Finding freq., Wavelentgh, Phase Velocity, and attenuation constant

In summary: Is your question about the solution for part (d) or about the original problem? In part (d), we are not solving for t, we are solving for the attenuation constant, alpha. The equation given in the problem is a general form for a wave function, so the sin term is not needed to solve for alpha.
  • #1
VinnyCee
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0

Homework Statement



Given...

[tex]v\left(z,\,t\right)\,=\,5\,e^{-\alpha\,z}\,sin\left(4\pi\,\times\,10^9\,t\,-\,20\pi\,z\right)[/tex]

where z is distance (m), find...

(a) Frequency

(b) Wavelength

(c) Phase Velocity

(d) At z = 2m, the amplitude is 1 [V], Find the attenuation constant ([itex]\alpha[/itex]).

Homework Equations



[tex]f\,=\,\frac{1}{T}[/tex]

[tex]y\left(x,\,t\right)\,=\,A\,cos\left(\frac{2\pi\,t}{T}\,-\,\frac{2\pi\,x}{\lambda}\,+\,\phi_0\right)[/tex]

[tex]u_p\,=\,f\,\lambda[/tex]

The Attempt at a Solution

(a)

Using the first term ([itex]\frac{2\pi\,t}{T}[/itex]) in the argument to the cosine in the general form above...

[tex]\frac{2\pi}{T}\,=\,4\pi\,\times\,10^9\,\,\longrightarrow\,\,T\,=\,\frac{2\pi}{4\pi\,\times\,10^9}\,=\,0.5\,\times\,10^{-9}[/tex]

[tex]f\,=\,\frac{1}{T}\,=\,\frac{1}{0.5\,\times\,10^{-9}}\,=\,2\,\times\,10^9\,=\,2\,Ghz[/tex](b)

Using the second term ([itex]-\,\frac{2\pi\,x}{\lambda}[/itex]) in the argument to the cosine in the general form above...

[tex]\frac{2\pi}{\lambda}\,=\,20\pi\,\,\longrightarrow\,\,\lambda\,=\,\frac{2\pi}{20\pi}\,=\,\frac{1}{10}\,=\,0.1\,m[/tex](c)

[tex]u_p\,=\,f\,\lambda\,=\,\left(2\,\times\,10^9\right)\,(0.1)\,=\,200,000,000\,\frac{m}{s}[/tex](d)

[tex]1\,=\,5\,e^{-2\,\alpha}\,sin\left(4\pi\,\times\,10^9\,t\,-\,40\pi\right)[/tex]

[tex]5\,e^{-2\alpha}\,=\,1\,\,\longrightarrow\,\,-2\alpha\,=\,ln\left(\frac{1}{5}\right)\,\,\longrightarrow\,\,\alpha\,=\,0.8047[/tex]

Right?
 
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  • #2
How did you eliminate 't' in part d?
 
  • #3
There is no "t" in (d). The "attenuation" constant is the rate at which the magnitude of the wave degrades- and that depends entirely upon the coefficient of the cosine term, [itex]5e^{-\alpha z}[/itex]. And here, we are given that z= 2.
 
  • #4
What about the sin term. 1= 5xexp(-alpha x z) x sin term which contains t?
 

What are frequency, wavelength, phase velocity, and attenuation constant?

Frequency is the number of cycles or oscillations of a wave that occur in one second. Wavelength is the distance between two consecutive peaks or troughs of a wave. Phase velocity is the speed at which a specific point on a wave travels. Attenuation constant is the rate at which a wave's amplitude decreases over distance.

How do you find the frequency of a wave?

The frequency of a wave can be found by dividing the speed of the wave by its wavelength. This can be represented by the equation f = v/λ, where f is frequency, v is speed, and λ is wavelength.

What is the relationship between wavelength and frequency?

Wavelength and frequency are inversely proportional. This means that as the frequency of a wave increases, its wavelength decreases, and vice versa.

How does phase velocity differ from group velocity?

Phase velocity is the speed at which a specific point on a wave travels, while group velocity is the speed at which the overall shape or envelope of a wave travels. In some cases, the two velocities may be the same, but in other cases, they may differ.

How does attenuation constant affect the propagation of a wave?

The attenuation constant determines how quickly a wave's amplitude decreases over distance. A higher attenuation constant means the wave will lose its energy and amplitude faster, resulting in a shorter propagation distance. This is important to consider in communication systems to ensure the signal reaches its intended destination with enough strength.

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