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I need to buy a DC to AC power inverter. Which type should I buy?

2007-02-05 17:34:31 · 2 answers · asked by will m 1 in Science & Mathematics Physics

2 answers

The most efficient inverter is a square wave , not sine wave generator. The reason is that at any given moment, it's either almost perfectly conducting or insulating. Either way, hardly any power is wasted. Most of the time a sine wave generator is partly conducting, which means that some of the DC power going in is wasted as heat, like in a resistor. But square wave inverters have some drawbacks. Not all appliances like them. Draw up a list of which appliances you're going to use it with and ask their manufacturers if a square wave supply is OK. The other drawback to square wave power is that it contains the fundamental frequency and also high frequency harmonics, which cause radio interference. This can cause problems with nearby TVs, radios, cellphones and computers. You can do things to reduce this problem, but it's a complicated subject. My advice is, if the amount of power you need is fairly small, so a little bit of wastage is acceptable, use sine. If power is at a premium, like if you're using a car battery out in the bush, and you haven't got fussy neighbours or temperamental electronic gear around, use square.

2007-02-05 17:58:23 · answer #1 · answered by zee_prime 6 · 0 0

a square wave consists of a fundamental sine wave component of the same frequency, together with odd harmonics. A 600 Hz square wave thus consists of 600 Hz, 1800Hz, 3000 Hz, ... etc. components.

Apply the LP filter. The only frequency component of the square wave signal lying within the filter passband is the 600 Hz fundamental, and you should therefore find that the filter output is sinusoidal, with a frequency (600 Hz) matching that of the square wave itself.

Now generate a 330 Hz square wave signal, and apply the LP filter as before. The third harmonic component of the square wave, at a frequency of 990 Hz, now lies within the filter passband as well as the 330 Hz fundamental. You should find that the filter output resembles a sine wave at the same frequency (330 Hz) as the square wave, but now with an added third harmonic (with an amplitude only one-third that of the fundamental) which produces distinctive dips in the peaks of the waveform, and also makes the zero-crossings of the waveform slightly steeper than for a pure sine wave.

Now, without altering the input signal (the 330 Hz square wave should be displayed in the upper graph), apply the BP (band pass) filter. This time, only the third harmonic component at 990 Hz lies within the filter passband (around 1000 Hz), and the filter should therefore select out the 990 Hz sine wave component of the square wave. This can be verified by confirming that there are three cycles of the sine wave shown in the lower (output signal) graph for each cycle of the square wave plotted input in the upper graph.

2007-02-05 20:04:04 · answer #2 · answered by brady ewart 3 · 0 0

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