ampliude= 240volts fundamental sine wave=50hz. 2nd harmonic=100hz---> 30% of 100hz is = 30 * 240/100 = 72 3rd harmonic= 150hz--> 30% of 150 hz is = 20 * 240/100= 48 5th harmonic= 250hz--> 10% 0f 250hz is = 10 *240/100 =24 now To inject the harmonics to fundamental wave we get harmonic distortion 240 + 72 + 48 +24 =384 volts. now we have to filter the harmonic distortion of 384 volts to recover the 240 volts to filter this harmonic distortion we have to use IIR filter order=2 sampling freq =1000 pass band =45 stop band =65 so by above specifications we get the filter coefficients reverse coeeficients= -3.4863 , 4.7565 , -3.0021 , 0.7432 farward coefficients= 0.0112 , 0.000 , -0.0223 , 0.0000 , these are the coefficients i am getting , sir now i need to recover the fundamental 240volts from the distortion of 384 volts using this filter and filter coefficients. i am stucked here to filter the distortion,, kindly reply to my mail.
FILTERING THE DISTORTION
Started by ●May 24, 2011
Reply by ●May 24, 20112011-05-24
On May 24, 12:51�pm, "rashmi venugopal" <rashu32003@n_o_s_p_a_m.gmail.com> wrote:> ampliude= 240volts > fundamental sine wave=50hz. > 2nd harmonic=100hz---> 30% of 100hz is �= 30 * 240/100 = 72Why are you computing the amplitudes of the harmonics? They are totally irrelevant. Rune
Reply by ●May 24, 20112011-05-24
>On May 24, 12:51=A0pm, "rashmi venugopal" ><rashu32003@n_o_s_p_a_m.gmail.com> wrote: >> ampliude=3D 240volts >> fundamental sine wave=3D50hz. >> 2nd harmonic=3D100hz---> 30% of 100hz is =A0=3D 30 * 240/100 =3D 72 > >Why are you computing the amplitudes of the harmonics? >They are totally irrelevant. > >Rune >Exactly. Furthermore, with a pass band of 45 Hz, your output after filtering will be attenuated (assuming unity gain across the pass band) since the fundamental is 50 Hz. A full power pass band of 50 Hz and a stop band of 100 Hz will give you the simplest filter computation. Mark
Reply by ●May 24, 20112011-05-24
On 24.5.11 1:55 , Rune Allnor wrote:> On May 24, 12:51 pm, "rashmi venugopal" > <rashu32003@n_o_s_p_a_m.gmail.com> wrote: >> ampliude= 240volts >> fundamental sine wave=50hz. >> 2nd harmonic=100hz---> 30% of 100hz is = 30 * 240/100 = 72 > > Why are you computing the amplitudes of the harmonics? > They are totally irrelevant. > > RuneHere in the far North, we call one as clueless to be outside like a snowman ... If the purpose of the OP is to clean up a 240 V 50 Hz power feed, maybe a ferroresonant transformer is the answer. To the OP: The harmonics do not combine in the way you calculate. To verify, compute the harmonics for a 240 V square wave, which is for sure 240 V each way, not anything more. -- Tauno Voipio
Reply by ●May 24, 20112011-05-24
On May 24, 6:51�am, "rashmi venugopal" <rashu32003@n_o_s_p_a_m.gmail.com> wrote:> ampliude= 240volts > fundamental sine wave=50hz. > 2nd harmonic=100hz---> 30% of 100hz is �= 30 * 240/100 = 72 > > 3rd harmonic= 150hz--> 30% of 150 hz is = 20 * 240/100= 48 > > 5th harmonic= 250hz--> 10% 0f 250hz is = 10 *240/100 =24 > > now �To inject the harmonics to fundamental wave we get harmonic > distortion > > 240 + 72 + 48 +24 �=384 volts.There's an assumption here about the phases that seems unwarranted. In any case, what does this amplitude tell you?> now we have to filter the harmonic distortion of 384 volts to recover the > 240 volts > > to filter this harmonic distortion we have to use IIR filterWho decided that?> order=2 > sampling freq =1000 > pass band =45 > stop band =65That's the wrong filter. You want to pass all of your 50 Hz, so the passband has to be a bit greater than that. You may be thinking ofr a bandpass filter, with the lower edge at 45 Hz and the upper edge at 65 Hz, but even that is not a good way. A bandpass filter isn't needed and the passband is centered at 55 Hz.> so by above specifications we get the filter coefficients > > reverse coeeficients= -3.4863 �, 4.7565 , -3.0021 , 0.7432 > farward coefficients= 0.0112 , �0.000 , -0.0223 , 0.0000 , > > these are the coefficients i am getting , > > sir now i need to recover the fundamental 240volts from the distortion of > 384 volts using this filter and filter coefficients. > > i am stucked here to filter the distortion,, > kindly reply to my mail.Before you get to this particular case, you need to learn how to determine the response of a filter to an input. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●May 24, 20112011-05-24
Rashmi, Normally I don't "top post" but I'm going to intersperse comments makred with *** below: On 5/24/2011 3:51 AM, rashmi venugopal wrote:> ampliude= 240volts > fundamental sine wave=50hz. > 2nd harmonic=100hz---> 30% of 100hz is = 30 * 240/100 = 72 > > 3rd harmonic= 150hz--> 30% of 150 hz is = 20 * 240/100= 48 > > 5th harmonic= 250hz--> 10% 0f 250hz is = 10 *240/100 =24 > >***I think you mean 30% of 50Hz above... etc. That's were the 240 comes from, right?> > now To inject the harmonics to fundamental wave we get harmonic > distortion > > 240 + 72 + 48 +24 =384 volts.***This ain't DC. You can't just add the rms voltage values to get a new rms voltage. Consider: just what *is* "rms"? There is an essential phase relationship that has to be accounted for to do this. And anyway, why do you care about this one thing? I don't see a use for it.> > now we have to filter the harmonic distortion of 384 volts to recover the > 240 volts > > to filter this harmonic distortion we have to use IIR filter***You don't have to use an IIR filter - so don't delude yourself into that thinking. But, if the professor says you must to suit his/her purposes, then you must.> > order=2 > sampling freq =1000 > pass band =45 > stop band =65***As Jerry points out, the pass band and stop band frequencies make no sense if the fundamental of interest is at 50Hz.> > > so by above specifications we get the filter coefficients > > reverse coeeficients= -3.4863 , 4.7565 , -3.0021 , 0.7432 > farward coefficients= 0.0112 , 0.000 , -0.0223 , 0.0000 , > > these are the coefficients i am getting ,***I won't compute or dispute these but if they are based on the specs above then they surely won't be what you need.> > sir now i need to recover the fundamental 240volts from the distortion of > 384 volts using this filter and filter coefficients.***Are we dealing with power here or just a signal that represents the voltages on a power line? Probably makes a difference in how you implement any filter!! ***Here is what you need to do with a signal: 1) implement the filter. It's an IIR digital filter so the data has to be sampled. 2) apply the sampled signal to the filter 3) the output of the filter is the desired filtered signal.> > i am stucked here to filter the distortion,, > kindly reply to my mail.***See above. Fred
Reply by ●May 24, 20112011-05-24
On 05/24/2011 03:55 AM, Rune Allnor wrote:> On May 24, 12:51 pm, "rashmi venugopal" > <rashu32003@n_o_s_p_a_m.gmail.com> wrote: >> ampliude= 240volts >> fundamental sine wave=50hz. >> 2nd harmonic=100hz---> 30% of 100hz is = 30 * 240/100 = 72 > > Why are you computing the amplitudes of the harmonics? > They are totally irrelevant.They are relevant if one must filter down to a total error specification, and then they are used to specify one's filter -- but the OP doesn't mention such a spec. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 24, 20112011-05-24
On 05/24/2011 03:51 AM, rashmi venugopal wrote:> ampliude= 240volts > fundamental sine wave=50hz. > 2nd harmonic=100hz---> 30% of 100hz is = 30 * 240/100 = 72 > > 3rd harmonic= 150hz--> 30% of 150 hz is = 20 * 240/100= 48 > > 5th harmonic= 250hz--> 10% 0f 250hz is = 10 *240/100 =24 >My, what strong harmonics we have.> > now To inject the harmonics to fundamental wave we get harmonic > distortion > > 240 + 72 + 48 +24 =384 volts.As pointed out, first, your voltage is almost certainly specified as RMS, and second, the harmonics don't necessarily add in phase. So that's a nonsensical calculation.> > now we have to filter the harmonic distortion of 384 volts to recover the > 240 volts > > to filter this harmonic distortion we have to use IIR filterOnly if the prof says. There's a lot of different ways to skin this particular cat.> order=2 > sampling freq =1000 > pass band =45 > stop band =65As mentioned, this is an odd filter specification. It looks like something that might be done by someone who wants to pass both 50 and 60Hz, but is naive about just how filters work (hint: are they flat on top? Particularly when it's a bandpass filter with just two poles?)> so by above specifications we get the filter coefficients > > reverse coeeficients= -3.4863 , 4.7565 , -3.0021 , 0.7432 > farward coefficients= 0.0112 , 0.000 , -0.0223 , 0.0000 , > > these are the coefficients i am getting ,Which is interesting, because a 2nd-order filter would only have three coefficients in the denominator polynomial, and you have three -- four if there's an assumed one. Moreover, your forward coefficients are not those of a pure bandpass filter. So -- how did you _get_ those coefficients, and what transfer function do they specify?> sir now i need to recover the fundamental 240volts from the distortion of > 384 volts using this filter and filter coefficients.Write the filter transfer function on a piece of rock. Use that as a paperweight while you do the computation all over again, properly and on paper.> i am stucked here to filter the distortion,,Make up a filter and use it?> kindly reply to my mail.You seem to be confused as to the difference between a USENET post and mail, or you are demanding a private reply to a public and unpaid plea. Why? -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 24, 20112011-05-24
>ampliude= 240volts >fundamental sine wave=50hz. >2nd harmonic=100hz---> 30% of 100hz is = 30 * 240/100 = 72 > >3rd harmonic= 150hz--> 30% of 150 hz is = 20 * 240/100= 48 > >5th harmonic= 250hz--> 10% 0f 250hz is = 10 *240/100 =24 > > > >now To inject the harmonics to fundamental wave we get harmonic >distortion > >240 + 72 + 48 +24 =384 volts. > >now we have to filter the harmonic distortion of 384 volts to recover the >240 volts > >to filter this harmonic distortion we have to use IIR filter > >order=2 >sampling freq =1000 >pass band =45 >stop band =65 > > >so by above specifications we get the filter coefficients > >reverse coeeficients= -3.4863 , 4.7565 , -3.0021 , 0.7432 >farward coefficients= 0.0112 , 0.000 , -0.0223 , 0.0000 , > >these are the coefficients i am getting , > >sir now i need to recover the fundamental 240volts from the distortion of >384 volts using this filter and filter coefficients. > >i am stucked here to filter the distortion,, >kindly reply to my mail.Why are you stuck? You keep asking roughly the same question, but choose to ignore the answers. Steve
Reply by ●June 1, 20112011-06-01
sorry for the late reply as i was not feeling well i could not check replies. ampliude= 240volts>>fundamental sine wave=50hz. >>2nd harmonic=100hz---> 30% of 100hz is = 30 * 240/100 = 72 >> >>3rd harmonic= 150hz--> 30% of 150 hz is = 20 * 240/100= 48 >> >>5th harmonic= 250hz--> 10% 0f 250hz is = 10 *240/100 =24 >> >> >> >>now To inject the harmonics to fundamental wave we get harmonic >>distortion >> >>240 + 72 + 48 +24 =384 volts. >> >>now we have to filter the harmonic distortion of 384 volts to recoverthe>>240 volts >> >>to filter this harmonic distortion we have to use IIR filter >> >>order=2 >>sampling freq =1000 >>pass band =45 >>stop band =65 >> >> >>so by above specifications we get the filter coefficients >> >>reverse coeeficients= -3.4863 , 4.7565 , -3.0021 , 0.7432 >>farward coefficients= 0.0112 , 0.000 , -0.0223 , 0.0000 , >> >>these are the coefficients i am getting , >> >>sir now i need to recover the fundamental 240volts from the distortionof>>384 volts using this filter and filter coefficients. >> >>i am stucked here to filter the distortion,, >>kindly reply to my mail. > >Why are you stuck? You keep asking roughly the same question, but chooseto>ignore the answers. > >Steve > >
