Is AC current analogous to sound waves?

AI Thread Summary
AC current can be likened to sound waves as both involve the transfer of energy through compressional waves. In a conductor connected to a sinusoidal voltage source, charge density oscillates, creating a compressional wave that reflects at the open circuit, forming standing waves with nodes and points of maximum displacement. The behavior of these waves changes with harmonics, where the number of nodes and displacement points increases. In antenna design, energy is also carried by surrounding fields, allowing for current flow even in a vacuum. Understanding the relationship between standing waves and antenna dimensions is crucial, as standing waves enhance amplitude, which is essential for effective RF power flow.
CraigH
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It is really hard to picture current as a wave. The best explanation I have had so far is that the energy is transferred along the wire as a longitudinal wave of compressions of charge density, just like a sound wave is a longitudinal compression of mass density.

so consider an open circuit connected to a sinusoidal voltage source. the charge's inside this conductor accelerate back and forward due to the force from the electric field. This creates a compressional wave of charge density through the conductor.

when this wave reaches the end of the conductor (because it is an open circuit), it will be reflected but upside down. This will create a standing wave.

So inside this conductor are points where the charge is not moving (the nodes) and in between the nodes there are points where the charge is oscillating with the maximum displacement.

For the fundamental standing wave there will be 1 node (at the end of the open circuit as the charge cannot oscillate out of the conductor) and 1 maximum point of displacement.
For the first harmonic, 2 nodes and 2 points of maximum displacement, for the second harmonic 3 etc...

Is this all correct? I am currently writing a report on antenna design and I need to know if I have the basics fully understood.

Thanks!
 
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If you're dealing with RF and antenna design it's not that simple. The open circuit can conduct 'current' even in a vacuum because the energy is carried in the fields that surround the wire or antenna with the conductors charge carriers creating a low-loss path (near field) for that energy.

For a simple electromagnetic radiation example with a quarter wave antenna element you have a standing wave at the (low impedance) feed point with high current (B field) and low voltage (E field) , as the wave moves to the end of the antenna the relationship of voltage and current changes causing the impedance to change (higher impedance) until we get to the end of antenna where current is low and voltage is high. We have the largest (far field) RF power flow away from the antenna near the region of the antenna where the impedance is also close to the reactive impedance of space around the antenna.

http://en.wikipedia.org/wiki/Near_and_far_field
http://en.wikipedia.org/wiki/Electromagnetic_radiation
http://www.moonraker.com.au/techni/marconi1.jpg
 
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Thanks for answering, but to be honest I still do not understand this. Am I at least right in saying that at certain frequencies there will be a standing wave, and between these will be traveling waves or partially standing waves, and that the frequency at which these standing waves are formed at are determined by the dimensions of the antenna?
We want standing waves because there amplitude is much higher.

If this is correct I'll be happy. I'm giving up on trying to understand the mechanism of the formation of these standing waves. The jump between A level physics and undergrad electrical engineering is to big when it comes to antennas in my opinion. Before this module all I had covered was basic wave theory, and electric and magnetic fields separately. Now suddenly they have all been thrown together and nothings been explained.
 
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