Homework 2 - City Tech OpenLab
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Juan Carlos Torres Homework 2 1-1 Calculate the free-space wavelength in meters for the following frequencies: a) 2KHz π 3π₯108 π/π π= = = π. πππππ π 3 π 2π₯10 π»π§ b) 200KHz π 3π₯108 π/π π= = = π. πππππ π π 2π₯105 π»π§ c) 20MHz π 3π₯108 π/π π= = = πππ π 20π₯106 π»π§ d) 2GHz π= π 3π₯108 π/π = = π. πππ π 2π₯109 π»π§ 1-2 Calculate the free-space wavelength in meters for the following frequencies: a) 80KHz π= π 3π₯108 π/π = = πππππ π 80π₯103 π»π§ π= π 3π₯108 π/π = = ππ. ππ π 8π₯108 π»π§ b) 8MHz c) 800MHz π 3π₯108 π/π π= = = π. ππππ π 800π₯106 π»π§ Juan Carlos Torres d) 8GHz π 3π₯108 π/π π= = = ππ. ππππ−π π π 8π₯109 π»π§ 1-3 Calculate the free-space wavelength in miles for a frequency of 400 Hz π= π 186000ππ/π = = πππππ π 400π»π§ 1-4 Calculate the free-space wavelength in miles for a frequency of 1.5KHz π= π 186000ππ/π = = πππππ π 1.5π₯103 π»π§ 1-5 A sinusoidal signal has a free-space wavelength of 80m. Calculate the frequency. π 3π₯108 π/π π= = = π. πππ²π―π π 80 π π= π 1-6 A sinusoidal signal has a free-space wavelength of 6m. Calculate the frequency. π= π π π= π 3π₯108 π/π = = πππ΄π―π π 6 1-7 A digital signal utilizes pulses whose minimum widths are about 3ns. Assuming the speed of light, determine the longest lengths of wire-pairs that can be allowed based on the 10% rule. π‘1 = 0.1 π₯ 3ππ = 0.3ππ π‘1 = π = π π = π‘1 π₯ π = (3π₯108 π/π )π₯ (0.3π₯10−9 π/π ) = π. πππ 1-8 The longest connecting wires in a digital system are about 20cm. Assuming the speed of light, determine the shortest acceptable pulse width based on the 10% rule. π‘1 = π 0.2π = = 66.6π₯10−8 π = ππππ π 3π₯108 π/π Juan Carlos Torres π‘1 = 10% π₯ π€ = π‘1 66.6π₯10−8 π π€= = = πππππ 10% 0.1 1-9 A communication system operates at a frequency of 800MHz. Assuming the speed of light, determine the length of connecting line that could be used without considering frequency-domain effects based on the 10% rule. π= π 3π₯108 π/π = = πππππ π 800π₯106 π»π§ 1-10 The length of a connecting cable between two points in a radio-frequency system is 50cm. Assuming the speed of light, determine the highest operating frequency that should be used without considering frequency-domain effects based on the 10% rule. π π= π π 3π₯108 π/π π= = = ππππ΄π―π π 0.5 1-11 In a coil, a current of 100mA results in a magnetic flux of 50μWb. Determine the inductance. π»= π½ 50πππ 50π₯10−6 ππ = = = π. ππ― πΌ 100ππ΄ 100π₯10−3 π΄ 1-12 A current of 4mA is flowing in a 20μH coil. Determine the magnetic flux. π»= π½ πΌ π½ = π»π₯πΌ = (20π₯10−6 π»)π₯(4π₯10−3 π΄) = ππππΎπ 1-13 In a capacitor, a voltage of 20 V results in charges storages of 5μC. Determine the capacitance. πΉ= πβππππ(π) 5π₯10−6 πΆππ = = πππππ π£πππ‘πππ (π) 20π 1-14 A capacitance of 40μF is charged to a voltage of 12V. Determine the electric charge. πΉ= πβππππ(π) π£πππ‘πππ (π) π = πΉ π₯ π = (40π₯10−6 πΉ)π₯(12π) = ππππͺππ 1-15 A lossless transmission line has an inductance of 320 nH/m and a capacitance of 57pF/m. Determine the characteristic impedance. Juan Carlos Torres πΏ 320π₯10−9 π»/π π π = √ = √ = √5614 = ππ. πππ πΆ 57π₯10−12 πΉ/π 1-16 A lossless transmission line has an inductance of 1.2 μH/m and a capacitance of 15pF/m. Determine the characteristic impedance. πΏ 1.2π₯10−9 π»/π π π = √ = √ = √80000 = πππ. πππ πΆ 57π₯10−12 πΉ/π 1-17 The dielectric constant of mica is 6. Determine the permittivity. ∈=∈π π₯ ∈° = 6 π₯ (8.842π₯10−12 πΉ/π) = ππ. πππ πππ−π π/π 1-18 The permittivity of a material is 14x10-12F/m. Determine the dielectric constant. ∈=∈π π₯ ∈° ∈ 14π₯10−12 πΉ/π ∈π = = = π. ππ ∈° 8.842π₯10−12 πΉ/π 1-19 The relative permeability of nickel is 800. Determine the actual permeability. π = ππ π₯ ππ = 800 π₯ (4ππ₯10−7 π»/π) = πππ―/π 1-20 The permeability of a ferromagnetic material is 10-4H/m. Determine the relative permeability. π = π π π₯ ππ ππ = π 10−4 π»/π = = πππ. ππ π° 4ππ₯10−7 π»/π 1-21 Determine the velocity of propagation of the transmission line of Problem 1-15. π= 1 √πΏπ₯πΆ = 1 √(320π₯10−9 π»/π)π₯(57π₯10−12 πΉ/π) = π. ππππππ π/π 1-22 Determine the velocity of propagation of the transmission line of Problem 1-16. π= 1 √πΏπ₯πΆ = 1 √(1.2π₯10−6 π»/π)π₯(15π₯10−12 πΉ/π) = π. ππππππ π/π Juan Carlos Torres 1-23 The dielectric constant in a transmission line is 4.7, and μ=μo. Determine the velocity of propagation. ∈=∈π π₯ ∈° = 4.7 π₯ (8.842π₯10−12 πΉ/π) = 41.55 π₯10−2 πΉ/π π= π = ππ = (4ππ₯10−7 π»/π) 1 1 1 1 = = = √ππ₯π √(4ππ₯10−7 π»/π)π₯(41.55 π₯10−2 πΉ/π) √5.22 π₯10−17 7.22 π₯10−9 = π. ππππππ π/π 1-24 The dielectric constant in a transmission line 3, and μ=μo. Determine the velocity of propagation. ∈=∈π π₯ ∈° = 3 π₯ (8.842π₯10−12 πΉ/π) = 26.52 π₯10−2 πΉ/π π= π = ππ = (4ππ₯10−7 π»/π) 1 1 1 1 = = = √ππ₯π √(4ππ₯10−7 π»/π)π₯(26.52 π₯10−2 πΉ/π) √3.33 π₯10−17 5.77 π₯10−9 = π. ππππππ π/π 1-25 A coaxial cable has the following specifications: Ro=73β¦, and velocity of propagation is 2.1x108m/s. Determine L and C. πΏ= πΆ= π π 73Ω = = π. ππ πππ−π π―/π 8 π 2.1 π₯10 π/π 1 1 = = ππ. ππ πππ−ππ π/π π π π 73Ω x (2.1 π₯108 π/π ) 1-26 A transmission line has the following specifications: Ro=150β¦, and velocity of propagation is 0.8. Determine L and C. ππ = π π π = ππ π₯ π = 0.8 π₯ (3π₯108 ) = 2.4π₯108 π/π πΏ= πΆ= π π 150Ω = = π. ππ πππ−π π―/π π 2.4 π₯108 π/π 1 1 = = ππ. ππ πππ−ππ π/π 8 π π π 150Ω x (2.4 π₯10 π/π ) Juan Carlos Torres 1-27 The lower end of the commercial amplitude-modulation (AM) band is about 550KHz. AM stations use "quarter-wave" vertical antennas whose lengths are 0.25λ. Determine the length in meters of a vertical antenna operating at the lower end. π= π π 3π₯108 π/π = 545.45π 550π₯103 π»π§ 0.25π = 0.25π = 545.45π π= 545.45π = πππππ 0.25 1-28 The upper end of the commercial AM band referred to in Problem 1-27 is about 1610KHz. Determine the length in meters of a vertical antenna operating at the upper end. 3π₯108 π/π 0.25π = = 186.3π 1610π₯103 π»π§ π π= π 0.25π = 186.3π π= 186.3π = ππππ 0.25 1-29 One popular simple antenna is the "half-wave" horizontal antenna whose theoretical length is 0.5λ at the operating frequency. In practice, however, the antenna is usually shorted by about 5% due to end effects. Determine the practical length in meters for a half-wave antenna to provide optimum reception at the lower end of the commercial FM band, which is about 88 MHz π= π π 0.5π = 186.3π π = 3.4π π₯ 0.05 = 0.17 π= 3π₯108 π/π = 3.4π 88π₯103 π»π§ π= 186.3π = 372.6π 0.5 π = 3.4π π₯ 0.17 = π. πππ 1-30 Based on the discussion of Problem 1-29, determine the practical length in meters for a half-wave antenna to provide optimum reception at the upper end of the FM band, which is about 108MHz. π= π π π= 3π₯108 π/π = 2.77π 108π₯103 π»π§ π = 5.55π π₯ 0.05 = 0.27 π= 2.77 = 5.55π 0.5 π = 5.55π π₯ 0.27 = π. πππ Juan Carlos Torres 1-31Show that the free-space velocity of light in feet/second is very close to c=982x106ft/s. 3π₯108 π π = 3π₯108 π/π 1ππ‘ = π π₯ 3.2808 ? ππ‘ = π 1π 1-32 Show that the free-space wavelength in feet can be expressed as π(ππ‘) = π 982 = π π(ππ»π§) π 984.25197π₯106 ππ‘ 10−6 πππ. ππππ π= = π₯ = π π(π»π§) 106 π(π΄π) You may use the result of Problem 1-31 = π. ππππππππ
