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Chapter 10: Problem 31

In order to compare the relative sharpness of the maxima and minima of a standing wave, assume a \(\operatorname{load} z_{L}=4+j 0\) is located at \(z=0 .\) Let \(|V|_{\min }=1\) and \(\lambda=1 \mathrm{~m}\). Determine the width of the \((a)\) minimum where \(|V|<1.1 ;(b)\) maximum where \(|V|>4 / 1.1\).

Short Answer

Expert verified
Answer: The approach involves calculating the reflection coefficient and standing wave ratio (SWR) for the given wave, then finding the positions of the minima and maxima using their respective formulas for the specific conditions given in cases (a) and (b). Finally, calculate the width by finding the difference between two consecutive positions of minima or maxima that meet the given criteria for each case. Note that exact numeric values can't be determined since the value of \(Z_0\) is unknown, but general expressions can be obtained for the widths.

Step by step solution

01

Calculate the reflection coefficient and SWR

The reflection coefficient \(\Gamma\) can be calculated using the formula \(\Gamma = \frac{z_L - Z_0}{z_L + Z_0}\), where \(Z_0\) is the characteristic impedance of the transmission line. Without knowing the value of \(Z_0\), we can't determine the exact value of \(\Gamma\). However, we can work with a general expression of \(\Gamma\): \(\Gamma = \frac{4 - Z_0}{4 + Z_0}\), since the load impedance is \(z_L = 4+j0\). Next, we need to calculate the standing wave ratio (SWR) using the reflection coefficient. The SWR can be found using the formula \(SWR = \frac{1 + |\Gamma|}{1 - |\Gamma|}\).
02

Find the positions of the minimum and maximum

Using the given information that \(|V|_{\min} = 1\) and \(\lambda = 1m\), we can find the positions for minimum and maximum. At the minima, \(|V| = \frac{1}{SWR}\) and at the maxima, \(|V| = SWR\). For the case \((a)\), we have, \(|V|<1.1\), which means \(\frac{1}{SWR}<1.1\). To find the position where such minima happen, use the formula \(z = \frac{\lambda}{2} * m\), where \(m\) is an integer. For the case \((b)\), we have, \(|V| > \frac{4}{1.1}\), which means \(SWR > \frac{4}{1.1}\). To find the position where such maxima happen, use the formula \(z = \frac{\lambda}{4} * (2n + 1)\), where \(n\) is an integer.
03

Calculate the width for minima and maxima

Now we have the positions for minima and maxima, we need to find their width. The width is the difference between two consecutive positions of minima or maxima that meet the given criteria. For case \((a)\), calculate \(z_1\) and \(z_2\) of the positions where \(\frac{1}{SWR}<1.1\), and find the difference in their positions: \(width = z_2 - z_1\). For case \((b)\), calculate \(z_1\) and \(z_2\) of the positions where \(SWR > \frac{4}{1.1}\), and find the difference in their positions: \(width = z_2 - z_1\). In both cases, we can't find the exact numeric values since the value of \(Z_0\) is unknown. However, we can represent the widths with the general expressions obtained in steps 2 and 3 for each case.

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Most popular questions from this chapter

Two lossless transmission lines having different characteristic impedances are to be joined end to end. The impedances are \(Z_{01}=100 \Omega\) and \(Z_{03}=25 \Omega\). The operating frequency is \(1 \mathrm{GHz}\). \((a)\) Find the required characteristic impedance, \(Z_{02}\), of a quarter-wave section to be inserted between the two, which will impedance-match the joint, thus allowing total power transmission through the three lines. \((b)\) The capacitance per unit length of the intermediate line is found to be \(100 \mathrm{pF} / \mathrm{m}\). Find the shortest length in meters of this line that is needed to satisfy the impedance-matching condition. ( \(c\) ) With the three-segment setup as found in parts \((a)\) and \((b)\), the frequency is now doubled to \(2 \mathrm{GHz}\). Find the input impedance at the line-1-to-line- 2 junction, seen by waves incident from line \(1 .(d)\) Under the conditions of part \((c)\), and with power incident from line 1 , evaluate the standing wave ratio that will be measured in line 1 , and the fraction of the incident power from line 1 that is reflected and propagates back to the line 1 input.

The parameters of a certain transmission line operating at \(\omega=6 \times 10^{8} \mathrm{rad} / \mathrm{s}\) are \(L=0.35 \mu \mathrm{H} / \mathrm{m}, C=40 \mathrm{pF} / \mathrm{m}, G=75 \mu \mathrm{S} / \mathrm{m}\), and \(R=17 \Omega / \mathrm{m}\). Find \(\gamma, \alpha, \beta, \lambda\), and \(Z_{0}\)

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 \(z_{\text {in }}=2+j 1 ?\)

The characteristic admittance \(\left(Y_{0}=1 / Z_{0}\right)\) of a lossless transmission line is \(20 \mathrm{mS}\). The line is terminated in a load \(Y_{L}=40-j 20 \mathrm{mS}\). Use the Smith chart to find \((a) s ;(b) Y_{\text {in }}\) if \(l=0.15 \lambda ;(c)\) the distance in wavelengths from \(Y_{I}\) to the nearest voltage maximum.

A lossless \(75-\Omega\) line is terminated by an unknown load impedance. VSWR of 10 is measured, and the first voltage minimum occurs at \(0.15\) wavelengths in front of the load. Using the Smith chart, find \((a)\) the load impedance; \((b)\) the magnitude and phase of the reflection coefficient; \((c)\) the shortest length of line necessary to achieve an entirely resistive input impedance.
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