- #1

FOIWATER

Gold Member

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## Main Question or Discussion Point

I recently did an experiment where I had a T-type attenuator.

It was connected to a 50 ohm load, and as such I used a function generator with an internal impedance of 50 ohms. Doing the correct calculations, I found the source and load impedance were matched to the attenuator at 50 ohms.

However, removing the power supply with internal impedance for a DC supply with NO internal impedance, and mismatching the circuit, had NO effect on the insertion loss.

I was asked a question, that what if we did not match the source to the attenuator? what would happen? Some student said it would keep the ratio of voltage into the attenuator to voltage out ratio the same. If so, what is the point of the matching in the first place? Surely it has to change it.

All I was ever told was that the matching is required so the waveform not be distorted, purely attenuated.

Perhaps because the resistance is not different, V^2/R on the output will no long be divided by V^2/50 on the input.... and we have more power loss than simply attenuation?

But then again.. this change in resistance obviously has an effect on Vin and ultimately Vout of the attenuator. Some one also said yes, but the ratio to Vout over Vin will NOT change, meaning according to -20log(Vout/Vin), the insertion loss does not change.

I really really do not understand.

Am I correct in saying that unless the source is matched, you cannot attain some intended insertion loss? Or at least, attenuation losses are smaller since some power is lost?.

It was connected to a 50 ohm load, and as such I used a function generator with an internal impedance of 50 ohms. Doing the correct calculations, I found the source and load impedance were matched to the attenuator at 50 ohms.

However, removing the power supply with internal impedance for a DC supply with NO internal impedance, and mismatching the circuit, had NO effect on the insertion loss.

I was asked a question, that what if we did not match the source to the attenuator? what would happen? Some student said it would keep the ratio of voltage into the attenuator to voltage out ratio the same. If so, what is the point of the matching in the first place? Surely it has to change it.

All I was ever told was that the matching is required so the waveform not be distorted, purely attenuated.

Perhaps because the resistance is not different, V^2/R on the output will no long be divided by V^2/50 on the input.... and we have more power loss than simply attenuation?

But then again.. this change in resistance obviously has an effect on Vin and ultimately Vout of the attenuator. Some one also said yes, but the ratio to Vout over Vin will NOT change, meaning according to -20log(Vout/Vin), the insertion loss does not change.

I really really do not understand.

Am I correct in saying that unless the source is matched, you cannot attain some intended insertion loss? Or at least, attenuation losses are smaller since some power is lost?.

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