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Also, not all radios are designed for a 50 ohm load. In particular, older tube radios typically have a variable output network, so they will work with a whole range of loads. Also practical coax is an invention only a few decades old; prior to that the most common feedline was some kind of balanced feedline typically with a higher impedance.

Also, not all radios are designed for a 50 ohm load. In particular, older tube radios typically have a variable output network, so they will work with a whole range of loads. And although coax has been around since the mid-19th century, it wasn't really until after WWII that it became available to regular folk. Prior to that the most common feedline was some kind of balanced feedline typically with a higher impedance.

There are a lot of topics in this question, so let's take them one at a time.

I figure the transmitter has a 50-ohm pure resistive output

Not necessarily. You're probably arriving at this conclusion based on the maximum power transfer theorem. Which of these circuits delivers more power to the load resistor?

^{simulate this circuit – Schematic created using CircuitLab}

What we can say is the manufacturer has designed the transmitter expecting a 50 ohm load, and they've also designed it to minimize their cost, thus maximizing their profit. This means when the load is 50 ohms the transmitter will be able to make its rated power while staying within the ratings of the components of the transmitter, the finals especially. But there won't be much margin for deviation, because that would increase costs.

When the load isn't 50 ohms this might subject the finals to too much current, voltage, or power. If we're lucky this means the radio just reduces power. If we're unlucky the radio doesn't reduce power, and the finals are damaged.

So we want the transmitter to see 50 ohms so it can operate as designed.

I think it's usually the antenna itself that actually needs to be tuned

Usually, but not always. Most radios these days are designed for a 50 ohm load, and 50 ohm coax is very popular. But there is also 75 ohm coax, and there are balanced feedlines between 200 and 600 ohms which are readily available.

Also, not all radios are designed for a 50 ohm load. In particular, older tube radios typically have a variable output network, so they will work with a whole range of loads. Also practical coax is an invention only a few decades old; prior to that the most common feedline was some kind of balanced feedline typically with a higher impedance.

my guess is that the transmitter sees a "perfect" match and will have no complaints but that there are reflections from the antenna port going back and forth between the tuner and the antenna.

This is exactly right. The reflected power from the antenna encounters the tuner, and the tuner (if you've managed to adjust it so the transmitter sees a 1:1 SWR) re-reflects that power back at the antenna. The consequence of these extra reflections is usually (but counter-intuitively, not always!) additional loss in the feedline.

i think this would create a power bottleneck where the transmitter could put out more power but the tuner/cable/antenna combo can't "absorb" that much power.

Not usually. Under normal circumstances, the feedline and antenna are linear systems, which means waves can be superimposed indefinitely. What happens which each reflection is independent of what's happening with other reflections at the same time.

However, linearity does break down eventually for just about any real system. To give one example, that additional loss in the feedline causes the feedline to become warmer. At some point it will become warm enough that the dielectric may melt, and the center conductor will short to the shield. Or, the voltage can get high enough to arc through the coax dielectric. It's pretty much impossible to get to this point with 100W and LMR-400, but 2kW and some really terrible RG-58 might. For broadcast stations in the megawatts, it's definitely a concern.

In summary, on purely theoretical grounds it usually is better to put the tuner at the antenna end of the feedline. However this requires making it weatherproof, and having some mechanism to remotely operate it, which makes it more expensive and difficult to install. Those downsides may or may not outweigh the benefit, depending on circumstances and priorities.

There are a lot of topics in this question, so let's take them one at a time.

I figure the transmitter has a 50-ohm pure resistive output

Not necessarily. You're probably arriving at this conclusion based on the maximum power transfer theorem. Which of these circuits delivers more power to the load resistor?

^{simulate this circuit – Schematic created using CircuitLab}

What we can say is the manufacturer has designed the transmitter expecting a 50 ohm load, and they've also designed it to minimize their cost, thus maximizing their profit. This means when the load is 50 ohms the transmitter will be able to make its rated power while staying within the ratings of the components of the transmitter, the finals especially. But there won't be much margin for deviation, because that would increase costs.

When the load isn't 50 ohms this might subject the finals to too much current, voltage, or power. If we're lucky this means the radio just reduces power. If we're unlucky the radio doesn't reduce power, and the finals are damaged.

So we want the transmitter to see 50 ohms so it can operate as designed.

I think it's usually the antenna itself that actually needs to be tuned

Usually, but not always. Most radios these days are designed for a 50 ohm load, and 50 ohm coax is very popular. But there is also 75 ohm coax, and there are balanced feedlines between 200 and 600 ohms which are readily available.

Also, not all radios are designed for a 50 ohm load. In particular, older tube radios typically have a variable output network, so they will work with a whole range of loads. Also practical coax is an invention only a few decades old; prior to that the most common feedline was some kind of balanced feedline typically with a higher impedance.

my guess is that the transmitter sees a "perfect" match and will have no complaints but that there are reflections from the antenna port going back and forth between the tuner and the antenna.

This is exactly right. The reflected power from the antenna encounters the tuner, and the tuner (if you've managed to adjust it so the transmitter sees a 1:1 SWR) re-reflects that power back at the antenna. The consequence of these extra reflections is usually (but counter-intuitively, not always!) additional loss in the feedline.

i think this would create a power bottleneck where the transmitter could put out more power but the tuner/cable/antenna combo can't "absorb" that much power.

Not usually. Under normal circumstances, the feedline and antenna are linear systems, which means waves can be superimposed indefinitely. What happens is that each reflection is independent of what's happening with other reflections at the same time.

However, linearity does break down eventually for just about any real system. To give one example, that additional loss in the feedline causes the feedline to become warmer. At some point it will become warm enough that the dielectric may melt, and the center conductor will short to the shield. Or, the voltage can get high enough to arc through the coax dielectric. It's pretty much impossible to get to this point with 100W and LMR-400, but 2kW and some really terrible RG-58 might. For broadcast stations in the megawatts, it's definitely a concern.

In summary, on purely theoretical grounds it usually is better to put the tuner at the antenna end of the feedline. However this requires making it weatherproof, and having some mechanism to remotely operate it, which makes it more expensive and difficult to install. Those downsides may or may not outweigh the benefit, depending on circumstances and priorities.

There are a lot of topics in this question, so let's take them one at a time.

I figure the transmitter has a 50-ohm pure resistive output

Not necessarily. You're probably arriving at this conclusion based on the maximum power transfer theorem. Which of these circuits delivers more power to the load resistor?

^{simulate this circuit – Schematic created using CircuitLab}

What we can say is the manufacturer has designed the transmitter expecting a 50 ohm load, and they've also designed it to minimize their cost, thus maximizing their profit. This means when the load is 50 ohms the transmitter will be able to make its rated power while staying within the ratings of the components of the transmitter, the finals especially. But there won't be much margin for deviation, because that would increase costs.

When the load isn't 50 ohms this might subject the finals to too much current, voltage, or power. If we're lucky this means the radio just reduces power. If we're unlucky the radio doesn't reduce power, and the finals are damaged.

So we want the transmitter to see 50 ohms so it can operate as designed.

I think it's usually the antenna itself that actually needs to be tuned

Usually, but not always. Most radios these days are designed for a 50 ohm load, and 50 ohm coax is very popular. But there is also 75 ohm coax, and there are balanced feedlines between 200 and 600 ohms which are readily available.

Also, not all radios are designed for a 50 ohm load. In particular, older tube radios typically have a variable output network, so they will work with a whole range of loads. Also coax is an invention only a few decades old; prior to that the most common feedline was some kind of balanced feedline typically with a higher impedance.

my guess is that the transmitter sees a "perfect" match and will have no complaints but that there are reflections from the antenna port going back and forth between the tuner and the antenna.

This is exactly right. The reflected power from the antenna encounters the tuner, and the tuner (if you've managed to adjust it so the transmitter sees a 1:1 SWR) re-reflects that power back at the antenna. The consequence of these extra reflections is usually (but counter-intuitively, not always!) additional loss in the feedline.

i think this would create a power bottleneck where the transmitter could put out more power but the tuner/cable/antenna combo can't "absorb" that much power.

Not usually. Under normal circumstances, the feedline and antenna are linear systems, which means waves can be superimposed indefinitely. What happens which each reflection is independent of what's happening with other reflections at the same time.

However, linearity does break down eventually for just about any real system. To give one example, that additional loss in the feedline causes the feedline to become warmer. At some point it will become warm enough that the dielectric may melt, and the center conductor will short to the shield. Or, the voltage can get high enough to arc through the coax dielectric. It's pretty much impossible to get to this point with 100W and LMR-400, but 2kW and some really terrible RG-58 might. For broadcast stations in the megawatts, it's definitely a concern.

In summary, on purely theoretical grounds it usually is better to put the tuner at the antenna end of the feedline. However this requires making it weatherproof, and having some mechanism to remotely operate it, which makes it more expensive and difficult to install. Those downsides may or may not outweigh the benefit, depending on circumstances and priorities.

There are a lot of topics in this question, so let's take them one at a time.

I figure the transmitter has a 50-ohm pure resistive output

Not necessarily. You're probably arriving at this conclusion based on the maximum power transfer theorem. Which of these circuits delivers more power to the load resistor?

^{simulate this circuit – Schematic created using CircuitLab}

What we can say is the manufacturer has designed the transmitter expecting a 50 ohm load, and they've also designed it to minimize their cost, thus maximizing their profit. This means when the load is 50 ohms the transmitter will be able to make its rated power while staying within the ratings of the components of the transmitter, the finals especially. But there won't be much margin for deviation, because that would increase costs.

When the load isn't 50 ohms this might subject the finals to too much current, voltage, or power. If we're lucky this means the radio just reduces power. If we're unlucky the radio doesn't reduce power, and the finals are damaged.

So we want the transmitter to see 50 ohms so it can operate as designed.

I think it's usually the antenna itself that actually needs to be tuned

Usually, but not always. Most radios these days are designed for a 50 ohm load, and 50 ohm coax is very popular. But there is also 75 ohm coax, and there are balanced feedlines between 200 and 600 ohms which are readily available.

Also, not all radios are designed for a 50 ohm load. In particular, older tube radios typically have a variable output network, so they will work with a whole range of loads. Also practical coax is an invention only a few decades old; prior to that the most common feedline was some kind of balanced feedline typically with a higher impedance.

my guess is that the transmitter sees a "perfect" match and will have no complaints but that there are reflections from the antenna port going back and forth between the tuner and the antenna.

This is exactly right. The reflected power from the antenna encounters the tuner, and the tuner (if you've managed to adjust it so the transmitter sees a 1:1 SWR) re-reflects that power back at the antenna. The consequence of these extra reflections is usually (but counter-intuitively, not always!) additional loss in the feedline.

i think this would create a power bottleneck where the transmitter could put out more power but the tuner/cable/antenna combo can't "absorb" that much power.

Not usually. Under normal circumstances, the feedline and antenna are linear systems, which means waves can be superimposed indefinitely. What happens which each reflection is independent of what's happening with other reflections at the same time.

However, linearity does break down eventually for just about any real system. To give one example, that additional loss in the feedline causes the feedline to become warmer. At some point it will become warm enough that the dielectric may melt, and the center conductor will short to the shield. Or, the voltage can get high enough to arc through the coax dielectric. It's pretty much impossible to get to this point with 100W and LMR-400, but 2kW and some really terrible RG-58 might. For broadcast stations in the megawatts, it's definitely a concern.

In summary, on purely theoretical grounds it usually is better to put the tuner at the antenna end of the feedline. However this requires making it weatherproof, and having some mechanism to remotely operate it, which makes it more expensive and difficult to install. Those downsides may or may not outweigh the benefit, depending on circumstances and priorities.