EE313 HW3 Solutions
Problem 2-1
Determine the period given the frequency of the waveform.
a) 𝑓 = 2 𝑀𝐻𝑧
𝑇=
1
1
*𝑓 = 𝑇 => 𝑇 = 𝑓
1
= 0.5𝑥10−6 = 𝟎. 𝟓 µ𝒔
2𝑥106
b) 𝑓 = 500 𝑘𝐻𝑧
1
𝑇=
= 2𝑥10−6 = 𝟐 µ𝒔
500𝑥103
c) 𝑓 = 4.27 𝑀𝐻𝑧
1
𝑇=
= 0.234𝑥10−6 = 𝟎. 𝟐𝟑𝟒 µ𝒔 = 𝟐𝟑𝟒 𝒏𝒔
4.27𝑥106
d) 𝑓 = 17 𝑀𝐻𝑧
1
𝑇=
= 0.0588𝑥10−6 = 𝟎. 𝟎𝟓𝟖𝟖 µ𝒔 = 𝟓𝟖. 𝟖 𝒏𝒔
17𝑥106
Determine the frequency given the period of the waveform.
e) 𝑇 = 2 µ𝑠
1
𝑇=
= 0.5𝑥106 = 𝟎. 𝟓 𝑴𝑯𝒛 = 𝟓𝟎𝟎 𝒌𝑯𝒛
2𝑥10−6
f) 𝑇 = 100 µ𝑠
1
𝑇=
= 0.01𝑥106 = 𝟏𝟎 𝒌𝑯𝒛
−6
100𝑥10
g) 𝑇 = 0.75 𝑚𝑠
1
𝑇=
= 1.333𝑥103 = 𝟏. 𝟑𝟑𝟑 𝒌𝑯𝒛
0.75𝑥10−3
h) 𝑇 = 1.5 µ𝑠
1
𝑇=
= 0.667𝑥106 = 𝟔𝟔𝟕 𝒌𝑯𝒛
1.5𝑥10−6
Problem 2-2
Draw the Serial and Parallel waveforms. Calculate how long it will take to transmit if
𝐶𝐶𝑝𝑝 = 2𝑀𝐻𝑧.
a) 45B16 = {4} {5} {𝐵} = {0100} {0101} {1011}
Serial
𝐶𝐶𝑝𝑝
𝑆𝑆𝑜𝑜
One
Clock
Cycle
1
0
1
1
0
1
0
1
1
0
1
0
0
0
1
LSB
MSB
1
= 0.5𝑥10−6 = 𝟎. 𝟓 µ𝒔
2𝑥106
Duration of Transmission: 0.5 µs x 12 clock-cycles = 6 µs
𝑇=
Parallel
One Clock
Cycle
𝐶𝐶𝑝𝑝
LSB 20
21
22
MSB 23
1
0
1
0
1
0
1
0
1
0
0
1
1
0
1
0
0
0
1
1
1
0
0
1
= 0.5𝑥10−6 = 𝟎. 𝟓 µ𝒔
2𝑥106
Duration of Transmission:
0.5 µs x 3 clock-cycles = 1.5 µs
𝑇=
b) A3C16 = {𝐴} {3} {𝐶𝐶} = {1010} {0011} {1100}
Serial
𝐶𝐶𝑝𝑝
𝑆𝑆𝑜𝑜
One
Clock
Cycle
1
0
1
0
0
1
0
1
1
1
0
0
0
LSB
1
0
1
MSB
1
= 0.5𝑥10−6 = 𝟎. 𝟓 µ𝒔
6
2𝑥10
Duration of Transmission: 0.5 µs x 12 clock-cycles = 6 µs
𝑇=
Parallel
One Clock
Cycle
𝐶𝐶𝑝𝑝
LSB 20
21
22
MSB 23
1
0
1
0
1
0
1
0
1
0
0
1
0
0
1
1
1
0
0
1
0
1
1
= 0.5𝑥10−6 = 𝟎. 𝟓 µ𝒔
2𝑥106
Duration of Transmission:
0.5 µs x 3 clock-cycles = 1.5 µs
𝑇=
Problem 2-3
a) 𝟑𝟑𝟏𝟎
Divide by Desired Base
33 ÷ 2 = 16
16 ÷ 2 = 8
8÷2=4
4÷2=2
2÷2=1
1÷2=0
𝟑𝟑𝟏𝟎 = 𝟎𝟎𝟏𝟎 𝟎𝟎𝟎𝟏𝟐
𝑇=
Remainder times Base
0.5 𝑥 2
0.0 𝑥 2
0.0 𝑥 2
0.0 𝑥 2
0.0 𝑥 2
0.5 𝑥 2
=
=
=
=
=
=
Output
1 (LSB)
0
0
0
0
1 (MSB)
1
= 0.27𝑥10−6 = 𝟎. 𝟐𝟕 µ𝒔
3.7𝑥106
Duration of Transmission: 0.27 µs x 8 clock-cycles = 2.16 µs
b)
𝑇𝑖𝑚𝑒 𝑜𝑜𝑓 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑜𝑛
𝑇𝑖𝑚𝑒 𝑜𝑜𝑓 𝑂𝑛𝑒 𝐶𝑙𝑜𝑜𝑐𝑘−𝐶𝑦𝑐𝑙𝑒
1.21
= 0.27 = 4.48
This tells us that the transmission has already transmitted the 4th Bit and is almost half
way through transmitting the 5th Bit. Starting with the LSB the 5th Bit is “0” therefore at
1.21 µs the serial line is Low.
Problem 2-4
a) 7-Bits make up each character in ASCII code and since transmission is 8-Bit Parallel it will
take 3 clock-cycles to transmit $14.
1
𝑇=
= 0.125𝑥10−6 = 𝟎. 𝟏𝟐𝟓 µ𝒔
8𝑥106
Duration of Transmission: 0.125 µs x 3 clock-cycles = 0.375 µs
b) For $78.18 it takes 6 clock-cycles.
1
𝑇=
= 0.24𝑥10−6 = 𝟎. 𝟐𝟒 µ𝒔
4.17𝑥106
Duration of Transmission: 0.24 µs x 6 clock-cycles = 1.44 µs