How can I analyze the output waveform at various points in a RF receiver model with and without a shunt resistor in SimRF 3.0 (R2010b) ?

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MathWorks Support Team 2011-5-10

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I have two RF receiver system models both modeled in SimRF, the following blocks are used in the in the receiver forward path:

1. LNA

2. Mixer

3. SimRF output Port

In the model 'example_shunt.mdl':

- A shunt resistor is used after the mixer.

In the model 'example_no_shunt.mdl':

- No shunt resistor is used after the mixer.

In both models the voltages going into the mixer (out of the LNA) is 3.5313. In the first model with the shunt resistor, the voltage output of the mixer is 3.5313, the voltage output of the mixer (without the shunt resistor) is 7.06237.

I have attached the models to the solution, namely:

example_shunt.mdl

example_no_shunt.mdl

Run each of the models to see the outputs on the Scope block that contains the signals from LNA, Mixer and the SimRF Output ports.

The voltage is doubled without the shunt resistor, in the model. I would think that the voltage should be the same at the output as the input to the mixer since the mixer has 0 gain.

Intuitively, I think that with a shunt resistor in the forward path, there should be difference in the voltage from LNA to the Mixer output by a factor of 2 (by maximum power transfer theorem). However, in this example the outputs are not so. Whereas in the absence of a shunt resistor, the output across LNA, mixer and the SimRF output port should be same.

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Answer 1

MathWorks Support Team 2011-5-10

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https://ww2.mathworks.cn/matlabcentral/answers/95307-how-can-i-analyze-the-output-waveform-at-various-points-in-a-rf-receiver-model-with-and-without-a-sh#answer_104659

The gain of the amplifier and mixer is defined under matching condition. So when the load impedance matches the output impedance of the amplifier, you will see a maximum power transfer and the entire power/voltage at the input will be available at the output (in case of unity gain).

In example 'example_shunt.mdl', this is what we observe when you measure the output across a 50 ohm resistance. The amplifiers output impedance is also 50 ohms. So there is a match.

In the second case 'example_no_shunt.mdl', the load impedance is infinity i.e an open circuit. So in this case the output voltage measured is the open circuit voltage. From a voltage number point of view, the first case has the output resistance and the load resistance forming a voltage divider, while in the second case there is not voltage division. So we would see an additional factor of 2 in the output voltage.