Module: TM111
Assignment: TMA 03
, Question 1
a.
The corresponding frequency of 460μs in kHz is:-
Given :- Period (t) = 460μs = 460x10−6 s
Frequency (f) = 1 / t
= x 10−6
= 2.
=2.17 kHz (3 s.f)
If the frequency is reduced by a factor of 4.25, the new period value
will be: -
f 2.173913043
New frequency (f) = =¿
4 ⋅ 25 4 ⋅25
= 5.115089514
= 5.11 μs (to 3 s.f)
b.
IP address 187.31.247.11 converted to binary is: -
Decimal Binary
187 – 128 = 59 1
59 – 64 0
59 – 32 = 27 1
27 – 16 = 11 1
11 – 8 = 3 1
3-4 0
3–2=1 1
1 – 1= 0 1
2|Page
Assignment: TMA 03
, Question 1
a.
The corresponding frequency of 460μs in kHz is:-
Given :- Period (t) = 460μs = 460x10−6 s
Frequency (f) = 1 / t
= x 10−6
= 2.
=2.17 kHz (3 s.f)
If the frequency is reduced by a factor of 4.25, the new period value
will be: -
f 2.173913043
New frequency (f) = =¿
4 ⋅ 25 4 ⋅25
= 5.115089514
= 5.11 μs (to 3 s.f)
b.
IP address 187.31.247.11 converted to binary is: -
Decimal Binary
187 – 128 = 59 1
59 – 64 0
59 – 32 = 27 1
27 – 16 = 11 1
11 – 8 = 3 1
3-4 0
3–2=1 1
1 – 1= 0 1
2|Page
