Why are harmonics considered waste?

Discussion in 'General Electronics Chat' started by shespuzzling, Apr 19, 2011.

  1. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
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    Hi,

    I'm just trying to get a qualitative understanding of why harmonics are considered waste. If, for example, a non-sinusoidal voltage waveform was applied to a purely resistive element, would the harmonics that exist in the waveform be wasted?

    Why does only the fundamental component of a waveform carry the real power? Is that always the case, no matter what the load is?

    What about if the voltage and current waveforms are identical but non-sinusoidal, are the harmonic components still not doing any work?

    If somebody can answer I'd really appreciate it or provide me with a resource so I can look into it myself. So far, everything I've read simply states facts about harmonics but doesn't go into "why".

    Thanks.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Well, they're not always waste. Consider an ideal square wave; it consists of the fundamental frequency, plus ALL of the odd harmonics of the fundamental frequency.

    Most of the power would be in the fundamental frequency; the odd harmonics would make up the rest.

    Have a look at this animated .gif from Wikipedia:
    http://upload.wikimedia.org/wikipedia/commons/f/f8/SquareWave.gif
    It's the second image down in this entry:
    http://en.wikipedia.org/wiki/Square_wave
    It shows how the waveform better approaches an ideal square wave as higher odd harmonic content is added. This implies that square waves require a great deal of bandwidth.

    If the load for an ideal square wave were purely resistive, like a heater for example, the voltage and current waveforms would be identical, and the harmonic components would be "doing useful work".
     
  3. Adjuster

    Well-Known Member

    Dec 26, 2010
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    This depends on the application. You are probably referring to AC power distribution systems, but the situation is quite different in some other cases, e.g. a sound reproduction system.

    Here harmonics can and do carry real power. If they represent desirable sound content they may be useful, although if they are the result of unwanted non-linearity they may be a nuisance.
     
  4. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
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    Yes, I am talking about power distribution systems, thanks for pointing out that distinction.

    I understand how singals are broken down into harmonics, but when you look at these signals as a whole, I can't see why they would, for example, cause more heat dissipation in a transformer or cause a load to have a lower power factor (i.e. the harmonics aren't doing any real work).

    If you have two current sources (for example), each with the same RMS value but one is perfectly sinusoidal and the other contains harmonics, and these signals are applied to identical circuits, why is the real power output different in each case?
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You asked a question in your original post that contains the answer.

    It's a question of whether the current waveform is the same as the voltage waveform. In a power distribution system, the voltage waveform is sinusoidal (as least the intention is that it should be; in the real world the waveform is somewhat distorted) but the current waveform depends on the load.

    If the voltage waveform is a pure sinusoid, and if the current waveform is not the same as the voltage waveform, the current harmonics deliver no power to the load, but they cause I squared R losses in the distribution hardware (wires, transformers, etc.).

    When the load is a resistor, such as a heater or incandescent light bulb, the current waveform is identical to the voltage waveform, and any harmonics will deliver power to the (resistive) load.

    Of course, when the load is resistive there are only current harmonics if the applied voltage has harmonics (is not a pure sinusoid). But if the applied voltage waveform is not a pure sinusoid, and the load is a resistor, there will be current harmonics due to the distortion of the applied voltage waveform, and those harmonics will deliver power to the load.
     
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  6. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
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    Okay thanks! Just to clarify the below statement:

    "If the voltage waveform is a pure sinusoid, and if the current waveform is not the same as the voltage waveform, the current harmonics deliver no power to the load, but they cause I squared R losses in the distribution hardware (wires, transformers, etc.)."

    Is no power delivered because multiplying a pure sinusoid with other sinusoids at diferent frequencies gives you an average value of 0?

    And in a purely resistive load, is the logic that the current would have to be identical to the voltage, so even if the voltage was distorted, the current would be distorted in the exact same way. So when figuring out power, each harmonic current has a corresponding harmonic voltage and the power has some positive average value.

    So why do I hear so often that power is only carried by the fundamental component? In a resistive load this seems to be not so. When would this be true? Is it just for non-linear loads where we can assume the voltage is almost a pure sinusoid but the current is distorted?
     
  7. careless_monkey

    New Member

    Jan 11, 2011
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    I'm confused with this statement here. What does bandwidth exactly mean? How does it require more bandwidth?
     
  8. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
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    I think he means that there are a large number of frequencies in the square wave...bandwidth describes the range of frequencies present in a given signal.
     
  9. careless_monkey

    New Member

    Jan 11, 2011
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    Oh ok.
    What about a sinusoidal wave then? Does it many frequencies or just one?
     
  10. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
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    If it's a perfect sinusoid, then it's just one frequency. Any signal that is not a perfect sinusoid (i.e. a square wave) can be decomposed into a summation (often times infinite) of sinusoids of different frequencies. This is Fourier Analysis.
     
  11. careless_monkey

    New Member

    Jan 11, 2011
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    Perfect. Thanks to clear that out for me.
     
  12. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
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    just re-posting my last question in case it got lost in the shuffle.....

     
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As an earlier poster pointed out the supply authorities attempt to provide a pure sine wave supply. If the AC supply voltage is distorted for any reason then there will be potentially harmonic losses in the load. It's unlikely those losses would equate to useful work - say in an AC motor. In the case of a purely resistive load - say a simple resistive heating element furnace - then the harmonic currents will produce produce heat in the load. You might argue that is a "useful" outcome since the sole purpose is to produce heat.

    However power supply authorities are highly averse to allowing harmonic content in the supply voltage and will take steps to reduce or eliminate that possibility. If a consumer produces voltage harmonics at their point of common coupling to the AC grid they may get a warning to desist or be required to take steps themselves to eliminate them.

    With the increasing use of power electronic switching technologies the issue of electromagnetic compatibility of power networks with other electronic systems such as telecommunication networks has become the basis of international standards and legislated compliance regulations on harmonic generation.
     
  14. shespuzzling

    Thread Starter Active Member

    Aug 13, 2009
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    This is what I am unsure of. So there's some kind of distortion in the voltage or current waveforms in a system. An AC motor (for example) now sees this distorted wave. The harmonics contained in the signal are losses in the motor because _________ ?
     
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    As I explained in post #5:

    "It's a question of whether the current waveform is the same as the voltage waveform. In a power distribution system, the voltage waveform is sinusoidal (as least the intention is that it should be; in the real world the waveform is somewhat distorted) but the current waveform depends on the load."

    If a distorted voltage waveform is applied to a pure resistive load, the current waveform will automatically be the same as the current waveform, and any harmonics will deliver power to the (resistive) load.

    If the load has any reactive components, the impedance it presents to the applied voltage waveform will vary with frequency, and if the applied voltage is distorted (contains harmonics) then the harmonics in the current will not all be in the same proportions that they are in the voltage (because the impedance of a reactance, inductive or capacitive varies with frequency the higher frequency harmonics will see a different impedance than the lower frequency harmonics) and the waveshape of the current will be different than the waveshape of the voltage.

    When the waveshape of a voltage is different than the waveshape of the associated current, the product of the RMS voltage times the RMS current (volt-amperes) is greater than the time averaged integral of the product of the instantaneous voltage and instantaneous current (real power).

    There is a mathematical theorem that expresses this potential inequality of volt-amperes and real power:

    http://mathworld.wolfram.com/SchwarzsInequality.html

    If the functions ψ1 and ψ2 represent the instantaneous voltage and current, then consider taking the square root of the left and right sides of the inequality on line (2). The left side is then the real power and the right side is the quantity volt-amperes (the quantities on the right become the RMS values of voltage and current after taking the square root). The inequality becomes an equality if the waveshape of ψ1 and ψ2 are identical; otherwise the inequality holds; if the waveshapes are different, the real power is always less than the volt-amperes.
     
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