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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
| + | == Define a Periodic CT signal and compute its Fourier series coefficients == | ||
| − | == 1 == | + | |
| + | Consider the following CT signal: | ||
| + | |||
| + | |||
| + | x(t) such that | ||
| + | |||
| + | <math> ak = \frac{1}{T} \int_{0}^{T} x(t) * e^{-j*k} * \frac{2*\pi}{T} *dt </math> | ||
| + | |||
| + | |||
| + | |||
| + | <math> x(t) = cos(2* \pi * t) * cos(4* \pi * t) </math> | ||
| + | |||
| + | |||
| + | <math> = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} </math> | ||
| + | |||
| + | |||
| + | <math> = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}}{4} + \frac{e^{-j*6*\pi*t}}{4} </math> | ||
| + | |||
| + | |||
| + | K = 3 | ||
| + | |||
| + | K= -1 | ||
| + | |||
| + | K= 1 | ||
| + | |||
| + | K= -3 | ||
| + | |||
| + | |||
| + | <math> a^{3} = \frac{1}{4} </math> | ||
| + | |||
| + | <math> a^{-1} = \frac{1}{4} </math> | ||
| + | |||
| + | <math> a^{1} = \frac{1}{4} </math> | ||
| + | |||
| + | <math> a^{-3} = \frac{1}{4} </math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:58, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Define a Periodic CT signal and compute its Fourier series coefficients
Consider the following CT signal:
x(t) such that
$ ak = \frac{1}{T} \int_{0}^{T} x(t) * e^{-j*k} * \frac{2*\pi}{T} *dt $
$ x(t) = cos(2* \pi * t) * cos(4* \pi * t) $
$ = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} $
$ = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}}{4} + \frac{e^{-j*6*\pi*t}}{4} $
K = 3
K= -1
K= 1
K= -3
$ a^{3} = \frac{1}{4} $
$ a^{-1} = \frac{1}{4} $
$ a^{1} = \frac{1}{4} $
$ a^{-3} = \frac{1}{4} $
