
#1
February 19th, 2019,
11:58
Good morning,
I am working on a problem where I am finding the 4th Coefficient in a sample of 4 discrete time Fourier Series coefficients. I got the sum but now I have to solve for a_3 which consists of a real and imaginary part. Any assitance on how to solve for the a_3? Thank you.
$a_k = \{3, 12j, 1, ?\}$
Step 1: $(12j)e^{j*.5\pi*n} +a_3 e ^ {(j*.5\pi*n)} + 3 + (1)^{n+1} $
Step 2: $[(12j)(\cos \frac\pi2 n + j \sin \frac\pi2n) + a_3 (\cos \frac\pi2nj \sin \frac\pi2n)]$
Last edited by Klaas van Aarsen; February 20th, 2019 at 03:26.
Reason:
Correct LaTeX Code

February 19th, 2019 11:58
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#2
February 20th, 2019,
03:31
Originally Posted by
Tan Tom
Good morning,
I am working on a problem where I am finding the 4th Coefficient in a sample of 4 discrete time Fourier Series coefficients. I got the sum but now I have to solve for a_3 which consists of a real and imaginary part. Any assitance on how to solve for the a_3? Thank you.
$a_k = \{3, 12j, 1, ?\}$
Step 1: $(12j)e^{j*.5\pi*n} +a_3 e ^ {(j*.5\pi*n)} + 3 + (1)^{n+1} $
Step 2: $[(12j)(\cos \frac\pi2 n + j \sin \frac\pi2n) + a_3 (\cos \frac\pi2nj \sin \frac\pi2n)]$
Hi Tan Tom, welcome to MHB!
If I understand correctly, you have
$$(12j)e^{j\frac\pi 2 n} +a_3 e ^ {(j\frac\pi 2n)} + 3 + (1)^{n+1}=sum$$
for some known $sum$.
We can rewrite it as:
$$a_3 e ^ {(j\frac\pi 2 n)}=sum(12j)e^{j\frac \pi 2n}  3  (1)^{n+1}\\
a_3 =\big[sum(12j)e^{j\frac \pi 2n}  3  (1)^{n+1}\big]e ^ {j\frac\pi 2 n}$$
Is that what you're looking for, or am I misunderstanding something?

#3
November 10th, 2019,
08:44
Before you can "solve" you have to have an equation! What is that supposed to be equal to? Klaas van Aarsen is assuming it is to be equal to some number he is calling "sum". Is that correct?