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- Thread starter HasuChObe
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uart

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Hah; Of course I try to calculate field energy with only the E field :/ Good answer. Should be further along now :P Any inputs on the geometry changes from a smaller to larger impedance? I've concluded that the cross section of the dielectric part has to get bigger, but I could be wrong again -.-

- #4

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Are you assuming your ideal source has a 50 ohm output impedance? To solve the mismatch equation at the transmission line impedance step, you have to consider both the voltage V and the current I at the mismatch. For a forward (F) and reflected (R) signal you have 3 equations for the voltages and currents at the mismatch (Z_{1}=50 ohms, and Z_{2}=100 ohms). This is easily simulated in SPICE.

V_{F1} = +Z_{1}·I_{F1}

V_{F2} = +Z_{2}·I_{F2},

V_{R1} = -Z_{1}·I_{R1}

where F = forward, R = reflected.

[added] The minus sign comes from the direction of the Poynting vector.

Bob S

V

V

V

where F = forward, R = reflected.

[added] The minus sign comes from the direction of the Poynting vector.

Bob S

Last edited:

- #5

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Here in thumbnail is a simulation of the mismatch using LTSpice. A back-terminated voltage source drives a 50-ohm transmission line, which is connected to a 100-ohm transmission line, terminated at the end. For 1 volt in,

V_{F1} = 1 volt

V_{R1} = 0.333 volts

V_{F2} = 1.333 volts

I_{F1} = +20 mA

I_{R1} = - 6.667 mA (note minus sign)

I_{F2} = +13.333 mA

So V_{F1} + V_{R1} = V_{F2}

and I_{F1} + I_{R1} = I_{F2}

Bob S

V

V

V

I

I

I

So V

and I

Bob S

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