Continuous Time Fourier Series of cosine equation

AI Thread Summary
The discussion revolves around finding the Continuous Time Fourier Series (CTFS) of the signal 2*cos(100*pi(t - 0.005)), with a fundamental period of T = 1/50. The initial solution derived is Cx[k] = (delta[k - 1] + delta[k + 1])*e^(-j*0.5*pi*k). However, the final solution in the manual simplifies this to Cx[k] = j*(delta[k - 1] + delta[k + 1]). Participants are seeking clarification on the transition from the first to the second expression. The focus is on understanding the simplification step involved in the CTFS calculation. The thread highlights the importance of mastering CTFS properties for accurate signal analysis.
owtu
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Homework Statement


Using the CTFS table of transforms and the CTFS properties, find the CTFS harmonic function of the signal

2*cos(100*pi(t - 0.005))

T = 1/50

Homework Equations



To = fundamental period

T = mTo

cos(2*pi*k/To) ----F.S./mTo---- (1/2)(delta[k-m] + delta[k+m])

The Attempt at a Solution



using the table of pairs and CTFS properties, the solution I arrived to was:

1) Cx[k] = (delta[k - 1] + delta[k+1])*e^(-j*0.5*pi*k)

My solution is right but the final solution from the solution manual there's one more step to simplify it even more:

2) Cx[k] = j*(delta[k - 1] + delta[k+1])How did they go from (1) to (2) ?
 
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