Understanding Smith Chart by Learning Examples and Questions

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Examples and Questions:

This article will be dedicated to examples and questions to find out how much you really know about Smith chart and enhance your understanding of this great chart.

But you should learn all Basic Parameters, Equations, and Plots first before continuing reading further here.

We’ll work on examples and then ask questions.

 

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Obtain \(Γ_r\) and \(Γ_i\) from given \(r\) and \(x\):

\(z\), normalized load impedance,

$$z={Z_L\over Z_0}={{R_L+jX_L}\over {Z_0}}=r+jx…..(1)$$

Given \(Z_L\), and find \(Γ\), reflection coefficient,

$$Γ=Γ_r+jΓ_i={{Z_L-Z_0}\over {Z_L+Z_0}}={{z-1}\over {z+1}}…..(2)$$

Or, obtain \(Γ_r\) and \(Γ_i\) directly from given \(r\) and \(x\):

$${Γ_r}={{r^2-1+x^2}\over {{(r+1)^2}+x^2}}…..(3)$$

$${Γ_i}={2x\over {{(r+1)^2}+x^2}}…..(4)$$

 

Fig. 1   \(Γ\) and impedance normalization

 

Fig. 2   \(z\) circles and \(Γ\)

 

Example #1:

Q.  Point A in Fig. 2, if the characteristic impedance \(Z_0\) is 50 ohms, and the load impedance \(Z_L={R_L+jX_L}=50+j50\),

calculate 1) \(z\), and 2) \(Γ\)?

Ans.

1)   $$z=r+jx={Z_L\over Z_0}={{50+j50}\over 50}=1.0+j1.0$$

So, \(r=1.0\) and \(x=1.0\).

2)   $$Γ={{Z_L-Z_0}\over {Z_L+Z_0}}={{(50+j50)-50}\over {(50+j50)+50}}$$

$$={{j50}\over {100+j50}}={{(j50)(100-j50)}\over {(100+j50)(100-j50)}}$$

$$=0.2+j0.4=Γ_r+jΓ_i$$

Or,

$$Γ={{z-1}\over {z+1}}={{1.0+j1.0-1}\over {1.0+j1.0+1}}$$

$$={{j1.0}\over {2.0+j1.0}}=0.2+j0.4$$

So, \({Γ_r}=0.2\), \({Γ_i}=0.4\)

Or, applying Equations (3) & (4),

$${Γ_r}={{r^2-1+x^2}\over {{(r+1)^2}+x^2}}={{1.0^2-1+1.0^2}\over {{(1.0+1)^2}+1.0^2}}=0.2$$

$${Γ_i}={2x\over {{(r+1)^2}+x^2}}={{2\times 1.0}\over {{(1.0+1)^2}+1.0^2}}=0.4$$

The answers match with the plot in Fig. 2.

 

Question #1:

Q. Point B in Fig. 2, calculate \(Γ\) if the characteristic impedance \(Z_0\) is 75 ohms, and the load impedance \(Z_L=15+j45\).

Ans.  \(Γ=-0.333+j0.667\)

 

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Obtain \(r\) and \(x\) from given \(Γ_r\) and \(Γ_i\):

For any \(Γ\), there is a corresponding \(z\) and we can find \(r\) and \(x\) by using these 2 equations:

$$r={{1-{Γ_r}^2-{Γ_i}^2}\over {({1-{Γ_r}})^2+{Γ_i}^2}}…..(5)$$

$$x={{2{Γ_i}}\over {({1-{Γ_r}})^2+{Γ_i}^2}}…..(6)$$

 

Fig. 2   \(z\) circles and \(Γ\)

Example #2:

Q.  Point C in Fig. 2, \(Z_0=50\), and \(Γ=-0.32-j0.38\),

calculate 1) \(z\), and 2) \(Z\)?

Ans.

1)

Equation (5),

$$r={{1-{Γ_r}^2-{Γ_i}^2}\over {({1-{Γ_r}})^2+{Γ_i}^2}}$$

$$={{1-{(-0.32)}^2-{(-0.38)}^2}\over {({1-{(-0.32)}})^2+{(-0.38)}^2}}=0.4$$

Equation (6),

$$x={{2{Γ_i}}\over {({1-{Γ_r}})^2+{Γ_i}^2}}$$

$$={{2\times (-0.38)}\over {({1-{(-0.32)}})^2+{(-0.38)}^2}}=-0.4$$

So,

\(z=0.4-j0.4\)

As it is showed in Fig. 2.

2)

\({R_L}={0.4 \times 50}=20 Ω\)

\({X_L}={-0.4 \times 50}=-20 Ω\)

So,

\(Z_L={20-j20}\)

 

Question #3:

Q.  Point D in Fig. 2, \(Z_0=50\), and \(Γ=0.25-j0.25\),

calculate \(z\).

 

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Fig. 3   \(z\) to \(y\) conversion

Fig. 4   Full Smith chart, \(z\) & \(y\) circles

 

Every impedance \(Z\) has a corresponding admittance \(Y\) and,

\(Y={1/Z}\), also, after normalization,

\(y={1/z}\)

 

Fig. 5   \(z\) & \(y\) conversions

 

Obtain \(g\) and \(b\) from given \(r\) and \(x\):

If \(z=r+jx\) and \(y=g+jb\), then,

$$y=g+jb={1\over z}={1\over {r+jx}}$$

And,

$$g={r\over {r^2+x^2}}…..(7)$$

$$b={-x\over {r^2+x^2}}…..(8)$$

 

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Obtain \(r\) and \(x\) from given \(g\) and \(b\):

Conversely,

$$z=r+jx={1\over y}={1\over {g+jb}}$$

And,

$$r={g\over {g^2+b^2}}…..(9)$$

$$x={-b\over {g^2+b^2}}…..(10)$$

 

Example #4:

Q.  Point A in Fig. 5, \(z=0.4-j0.3\),

calculate 1) \(y\), and 2) if \(Z_0\) is 50Ω, what is \(Y\)?

Ans.

1)

\(r=0.4\) and \(x=-0.3\)

Equation (7), $$g={r\over {r^2+x^2}}={0.4\over {0.4^2+(-0.3)^2}}=1.6$$

Equation (8), $$b={-x\over {r^2+x^2}}={0.3\over {0.4^2+(-0.3)^2}}=1.2$$

As showed in Fig. 5.

2)

\(Y=1/Z\), and \({Y_0}={1/{Z_0}}={1/50}=0.02\).

Since \(y={Y/{Y_0}}\), so \(Y={Y_0}\times y\).

$$Y={{Y_0}\times (g+jb)}={0.02\times (1.6+j1.2)}$$

$$={0.032+j0.024}$$

Example #5:

Q. Point B in Fig. 5, \(y=1.0-j1.0\), calculate \(z\)?

Ans.

Equation (9),

$$r={g\over {g^2+b^2}}={1.0\over {1.0^2+(-1.0)^2}}=0.5$$

Equation (10),

$$x={-b\over {g^2+b^2}}={1.0\over {1.0^2+(-1.0)^2}}=0.5$$

So, \(z=0.5+j0.5\).

As showed in Fig. 5.

 

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Obtain \(Γ_r\) and \(Γ_i\) from given \(g\) and \(b\):

$${Γ_r}={{1-g^2-b^2}\over {{(g+1)^2}+b^2}}…..(11)$$

$${Γ_i}={-2b\over {{(g+1)^2}+b^2}}…..(12)$$

 

Example #6:

Q. Point C in Fig. 5, \(y=0.6+j0.2\), calculate \(Γ\)?

Ans.

$${Γ_r}={{1-g^2-b^2}\over {{(g+1)^2}+b^2}}$$

$$={{1-0.6^2-0.2^2}\over {{(0.6+1)^2}+0.2^2}}=0.231$$

 

$${Γ_i}={-2b\over {{(g+1)^2}+b^2}}$$

$$={{-2\times 0.2}\over {{(0.6+1)^2}+0.2^2}}=-0.154$$

So, \(Γ=0.231-j0.154\)

As showed in Fig. 5.

 

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Obtain \(g\) and \(b\) from given \(Γ_r\) and \(Γ_i\):

$$g={{1-{Γ_r}^2-{Γ_i}^2}\over {1+{Γ_r}^2+{2Γ_r}+{Γ_i}^2}}…..(13)$$

$$b={{-2{Γ_i}}\over {1+{Γ_r}^2+{2Γ_r}+{Γ_i}^2}}…..(14)$$

Example #7:

Q. Point D in Fig. 5, \(Γ=0.4+j0.2\), calculate \(y\)?

Ans.

$$g={{1-{Γ_r}^2-{Γ_i}^2}\over {1+{Γ_r}^2+{2Γ_r}+{Γ_i}^2}}$$

$$={{1-0.4^2-0.2^2}\over {1+0.4^2+{2\times 0.2}+0.2^2}}=0.4$$

 

$$b={{-2{Γ_i}}\over {1+{Γ_r}^2+{2Γ_r}+{Γ_i}^2}}$$

$$={{-2\times 0.2}\over {1+0.4^2+2\times 0.4+0.2^2}}=-0.2$$

So, \(y=0.4-j0.2\)

As showed in Fig. 5.

 

Now, you have learned all basics of Smith chart and you know the chart is consisted of 3 very basic parameters, \(Γ, z, y\), and they can be converted among each other based on a few sophisticated equations.

With any one of these 3 parameters given, you can read the other 2 in the chart simultaneously with a very reasonable accuracy.

You don’t need to remember those equations by heart but you should know how to apply them without any difficulty, whenever you need to use it to conveniently solve impedance and reflection issues.

If you can follow those 6 examples easily, then you are good to go to answer the questions below, and, once you get them done correctly, you can continue to learn the most exciting application of Smith chart, impedance matching.

You’ll learn a unique way of impedance matching using Smith chart, and you will also be directed to where you can get a powerful spreadsheet which will help you get the matching job solved within a fraction of a second with very minimal effort.

However, do not force yourself ahead to the next article if you have difficulty to answer the following questions. You should go back to review this article-Basics, Parameters, Equations, and Plots, then come back to practice on this article again.

After that, you will be well prepared for learning impedance matching.

Let’s charge ahead to answer these simple questions below.

 

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Fig. 6   Smith chart

 

Unless otherwise stated, \(Z_0=50Ω\).

Question #3:

If \(Γ=-0.4+j0.25\), then what are:

  1.  \(z\)?
  2.  \(y\)?
  3.  \(Z\)?
  4.  \(Y\)?
  5.  Location on the Smith chart?

Ans.

  1.  \(z=0.384+j0.247\)
  2.  \(y=1.842-j1.185\)
  3.  \(Z=19.22+j12.36\)
  4.  \(Y=0.037-j0.024\)
  5.  Point A in Fig. 7.

 

Question #4:

If \(Z=85-j35\), then what are:

  1.  \(Y\)?
  2.  \(Γ\)?
  3.  Location on the Smith chart?

Ans.

  1.  \(Y=0.01+j0.004\)
  2.  \(Γ=0.306-j0.18\)
  3.  Point B in Fig. 7.

 

Question #5:

If \(Y=0.04+j0.028\), then what are:

  1.  \(Z\)?
  2.  \(Γ\)?
  3.  Location on the Smith chart?

Ans.

  1.  \(Z=16.78-j11.74\)
  2.  \(Γ=-0.453-j0.255\)
  3.  Point C in Fig. 7.

 

Question #6:

If \(z=1.4+j1.2\), then, read directly from the Smith chart without using equations, what are the approximate values of:

  1.  \(Γ\)?
  2.  \(y\)?
  3.  Location on the Smith chart?

Ans.

  1.  \(Γ=0.33+j0.33\)
  2.  \(y=0.41-j0.35\)
  3.  Point D in Fig. 7.

 

Fig. 7   Q & A Smith chart

 

 

‘Note: This is an article written by an RF engineer who has worked in this field for over 40 years. Visit ABOUT to see what you can learn from this blog.’

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