Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω

A transmission line with a characteristic impedance \[ Z_0 = 100\,\Omega \] is terminated with a load impedance \[ Z_L = (75 + j40)\,\Omega \] The load is to be matched to the line using a double shunt stub tuner where the spacing between the two stubs is \[ \frac{\lambda}{8} \] Assuming the first stub is placed at a distance \[ d_1 = 0.2\lambda \] from the load, determine the required lengths of both stubs \[ (s_1 \text{ and } s_2) \] for:
  • Short-circuited stubs
  • Open-circuited stubs

GIven:

\[
Z_L = 75 + j40\,\Omega
\]

\[
Z_0 = 100\,\Omega
\]

Step 1: Normalized the load impedance by dividing it by characteristic impedance of line. Plot Zn in the Smith Chart. Construct SWR circle for load line recording the wave length at Yn.

\[
z_N = \frac{Z_L}{Z_0}
\]

\[
z_N = \frac{75 + j40}{100}
\]

\[
z_N = \frac{75}{100} + j\frac{40}{100}
\]

\[
z_N = 0.75 + j0.4
\]

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

SWR = 1.7:1

 

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

Yn = 1.0+j0.55

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

WTGb = 0.349 λ

Step2: Construct λ/8 Circle

The Spacing circle cuts the stub circle at point c and d. Join c and d with the prime circle. Starting at the wavelength reading at the Yn. Move clockwise around the wavelength scale so that the line ends up anywhere between the dashed line C and D. Line C and D describes an arc between two radii that. Point C and D are the points that intersects both SWR and Spacing circle

Draw a straight line from origin to 90 degrees

numerical5doublestub8zl75-j40zo50-6 

 

 

Draw a bisector line

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

Draw a circle with bisector line as pivot point

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

Step 3: Since the distance of 1st stub is not given, we can assume a distance d1 that is d1 distance apart from the load overlapped within the spacing circle or we could assume a point E, and based on WTGE distance can be calculated. For point E, the point should be inside the overlapping region between 1st SWR and spacing Circle.  

Overlapping Region:

       The region where the SWR circle and the spacing circle intersect.

       Any point within this overlapping region satisfies both the SWR and the spacing requirements.

Arbitrarily Choosing Point E:

       Within the overlapping region, you can choose any point for the stub placement.

       This point is chosen based on practical considerations or ease of implementation.

For Ease of implementation, Point A (Zn) lies between Spacing Circle and on SWR circle, we chose arbitrarily Point E somewhere between Point A and Point C.

Let us consider a distance to the WTGE as 0.09 λ so that the distance d1 can be calculated. The placement distance of 1st Stub from the load is equal to the clockwise distance from Point B To Point E.

Point E is

\[
y_E = 0.73 + j0.375
\]

\[
D_1 = \text{WTG}_E + \left(0.5\lambda - \text{WTG}_B\right)
\]

\[
D_1 = 0.09\lambda + \left(0.5\lambda - 0.349\lambda\right)
\]

\[
D_1 = 0.09\lambda + 0.151\lambda
\]

\[
D_1 = 0.241\lambda
\]

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

Step 4: Follow the reactance circle through point E (0.73) in the direction of a smaller reactance. In the case, move left to the point C at the edge of spacing circle and note the coordinates i.e. move anticlockwise towards the edge of the spacing circle. Label the point as F.

 

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

Zf = 0.73+j0.04 ohm

Step 5: Find the difference in reactance between point F and E.

\[
\text{The amount of susceptance that must be cancelled is}
\]

\[
X_E - X_F
\]

\[
= j0.375 - j0.04
\]

\[
= j0.335
\]

\[
\therefore \text{Required susceptance to be cancelled} = j0.335
\]

\[
\text{To cancel } +j0.335,\ \text{locate the point } -j0.335
\]

\[
\text{on the susceptance circle and mark it as Point H.}
\]

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

Also find WTGH

numerical5doublestub8zl75-j40zo50-7 

WTGH = 0.448 λ

Step 6: The Stub length is found in the same way as single stub matching. The length of the 1st Stub is calculated as  

\[
\text{For the open-circuited stub, measure clockwise from the zero-admittance point.}
\]

\[
L_{1O} = 0.448\lambda
\]

\[
\text{For the short-circuited stub, measure clockwise from the infinite-admittance point.}
\]

\[
L_{1S} = 0.448\lambda - 0.25\lambda
\]

\[
L_{1S} = 0.198\lambda
\]

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

Step 7: From Point F, again draw 2nd SWR Circle.  Using the point F as a circumference location and prime center of the short as a pivot point. Construct a second SWR circle

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

SWR2 = 1.4:1

Step 8: From point F, move around the 2nd SWR circle in clockwise direction until reaching to R=1 circle on the inside of the spacing circle, note the edge as G.

 

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

ZG = 1+j0.35

Step 9: Find the point -j0.35to cancel j0.35 and note it as Point I

 

Numerical5:DoubleStub(λ/8):Zl:75 + j40Ω,Zo=50Ω 

Now draw a line to that point

numerical5doublestub8zl75-j40zo50-8 

WTGI = 0.446 λ

Step 9: The Stub length is found in the same way as single stub matching. The length of the 2nd Stub is calculated as  

\[
\text{For the open-circuited stub, measure clockwise from the zero-admittance point.}
\]

\[
L_{2O} = 0.446\lambda
\]

\[
\text{For the short-circuited stub, measure clockwise from the infinite-admittance point.}
\]

\[
L_{2S} = 0.446\lambda - 0.25\lambda
\]

\[
L_{2S} = 0.196\lambda
\]numerical5doublestub8zl75-j40zo50-9

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