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 \(\phi\).

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.

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