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Audio Signal Conversion

Hi Guys, This might be a bit of a different question. I have a sound generator for my model railroad. It generate the sounds of locomotives as I operate them. It is triggered by voltage changes as I move the rheostat throttle. This power pack has screw terminals. The instructions say to run the two screw terminals by wire to a 4 inch 8 ohm speaker. What I need to do is to run the wires from the two screw terminals to my mix board so I can take control of the sounds and shape them E.Q. wise and then amplify without distortion. The specs on the speaker out of the power pack says "1 Watt Audio Power into 8 ohms impedance 65db signal to noise ratio". Are you aware of a converter box or something that would work well for this? This is outdated technology but I want to bring out the best of what is there. I thought I would ask you folks as you have never steered me wrong in the past.
Thanks, Tim
 

Harald Kapp

Moderator
Moderator
Welcome to EP.
1 W into 8 Ohm is equivalent to ~2.8 V (rms) as per P = V²/R -> V = sqrt(P × R)
Line level is on the order of 1.2 V (rms) (Wikipedia) and impedance of your mixer is way higher than 8 Ω, typically around 10 kΩ.
You need to scale down the output voltage of your signal generator by a factor of 2.3 (2.8 V / 1.2 V). For simplicity, let's make that a factor of 3 (you're going to amplify the signal after the mixer anyway. Thus a 20 kΩ resistor between the signal generator and the mixer should do the job. For better symmetry use 2 × 10 kΩ resistors in this configuration:
upload_2021-2-19_7-25-18.png
 
Thanks for the reply. -10db is where I want to end up. That way I can pack a punch with minimal distortion. This sound generator was produced in 1993 and although they were using pulse width modulation for their capture, it still could use some reinforcement. Thanks again.
 

Harald Kapp

Moderator
Moderator
-10 dB is equivalent to a reduction in voltage by a factor of 0.3. Multiply by 1/2.3 to account for the adaptation to line level and you get a total og 0.13. make that 0.1 for some headroom.That means with 10 kΩ input impedance the series resistance (R1+R2) needs to b 9 kΩ (voltage divider: Vout/Vin = R3/(R1+R2+R3). Splitting 9 kΩ equally between R1 and R2 makes 4.5 kΩ for each of them. A standard value for resistors would be 4.7 kΩ. That should bring you near the desired -10 dB.
If you want to add the flexibility of adjustment, replace one of R1 or R2 (not both) by a 10 kΩ potentiometer connected as a rheostat.
 
Thank you for your reply. This is why I love you guys! You explain what is needed and what is involved theory wise. You see, I am a model railroader and an amatuer sound engineer. I have a home studio and lots of audio equipment laying around. I purchased a power pack for the trains that syncs PWM sounds with the trains. It does this by way of applied voltage to the track. So, the pack has two screw terminals that are labeled "speaker out". The manual wants you to run wires to a 8 ohm 4" speaker. Having done this, I recognized that the sound isn't half bad. Oh ya, I get to blow the whistle or horn depending on if I am driving a steam engine or a diesel. Anyway, I thought that I would be able to find a attenuator box that would step the signal down to line-level (-10db). However, while there are numerous boxes that step up the signal, there aren't any advertised to do a step-down. You guys gave me the process written out. That, in turn, lead me to an audio rack-mount signal processor that, while not advertised as such, all the components are there to do a step-down. At least it looks that way to me. I am having some engineers I have worked with check this box out. If it does the trick, great! If not, it will be time to get the soldering iron and some circuit board out and follow your collective advice further. I thank you guys for being here and for your willingness to help. Tim
 
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