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What are the respective advantages of using either an open circuit stub or a short circuit stub?

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  • \$\begingroup\$ Well, given that they do different things, you would pick the one that does what you need. Seriously, this question is far too broad. You'll have to be a lot more specific about your application. \$\endgroup\$ – Dave Tweed Jan 12 '17 at 0:45
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The application dictates when you want to use one or another. When matching the load impedance to the line's, you want to eliminate the imaginary part of the load's impedance and that's what the stub would help you do.

Consider a load with an impedance expressed as \$z_L=A+jB\$ and for simplicity, let's say the line characteristic impedance is \$Z_o=50\$. In order for you to match those two, you have to add something to the load so that the imaginary part gets cancelled out.

Adding a parallel stub and taking advantage of using admittance instead impedance to simplify the math, you can then find the admittance of the load, \$y_L=C+jD\$ and find a corresponding value for the admittance of the stub so that when you add \$y_L\$ and \$y_{\text{stub}}\$ you eliminate the imaginary part.

A short circuited stub has an admittance in the form: \$y_{\text{short}} = -jX\$ and an open stub has an admittance in the form: \$y_{\text{open}} = -jY\$. Then, if for example, the imaginary part of the load's admittance is positive, you want to use a shorted stub in parallel to cancel out the imaginary part.

Keep in mind, though, that stub matching helps you with the imaginary part only. You may have cases where you need to adjust the real part too. For those cases, in conjunction with using a stub, you need to add some 'length' between the load and the stub in order to fully match.

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