Fourier Series Reduced Form: Phase Angle and Spectra
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Im very confused regarding how to determine the angle on the reduced or harmonic form representation of the Fourier series. Some books state the following:
$$f(t)=F_0+sum_n=1^infty |F_n |cos(nomega t +theta)$$
$$F_0=a_0$$
$$|F_n |=sqrta_n^2+b_n^2$$
$$theta =arctanleft(frac-b_na_nright) $$
Now, say you want to plot the phase spectra of a square wave, with the following Fourier series:
$$frac4Api sum_n=1^infty frac12n-1sin((2n-1)omega t) ,,,,,,,,mathbf(1)$$
$$mathbf A text is the amplitude of the signal$$
then:
$$F_0=a_0=0$$
$$|F_n |=sqrta_n^2+b_n^2=frac4Apifrac12n-1$$
$$theta =arctanleft(frac-b_na_nright)=arctanleft(frac-frac4Apifrac12n-10right)=-90^circ ,,,,,,,,mathbf(2a)$$
Its phase spectrum is the following:
And the Fourier series (equation $mathbf(1)$ ) of the square waveform represented in its harmonic form is:
$$frac4Api sum_n=1^infty frac12n-1cosleft((2n-1)omega t -90^circright) ,,,,,,,,mathbf(2b)$$
Now here is where I get confused: other books state that the reduced or harmonic form representation is as follows (notice the negative sign on the angle):
$$f(t)=F_0+sum_n=1^infty |F_n |cos(nomega t colorredmathbf-theta)$$
$$F_0=a_0$$
$$|F_n |=sqrta_n^2+b_n^2$$
$$theta =arctanleft(fracb_na_nright) $$
So for the same square wave of equation $mathbf(1)$ from the example above, the harmonic representation with the negative sign should be:
$$F_0=a_0=0$$
$$|F_n |=sqrta_n^2+b_n^2=frac4Apifrac12n-1$$
$$theta =arctanleft(frac-b_na_nright)=arctanleft(fracfrac4Apifrac12n-10right)=90^circ ,,,,,,,,mathbf(3a)$$
Thus:
$$frac4Api sum_n=1^infty frac12n-1cosleft((2n-1)omega t -90^circright) ,,,,,,,,mathbf(3b)$$
Notice that equation $mathbf(3b)$ is exactly the same as $mathbf(2b)$ but the angle $theta$ in $mathbf(3a)$ has a positive sign while the angle $theta$ in $mathbf(2a)$ has a negative sign.
So the phase spectrum using the angle calculated in $mathbf(3a)$ is:
Which is the mirror image of the phase spectrum found using $mathbf(2a)$
So bottom line: What is the correct way or the standard for the harmonic representation (plus sign or negative angle sign) and how should one decide the angle sign to plot the phase spectra ?
fourier-analysis fourier-series
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Im very confused regarding how to determine the angle on the reduced or harmonic form representation of the Fourier series. Some books state the following:
$$f(t)=F_0+sum_n=1^infty |F_n |cos(nomega t +theta)$$
$$F_0=a_0$$
$$|F_n |=sqrta_n^2+b_n^2$$
$$theta =arctanleft(frac-b_na_nright) $$
Now, say you want to plot the phase spectra of a square wave, with the following Fourier series:
$$frac4Api sum_n=1^infty frac12n-1sin((2n-1)omega t) ,,,,,,,,mathbf(1)$$
$$mathbf A text is the amplitude of the signal$$
then:
$$F_0=a_0=0$$
$$|F_n |=sqrta_n^2+b_n^2=frac4Apifrac12n-1$$
$$theta =arctanleft(frac-b_na_nright)=arctanleft(frac-frac4Apifrac12n-10right)=-90^circ ,,,,,,,,mathbf(2a)$$
Its phase spectrum is the following:
And the Fourier series (equation $mathbf(1)$ ) of the square waveform represented in its harmonic form is:
$$frac4Api sum_n=1^infty frac12n-1cosleft((2n-1)omega t -90^circright) ,,,,,,,,mathbf(2b)$$
Now here is where I get confused: other books state that the reduced or harmonic form representation is as follows (notice the negative sign on the angle):
$$f(t)=F_0+sum_n=1^infty |F_n |cos(nomega t colorredmathbf-theta)$$
$$F_0=a_0$$
$$|F_n |=sqrta_n^2+b_n^2$$
$$theta =arctanleft(fracb_na_nright) $$
So for the same square wave of equation $mathbf(1)$ from the example above, the harmonic representation with the negative sign should be:
$$F_0=a_0=0$$
$$|F_n |=sqrta_n^2+b_n^2=frac4Apifrac12n-1$$
$$theta =arctanleft(frac-b_na_nright)=arctanleft(fracfrac4Apifrac12n-10right)=90^circ ,,,,,,,,mathbf(3a)$$
Thus:
$$frac4Api sum_n=1^infty frac12n-1cosleft((2n-1)omega t -90^circright) ,,,,,,,,mathbf(3b)$$
Notice that equation $mathbf(3b)$ is exactly the same as $mathbf(2b)$ but the angle $theta$ in $mathbf(3a)$ has a positive sign while the angle $theta$ in $mathbf(2a)$ has a negative sign.
So the phase spectrum using the angle calculated in $mathbf(3a)$ is:
Which is the mirror image of the phase spectrum found using $mathbf(2a)$
So bottom line: What is the correct way or the standard for the harmonic representation (plus sign or negative angle sign) and how should one decide the angle sign to plot the phase spectra ?
fourier-analysis fourier-series
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Im very confused regarding how to determine the angle on the reduced or harmonic form representation of the Fourier series. Some books state the following:
$$f(t)=F_0+sum_n=1^infty |F_n |cos(nomega t +theta)$$
$$F_0=a_0$$
$$|F_n |=sqrta_n^2+b_n^2$$
$$theta =arctanleft(frac-b_na_nright) $$
Now, say you want to plot the phase spectra of a square wave, with the following Fourier series:
$$frac4Api sum_n=1^infty frac12n-1sin((2n-1)omega t) ,,,,,,,,mathbf(1)$$
$$mathbf A text is the amplitude of the signal$$
then:
$$F_0=a_0=0$$
$$|F_n |=sqrta_n^2+b_n^2=frac4Apifrac12n-1$$
$$theta =arctanleft(frac-b_na_nright)=arctanleft(frac-frac4Apifrac12n-10right)=-90^circ ,,,,,,,,mathbf(2a)$$
Its phase spectrum is the following:
And the Fourier series (equation $mathbf(1)$ ) of the square waveform represented in its harmonic form is:
$$frac4Api sum_n=1^infty frac12n-1cosleft((2n-1)omega t -90^circright) ,,,,,,,,mathbf(2b)$$
Now here is where I get confused: other books state that the reduced or harmonic form representation is as follows (notice the negative sign on the angle):
$$f(t)=F_0+sum_n=1^infty |F_n |cos(nomega t colorredmathbf-theta)$$
$$F_0=a_0$$
$$|F_n |=sqrta_n^2+b_n^2$$
$$theta =arctanleft(fracb_na_nright) $$
So for the same square wave of equation $mathbf(1)$ from the example above, the harmonic representation with the negative sign should be:
$$F_0=a_0=0$$
$$|F_n |=sqrta_n^2+b_n^2=frac4Apifrac12n-1$$
$$theta =arctanleft(frac-b_na_nright)=arctanleft(fracfrac4Apifrac12n-10right)=90^circ ,,,,,,,,mathbf(3a)$$
Thus:
$$frac4Api sum_n=1^infty frac12n-1cosleft((2n-1)omega t -90^circright) ,,,,,,,,mathbf(3b)$$
Notice that equation $mathbf(3b)$ is exactly the same as $mathbf(2b)$ but the angle $theta$ in $mathbf(3a)$ has a positive sign while the angle $theta$ in $mathbf(2a)$ has a negative sign.
So the phase spectrum using the angle calculated in $mathbf(3a)$ is:
Which is the mirror image of the phase spectrum found using $mathbf(2a)$
So bottom line: What is the correct way or the standard for the harmonic representation (plus sign or negative angle sign) and how should one decide the angle sign to plot the phase spectra ?
fourier-analysis fourier-series
Im very confused regarding how to determine the angle on the reduced or harmonic form representation of the Fourier series. Some books state the following:
$$f(t)=F_0+sum_n=1^infty |F_n |cos(nomega t +theta)$$
$$F_0=a_0$$
$$|F_n |=sqrta_n^2+b_n^2$$
$$theta =arctanleft(frac-b_na_nright) $$
Now, say you want to plot the phase spectra of a square wave, with the following Fourier series:
$$frac4Api sum_n=1^infty frac12n-1sin((2n-1)omega t) ,,,,,,,,mathbf(1)$$
$$mathbf A text is the amplitude of the signal$$
then:
$$F_0=a_0=0$$
$$|F_n |=sqrta_n^2+b_n^2=frac4Apifrac12n-1$$
$$theta =arctanleft(frac-b_na_nright)=arctanleft(frac-frac4Apifrac12n-10right)=-90^circ ,,,,,,,,mathbf(2a)$$
Its phase spectrum is the following:
And the Fourier series (equation $mathbf(1)$ ) of the square waveform represented in its harmonic form is:
$$frac4Api sum_n=1^infty frac12n-1cosleft((2n-1)omega t -90^circright) ,,,,,,,,mathbf(2b)$$
Now here is where I get confused: other books state that the reduced or harmonic form representation is as follows (notice the negative sign on the angle):
$$f(t)=F_0+sum_n=1^infty |F_n |cos(nomega t colorredmathbf-theta)$$
$$F_0=a_0$$
$$|F_n |=sqrta_n^2+b_n^2$$
$$theta =arctanleft(fracb_na_nright) $$
So for the same square wave of equation $mathbf(1)$ from the example above, the harmonic representation with the negative sign should be:
$$F_0=a_0=0$$
$$|F_n |=sqrta_n^2+b_n^2=frac4Apifrac12n-1$$
$$theta =arctanleft(frac-b_na_nright)=arctanleft(fracfrac4Apifrac12n-10right)=90^circ ,,,,,,,,mathbf(3a)$$
Thus:
$$frac4Api sum_n=1^infty frac12n-1cosleft((2n-1)omega t -90^circright) ,,,,,,,,mathbf(3b)$$
Notice that equation $mathbf(3b)$ is exactly the same as $mathbf(2b)$ but the angle $theta$ in $mathbf(3a)$ has a positive sign while the angle $theta$ in $mathbf(2a)$ has a negative sign.
So the phase spectrum using the angle calculated in $mathbf(3a)$ is:
Which is the mirror image of the phase spectrum found using $mathbf(2a)$
So bottom line: What is the correct way or the standard for the harmonic representation (plus sign or negative angle sign) and how should one decide the angle sign to plot the phase spectra ?
fourier-analysis fourier-series
fourier-analysis fourier-series
edited Jan 8 '15 at 0:21
dustin
6,64492966
6,64492966
asked Jan 7 '15 at 23:06
S.s.
12617
12617
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add a comment |Â
1 Answer
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up vote
0
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The two spectra should be the same, as (2b) and (3b) indicate. These two equations should lead you to the same result.
You compare (2a) with (3a). That's wrong because you don't take into account the sign before $theta$ of the initial equation.
Im not conviced that is actually true, because by the different definitions, a negative angle on 3b implies that the angle is positive (comparing it with the definition that uses a negative angle), and a negative angle on 2b implies the angle is in fact negative (comparing it with the definition that uses the positive angle)
â S.s.
Jan 8 '15 at 3:39
(2b) and (3b) are exactly the same. So their spectra should be the same too.
â Joker
Jan 11 '15 at 18:52
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The two spectra should be the same, as (2b) and (3b) indicate. These two equations should lead you to the same result.
You compare (2a) with (3a). That's wrong because you don't take into account the sign before $theta$ of the initial equation.
Im not conviced that is actually true, because by the different definitions, a negative angle on 3b implies that the angle is positive (comparing it with the definition that uses a negative angle), and a negative angle on 2b implies the angle is in fact negative (comparing it with the definition that uses the positive angle)
â S.s.
Jan 8 '15 at 3:39
(2b) and (3b) are exactly the same. So their spectra should be the same too.
â Joker
Jan 11 '15 at 18:52
add a comment |Â
up vote
0
down vote
The two spectra should be the same, as (2b) and (3b) indicate. These two equations should lead you to the same result.
You compare (2a) with (3a). That's wrong because you don't take into account the sign before $theta$ of the initial equation.
Im not conviced that is actually true, because by the different definitions, a negative angle on 3b implies that the angle is positive (comparing it with the definition that uses a negative angle), and a negative angle on 2b implies the angle is in fact negative (comparing it with the definition that uses the positive angle)
â S.s.
Jan 8 '15 at 3:39
(2b) and (3b) are exactly the same. So their spectra should be the same too.
â Joker
Jan 11 '15 at 18:52
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The two spectra should be the same, as (2b) and (3b) indicate. These two equations should lead you to the same result.
You compare (2a) with (3a). That's wrong because you don't take into account the sign before $theta$ of the initial equation.
The two spectra should be the same, as (2b) and (3b) indicate. These two equations should lead you to the same result.
You compare (2a) with (3a). That's wrong because you don't take into account the sign before $theta$ of the initial equation.
answered Jan 8 '15 at 0:09
Joker
1223
1223
Im not conviced that is actually true, because by the different definitions, a negative angle on 3b implies that the angle is positive (comparing it with the definition that uses a negative angle), and a negative angle on 2b implies the angle is in fact negative (comparing it with the definition that uses the positive angle)
â S.s.
Jan 8 '15 at 3:39
(2b) and (3b) are exactly the same. So their spectra should be the same too.
â Joker
Jan 11 '15 at 18:52
add a comment |Â
Im not conviced that is actually true, because by the different definitions, a negative angle on 3b implies that the angle is positive (comparing it with the definition that uses a negative angle), and a negative angle on 2b implies the angle is in fact negative (comparing it with the definition that uses the positive angle)
â S.s.
Jan 8 '15 at 3:39
(2b) and (3b) are exactly the same. So their spectra should be the same too.
â Joker
Jan 11 '15 at 18:52
Im not conviced that is actually true, because by the different definitions, a negative angle on 3b implies that the angle is positive (comparing it with the definition that uses a negative angle), and a negative angle on 2b implies the angle is in fact negative (comparing it with the definition that uses the positive angle)
â S.s.
Jan 8 '15 at 3:39
Im not conviced that is actually true, because by the different definitions, a negative angle on 3b implies that the angle is positive (comparing it with the definition that uses a negative angle), and a negative angle on 2b implies the angle is in fact negative (comparing it with the definition that uses the positive angle)
â S.s.
Jan 8 '15 at 3:39
(2b) and (3b) are exactly the same. So their spectra should be the same too.
â Joker
Jan 11 '15 at 18:52
(2b) and (3b) are exactly the same. So their spectra should be the same too.
â Joker
Jan 11 '15 at 18:52
add a comment |Â
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