lecture 18 – 24: smith chart
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Lecture 18 – 24: Smith Chart. Instructor: Engr. Zuneera Aziz Course: Microwave Engineering. Introduction. Invented by Phillip H. Smith It is a graphical ‘ nomogram ’ Used for solving Transmission line problems and matching circuits - PowerPoint PPT PresentationTRANSCRIPT
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Lecture 18 – 24: Smith ChartInstructor: Engr. Zuneera AzizCourse: Microwave Engineering
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Introduction•Invented by Phillip H. Smith•It is a graphical ‘nomogram’•Used for solving Transmission line
problems and matching circuits•Used to represent parameters such as:
Impedance, Admittance, Reflection coefficients, S parameters, etc.
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Introduction• Plotted on the complex reflection coefficient
plane in two dimensions and is scaled in normalized impedance, normalized admittance or both. (Z Smith Charts, Y Smith Charts or YZ Smith Charts respectively)
• Transmission lines having differing values of Zo all behave the same, as far as their normalized impedance properties are concerned.
• The smith chart may also be used for lumped element matching and analysis problems
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Introduction• The Smith chart is a graphical calculator that
allows the relatively complicated mathematical calculations, which use complex algebra and numbers, to be replaced with geometrical constructs, and it allows us to see at a glance what the effects of altering the transmission line (feed) geometry will be. If used regularly, it gives the practitioner a really good feel for the behavior of transmission lines and the wide range of impedance that a transmitter may see for situations of moderately high mismatch (VSWR).
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Normalized impedance at any point along the transmission line• The Smith chart lets us relate the complex dimensionless
number gamma at any point P along the line, to the normalized load impedance zL = ZL/Zo which causes the reflection, and also to the distance we are from the load in terms of the wavelength of waves on the line.
• We can then read off the normalized impedance zL at the point P along the line, where the actual impedance Zo is the local ratio of total voltage to total current taking into account the phase angles as well as the sizes. This impedance is what a generator would "see" if we cut the line at this point P and connected the remaining transmission line and its load to the generator terminals.
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Normalized impedance at any point along the transmission line•Why should the impedance we see vary
along the transmission line? Well, the impedance we measure is really the total voltage on the line (formed from the sum of forward and backward wave voltages) divided by the total current on the line (formed by the sum of forward and backward wave currents).
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So this is how a Smith Chart Looks Like!
http://sss-mag.com/pdf/smithchart.pdf
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This is what you can find from a Smith Chart•Reflection Coefficient•VSWR•Transmission Coefficient•Load Impedance•Admittance•Input Impedance•Lmin and Lmax
•and even more …
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Reflection Coefficient•The smith chart can be thought of as a
circle (with radius equal to 1)•This circle is the circle for the Reflection
Coefficient Γ•The reflection coefficient Γ will possess
values between 0 and 1 (0 for no reflection and 1 for complete reflection)
•So that means that the Γ should be plotted inside the unit circle
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Reflection Coefficient•The center of this unit circle is the place
where Γ is 0 (ideally this is what we want!)
•At any point ON the unit circle, Γ = 1 and that means that all the power is reflected back from the load (Do we want that? NO!)
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A few starter examples•Γ = 0.5•Γ = -0.3 + j 0.4•Γ = -j
•What will be the load impedance if Zo=50 ohms for the above values of Γ?
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The Answers:•For Γ = 0.5
For ZL:
Which corresponds to ZL = 150 Ohms
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The Answers:•For Γ = -0.3 + j 0.4
ZL = 20.27 + j 21.62 Ohms
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The Answers:•For Γ = -j
ZL = -j 50 Ohms
A purely capacitive load!
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Concept of Normalized Impedances•Normalized Load Impedance•To make the Smith Chart more general
and independent of the characteristic impedance Z0 of the transmission line, we will normalize the load impedance ZL by Z0 for all future plots:
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Constant Resistance Circles•Suppose we have a normalized load
impedance given by:
zL= R + j Y
•Y is NOT the admittance here, Y is any real number here.
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Constant Resistance Circles• So for different values of
Resistance you the following circles in the smith chart. For any one circle, the resistance will be constant but the reactance (the imaginary part) is going to change on the circumference of that particular constant resistance circle.
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Constant Reactance Curves• Now lets hold constant the
reactance (Y) and vary R. This will result in producing the following curves on the Smith Chart. So for a given curve, the reactance will be same but as we move ON the curve, we’ll have different values of Resistance.
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Vary Reactance on Resistance Circle• In the following example,
we have taken 2 values of Resistance (R=1.0, 0.3)
• For R=1.0 we have taken X=0 (zL=1+j 0)
• For R=0.3, is taken as 0, 3 and -0.4. Corresponding to zL=0.3 zL=0.3+ j 3 zL=0.3-j 0.4
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Vary Resistance on a Reactance Curve• Now here we’re taking
constant reactance of 1.0 and different resistances (R=0.3 and 2.o) implying to zL=0.3 + j 1.0 and zL=2.0 + j 1.0)
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Zero Reactance• An important curve is
given by Im[zL]=0. That is, the set of all impedances given by zL = R, where the imaginary part is zero and the real part (the resistance) is greater than or equal to zero. The result is shown in Figure.
• It is understood that for Short Circuit zL=0 and for Open Circuit zL= ∞
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Zero Reactance•So the three special cases are:•zL=0+j 0, Γ =1•zL=1+j 0, Γ =0•zL=∞+ j 0, Γ =1
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Transmission Coefficients
incident
dtransmitte
VVT 1
• The transmission coefficient is given by:
• The inner most axis denotes the Transmission coefficient angle
• The transmission could be easily plotted with the help of reflection coefficient (taking the same point) and then calculating the magnitude of the transmission coefficient with the help of the linear scale
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Impedance• The first thing that should be done is the Normalization of
the given impedance from Zo, in the following manner:
• Check the point where the Resistance Circle and the Reactance Curve intersect
• Negative Reactance implies to being Capacitive
• Positive Reactance implies to being Inductive
o
imaginaryrealL Z
jZZz
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Impedance - Example•So lets suppose I have a load
impedance ZL = 50 + j 50 ohms, the Characteristic impedance Zo= 50 ohms
•After I normalize it: zL= 1 + j 1
•Now let’s plot this on the Smith Chart
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Impedance – Example zL=1+j 1
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Impedance – Example 2•Do the same thing for ZL=50 - j 50
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Impedance – Example 2zL=1- j 1
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Finding Reflection coefficient with the help of zL• Find the angle of
reflection coefficient by drawing a line from the center of the smith chart to the circumference, crossing the zL point.
• The axis will show you the angle
• For magnitude, scale the point from center of the smith chart to zL and arc it down on the linear scale of reflection coefficient
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Admittance•Lets begin with the previous example:•zL=1+j1 (normalized impedance of the
load)•yL = 1/zL•The method is called ‘translating through
the center of the smith chart’
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Plotting Admittance
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Plotting Admittance• Now copy the distance of
zL and plot it on the line from the center of the smith chart.
• So the yL comes out to be : 0.5 – j 0.5 (in normalized form)
• To find YL in mhos, multiply yL by Yo which is given by Yo = 1/Zo
• So for this example: YL (in mhos) = 0.01 – j 0.01 mhos
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Special cases of Impedance•We know that ZOC = ∞•And ZSC = 0
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Special Cases of Admittance•Just translate through the center!
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Special Cases of Reflection Coefficient•ΓOC= 1 ∠ 0°(Plot it
at zOC point)•ΓSC= 1 ∠ 180°
(Plot it at zSC point)
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Input Impedance - Zin
•Go ‘towards the generator’ in order to find input impedance. You are provided with load impedance and the length of the transmission line (outer most axis)
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Input Impedance - Zin
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Another way to find ZL• Now we’d like to find
the load impedance, provided we have the input impedance and line length
• Suppose Zin = 100 – j 100 ohms
• zin= 2 – j 2 ohms
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Another way to find ZL
•Now move towards the ‘load’ on the 2nd last axis
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Another way to find ZL
•Not really a complexity! •What if you get the length greater than
0.5 λ such as may be 2.2 λ, 0.61 λ•The maximum scale is 0.5 λ so you just
have to add up the wavelengths and move around the smith chart
•Negate the multiples of 0.5 λ from the given transmission length (if greater than 0.5 λ )
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VSWR•Draw a circle round the zL point•The value of resistance circle (easily read
at the 0 reactance line) represents the VSWR
•You can follow up with the linear scale, where you can find the value in dB too
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VSWR – Lmin, Lmax
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Example•A lossless transmission line is terminated
by a load ZL=10+j20 Ω. Find Γ, T, VSWR, lmin, lmax, YL and Zin if the line length is 4 cm and frequency of 3 GHz.
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Example - Solution•Step 1: Normalize the load impedance•zL=0.2+j0.4•Step 2: Calculate length of transmission
line in number of wavelengths
•So the transmission line length in number of wavelengths is:
cmmXX 101.0103103
fc
9
8
4.0104
cmcm
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Example - Solution• Step 3: plot zL
• Step 4: Draw a circle through the zL point• Step 5: Find out reflection coefficient• Step 6: Find VSWR (with the circle method and the
linear scale too, both give the same results)• Step 7: Find yL and then YL (translating through
the center of the smith chart)• Step 8: find T with the help of reflection coefficient• Step 9: find zin by rotating towards the generator
0.4 wavelengths