Ayouba Ouedraogo
Professor: Viviana
TCET 2220
Date; 09/26/2013
Homework#2
Problem#1.1
The free-space wavelength in meters for the following frequencies
a) f=2kh Λ =C/f =3×10^8m/s/2×10^3 =15×10^4m
b) f= 200khz
Λ =C/f =3×10^8m/s/2×10^5Hz =1500m
c) f= 20MHz
Λ =C/f =3×10^8m/s/2×10^7Hz =15m
d) f=2GHz
Λ =C/f =3×10^8m/s/2×10^9 =0.15m
Problem#1.2
The free-space wavelength in meters for the following frequencies
a) f=80kHz
Λ =C/f =3×10^8m/s/8×10^4Hz =3750m
b) f= 8MHz
Λ =C/f =3×10^8m/s/8×10^6Hz =37.5m
c) f=800MHz
Λ =C/f =3×10^10m/s/8×10^8 =0.375m
d) f= 8GHz
Λ =C/f =3×10^8m/s/8×10^9Hz =0.0375m
Problem#1.3
The free-space wavelength in miles for 400Hz
Λ =C/f =186,000mi/s/400Hz =465mi
Problem#1.4
The free-space wavelength in miles for f=1.5 kHz
Λ =C/f=186,000mi/s/1500Hz =124mi
Problem#1.5f
The frequency is
f=c/λ 3×10^8m/s/80m =375×10^4m
Problem#1.6
The frequency is
f =c/λ =3×10^8m/s/6m =5×10^7Hz
Problem#1.7
Based on the 10% rule t =0.3ns
The length d is
d =Ct (3×10^8)(0.3×10^-9) 0.09m
Problem#1.8
Based on 10% rule the longest propagation allows is 0.1x20cm =2cm =0.02m
Problem#1.9
The wavelength at a frequency of 800MHz is
Λ =C/f =3×10^8m/s/8×10^8 =0.375m
Problem# 1.10
The highest operating frequency is
f =c/λ =3×10^8m/s/0.5m =6×10^8Hz
Problem#1.11
The inductance is
Inductance = Flux/Current =5×10^-5Wb/Current =5×10^-5Wb/0.1A =500×10^-6H
Problem#12
The magnitude flux is
Flux = Inductance x Current
Flux = 20×10^-6H x 4×10^-3A =80×10^-9 Wb
Problem#1.13
The capacitance is
Capacitance = Charge/Voltage =5×10^-6C/20V =0.25×10^-6F
Problem#1.14
The electric charge is
Charge = Capacitance x Voltage = 40x 10^-6Fx 12V =480×10^-6C
Problem#1.15
The characteristic impedance
R0=√L/C =√320×10^-9H/m/57×10^-12F/m =74.93Ω
Problem#1.16
The characteristic impedance is
R0 =√L/C =√1.2×10^-6H/m/15×10^-12F/m =282.84Ω
Problem#1.17
The permittivity is
ɛ =ɛrɛ0 = (6×1)x10^-9/36×3.14 =0.053×10^-9F/m
Problem#1.18
The dielectric constant is
ɛr =ɛ/ɛ0 =(14×10^-12F/m):(1×10^-9F/m/36×3.14) =1.58
Problem#1.19
The actual permeability is
µ =µrµ0 =(800x4x3.140)x10^-7H/m =1.005×10^-3H/m
Problem#1.20
The relative permeability is
µr =µ/µ0 =(10^-4H/m):(4×3.14×10^-7H/m) =79.58
Problem#1.21
The velocity of propagation of the transmission line
V=1/√LC = 1/√320×10^-19x57x10^-12 =2.34×10^8m/s
Problem#1.22
The velocity of propagation of transmission line
V=1/√LC = 1/√(1.2×15)x10^-18 =0.235×10^9m/s
Problem#1.23
The velocity of propagation
V=1/√µɛ =1/√4×3.14×10^-7×4.7 =0.0169×10^7m/s
Problem#1.24
The velocity of propagation
V =1/√µɛ =1/√(4×3.14×3)x10^-7 =515m/s
Problem#1.25
The value of L
L =R0/V =73Ω/2.1×10^8m/s =34.76×10^-8Ω
Problem#1.26
The velocity is
V =0.8C =0.8(3×10^8) =2.4×10^8m/s
The value of L
L =R0/V =150Ω/2.4×10^8m/s =62.5×10^-8Ω
The value of C
C =1/R0V =1/150x24x10^8 =
2.77×10^-11Ω
Problem#1.27
The length in meters of the vertical antenna
Λ =C/f =L/0.25
C/f =L/0.25
L =0.25C/f =0.25x3x10^8m/s/550×10^3 =136.36m
Problem#1.28
The length in meters of a vertical antenna
L =0.25C/f =0.25x3x10^8m/s/1610×10^3Hz =4.65m
Problem#1.29
The length of antenna base on 5% shortened
L’ =0.5λx0.05 =0.025λ
L =0.5λ-0.025λ =0.47λ
The pratical length in meters
L =0.47C/t =0.47x3x10^8m/s/88×10^6hz =1.61m
Problem#1.30
The practical length in meter is
L =0.47C/f =0.47x3x10^8/108×10^6 =1.30m
Problem#1.31
The free-space velocity of light
3×10^8m/s x 0.9113ft/s/0.2778m/s =984×10^6ft/s
Problem#1.32
I show that the free-space wavelength in feet can be expressed
Λ (t) =984×10^6ft/s/f
