Unformatted text preview: 13.38 CHAPTER 13. TRANSMISSION LINES >> smith303(ZL, ZG)
where Z L and Z0 are the load and transmission line impedances, respectively. In
response, smith303 does the following:
i) draws a Smith Chart
ii) points out the starting point (i.e., I‘(z = 0) and Zn(z = 0)) on the
Smith Chart with a little circle
iii) draws the circle that shows the values of l"(z) and Z"(z) on the Smith
Chart as one moves back from the load on the transmission line. Solution strategy:
i) First choose the smallest length of 31 such that the normalized admit tance YMz = —€1) has a real part of unity.
ii) Then choose the smallest length of 22 such that the normalized ad— mittance Yn(z = —€2) has an imaginary part that exactly cancels the
imaginary part of Yn(z = —€1). a. Find the smallest length of E1 in terms of the wavelength such that Yn(z = —£1) has
a real part of unity. Use a Smith Chart to calculate £1 and include a printout of the
Smith Chart showing your work with your answer sheet. Note that the normalized
admittance is always diagonally opposite to the normalized impedance on a Smith
Chart. b. Find the smallest length of £2 in terms of the wavelength such that K,(z = —£2)
has an imaginary part that exactly cancels the imaginary part of Yn(z = —E1). Use
a Smith Chart to calculate E2 and include a. printout of the Smith Chart showing
your work with your answer sheet. Note that the normalized admittance is always
diagonally opposite to the normalized impedance on a Smith Chart. c. Now suppose that instead of using an opencircuit stub you use a shortcircuit
stub, as shown in the ﬁgure below. Figure 13.40: Shorted—stub matching of a transmission line to a load. Assuming that the length of 31 is the same as that calculated in part (a) above, ﬁnd
the smallest length of E2 in terms of the wavelength A such that Yn(z = —£2) has
an imaginary part that exactly cancels the imaginary part of Yn(z = —£1). Use the ...
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 Fall '06
 RANA
 Electromagnet

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