Hello I am confused about how EM shielding works, especially after checking out my VGA cable.
The end of the VGA cable has a metal chassis around the 15 pins, covered by a plastic coat, and 2 screws on the 2 sides. The 15 pins connect to the 15 wires inside the cable. The core wires are insulated by a layer of plastic coat and twisted together, and then shielded by a layer of aluminium foil, and then by a layer of braided aluminium, and then covered by the outside rubber coat. This is how it's made.
Now my assumption is that the shield (the foil+ braided aluminium) has to be insulated from the core of the signal wire otherwise it short circuit's it, or perhaps cause interference and act as a receiver antenna that would absorb the noise to destroy the signal inside the cable.
But that is not true, there is electric connection between the 15 pins, the end-chassis, and even the screws, therefore I assume that the shield itself is also connected to the signal wire. How is this even possible?
Can somebody explain how the electromagnetic shielding works for a VGA cable, because it sounds illogical to me how the inside signal is connected to the shield, I thought they have to be insulated from one another.
Especially how does this shield the inside of the cable from outside noise, doesn't this make the cable an ever bigger antenna if the metal surface is bigger?
And note that this is not a defect cable, I have checked it on 2 different cables, they are all the same and they all work perfectly fine with my monitor.