Sorry but the phase expression is incorrect because you do not know the actual sign of a[n] and b[n] which in general could be either positive or negative depending on the value of n. In complex form assuming a cosine series, A[n] = a[n]-j*b[n]. The argument must be calculated using the atat2(a[n], -b[n]) alroithm.
One way to think about why is it valid to combine both cosine and sine terms into one term (i.e. why it is valid to convert the trigonometric form to the amplitude-phase form of the Fourier series), is by remembering that it is possible to combine any number of sines and cosines of same frequency, and the result also has the same frequency. In our case, for a given value of n, the sine and the cosine terms of the trigonometric form will have the same frequency, hence we can combine them into one.
Not sure i understand correctly. Hopefully it helps. Trig identity cos(a+b) = cos(a)cos(b)-sin(a)sin(b). We have An*cos(nwt) + Bn*sin(nwt) in the expanded form, we could make An and Bn equivalent to An=Mn*cos(b), Bn=Mn*sin(b), where Mn=√(An^2+Bn^2), and b would be theta. Mn*cos(a+b) = Mn*cos(a)cos(b) - Mn*sin(a)sin(b). However, the expanded form has positive Bn in all its sine component, by making theta = -tan^-1(Bn/An), we achieve Mn*cos(a-b) = Mn*cos(a)cos(b) + Mn*sin(a)sin(b), exactly what we want.
sir how did we derived these equations . im confused because i dont understand why we used -jbn in the phase equation!!! if you can refer us some reading material to understand that it would be great.
Sorry but the phase expression is incorrect because you do not know the actual sign of a[n] and b[n] which in general could be either positive or negative depending on the value of n. In complex form assuming a cosine series, A[n] = a[n]-j*b[n]. The argument must be calculated using the atat2(a[n], -b[n]) alroithm.
One way to think about why is it valid to combine both cosine and sine terms into one term (i.e. why it is valid to convert the trigonometric form to the amplitude-phase form of the Fourier series), is by remembering that it is possible to combine any number of sines and cosines of same frequency, and the result also has the same frequency. In our case, for a given value of n, the sine and the cosine terms of the trigonometric form will have the same frequency, hence we can combine them into one.
Interesting concept.
Not sure i understand correctly. Hopefully it helps.
Trig identity cos(a+b) = cos(a)cos(b)-sin(a)sin(b).
We have An*cos(nwt) + Bn*sin(nwt) in the expanded form, we could make An and Bn equivalent to An=Mn*cos(b), Bn=Mn*sin(b), where Mn=√(An^2+Bn^2), and b would be theta.
Mn*cos(a+b) = Mn*cos(a)cos(b) - Mn*sin(a)sin(b).
However, the expanded form has positive Bn in all its sine component, by making theta = -tan^-1(Bn/An), we achieve Mn*cos(a-b) = Mn*cos(a)cos(b) + Mn*sin(a)sin(b), exactly what we want.
sir how did we derived these equations . im confused because i dont understand why we used -jbn in the phase equation!!! if you can refer us some reading material to understand that it would be great.
What about even functions where bn = 0 ? How do you find the phase?
You can use the sine inverse or cosine inverse
Makes sense!
Great. 🙂
Thanx!
You are welcome. 🙂
hi ,there , thank for the lesson
how can i contact u , i have some questions and thank for your time
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i want to complet my study in ur contry , and it be nice of u if give some information and thnx
Thanks sir
Sir please solve it for phase and magnitude
X(f)=z+4sin2pi t+ 2cospi t
Use the R formula
THANK MY FATHOAR
assumed bn = negative thus -bn
tan(theta) = -bn/an
-tan(theta)=bn/an
-theta = tan^-1(bn/an)
theta= -tan^-1(bn/an)
Just Remember trig i.e. SOH CAH TOA
come on, can u atleast derive the equation