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(15 points) Plot the following impedance values on the same admittance Smith chart, assuming Z. = 7522. (a) ZL = 15 - j2712 (b) ZL = 120 - 1202 (c) ZL = 45 + j3712 (d) ZL = 75 + j3012 . Please show all work and show on Admittance Smith Chart as well! Thank you! Show transcribed image text.Problem 4 (part (a) 20, part (b) 10, total 30 points) A lossless transmission line with characteristic impedance Z = 50 2 is terminated by a load Z, as shown in the figure below. (a) Eight different values of Zl correspond to eight points in an admittance Smith chart In the Smith chart, these eight points are denoted A to H.The load and source resistances are plotted on the Smith chart in Figure 10.10.4 (a) after choosing a normalization impedance of Z0 = 50Ω (and so rS = RS / Z0 = 0.5 and rL = RL / Z0 = 4). The normalized source impedance, rS, is Point A, and the normalized load impedance, rL, is Point C. The matching network must be lossless, which means that ...1/13 T 1 Syllabus, Preliminaries, The Smith Chart 2 92-104 2.1all, 2.5 1/15 R 2 The Normalized Impedance and Admittance Smith Chart, Impedance Matching Networks (Part 1) 2 104-124 2.8, 2.9all, 2.10, 2.12, 2.14 1/21 T 3 QUIZ 1, Impedance Matching Networks (Part 2) 2 125-141 2.13; 2.15all 1/23 R 4 Microstrip Matching Networks• The admittance Smith chart is the impedance Smith chart rotated 180 degrees. - We could use one Smith chart and flip the reflection coefficient vector 180 degrees when switching between a series configuration to a parallel configuration. Toward Generator Away From Generator ConstantConsider the load admittance yL = 0.3 + i*0.5 (then zL = 0.882 - i*1.471). The absolute value of is 0.6176, which can be found from equation [2] on the admittance Smith Chart page. The load yL and the constant VSWR circle (in black) is plotted in Figure 1: Figure 1. Load Admittance and the Corresponding Constant SWR Circle.Figure 1. Constant Resistance Circle for zL=1 on Smith Chart. In Figure 1, the outer blue ring represents the boundary of the smith chart. The black curve is a constant resistance circle: this is where all values of z1 = 1 + i*Y will lie on. Several points are plotted along this curve, z1 = 1, z1 = 1 + i*2, and zL = 1 - i*4.The admittance form of the Smith chart is used in the following example. Example 14.6.1. Single Stub Matching. In Fig. 14.6.6a, the load admittance Y L is to be matched to a transmission line having characteristic admittance Y o by means of a "stub" consisting of a shorted section of line having the same characteristic admittance Y o.Variables that can be used to accomplish the matching are ...What is the admittance of an inductor of inductance L at angular frequency ω? − ω L j jω L jω C Question 2 0 / 1 pts What point on the admittance Smith chart corresponds to a short-circuit load? (Give your answer in terms of Γ r and Γ i coordinates.)2/17/2010 Admittance and the Smith Chart 3/7 Jim Stiles The Univ. of Kansas Dept. of EECS numerical value as the impedance of the first load (i.e., y′=+1 j). Q: Where would this admittance value map onto the complex Γ plane? A: Start at the location zj′=1+ on the Smith Chart, and then rotate around the center 180D.You are now at theThese are two families of circles as and . Superposition of Smith Chart and its 180o ( ) rotated version is known as Combined Impedance-Admittance Smith Chart. is the complete revolution of Smith chart. is the half of Smith chart (180o ). The images of is in Smith chart. 2.3. Slotted Line This device is used to find as first .Figure 1. Admittances in Parallel Add. Since the admittance of an inductor is entirely imaginary (that is, Re [y_ind]=0), the result of adding an inductor in parallel is to change the susceptance of the antenna (load). That is, we are only altering the imaginary part of the antenna's admittance.Electrical Engineering questions and answers. Use the Smith chart to find the following quantities for the transmission line circuit shown in the accompanying figure: Make sure to to solve the All parts, specially part (g) and (h) please. (a) The SWR on the line. (b) The reflection coefficient at the load. (c) The load admittance.A Smith Chart is used to calculate: a. transmission line impedances. b. propagation velocity. c. optimum length of a transmission line. d. transmission line losses. View Answer: Answer: Option A. Solution: 20. Compared to a 300-ohm line, the loss of a 50-ohm cable carrying the same power:Admittance Smith chart • Alternative approach to solve parallel network elements is through 180⁰ rotated Smith chart • This rotated Smith chart is called admittance Smith chart or Y-Smith chart • The corresponding normalized resistances become normalized conductances & normalized reactances become normalized suceptances 0 00 RG r g Z GAre you looking for a cozy and affordable place to call home in Fort Smith, AR? Look no further than a duplex rental. Duplexes are a great option for those who want to enjoy the convenience of apartment living without breaking the bank.(g) Use the chart to determine the length of the stub. The imaginary normalized admittance values are found on the circle of zero conductance on the chart. On a commercial Smith chart one can use a printed scale to read the stub length in terms of wavelength. We assume here that the stub line has characteristic impedance Z0 as the main line. If ...The stub value is the amount of susceptance that is required to move the normalized load admittance (the center of the Smith chart) to point "A" on the constant magnitude circle. An open stub transmission line can be used to supply this value of susceptance. Its wavelength is defined by the amount of angular rotation from the open-circuit ...What is the admittance of an inductor of inductance L at angular frequency ω ? −ωLjjωLjωC Question 2 What point on the admittance Smith chart corresponds to a short-circuit load? (Give your answer in terms of Γr and Γi coordinates.) (Γr,Γi)=(0,0) (Γr,Γi)=(1,0) (Γr,Γi)=(0,−1)The very basic RULES of impedance matching are: Add a lossless element, capacitor or inductor, to get the real part of either impedance or admittance to be 1. Add the second lossless element to tune out the remaining imaginary part, reactance or susceptance, so the resultant impedance or admittance is a real number 1 ( z = 1 + j0 or …Consider the load admittance yL = 0.3 + i*0.5 (then zL = 0.882 - i*1.471). The absolute value of is 0.6176, which can be found from equation [2] on the admittance Smith Chart page. The load yL and the constant VSWR circle (in black) is plotted in Figure 1: Figure 1. Load Admittance and the Corresponding Constant SWR Circle.Figure 1. Smith chart If Z0 = 50 Ω, when there is matching (Z = Z 0) the normalized impedance at 50 Ω is 1 and it is the center of the Smith chart. The goal in the search of a matching network is to converge towards this point. The horizontal axis of the Smith chart represents pure resistors: at the left side, z = 0 (short circuit) and at the ...Solve the following problem using Smith chart . impedance. The input end of the 50Ω line is attached to the load end of a 75- -Ω lossless transmission line with another load . Z. L1 =150 +j150Ω. Find out the unknown load impedance on the 50-Ω lossless transmission line that can make the reflection coefficientThe combination of the reflection/transmission polar plots, nomographs, and the impedance and admittance Smith chart leads to the combined Smith chart (see Figure \(\PageIndex{9}\)). This color Smith chart is the preferred version for use in design and the separate impedance and admittance versions of the Smith chart are rarely used.

This occurs because the plot is an impedance plot. But the new point is, in fact, an admittance. Therefore, the value read on the chart has to be read ...The Imaginary Smith Chart was used in [12], [13] to design a narrowband match to the aperture admittance of an evanescent-mode open-ended waveguide. In [12] this resulted in a narrowband antenna ...• Smith chart . Trace Data Types. The data types generated by the VNA (real, imaginary, magnitude, phase) are used in the display graph to show the possible ways in which S-Parameter data can be represented. ... Displayed on a Smith chart graph as impedance or as admittance • Real and Imaginary. If simultaneous displays are required ...7. For a load impedance of ZL=60-j80. Design of 2 single-stub shunt tuning networks to match this load to a 50Ω line is to be done. What is the normalized admittance obtained so as to plot it on smith chart? a) 1+j b) 0.3+j0.4 c) 0.4+j0.3 d) 0.3-j0.4 View Answer

The stub value is the amount of susceptance that is required to move the normalized load admittance (the center of the Smith chart) to point "A" on the constant magnitude circle. An open stub transmission line can be used to supply this value of susceptance. Its wavelength is defined by the amount of angular rotation from the open-circuit ...Analyze RLC Components Using the Admittance Smith Chart and Examples by Dr. Steve Arar More Than Moore: Lux Semiconductors' System-on-Foil Shrinks Circuitsnormalized admittance and impedance. These are gotten from the smith Chart by switching. The impedance and admittance charts. For Ž the corresponding admittance ...…

Reader Q&A - also see RECOMMENDED ARTICLES & FAQs. Use the Smith chart to find the followin. Possible cause: The stub value is the amount of susceptance that is required to move the normalized l.