Hence,
the total Number of bits in 25 images = 2,097,152 * 25 = 52,428,800 bits
b.Ideally, however, doctors would like to use 512 × 512 × 8 bit
frames at 30 fps (frames per second). Ignoring possible compression
and overhead factors, what is the minimum channel capacity required
to sustain this full-motion ultrasound?
Answer:
Frame Rate = 30 fps. From the above question, we found that the no of
bits/image = 2,097,152 Hence, the minimum channel capacity required
to sustain full-motion ultrasound= 2,097,152*30 = 62,914,560 bits per
second
c.Suppose each full-motion study consists of 25 seconds of
frames. How many bytes
Problem 3.13 (page 98) – Previous Exam problem:Solution
a) (30 pictures/s) (480 x 500 pixels/picture) = 7.2 x 106 pixels/s
Each pixel can take on one of 32 values and can therefore be
represented by 5 bits:
B) We use the formula: C = B log2 (1 + SNR)
B = 4.5 x 106 MHz = bandwidth, and
SNR = 35 dB = 10^(35/10) = 3162, and therefore
C = 4.5 x 106 log2 (1 + 103.5) = 4.5 x 10^6 log2 (3163) = (4.5 x 10^6 x
11.63) = 52.335 x 10^6 bps.
c) Allow each pixel to have one of ten intensity levels and let each
pixel be one of three colors (red, blue, green) for a total of 10 x 3 = 30
levels for each pixel element.