Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 3 of 3
  1. MHB Apprentice

    Status
    Offline
    Join Date
    Feb 2019
    Posts
    1
    Thanks
    0 times
    Thanked
    0 times
    #1
    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.

    $a_k = \{3, 1-2j, -1, ?\}$

    Step 1: $(1-2j)e^{j*.5\pi*n} +a_3 e ^ {-(j*.5\pi*n)} + 3 + (-1)^{n+1} $

    Step 2: $[(1-2j)(\cos \frac\pi2 n + j \sin \frac\pi2n) + a_3 (\cos \frac\pi2n-j \sin \frac\pi2n)]$
    Last edited by Klaas van Aarsen; February 20th, 2019 at 03:26. Reason: Correct LaTeX Code

  2. # ADS
    Circuit advertisement
    Join Date
    Always
    Posts
    Many
     

  3. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    MHB Coder
    Klaas van Aarsen's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Leiden
    Posts
    8,403
    Thanks
    5,435 times
    Thanked
    14,615 times
    Thank/Post
    1.739
    Awards
    MHB Pre-University Math Award (2018)  

MHB Model Helper Award (2017)  

MHB Best Ideas (2017)  

MHB Analysis Award (2017)  

MHB Calculus Award (2017)
    #2
    Quote Originally Posted by Tan Tom View Post
    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.

    $a_k = \{3, 1-2j, -1, ?\}$

    Step 1: $(1-2j)e^{j*.5\pi*n} +a_3 e ^ {-(j*.5\pi*n)} + 3 + (-1)^{n+1} $

    Step 2: $[(1-2j)(\cos \frac\pi2 n + j \sin \frac\pi2n) + a_3 (\cos \frac\pi2n-j \sin \frac\pi2n)]$
    Hi Tan Tom, welcome to MHB!

    If I understand correctly, you have
    $$(1-2j)e^{j\frac\pi 2 n} +a_3 e ^ {-(j\frac\pi 2n)} + 3 + (-1)^{n+1}=sum$$
    for some known $sum$.

    We can rewrite it as:
    $$a_3 e ^ {-(j\frac\pi 2 n)}=sum-(1-2j)e^{j\frac \pi 2n} - 3 - (-1)^{n+1}\\
    a_3 =\big[sum-(1-2j)e^{j\frac \pi 2n} - 3 - (-1)^{n+1}\big]e ^ {j\frac\pi 2 n}$$
    Is that what you're looking for, or am I misunderstanding something?

  4. MHB Master
    MHB Math Helper

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    1,102
    Thanks
    318 times
    Thanked
    1,098 time
    #3
    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?

Similar Threads

  1. Imaginary numbers and real numbers?
    By Ilikebugs in forum Pre-Algebra and Algebra
    Replies: 1
    Last Post: February 7th, 2017, 22:47
  2. Replies: 1
    Last Post: October 8th, 2014, 01:04
  3. Complex numbers finding the Imaginary part
    By Drain Brain in forum Pre-Algebra and Algebra
    Replies: 7
    Last Post: June 14th, 2014, 12:10
  4. Replies: 6
    Last Post: September 8th, 2013, 11:57
  5. Separate imaginary from real
    By Amer in forum Analysis
    Replies: 2
    Last Post: October 5th, 2012, 10:07

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards