Question

g An electromagnetic wave is traveling on a transmission line. Its voltage is given by v(x, t) = 10 Exp[−γx] cos(15x − 109 t) where x is the distance (in meters) from the generator, and t is in seconds. (a) What is the frequency, wavelength and phase veolicty of the electromagnetic wave. (b) at x = 3 m, the wave amplitude is measured to be 4V . Determine γ

Answer 1

a)

f = 1.59*10⁸ Hz.

λ = 0.42 m

vp = 0.67*10⁸ m/s = 0.22 c

b)

γ = 0.31 1/m

Step-by-step explanation:

a)

• In a transmission line, voltage varies sinusoidally with the distance and time.
• We can express mathematically this relationship in a cosine form as follows:
• V (x,t) = A cos (kx-ωt + φ₀) (1)
• where A = amplitude of the wave (Volts)
• k = number of wave = 2*π/λ (being λ the wavelength of the wave)
• ω = angular frequency = 2*π*f (being f the frequency of the wave)
• φ₀ = phase of the wave.

• In our case, the expression for the voltage is as follows:
• v (x,t) = 10 V Exp [-γx] cos (15x-10⁹ t) (2)
• From (1) and (2) we find the following equalities:
• k = 15 = 2*π/λ (3)
• Solving for λ:

`\lambda = (2*\pi )/(k) =(2*\pi )/((15)1/m) = 0.42 m (4)`

• ω = 2*π*f = 10⁹ rad/sec (5)
• Solving for f:

`f = (\omega )/(2*\pi ) =(10e9rad/sec )/((2*\pi rad) = 1.59*e8 (6)`

• The phase velocity of the wave is just the product of the wavelength times the frequency, as follows:

`v_(phase) =\lambda * f = 0.42m * 1.59e8 1/sec = 0.67e8 m/s (7)`

b)

• We can express the amplitude at any time as follows:

`V = 10 V *e^(-\gamma*x) (8)`

• If we know that V= 4 V at x=3m, (8) becomes:

`10 V *e^(-\gamma*3m) = 4 V (9)`

• Taking ln on both sides, we can solve for γ as follows:

`-\gamma*3m* ln e = -\gamma*3m = ln ((4V)/(10V)) = ln 0.4\\\gamma = -(ln 0.4)/(3m) = 0.31 (1/m) (10)`

γ = 0.31 1/m