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Homework - Microwave Engineering
HW I.1 Locate the following load impedances terminating a 50Ω T-line:
(a) ZL = 0 (a short circuit),
(b) ZL = ∞ (an open circuit),
(c) ZL = 100 + j100 Ω,
(d) ZL = 100 - j100 Ω, and
(e) ZL = 50 Ω.
HW I.2 A 0.334λ - long ZO = 50 Ω T - line is terminated in a load ZL = 100 - j100 Ω. Use the Smith Chart to find:
(c) Zin, and
(d) the distance from the load to the first voltage minimum.
HW I.3 Suppose in Example 2.1 that the 50 Ω coaxial air line extends all the way from 0 cm scale location to the location of the termination. What is the shortest length this extension can be?
HW I.1 For the graphical representations, see Figure 1 below. We know that ZC = 50 Ω.
(a) ZL = 0 => see the blue color;
(b) ZL = ∞ (an open circuit) => see the red color;
(c) ZL = 100 + j100 Ω => the normalized impedance zL = 2 + j2 Ω; see the magenta color;
(d) ZL = 100 - j100 Ω => the normalized impedance zL = 2 - j2 Ω; we move on the Smith Chart in the clockwise direction; see the green color;
(e) ZL = 50 Ω => zL = 1 Ω; we are now in the center of the Smith Chart; see the orange color.
Figure 1. The representations on the Smith Chart for HW I.1
(a) ZL = 100 + j100 Ω and ZC = 50 Ω => the normalized impedance is zL = 2 + j2 Ω; we now use the sub point (c) from exercise HW I.1.
In order to find | ΓL | (the magnitude of the reflection coefficient), we have to measure the distance from the center of the Smith Chart to the point (c), and divide it to the radius of the big Smith circle; we obtain | ΓL | = 0.62. The angle of the reflection coefficient is read from the graph: θΓ = 30o.
For the graphical representation, see Figure 2 below:
Figure 2. Graphical representation for determining ΓL
(b) We now draw the constant - | ΓL | circle. The voltage standing wave ratio VSWR can be determined by reading the value of r at the θΓ = 0o crossing for the constant - | ΓL | circle:
z = r = VSWR = 4.2
(c) We still use the constant - | ΓL | circle. The starting point is at 0.208 λ. Adding the length of the line 0.334λ, we obtain 0.542λ; normalized to the interval (0 λ, 0.5 λ), we obtain 0.042λ. At this wavelength, we consider the intersection point with the constant - | ΓL | circle => zin = 0.24 + j0.26 Ω; de-normalizing, we obtain the input impedance:
Zin = (0.24 + j0.26) x 50 Ω = 12 + j13 Ω.
For the graphical representation of (b) and (c), see Figure 3 below:
Figure 3. Graphical representation of the input impedance determination
d) We have: ZL = 100 + j100 Ω and ZC = 50 Ω => the normalized impedance is zL = 2 + j2 Ω => we read on the Smith Chart 0.208 λ (from Figure 3). Considering the same scale as in Example 2.1, we have 30cm/λ. Thus, the distance from the load to the first voltage minimum is:
d = 0.208λ x 30 cm/λ = 6.24 cm.
HW I.3 At the given frequency of the signal, the coaxial air line should contain at least a minimum and a maximum; thus, the minimum length of the wire should be λ/2
=> dmin = 30 cm/λ x λ/2 = 15cm.
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