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Q&A Electric circuit with antenna

Antennas are not lumped-parameter systems. They may be open circuits or shorts at DC, but this changes as the frequency goes up. They aren't just a bunch of wires anymore at the intended operatin...

posted 3y ago by Olin Lathrop‭  ·  edited 3y ago by Olin Lathrop‭

Answer
#3: Post edited by user avatar Olin Lathrop‭ · 2021-07-22T16:59:18Z (over 3 years ago)
  • Antennas are not lumped-parameter systems. They may be open circuits or shorts at DC, but this changes as the frequency goes up. They aren't just a bunch of wires anymore at the intended operating frequency.
  • Take a simple dipole as example. It's just two disconnected wires at DC, but that doesn't tell you anything useful about what happens at the optimum radiation frequency. If you apply a steady voltage, one wire becomes positive, and the other negative. However, if you keep changing that voltage, then charges must move back and forth.
  • If you change the voltage at the right rate so that it matches the speed of propagation in the wire, then you're exiting resonance. Charges slosh back and forth smoothly from one end to the other. Just as the charges get to one end, the driving voltage is changed so that they need to slosh to the other end again. You end up with a voltage profile of half a sine wave. The current profile is also half a sine wave, but with a 90&deg; phase shift.
  • The concept is similar to moving hour hand back and forth in a bathtub of water. If you do it at the right frequency, a little back and forth with your hand in the middle causes a lot of back and forth of all the water in the bathtub. Each time your hand goes back and forth, it adds a little energy to the system. That energy goes into increasing the amplitude of the water's movement. The wave peaks at each end get higher, and the volume of water that sloshes past the middle each cycle gets larger.
  • The height of the water is like the voltage on the dipole, and volume that moves each cycle is like the current.
  • The dipole has an additional trick that the bathtub doesn't. The charges sloshing back and forth is a current, which creates a magnetic field. The voltage going up and down creates an electric field. When these are matched at the right phase and relative amplitudes, then energy is propagated into space.
  • Since energy is radiated from the antenna, it has to be replaced if a signal is transmitted continuously. The appears to the driving circuit as the antenna having a resistive component. In fact, at the optimum radiation frequency, the antenna ideally looks like a pure resistance to the driving circuit. That resistance happens to be 75 &Omega; for a simple half wave dipole. Other antennas have other characteristic impedances.
  • Antennas are not lumped-parameter systems. They may be open circuits or shorts at DC, but this changes as the frequency goes up. They aren't just a bunch of wires anymore at the intended operating frequency.
  • Take a simple dipole as example. It's just two disconnected wires at DC, but that doesn't tell you anything useful about what happens at the optimum radiation frequency. If you apply a steady voltage, one wire becomes positive, and the other negative. However, if you keep changing that voltage, then charges must move back and forth.
  • If you change the voltage at the right rate so that it matches the speed of propagation in the wire, then you're exiting resonance. Charges slosh back and forth smoothly from one end to the other. Just as the charges get to one end, the driving voltage is changed so that they need to slosh to the other end again. You end up with a voltage profile of half a sine wave. The current profile is also half a sine wave, but with a 90&deg; phase shift.
  • The concept is similar to moving your hand back and forth in a bathtub of water. If you do it at the right frequency, a little back and forth with your hand in the middle causes a lot of back and forth of all the water in the bathtub. Each time your hand goes back and forth, it adds a little energy to the system. That energy goes into increasing the amplitude of the water's movement. The wave peaks at each end get higher, and the volume of water that sloshes past the middle each cycle gets larger.
  • The height of the water is like the voltage on the dipole, and volume that moves each cycle is like the current.
  • The dipole has an additional trick that the bathtub doesn't. The charges sloshing back and forth is a current, which creates a magnetic field. The voltage going up and down creates an electric field. When these are matched at the right phase and relative amplitudes, then energy is propagated into space.
  • Since energy is radiated from the antenna, it has to be replaced if a signal is transmitted continuously. This appears to the driving circuit as the antenna having a resistive component. In fact, at the optimum radiation frequency, the antenna ideally looks like a pure resistance to the driving circuit. That resistance happens to be 75 &Omega; for a simple half wave dipole. Other antennas have other characteristic impedances.
#2: Post edited by user avatar Olin Lathrop‭ · 2021-07-22T12:50:07Z (over 3 years ago)
  • Antennas are not lumped-parameter systems. They may be open circuits or shorts at DC, but this changes as the frequency goes up. They aren't just a bunch of wires anymore at the intended operating frequency.
  • Take a simple dipole as example. It's just two disconnected wires at DC, but that doesn't tell you anything useful about what happens at the optimum radiation frequency. If you apply a steady voltage, one wire becomes positive, and the other negative. However, if you keep changing that voltage, then charges must move back and forth.
  • If you change the voltage at the right rate so that it matches the speed of propagation in the wire, then you're exiting resonance. Charges slosh back and forth smoothly from one end to the other. Just as the charges get to one end, the driving voltage is changed so that they need to slosh to the other end again. You end up with a voltage profile of half a sine wave. The current profile is also half a sine wave, but with a 90&deg; phase shift.
  • The concept is similar to moving hour hand back and forth in a bathtub of water. If you do it at the right frequency, a little back and forth with your hand in the middle causes a lot of back and forth of all the water in the bathtub. Each time your hand goes back and forth, it add a little energy to the system. That energy goes into increasing the amplitude of the water's movement. The wave peaks at each end get higher, and the volume of water that sloshes past the middle each cycle gets larger.
  • The height of the water is like the voltage on the dipole, and volume that moves each cycle is like the current.
  • The dipole has an additional trick that the bathtub doesn't. The charges sloshing back and forth is a current, which creates a magnetic field. The voltage going up and down creates an electric field. When these are matched at the right phase and relative amplitudes, then energy is propagated into space.
  • Since energy is radiated from the antenna, it has to be replaced if a signal is transmitted continuously. The appears to the driving circuit as the antenna have a resistive component. In fact, at the optimum radiation frequency, the antenna ideally looks like a pure resistance to the driving circuit. That resistance happens to be 75 &Omega; for a simple half wave dipole. Other antennas have other characteristic impedances.
  • Antennas are not lumped-parameter systems. They may be open circuits or shorts at DC, but this changes as the frequency goes up. They aren't just a bunch of wires anymore at the intended operating frequency.
  • Take a simple dipole as example. It's just two disconnected wires at DC, but that doesn't tell you anything useful about what happens at the optimum radiation frequency. If you apply a steady voltage, one wire becomes positive, and the other negative. However, if you keep changing that voltage, then charges must move back and forth.
  • If you change the voltage at the right rate so that it matches the speed of propagation in the wire, then you're exiting resonance. Charges slosh back and forth smoothly from one end to the other. Just as the charges get to one end, the driving voltage is changed so that they need to slosh to the other end again. You end up with a voltage profile of half a sine wave. The current profile is also half a sine wave, but with a 90&deg; phase shift.
  • The concept is similar to moving hour hand back and forth in a bathtub of water. If you do it at the right frequency, a little back and forth with your hand in the middle causes a lot of back and forth of all the water in the bathtub. Each time your hand goes back and forth, it adds a little energy to the system. That energy goes into increasing the amplitude of the water's movement. The wave peaks at each end get higher, and the volume of water that sloshes past the middle each cycle gets larger.
  • The height of the water is like the voltage on the dipole, and volume that moves each cycle is like the current.
  • The dipole has an additional trick that the bathtub doesn't. The charges sloshing back and forth is a current, which creates a magnetic field. The voltage going up and down creates an electric field. When these are matched at the right phase and relative amplitudes, then energy is propagated into space.
  • Since energy is radiated from the antenna, it has to be replaced if a signal is transmitted continuously. The appears to the driving circuit as the antenna having a resistive component. In fact, at the optimum radiation frequency, the antenna ideally looks like a pure resistance to the driving circuit. That resistance happens to be 75 &Omega; for a simple half wave dipole. Other antennas have other characteristic impedances.
#1: Initial revision by user avatar Olin Lathrop‭ · 2021-07-22T12:06:08Z (over 3 years ago) 
Antennas are not lumped-parameter systems.  They may be open circuits or shorts at DC, but this changes as the frequency goes up.  They aren't just a bunch of wires anymore at the intended operating frequency.

Take a simple dipole as example.  It's just two disconnected wires at DC, but that doesn't tell you anything useful about what happens at the optimum radiation frequency.  If you apply a steady voltage, one wire becomes positive, and the other negative.  However, if you keep changing that voltage, then charges must move back and forth.

If you change the voltage at the right rate so that it matches the speed of propagation in the wire, then you're exiting resonance.  Charges slosh back and forth smoothly from one end to the other.  Just as the charges get to one end, the driving voltage is changed so that they need to slosh to the other end again.  You end up with a voltage profile of half a sine wave.  The current profile is also half a sine wave, but with a 90&deg; phase shift.

The concept is similar to moving hour hand back and forth in a bathtub of water.  If you do it at the right frequency, a little back and forth with your hand in the middle causes a lot of back and forth of all the water in the bathtub.  Each time your hand goes back and forth, it add a little energy to the system.  That energy goes into increasing the amplitude of the water's movement.  The wave peaks at each end get higher, and the volume of water that sloshes past the middle each cycle gets larger.

The height of the water is like the voltage on the dipole, and volume that moves each cycle is like the current.

The dipole has an additional trick that the bathtub doesn't.  The charges sloshing back and forth is a current, which creates a magnetic field.  The voltage going up and down creates an electric field.  When these are matched at the right phase and relative amplitudes, then energy is propagated into space.

Since energy is radiated from the antenna, it has to be replaced if a signal is transmitted continuously.  The appears to the driving circuit as the antenna have a resistive component.  In fact, at the optimum radiation frequency, the antenna ideally looks like a pure resistance to the driving circuit.  That resistance happens to be 75 &Omega; for a simple half wave dipole.  Other antennas have other characteristic impedances.