# EE313 HW3 Solutions Problem 2-1

```EE313 HW3 Solutions
Problem 2-1
Determine the period given the frequency of the waveform.
a) π = 2 ππ»π§
π=
1
1
*π = π =&gt; π = π
1
= 0.5π₯10−6 = π. π &micro;π
2π₯106
b) π = 500 ππ»π§
1
π=
= 2π₯10−6 = π &micro;π
500π₯103
c) π = 4.27 ππ»π§
1
π=
= 0.234π₯10−6 = π. πππ &micro;π = πππ ππ
4.27π₯106
d) π = 17 ππ»π§
1
π=
= 0.0588π₯10−6 = π. ππππ &micro;π = ππ. π ππ
17π₯106
Determine the frequency given the period of the waveform.
e) π = 2 &micro;π
1
π=
= 0.5π₯106 = π. π π΄π―π = πππ ππ―π
2π₯10−6
f) π = 100 &micro;π
1
π=
= 0.01π₯106 = ππ ππ―π
−6
100π₯10
g) π = 0.75 ππ
1
π=
= 1.333π₯103 = π. πππ ππ―π
0.75π₯10−3
h) π = 1.5 &micro;π
1
π=
= 0.667π₯106 = πππ ππ―π
1.5π₯10−6
Problem 2-2
Draw the Serial and Parallel waveforms. Calculate how long it will take to transmit if
πΆπΆππ = 2ππ»π§.
a) 45B16 = {4} {5} {π΅} = {0100} {0101} {1011}
Serial
πΆπΆππ
ππππ
One
Clock
Cycle
1
0
1
1
0
1
0
1
1
0
1
0
0
0
1
LSB
MSB
1
= 0.5π₯10−6 = π. π &micro;π
2π₯106
Duration of Transmission: 0.5 &micro;s x 12 clock-cycles = 6 &micro;s
π=
Parallel
One Clock
Cycle
πΆπΆππ
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
= 0.5π₯10−6 = π. π &micro;π
2π₯106
Duration of Transmission:
0.5 &micro;s x 3 clock-cycles = 1.5 &micro;s
π=
b) A3C16 = {π΄} {3} {πΆπΆ} = {1010} {0011} {1100}
Serial
πΆπΆππ
ππππ
One
Clock
Cycle
1
0
1
0
0
1
0
1
1
1
0
0
0
LSB
1
0
1
MSB
1
= 0.5π₯10−6 = π. π &micro;π
6
2π₯10
Duration of Transmission: 0.5 &micro;s x 12 clock-cycles = 6 &micro;s
π=
Parallel
One Clock
Cycle
πΆπΆππ
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
= 0.5π₯10−6 = π. π &micro;π
2π₯106
Duration of Transmission:
0.5 &micro;s x 3 clock-cycles = 1.5 &micro;s
π=
Problem 2-3
a) ππππ
Divide by Desired Base
33 &divide; 2 = 16
16 &divide; 2 = 8
8&divide;2=4
4&divide;2=2
2&divide;2=1
1&divide;2=0
ππππ = ππππ πππππ
π=
Remainder times Base
0.5 π₯ 2
0.0 π₯ 2
0.0 π₯ 2
0.0 π₯ 2
0.0 π₯ 2
0.5 π₯ 2
=
=
=
=
=
=
Output
1 (LSB)
0
0
0
0
1 (MSB)
1
= 0.27π₯10−6 = π. ππ &micro;π
3.7π₯106
Duration of Transmission: 0.27 &micro;s x 8 clock-cycles = 2.16 &micro;s
b)
ππππ πππ πππππ πππ π ππππ
ππππ πππ πππ πΆπππππ−πΆπ¦πππ
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 &micro;s 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
π=
= 0.125π₯10−6 = π. πππ &micro;π
8π₯106
Duration of Transmission: 0.125 &micro;s x 3 clock-cycles = 0.375 &micro;s
b) For \$78.18 it takes 6 clock-cycles.
1
π=
= 0.24π₯10−6 = π. ππ &micro;π
4.17π₯106
Duration of Transmission: 0.24 &micro;s x 6 clock-cycles = 1.44 &micro;s
```