Determination of a signal from its Fourier Coefficients

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The discussion focuses on deriving a discrete-time signal x[n] from its Fourier coefficients ak, specifically ak = cos(k*Pi/4) + sin(3*k*Pi/4) with a fundamental period N=12. The method proposed involves using Euler's formula to expand ak, but participants note that this approach becomes tedious when evaluating the sum multiple times. Suggestions include summing directly without conversion to Euler's form or simplifying the equation to identify patterns, particularly for specific values of k. Concerns are raised about the complexity of the process if the period increases, leading to the suggestion of creating a program to automate the calculations. Overall, the discussion emphasizes the need for more efficient methods to derive x[n] by hand.
ColdStart
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letsa say i have an ak = cos ( k*Pi/4) + sin(3*k*Pi/4), the signal is discrete time, fundamental period N=12.

the way i would derive its x[n] is.. Sum(k=0, to 11 of: 0.5*exp(j*k*Pi/4) *exp(j*k*w*n) + 0.5*exp(-j*k*Pi/4) *exp(j*k*w*n) + (1/2*j)*exp(j*k*3*Pi/4) *exp(j*k*w*n) - (1/2*j)*exp(-j*k*3*Pi/4) *exp(j*k*w*n)

In other words, i expanded ak using eulers, and put it into the formula for finding x[n]... however as u see it turns out to be tedious process to evaluate this sum 12 times..

i was wondering, what is the more effective and QUICK method getting of x[n] by hand?

thanks
 
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You mean X(t) right ?

Anyway, do not convert to euler as you see it looks terrible.

You can just sum it the way you see it.

Another choice would be to invest a little time to simplifying the equation.

For example whe k =4

cos ( k*Pi/4) = -1
sin (3*k*pi/4) =0
Usually this kind of patterns repeat.
 
no actually x[n] because its discrete time... wht if my period would be 40 or more? it would be more tedious then...
 
Well, perhaps you should look into creating a program to do the job. I personally do not know any mathematical acrobatics that can help you.
 
thanks anyway!
 
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