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A sinusoidal wave on a transmission line is specified by voltage and current in phasor form: $$V_{s}(z)=V_{0} e^{\alpha z} e^{j \beta z} \quad \text { and } \quad I_{s}(z)=I_{0} e^{\alpha z} e^{j \beta z} e^{j \phi}$$ where \(V_{0}\) and \(I_{0}\) are both real. (a) In which direction does this wave propagate and why? \((b)\) It is found that \(\alpha=0, Z_{0}=50 \Omega\), and the wave velocity is \(v_{p}=2.5 \times 10^{8} \mathrm{~m} / \mathrm{s}\), with \(\omega=10^{8} \mathrm{~s}^{-1}\). Evaluate \(R, G, L, C, \lambda\) and \(\phi\).

Short Answer

Expert verified
Answer: The values for the transmission line are as follows: - Direction of wave propagation: positive z-direction - R = 0 - G = 0 - L = 2 x 10^-9 H/m - C = 4 x 10^-12 F/m - λ = 2.5 m - φ = 0

Step by step solution

01

(a) Direction of wave propagation

To determine the direction of wave propagation, let's look at the expressions for \(V_s(z)\) and \(I_s(z)\). We know that if the wave is propagating in the positive \(z\) direction, the exponents in both expressions should be negative. Conversely, if the wave is propagating in the negative \(z\) direction, the exponents should be positive. In this case, the exponent in both expressions is: $$e^{\alpha z} e^{j \beta z}$$ Since \(\alpha z\) and \(j \beta z\) have the same sign, the wave is propagating in one direction. If the signs were opposite, the wave would propagate in the opposite direction. In this case, both exponents have the same sign. Therefore, the wave is propagating in the positive z-direction.
02

(b) Evaluating R, G, L, C, λ, and φ

First, we can use the given values for \(\alpha\), \(Z_0\), \(v_p\), and \(\omega\) to find the phase constant \(\beta\) using the formula for wave velocity: $$v_p = \frac{\omega}{\beta}$$ Now, solving for \(\beta\): $$\beta = \frac{\omega}{v_p} = \frac{10^8 s^{-1}}{2.5 \times 10^8 m/s} = \frac{2\pi}{\lambda}$$ So we can find \(\lambda\): $$\lambda = \frac{2\pi}{\beta} = 2.5 m$$ Next, we can use the transmission line formulas to determine \(R\), \(G\), \(L\), and \(C\). Since \(\alpha=0\), we can assume that there is no loss in the transmission line. Therefore, the primary constants (R, G, L, and C) can be determined as follows: $$R = 0 \,\text{since}\, \alpha = 0$$ $$G = 0 \,\text{since}\, \alpha = 0$$ Now, we can find L and C using the given characteristic impedance \(Z_0\) and the wave velocity \(v_p\): $$Z_0 = \sqrt{\frac{R + j\omega L}{G + j\omega C}} \Rightarrow 50\Omega = \sqrt{\frac{j\omega L}{j\omega C}}$$ From this equation, we have: $$\frac{L}{C} = \left(\frac{50\Omega}{\omega}\right)^2$$ And since the wave velocity \(v_p\) is given by: $$v_p = \frac{1}{\sqrt{LC}}$$ We can solve for L and C: $$L = \frac{50\Omega}{\omega} \cdot \left(\sqrt{\frac{1}{v_p^2 \cdot C}}\right)$$ $$C = \frac{1}{\omega \cdot 50\Omega} \cdot \left(\sqrt{\frac{1}{v_p^2 \cdot L}}\right)$$ Using the given values for \(\omega\) and \(v_p\), we get: $$L = 2 \times 10^{-9} H/m$$ $$C = 4 \times 10^{-12} F/m$$ Finally, to find the phase angle \(\phi\), we can look at the relation between \(V_s(z)\) and \(I_s(z)\): $$I_s(z) = \frac{V_s(z)}{Z_0} e^{j\phi}$$ Since \(V_s(z)\) and \(I_s(z)\) are given in terms of real values \(V_0\) and \(I_0\), their ratios are also real. Therefore, the phase angle \(\phi\) between the voltage and the current will be zero: $$\phi = 0$$ In conclusion, the values for the transmission line are as follows: - Direction of wave propagation: positive z-direction - R = 0 - G = 0 - L = 2 x 10^-9 H/m - C = 4 x 10^-12 F/m - λ = 2.5 m - φ = 0

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Propagation
Imagine a wave traveling along a string. It moves from one end to the other, carrying energy and information. In a transmission line, which could be a wire or a cable, wave propagation works similarly by transmitting signals in the form of electromagnetic waves. This concept is essential in telecommunications and power systems as it explains how signals move through cables.
  • Wave propagation can occur in positive or negative directions along the transmission line.
  • The direction is determined by the sign of the exponential term in the wave equations. Positive exponents suggest propagation in the negative direction, and vice versa.
  • Understanding wave propagation helps engineers design efficient communication systems.
In our exercise, the wave propagates in the positive z-direction, indicated by the positive exponential expressions in the voltage and current phasor equations. This means the wave travels forward along the line.
Characteristic Impedance
Characteristic impedance, denoted as \(Z_0\), is like the DNA of a transmission line. It defines the relationship between current and voltage in a wave traveling through the line, unaffected by line length or frequency changes.
  • The formula \(Z_0 = \sqrt{\frac{R + j\omega L}{G + j\omega C}}\) relates the primary constants of the line.
  • \(Z_0\) is significant for matching the line with connected devices to minimize reflections and energy loss.
  • If a line is matched to its characteristic impedance, signals are transmitted efficiently with minimal distortion.
In our scenario, \(Z_0\) is given as \(50 \Omega\), meaning that the line is designed to operate optimally with devices that have the same impedance.
Primary Constants
The primary constants of a transmission line are the foundational electrical properties that determine its behavior. These include resistance \(R\), conductance \(G\), inductance \(L\), and capacitance \(C\).
  • Resistance \(R\): Represents how much the line resists the flow of electrical current. In low-loss or ideal lines, \(R = 0\).
  • Conductance \(G\): Shows how well the line can conduct electrical signals. In our case, \(G = 0\), meaning electrical leakage is negligible.
  • Inductance \(L\): Indicates the line's property to store electrical energy in a magnetic field. Our exercise shows \(L = 2 \times 10^{-9} H/m\).
  • Capacitance \(C\): Represents the line's ability to store electrical energy in an electric field. Here, \(C = 4 \times 10^{-12} F/m\).
These constants define how signals behave on a transmission line. They influence wave speed, energy consumption, and signal integrity.
Phase Constant
The phase constant \(\beta\) is a crucial aspect of wave propagation along the transmission line. It describes the change in phase of the wave per unit length of the line.
  • The wave velocity \(v_p\) is given by \(v_p = \frac{\omega}{\beta}\), linking it with the angular frequency \(\omega\).
  • By knowing \(v_p = 2.5 \times 10^{8} m/s\) and \(\omega = 10^{8} s^{-1}\), you can calculate \(\beta\).
  • For our exercise, \(\beta = \frac{\omega}{v_p} = \frac{10^8 s^{-1}}{2.5 \times 10^8 m/s} = \frac{2\pi}{\lambda}\).
  • This enables us to determine the wavelength \(\lambda = 2.5 m\).
The phase constant is vital in assessing how signals behave as they travel, affecting timing and synchronization in communication systems.

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Most popular questions from this chapter

A sinusoidal voltage source drives the series combination of an impedance, \(Z_{g}=50-j 50 \Omega\), and a lossless transmission line of length \(L\), shorted at the load end. The line characteristic impedance is \(50 \Omega\), and wavelength \(\lambda\) is measured on the line. \((a)\) Determine, in terms of wavelength, the shortest line length that will result in the voltage source driving a total impedance of \(50 \Omega .(b)\) Will other line lengths meet the requirements of part \((a)\) ? If so, what are they?

The characteristic impedance of a certain lossless transmission line is \(72 \Omega\). If \(L=0.5 \mu \mathrm{H} / \mathrm{m}\), find \((a) C ;(b) v_{p} ;(c) \beta\) if \(f=80 \mathrm{MHz} .(d)\) The line is terminated with a load of \(60 \Omega\). Find \(\Gamma\) and \(s\).

In Figure \(10.39, R_{L}=Z_{0}\) and \(R_{g}=Z_{0} / 3\). The switch is closed at \(t=0\). Determine and plot as functions of time \((a)\) the voltage across \(R_{L} ;(b)\) the voltage across \(R_{g} ;(c)\) the current through the battery.

In the transmission line of Figure \(10.20, R_{g}=Z_{0}=50 \Omega\), and \(R_{L}=25 \Omega\). Determine and plot the voltage at the load resistor and the current in the battery as functions of time by constructing appropriate voltage and current reflection diagrams.

The normalized load on a lossless transmission line is \(2+j 1\). Let \(\lambda=20 \mathrm{~m}\) and make use of the Smith chart to find \((a)\) the shortest distance from the load to a point at which \(z_{\text {in }}=r_{\text {in }}+j 0\), where \(r_{\text {in }}>0 ;\) (b) \(z_{\text {in }}\) at this point. (c) The line is cut at this point and the portion containing \(z_{L}\) is thrown away. A resistor \(r=r_{\text {in }}\) of part \((a)\) is connected across the line. What is \(s\) on the remainder of the line? \((d)\) What is the shortest distance from this resistor to a point at which \(z_{\text {in }}=2+j 1 ?\)

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