IT 420
Spring 2019
Homework 1 Solutions
Note: Where applicable, units of the solution must be correct for full credit.
1.2 Which are peer layers?
1.3 If the unit exchanged at the network interface layer (layer 2) is called a frame and the unit exchanged at the internet layer (layer 3) is called a packet,
___B___ Process to process communication
___E___ Media
__ A___ Interacts with the user
___E____ Encodes data using physical phenomena, such as changing voltage
___C___ Host to host communication
___D____Sends a frame through a single local area network
___D___ Uses physical (MAC) addresses in the header
___C___ Uses IP addresses in the header
___B___ Uses protocol ports in the header
Physical Layer Bytes = A + B + C + D + E
% overhead = number bytes overhead data in physical layer * 100%
total number of physical layer bytes
= ((B + C + D + E) /( A + B + C + D + E )) * 100%
Transport Layer Packet = A + B
% overhead = number bytes overhead data in transport layer * 100%
total number of transport layer bytes
= (B/( A + B)) * 100%
5.1 A signal has 100 possible levels. This signal is
5.2 As the period of a sine wave increases, the frequency __________________ .
5.3 A possible unit for analog bandwidth is bits per second.
5.4 A simple sine wave is created by adding multiple composite signals together.
__D____ The difference between the high frequency and the low frequency
__B____ The number of cycles per second
__C____ The absolute value of the peak value
__A_____ The time for one cycle
s(t) =3sin(2π5t + π/2)
s(t) = 4sin (2πt + π)
Top: 3π/2
Middle: 0
Bottom: π
T = 1 / f
= 1 = 1 = 0.2 x 10-6 sec = 0.2µsec
5 MHz 5 x106 cycles/sec cycle
f = 1 / T
= 1 cycle = 1 cycle = 0.25 x 103 cycles = 0.25 KHz
4 msec 4 x 10-3sec sec
f(t) = 2sin(2π5t) + 3sin(2π7t) + 4sin(2π9t)
BW = high frequency – low frequency
= 9 - 5
= 4 Hz
hihi
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