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Basic Adivce/tutoring sought
Greetings all
I hope I make sense, as I'm seeking some guidance through this, please. I'm trying to get to grips with some basic electronics. I know and have learnt the electronic math rules W = V x A ; V = IR , etc.. What I get stuck on is more "why would one use a capacitor in a circuit, why not a resistor?" I understand the functions that a part plays (sometimes not 100% correct) but generally eg: capacitor: takes in energy (voltage) and stores it, until it is released. Resistor: acts to slow down the voltage as like a form of friction. much like one lane of traffic compared to a 4 lane highway. But what I don't grasp is why do we place a resistor with a coil to make a tuning circuit, why not use a capacitor. I'm trying to grasp what is it about the resistor, that makes the two components work and thus form a tuning circuit.? For example: If I take an FM Receiver and Transmitter kit (wireless microphone kit or remote control kit), say I connect a 1 Watt power source to the Transmitter, (in place of the human voice on the microphone kit) what components will effect the Voltage and Amperage (v x a = W) in the receiver side, and should it be the same +/- 1 Watt and what would cause the decrease should there be a difference? I thank you for your assistance up front, Thanks Barry |
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Roy Ingham wrote:
Greetings all I hope I make sense, as I'm seeking some guidance through this, please. I'm trying to get to grips with some basic electronics. I know and have learnt the electronic math rules W = V x A ; V = IR , etc.. What I get stuck on is more "why would one use a capacitor in a circuit, why not a resistor?" Capacitors remember the past, resistors don't. The voltage across a capacitor is proportional to the total charge that has passed through it, since it last had zero volts across it. Or mathematically, V=Q/C. 1/C is just the constant of proportionality between voltage and charge. If you look at this as a process, in time, rather than as a result, you can say that the rate of change of the voltage across a capacitor is proportional to the current passing through it. I=C*(dv/dt), where dv/dt means rate of change of voltage with respect to time, with units like "volts per second". I understand the functions that a part plays (sometimes not 100% correct) but generally eg: capacitor: takes in energy (voltage) and stores it, until it is released. Resistor: acts to slow down the voltage as like a form of friction. much like one lane of traffic compared to a 4 lane highway. Think of a resistor as a device that enforces a fixed proportionality between voltage and current. Another way to look at ohms is to call them volts per ampere. Double the volts (electromotive force across the resistor) and the amperes through it (amount of charge per second) also doubles. But what I don't grasp is why do we place (replace?) a resistor with a coil to make a tuning circuit, why not use a capacitor. I'm trying to grasp what is it about the resistor, that makes the two components work and thus form a tuning circuit.? An inductor also remembers the past, but instead of having a voltage in proportion to total charge, it has a current proportional to total volt seconds. So it remembers how much voltage for how long has been applied to it, and that memory is represented by its current. A capacitor and inductor in combination act something like a spring and mass act, mechanically to produce a resonance. When the spring is at maximum distortion and the mass is changing directions, but momentarily at rest, all the energy is stored in the spring. But when the mass hits peak velocity and the spring is passing through its relaxed state, all the energy is stored in the kinetic energy of the mass. The energy sloshes back and forth between spring and mass, twice per cycle, and the value of spring stiffness and mass determine the resonant frequency. For example: If I take an FM Receiver and Transmitter kit (wireless microphone kit or remote control kit), say I connect a 1 Watt power source to the Transmitter, (in place of the human voice on the microphone kit) what components will effect the Voltage and Amperage (v x a = W) in the receiver side, and should it be the same +/- 1 Watt and what would cause the decrease should there be a difference? I'm not following the question. I think you are trying to combine several processes into one, and skipping too many steps. It is sort of like saying that if you can run a mile on one hamburger, how fast can you run if you eat a gallon of gasoline. The FM transmitter combines radio frequency energy with audio by varying the frequency of a fixed energy carrier in proportion to the amplitude of the audio signal. Replacing the audio with something else does not change the power of the carrier. The receiver separated the modulated carrier from all other frequencies (if it receives enough energy from the transmitter) and then discards all information about how strong the carrier is, and just responds to the frequency variations to recreate the audio signal in proportion to the frequency shifts. Doubling the power of the transmitter carrier allows this process to work over a longer distance, but does not change the volume of the audio at the receiver. I thank you for your assistance up front, Thanks Barry |
#3
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Firstly thank you for your reply it was helpful.
John Popelish wrote: .... Capacitors remember the past, resistors don't. .... An inductor also remembers the past, So then the an issue of why a capacitor and a inductor / coil are used in combination is due the time "remberance" that both components have. If a inductor was put with a resistor, then only the inductor would be aware of the time factor, while the resistor would not, correct? Thus: frequency, (the tuning circuit = inductor with capacitor) which is time based, requireds all or both components to be time aware, right? Do you know of an online source, that explains the time aspects of a capacitor. So far I have only come across the voltage storage principles of a capacitor, none mentioning the time rememberance factors? For example: If I take an FM Receiver and Transmitter kit (wireless microphone kit or remote control kit), say I connect a 1 Watt power source to the Transmitter, (in place of the human voice on the microphone kit) what components will effect the Voltage and Amperage (v x a = W) in the receiver side, and should it be the same +/- 1 Watt and what would cause the decrease should there be a difference? I'm not following the question. I think you are trying to combine several processes into one, and skipping too many steps. It is sort of like saying that if you can run a mile on one hamburger, how fast can you run if you eat a gallon of gasoline. I could be skipping steps, not sure? Ok, so let me rephrases this another way. Using a wireless microphone kit, with a tranmitter and receiver. Designed or should be used as follows: Human Voice - Transmitter Receiver - Tape Deck Using the wireless system I can now record what is said onto the tape deck. Now making an assumption, lets replace the Human Voice with 1 Watt power supply and the tape deck with a Light Blub, thus giving: 1 Watt Power - Transmitter Receive - Light Bulb Questions: 1) Would the Light blub be able to receive 1 watt of power, if not why? 2) If the light bulb was to receive power. How much would it receive, ans what would cause the loss if the power received is less than the 1 watt supplied.? Thank you for your input John, your insight was helpful. Barry |
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On Wed, 28 Sep 2005 18:29:49 +0000, Roy Ingham wrote:
.... I could be skipping steps, not sure? Ok, so let me rephrases this another way. Using a wireless microphone kit, with a tranmitter and receiver. Designed or should be used as follows: Human Voice - Transmitter Receiver - Tape Deck Using the wireless system I can now record what is said onto the tape deck. Now making an assumption, lets replace the Human Voice with 1 Watt power supply and the tape deck with a Light Blub, thus giving: 1 Watt Power - Transmitter Receive - Light Bulb Questions: 1) Would the Light blub be able to receive 1 watt of power, if not why? Not necessarily - only if the receiver is designed to provide that power to the load from its own power supply. The actual amount of "power" sent from the transmitter to the receiver is minuscule. http://www.google.com/search?q=%22radio+fundamentals%22 Good Luck! Rich |
#5
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Thanks Rich
Rich Grise wrote: Not necessarily - only if the receiver is designed to provide that power to the load from its own power supply. The actual amount of "power" sent from the transmitter to the receiver is minuscule. In the same way that I can put human voice into the transmitter (microphone) and get the human voice out of the receiver to the tape deck (or house monitors). Can I not "transport or send" power/electricity from a transmitter to a receiver (not linked with the power required to operate)? The 1 watt power source, is not the power source required to operate the transmitter, nor was I concerned about the power source(battery) to power the receiver. But rather the ability to transmit power from transmitter to receiver. I take it this is not possible? Thanks again for all your help and input Barry |
#6
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I read in sci.electronics.design that Roy Ingham
wrote (in ) about 'Basic Adivce/tutoring sought', on Wed, 28 Sep 2005: But rather the ability to transmit power from transmitter to receiver. I take it this is not possible? It's not exactly impossible, but it's so very difficult that it's not proved practical. Both Tesla and Yagi (he of the xylophone-like antenna) tried. In fact, Yagi invented his antenna for power-transmission experiments. There has been speculation about transmitting power as microwaves from solar-powered satellites to ground, but there are huge problems, not least of which is ensuring that the power beam doesn't fry the nearest city when the satellite's guidance system is zapped by a solar eruption. People have managed to steal a hundred watts or so from a nearby high-powered broadcasting station (think 50 kilowatts), but the receiving antenna had to be very large, and the people running the station could tell it was happening. so the FCC moved in rapidly to stop it. -- Regards, John Woodgate, OOO - Own Opinions Only. If everything has been designed, a god designed evolution by natural selection. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk |
#7
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On Wed, 28 Sep 2005 20:17:44 +0000, Roy Ingham wrote:
Thanks Rich Rich Grise wrote: Not necessarily - only if the receiver is designed to provide that power to the load from its own power supply. The actual amount of "power" sent from the transmitter to the receiver is minuscule. In the same way that I can put human voice into the transmitter (microphone) and get the human voice out of the receiver to the tape deck (or house monitors). Can I not "transport or send" power/electricity from a transmitter to a receiver (not linked with the power required to operate)? The 1 watt power source, is not the power source required to operate the transmitter, nor was I concerned about the power source(battery) to power the receiver. But rather the ability to transmit power from transmitter to receiver. I take it this is not possible? One of these sources can probably explain it better than I can: http://www.google.com/search?&q=wire...r+transmission Good Luck! Rich |
#8
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On Wed, 28 Sep 2005 20:17:44 GMT, Roy Ingham
wrote: Thanks Rich Rich Grise wrote: Not necessarily - only if the receiver is designed to provide that power to the load from its own power supply. The actual amount of "power" sent from the transmitter to the receiver is minuscule. In the same way that I can put human voice into the transmitter (microphone) and get the human voice out of the receiver to the tape deck (or house monitors). Can I not "transport or send" power/electricity from a transmitter to a receiver (not linked with the power required to operate)? --- Not generally, in a practical sense, considering that the energy in the transmitted beam spreads geometrically when it leaves the transmitting antenna and is, therefore not intercepted greatly by the receiving antenna. To a lesser degree, there's also atmospheric absorbtion to consider. -- John Fields Professional Circuit Designer |
#9
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On Wed, 28 Sep 2005 13:43:52 GMT, Roy Ingham
wrote: Greetings all I hope I make sense, as I'm seeking some guidance through this, please. I'm trying to get to grips with some basic electronics. I know and have learnt the electronic math rules W = V x A ; V = IR , etc.. What I get stuck on is more "why would one use a capacitor in a circuit, why not a resistor?" I understand the functions that a part plays (sometimes not 100% correct) but generally eg: capacitor: takes in energy (voltage) and stores it, until it is released. Resistor: acts to slow down the voltage as like a form of friction. much like one lane of traffic compared to a 4 lane highway. --- Picture this: You have a large tank full of water in which is submerged a bidirectional pump with its output connected to a hose. The other end of the hose is connected to a fitting on the bottom of an aquarium, such that the pump can either pump water into, or out of, the aquarium. The pressure that the pump generates when it's either filling or emptying the tank is analogous to voltage, The diameter and length of the hose is analogous to resistance, and the volume of water which can be pumped either way in a given amount of time is analogous to current. --- But what I don't grasp is why do we place a resistor with a coil to make a tuning circuit, why not use a capacitor. I'm trying to grasp what is it about the resistor, that makes the two components work and thus form a tuning circuit.? --- In the case of the resistor and capacitor (RC) or of the resistor and inductor,(RL) they provide a known attenuation of a signal as a function of the signal's frequency and either an RC or an RL could be used, depending on the application. For example, in the water analogy the aquarium is analogous to a capacitor, so the corresponding electrical circuit would be: (View with a fixed-pitch font like Courier) E1 | [R1] | +----E2 | [C1] | GND Where E1 is the pump pressure generated, in volts R1 is the resistance of the hose, in ohms E2 is the pressure on the bottom of the aquarium, in volts C1 is the capacitance of the aquarium, in farads, and GND is the pressure on the bottom of the empty aquarium, zero volts. Now let's say that we start with an empty aquarium and that when we start the pump it will generate a pressure of 1 volt and start pumping water through the hose, which has a resistance of 1 ohm, into the aquarium, which has a capacitance of 1 farad. Initially, the pressure at E2 will be 0 volts because the aquarium was empty, but as the aquarium fills up the voltage at E2 will rise because of the head of water being built up in the aquarium. If the aquarium is tall enough to allow the pressure at E2 to rise to 1 volt, then the pump will no longer be able to pump water into the aquarium because it will be pumping against a 1 volt head with its 1 volt of pressure so everything will be in equilibrium. What's important to realize is that it took time for the aquarium to fill up. There's a certain "time constant" associated with the system which is given by: t = RC Where t is the time it takes, in seconds, for the aquarium to fill up to the point where E2 is about 2/3 of E1, R is the resistance of the hose in ohms, and C is the capacitance of the aquarium in farads. If we plug in what we know we'll have: t = RC = 1R * 1F = 1s So in one second, with a pressure of 1V at E1, we'll have the aquarium about 2/3 full and a pressure of ~ 0.67V at E2. Now, if at the point where E2 rises to 0.67V we suddenly reverse the pump, it will take 0.67s to empty the aquarium. So, for an input frequency of 0.5Hz (1s fill, 1s empty) we have a peak output amplitude at E2 of 0.67V. However, if we only fill the aquarium for 1/2 second and empty it for 1/2 second, the voltage at E2 won't have enough time to rise to 0.67V, so you can see that the higher the input frequency, the lower the amplitude of the output signal. -- John Fields Professional Circuit Designer |
#10
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Roy Ingham wrote:
Firstly thank you for your reply it was helpful. John Popelish wrote: .... Capacitors remember the past, resistors don't. .... An inductor also remembers the past, So then the an issue of why a capacitor and a inductor / coil are used in combination is due the time "remberance" that both components have. This is why, together this pair is called a second order system. If a inductor was put with a resistor, then only the inductor would be aware of the time factor, while the resistor would not, correct? Correct, and this pair makes up a first order system. Thus: frequency, (the tuning circuit = inductor with capacitor) which is time based, requireds all or both components to be time aware, right? Frequency selective circuits can be made with just an RC or RL pair, but this kind of selectivity is just low pass (everything lower than some cut off frequency is passed through with minimal attenuation) or high pass (everything higher than some cut off frequency is passed with minimal attenuation) but it takes at least a second order system to produce resonance that act as a band pass filter. Do you know of an online source, that explains the time aspects of a capacitor. The search key words might be [capacitor capacitance tutorial]. E.G. http://www.electronics-tutorials.com...apacitance.htm So far I have only come across the voltage storage principles of a capacitor, none mentioning the time rememberance factors? And you may not find such a description. The mathematical way to say "remembrance factor" is to refer to a function of time. The voltage across a capacitor is a function of current passing through it and time. The rate of change of voltage across a capacitor is proportional to the current passing through it. Current is charge per time. Rate of change of voltage with respect to time is a function of time. If the present value involves time, there is some sense of history in the present value. For example: If I take an FM Receiver and Transmitter kit (wireless microphone kit or remote control kit), say I connect a 1 Watt power source to the Transmitter, (in place of the human voice on the microphone kit) what components will effect the Voltage and Amperage (v x a = W) in the receiver side, and should it be the same +/- 1 Watt and what would cause the decrease should there be a difference? I'm not following the question. I think you are trying to combine several processes into one, and skipping too many steps. It is sort of like saying that if you can run a mile on one hamburger, how fast can you run if you eat a gallon of gasoline. I could be skipping steps, not sure? Ok, so let me rephrases this another way. Using a wireless microphone kit, with a tranmitter and receiver. Designed or should be used as follows: Human Voice - Transmitter Receiver - Tape Deck Using the wireless system I can now record what is said onto the tape deck. Now making an assumption, lets replace the Human Voice with 1 Watt power supply and the tape deck with a Light Blub, thus giving: 1 Watt Power - Transmitter Receive - Light Bulb Questions: 1) Would the Light blub be able to receive 1 watt of power, if not why? In the voice receiver combination, the amplitude of the voice signal produces a proportional receiver output signal, even though the actual power in the receiver's output comes from its power supply, not from the voice. The voice provides a controlling signal that determines how the receiver's power supply is converted to a proportional copy of the original voice signal. So the conversion of 1 watt going into the transmitter has no particular form implied to make it an appropriate control signal at the receiver. If the transmitter is designed to modulate its carrier energy with a few millivolts from a microphone, and you replace the microphone output with a 1 watt audio tone (say, a speaker drive signal) at the very least, the transmitter would produce a distorted modulation and be received as a distorted (like a fuzzed electric guitar) version of the original tone. At worst, the large audio power to the transmitter input would destroy the modulation circuit. 2) If the light bulb was to receive power. How much would it receive, ans what would cause the loss if the power received is less than the 1 watt supplied.? Power leaves the transmitter in all directions into space. Only a tiny fraction of that energy is picked up by the receiver. That tiny signal is amplified (a more powerful copy is created that matches the instant by instant variations of the radiated wave) with power from the receiver's DC supply. In a similar way, power steering copies the movements you make with the steering wheel, only with much more force capability, using power from the engine via the power steering pump. The modulation (audio) information is recovered from that amplified copy of the radiated wave, and then that signal is probably again amplifier with more power from the receiver's supply before it comes out of the audio jack of the receiver. Thank you for your input John, your insight was helpful. Barry |
#11
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Roy Ingham wrote:
Greetings all I hope I make sense, as I'm seeking some guidance through this, please. I'm trying to get to grips with some basic electronics. I know and have learnt the electronic math rules W = V x A ; V = IR , etc.. What I get stuck on is more "why would one use a capacitor in a circuit, why not a resistor?" I understand the functions that a part plays (sometimes not 100% correct) but generally eg: capacitor: takes in energy (voltage) and stores it, until it is released. Resistor: acts to slow down the voltage as like a form of friction. much like one lane of traffic compared to a 4 lane highway. But what I don't grasp is why do we place a resistor with a coil to make a tuning circuit, why not use a capacitor. I'm trying to grasp what is it about the resistor, that makes the two components work and thus form a tuning circuit.? For example: If I take an FM Receiver and Transmitter kit (wireless microphone kit or remote control kit), say I connect a 1 Watt power source to the Transmitter, (in place of the human voice on the microphone kit) what components will effect the Voltage and Amperage (v x a = W) in the receiver side, and should it be the same +/- 1 Watt and what would cause the decrease should there be a difference? I thank you for your assistance up front, Thanks Barry You are probably biting off more than you can chew at this stage. The concept of tuned circuits are not something that a beginner should start with, esp if you don't fully understand how a capacitor works yet. Start off with basic DC theory which involves resistors and voltages, and then move on to basic AC theory with Resistor and Capacitor (RC) circuits. Stuff like RC time constants, how a capacitor lets AC through but not DC, that kind of stuff. Then learn about inductors, and Resistor and Inductor (RL) circuits, and then you might be ready to grasp LC tuned circuits and other more complex AC theory. All of these are basic "building blocks" which you combine to make a circuit that does something useful. I can recommend "Circuit Theory & Techniques" by Hans Goodman. Vol 1 and Vol 2. Dave |
#12
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John Woodgate wrote:
I read in sci.electronics.design that Roy Ingham wrote (in ) about 'Basic Adivce/tutoring sought', on Wed, 28 Sep 2005: But rather the ability to transmit power from transmitter to receiver. I take it this is not possible? It's not exactly impossible, but it's so very difficult that it's not proved practical. Both Tesla and Yagi (he of the xylophone-like antenna) tried. In fact, Yagi invented his antenna for power-transmission experiments. Actually it's done all the time with RFID systems, but maybe not in the ballpark that the OP imagined. There has been speculation about transmitting power as microwaves from solar-powered satellites to ground, but there are huge problems, not least of which is ensuring that the power beam doesn't fry the nearest city when the satellite's guidance system is zapped by a solar eruption. People have managed to steal a hundred watts or so from a nearby high-powered broadcasting station (think 50 kilowatts), but the receiving antenna had to be very large, and the people running the station could tell it was happening. so the FCC moved in rapidly to stop it. You need to be quite close to the transmitter, as the field falls off quickly. -- ciao Ban Bordighera, Italy |
#13
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On 2005-09-28, Roy Ingham wrote:
Greetings all I hope I make sense, as I'm seeking some guidance through this, please. I'm trying to get to grips with some basic electronics. I know and have learnt the electronic math rules W = V x A ; V = IR , etc.. What I get stuck on is more "why would one use a capacitor in a circuit, why not a resistor?" I understand the functions that a part plays (sometimes not 100% correct) but generally eg: capacitor: takes in energy (voltage) and stores it, until it is released. capacitors behave like springs Resistor: acts to slow down the voltage as like a form of friction. But what I don't grasp is why do we place a resistor with a coil to make a tuning circuit, why not use a capacitor. I'm trying to grasp what is it about the resistor, that makes the two components work and thus form a tuning circuit.? they don't a typical tuning circuit uses a capacitor and an a coil with a resistor and a coil you can make a filter circuit. For example: If I take an FM Receiver and Transmitter kit (wireless microphone kit or remote control kit), say I connect a 1 Watt power source to the Transmitter, (in place of the human voice on the microphone kit) 1 watt is way too much input to the microphone terminals, more like 1 microwatt. 1 watt into the power source (battery terminals) is quite a lot too, what components will effect the Voltage and Amperage (v x a = W) in the receiver side, the volume knob or if you mean the strength of the radio signal received the antenna, is probably the main consideration. and should it be the same +/- 1 Watt and what would cause the decrease should there be a difference? there will be a decrease, as the 1 watt radio signal spreads out to cover more area it gets weaker like the way a distant lamp gives less illumination. Bye. Jasen |
#14
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In the same way that I can put human voice into the transmitter
(microphone) and get the human voice out of the receiver to the tape deck (or house monitors). Can I not "transport or send" power/electricity from a transmitter to a receiver (not linked with the power required to operate)? The 1 watt power source, is not the power source required to operate the transmitter, nor was I concerned about the power source(battery) to power the receiver. But rather the ability to transmit power from transmitter to receiver. I take it this is not possible? it's possible, just not practical. a "crystal set" am receiver is totally powered by the received signal, they typically need a large antenna, and only output into a earpiece. Not necessarily - only if the receiver is designed to provide that power to the load from its own power supply. The actual amount of "power" sent from the transmitter to the receiver is minuscule. In the same way that I can put human voice into the transmitter (microphone) and get the human voice out of the receiver to the tape deck (or house monitors). Can I not "transport or send" power/electricity from a transmitter to a receiver (not linked with the power required to operate)? energy is sent, but the energy sent is one millionth or less of the energy used to power a typical receiver... the energy sent is just used to carry the information - the voice. The 1 watt power source, is not the power source required to operate the transmitter, nor was I concerned about the power source(battery) to power the receiver. But rather the ability to transmit power from transmitter to receiver. I take it this is not possible? if you put a louder input into the transmiiter you'll get a louder output from the receiver (within limits), but still the receiver is responsible for the energy in the output, the transmitter only for the content (information part) Bye. Jasen |
#15
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On Wed, 28 Sep 2005 17:06:32 -0500, John Fields
wrote: Now, if at the point where E2 rises to 0.67V we suddenly reverse the pump, it will take 0.67s to empty the aquarium. ^^^^^ Oops... 1s -- John Fields Professional Circuit Designer |
#16
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Wow!! so much input, thank you
I didn't even think about the radiation of the signal from the transmitter, thus making it obvious that the receiver can only get a part of the signal. For everyones input thank you. Barry |
#17
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David Jones gave very good advice- follow it and learn the basics. There is
a hell of a lot between understanding of V=IR and radiation of signals and the intermediate steps are important. -- Don Kelly @shawcross.ca remove the X to answer ---------------------------- -- Don Kelly @shawcross.ca remove the X to answer ---------------------------- "Roy Ingham" wrote in message .. . Wow!! so much input, thank you I didn't even think about the radiation of the signal from the transmitter, thus making it obvious that the receiver can only get a part of the signal. For everyones input thank you. Barry |
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