Home |
Search |
Today's Posts |
#31
![]() |
|||
|
|||
![]()
Yuri Blanarovich wrote:
They must be inteligent smileys, with glasses 8-) or divers 0-) or rich ones $-) I'm not very internet savvy. What is an intelligent smiley? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#32
![]() |
|||
|
|||
![]()
"Isn't a 50 ohm transmitter conjugately matched to a 50 ohm load
when fed through 1/2 wavelength of 450 ohm ladder-line?" ============== Heck yes, I'm assuming u mean 50 + j 0 Ohms. 73 de Jack, K9CUN |
#33
![]() |
|||
|
|||
![]()
Everybody keeps going on and on and on about 50-ohm transmitters when in
fact there's no such thing. |
#34
![]() |
|||
|
|||
![]()
Gene Fuller wrote:
"The maximum power transfer theorem describes the impact from change of the load impedance." That`s true. When impedances at a junction become matched, you can cut the matched line at any place and the impedances looking in opposite directions are conjugates of each other. It is not pathological to take full advantage of the theeorem. Conjugate matching is the same as d-c matching except that reactance must be neutralized not to get into the way of a-c maximum power transfer. W.L. Everitt is one of many authors who state the maximum power transfer theorem. I quoted Terman early in this thread. Here is a quote from King, Mimno, annd Wing on page 43 of "Transmission Lines, Antennas, and Wave Guides": If a dissipationless network is inserted between a constant-voltage generator of internal impedance ZG and a load of impedance ZR such that maximum power is delivered to the load, at every pair of terminals the impedances looking in opposite directions are conjugates of each other. There`s lots more. Read the book. Best regards, Richard Harrison, KB5WZI |
#35
![]() |
|||
|
|||
![]()
On Wed, 3 Mar 2004 15:48:38 +0000 (UTC), "Reg Edwards"
wrote: Everybody keeps going on and on and on about 50-ohm transmitters when in fact there's no such thing. Yes, you do keep going on and on about it, don't you? That and Q and SWR meter alternative names - all pretty lame stuff. |
#36
![]() |
|||
|
|||
![]()
Richard Harrison wrote:
If a dissipationless network is inserted between a constant-voltage generator of internal impedance ZG and a load of impedance ZR such that maximum power is delivered to the load, at every pair of terminals the impedances looking in opposite directions are conjugates of each other. Since dissipationless networks don't exist in reality, doesn't this imply that it is impossible to achieve a conjugate match at *all* points in a real-world system? - Not trying to be ornery, just realistic. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#37
![]() |
|||
|
|||
![]()
Old Ed wrote:
"Couild you explain this concept, and/or provide some references?" Suppose we adjust a variable d-c supply to full-scale indication on an external meter. Next, install a chopper (lossless on-off circuit) driven at a high frequency to produce a square wave interruption of the d-c with a 50% duty cycle, and insert the chopper contacts in series with the external meter. The chopper connects the external meter 50% of the time and disconnercts it 50% of the time. The meter reads 50% of full scale. Another way to reduce the scale reading to 50% is to insert a resistor in series with the meter. If it is an 0-1 ma meter and if it has an internal resistance of 1000 ohms, insertion of a 1000-ohm resistor in series with the meter will reduce the meter scale indication to 50%. The chopper as part of the meter source eliminates current to the meter 50% of the time. The resistor which has the same effect and produces the same scale indication as the chopper on the effective output current exacts its loss of 0.0005 amp x 0.5 volt or 0.25 milliwatt during 100% of the time. The chopper eliminates the power-losing resistor by substituting off-time in the power source. The source only supplies the power used by the load. With a resistor limiting power to the load, the source supplies its loss and the load power. The power in the load, a meter in our example, is the same using either the resistor or the chopper. The resistor is analogous to a Class-A amplifier. The chopper is analogous to a Class-C amplifier. The off-time has the same effective opposition to current to a load as a dissipative resistance. As the time-limited currented opposition to load current consumes no power, it is called a dissipationless resistance. Best regards, Richard Harrison, KB5WZI |
#38
![]() |
|||
|
|||
![]()
It is in the same class of parts with inductancless inductors,
and capacitanceless capacitors are in. With lossless resistance they can form a lossless, non-energy storing, extremely low/high Q, network. No need for tuning either! (sorry-just not serious today) "Old Ed" wrote in message hlink.net... "Lossless resistance?" Would that be zero resistance, or perhaps a negative resistance, as in the active part of a tunnel diode's V-I characteristic? I am a career EE, with a couple of graduate EE degrees; and this is something entirely new to me. Could you explain this concept, and/or provide some references? Thanks, Ed |
#39
![]() |
|||
|
|||
![]() "Richard Clark" wrote in message ... On Wed, 3 Mar 2004 15:48:38 +0000 (UTC), "Reg Edwards" wrote: Everybody keeps going on and on and on about 50-ohm transmitters when in fact there's no such thing. Yes, you do keep going on and on about it, don't you? That and Q and SWR meter alternative names - all pretty lame stuff. and on, and on, and on, and on... i knew this would be a good thread. |
#40
![]() |
|||
|
|||
![]()
John Smith wrote:
"It is in the same class of parts with inductanceless inductors, and capacitanceless capacitors are in." Lossless resistance makes volts per amp. Inductors and capacitors are lossless because they only store energy and give it all back. A lossless resistance stores no energy. Its action is immediate. An inductance or capacitance with zero reactance also stores no energy and makes no volts per amp as opposition. Best regards, Richard Harrison, KB5WZI |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
Complex Z0 [Corrected] | Antenna | |||
Derivation of the Reflection Coefficient? | Antenna | |||
The Cecilian Gambit, a variation on the Galilean Defense revisited | Antenna |