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## Regarding the matching of a signal attenuation to source/load.

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?.
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 Quote by FOIWATER 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.
OK so with the resistors you used what is the attenuation supposed to be ?
Are we talking audio freq's or RF ? what frequency specifically?
and what is the load .... specifically ?
just a 50 Ohm resistor or some piece of equip ?

 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.
this comment makes no sense ..... What power supply ?

 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.
if the inputs and outputs are not matched between source and attenuator and attenuator and load then the mismatch is just going to insert more loss ( attenuation) due to reflections of power at the mismatch points.

 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?.
with matched source and load impedances to the impedance of the attenuator you should be able to seethe expected attenuation of the signal. Obviously give or take a small variations due to resistor tolerences, non ideal impedances quoted for the source and load etc

Dave
 Recognitions: Gold Member 10dB insertion loss was what we wanted. Obviously I couldn't get exact resistors so I got about 10.38dB loss with the ones I had at hand 1KHz sine wave 2Vrms input from a DC source with no impedance connected. Gave me the same insertion loss as a function generator who's manufacturers specs said it had a 50 ohm internal impedance, what didn't make sense about it was it what I said or the results themselves. What I mean was, the DC power supply I was told by my lab technologist has no internal impedance or at least it is so considerably less than the 50 ohm load (just a resistor) it could be considered UNMATCHED. It in neither situation experienced more or less insertion loss. Maybe the difference of between 10.22dB (with DC (nonmatched)) and 10.38 or so with matching

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## Regarding the matching of a signal attenuation to source/load.

but that implies less insertion loss with no matching.

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 Quote by FOIWATER 1KHz sine wave 2Vrms input from a DC source with no impedance connecte
Hi FOIWATER. You confused me earlier by including discussion of a DC power supply while asking a question about impedance matching of a function generator, but now your statement of a "1KHz sine wave 2Vrms input from a DC source" has lost me totally.
 Recognitions: Gold Member haha, sorry, I mean AC There were two parts to the experiment. Initially, we were given parameters by which to design a T type signal attenuator with 10dB insertion loss, using a load resistor of 50 ohms. Got me? From here, we proceeded to connected a DC supply across the attenuator, at this point, the source was not matched to the attenutor circuit (I assume part of the experiment to force me to understand matchings importance) I was asked to calculate the attenuation loss. I did, and proceeded to the next part which asked me now to include a 50 ohm resistor in series with the source, on the input of the attenuator. Doing such, I calculated the insertion loss. Next, I was asked to replace the DC supply for an AC one, outputting 2Vrms sine wave into the attenuator. The manufacturer specs says 50 ohm internal impedance. I removed the 50 ohm resistor from attenuator input. I was asked to again record the insertion loss. A lab question asked for the importance of matching in this circuit. from my lab results, I see no importance. My results were all favourable. Understand or?...
 Recognitions: Gold Member Science Advisor doing this with DC and AC signals/voltages is pretty confusing firstly using a DC voltage connected to the attenuator, you are only going to be using Ohms law to work out voltage drops across the attenuator network and the resistive load But using an AC signal, be it audio or RF, is a whole different ball game as now you are moving into impedance matching and the effects of the attenuator on an AC signal I really dont see how bring DC into this experiment is helpful to understanding how the attenuator works Dave
 Yeah, 'Matching' means different things in different circumstances. The 50 ohms is not a DC resistance it is an impedance over a specific frequency range. Secondly the 50 ohms will be across the terminals or in. It is the characteristic transmission line impedance as seen by a source or load. Its purpose is to prevent reflections in the line which result in loss of energy due to standing waves. Reflections are generated at an electrical discontinuity such as a change of impedance. (Theoretically) no reflections are generated from a 50 ohm source feeding a 50 ohm line connected to a 50 ohm load, using 50 ohm connectors. The impedance is kept low because the reactance of the line contributes to the impedance increasingly as frequency increases, thus attenuating the higher frequency componeents more than the lower frequency ones and altering the waveshape. This is all about preserving the signal integrity and not about transferring power. A power supply on the other hand needs to do one of two things Either you need to transfer the maximum power to the load in which case the generator impedance (resistance if DC) need to equal the load impedance (resistance), in accordance with the maximum power theorem. Or you need a low series internal impedance (resistance if DC) within a voltage power supply so that it can maintain its voltage at a specified level as the current draw increases. This is because the and increasing proportion of the generator voltage is dropped across the internal series impedance as the current increases (Ohms law) by the potential divider effect.
 Recognitions: Gold Member thankyou studiot that's what I needed - sorry to the rest for not being clear - I did not make the experiment however, and I also am very confused about it.
 Recognitions: Gold Member However - I do not really understand the reflection of standing waves as you mentioned. I do understand that at higher frequencies the line reactance is increased and that will distort the signal. And yes - I see now that is what we are aiming for here.

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