Digital Fourier Transform(DFT) Fourier Series ο΅ Signal used in daily life are continuous or discrete in nature. Discrete signal are used mainly in digital communication. Signals may be periodic or aperiodic also in nature. For these type of signals, we can use Fourier analysis to extract the information. Periodic signal, which may be represented by a function f(X) called as Fourier series. Where we for extract Fourier Series ο΅ But in the case of aperiodic signals Fourier transform and inverse Fourier transform are used to analysis the signals. Fourier transform f(w) is in generally in frequency domain, whereas as inverse F(t) is in the real time domain. Fourier Series F(ω) F(ω) Periodic signals ω 2ω ω Aperiodic signals F(t) F(ω) Fourier Transform Inverse Fourier Transform t t ω ω F(x) Fourier Series Period a Complex continuous periodic function by fourier series. x F(π) Each terms contains frequency as a integral multiple of original complex wave frequency. πΆπ πΆπ πΆπ πΆπ π Fourier Series ∞ π π = ππ + πΆπ πππ πππ + π·π πππ πππ π=π ∞ π π = ππ + π=π π π·π = π» π πΆπ = π» π/π n=1,2,3,4……………… πππ π πππ π πΆπ ππ¨π¬ + π·π π¬π’π§ π π n=1,2,3,4……………… πππ π π π π¬π’π§( )π π π −π/π π/π πππ π π π π¬π’π§ π π π −π/π Fourier Series Using the Euler's relation, e=cosθ+isinθ ∞ we can modify the Fourier series as πππ ππ/π π π = π ππ π=−∞ The above summation of exponential summation series extends from negative infinity and the ππ is the Fourier coefficient .If we can found out ππ then the series is determined. we use the kronecker delta function to solve. π/π π π π=−π/π aπΉππ = Due to this property the exponential function with integer value n is orthogonal −πππ ππ/π πππ ππ/π π π πΉππ π, ππ π = π = π, ππ π ≠ π π/π . We will use this to determine ππ . −π/π π/π in nature π π πππ ππ/π π π=−∞ π π π π=−∞ −π/π ∞ πππ ππ − π π π(π) π/π ∞−π/π = π π πππ ππ − π π π(π)= πππ ππ/π πππ ππ/π π ππ ∞ ππ = ππΉππ ππ π=−∞ = πππ ∞ ππ = π πππ π ππ π=−∞ π π π/π π π π −πππ π π(π) = ππ −π/π Where ππ = ππ π/π is the wave number Fourier π π = π₯π’π¦ Transform π→∞ ∞ ππππ ππ/π ππ π=−∞ ∞ πππππ ππ π π = π₯π’π¦ π→∞ π=−∞ ∞ π π = π₯π’π¦ π→∞ π=−∞ ∞ Δk ππ π π π π(ππ ) π(ππ )= ππ π π = π₯π’π¦ π→∞ ππ =nΔk, Δk= ππ /π π=−∞ Δk ππ π ππ ππ π π ππ Δk dk ππ π π π = π π π(π) −∞ ππ ∞ ππ ππ Δk ππ ππ π(ππ ) = π₯π’π¦ π→∞ Δk ππ π π(ππ ) = π₯π’π¦ π→∞ ππ /π π π/π π π π−πππ π π(π) −π/π ∞ π(ππ ) = π π π −πππ π(π) −∞ ∞ dk ππ π π π = π π π(π) −∞ ππ ∞ π(π) = π π π −∞ −πππ π(π) Now let us consider a digital sequence that is periodic in nature. The sequence contain n datas and have a period N. Then we can express the amplitude of each data point as x[1], x[2]… x[n]. Since the time period for a periodic signal is finite, we can represent this as a summation of exponential series as did for Fourier series equation 1 Frequency Analysis of Discrete time signals . Exponential series contains frequency components as an integral multiple of its fundamental frequency 1/N. Then the difference between successive harmonics is 1/N Hz or 2pi/N radians. X[n] πΏ[π] = π[π]π πππ ππ/π΅ ,k=0,1,23………..(1) N(period) n data points n 2 1 t πππ ππ/π΅ π[π]π ,k=0,1,2…(1) X[n] ππ§ = figure x[n] is periodic in nature, so X[n]=x[n+rN], where r is an integer N(period) n data points n 2 1 t Fourier series of a continuous periodic signals contains frequency up to a infinite range in the summation series but for Fourier series of discrete periodic signals only frequency in the range (0,1,…N) is needed. The proof is that the exponential function is identical for the kth and (k+N)th frequency. That is, π πππ ππ/π΅ =π πππ (π+π΅)π/π΅ π΅−π π[π]ππππ ππ/π΅ ………………..2 X[n]= π To find out x[k], we multiply πππ ππ/π΅ equation 2 with π on both sides and summing n=0 to N-1. π΅−π π΅−π π΅−π −πππ ππ/π΅ π[π] π = πππ (π−π)π/π΅ π[π]π π=π π=π π=π Recall the exponential summation of complex function as given in below π΅−π πππ ππ/π΅ π π=π π΅ ππ π = π, ±π΅, ±ππ΅ = π ππ πππππππππ π΅−π πππ (π−π)π/π΅ π π΅ ππ π − π = π, ±π΅, ±ππ΅ = π ππ πππππππππ π=π π΅−π π[π] π −πππ ππ/π΅ = π=π π΅−π π΅−π πππ (π−π)π/π΅ π[π]π π=π π=π If k-r=0,then k=r and the series is modified π = πΏ[π] π΅ π΅−π π[π] π π=π −πππ ππ/π΅ k=0,1,2…N1…….(2) π΅−π π[π] π=π −πππ ππ/π΅ π π΅−π π΅−π = πππ (π−π)π/π΅ π[π]π π=π π=π π΅−π π΅−π πππ (π−π)π/π΅ π[π]π π=π π΅−π + π[π]π π=π π[π΅ − π]π +…….. π=π πππ (π−π)π/π΅ πππ (π΅−π−π)π/π΅ If k-r=0, only this term is ≠0 π΅−π π΅−π π[π] π=π π΅−π π=π π΅−π π[π]π πππ (π−π)π/π΅ + π[π]π −πππ ππ/π΅ = π πππ (π−π)π/π΅ +…. π[π]ππππ (π−π)π/π΅ +…. π=π π=π If k-r=0,then k=r and then π πΏ[π] = π΅ π΅−π −πππ ππ/π΅ π[π] π π=π k=0,1,2…N- X[k]N π πΏ[π] = π΅ π΅−π −πππ ππ/π΅ π[π] π π=π k=0,1,2…..N-1…………….(2) DFTS can be represented by equation 2 and corresponding Fourier coefficient X[k] gives its frequency domain spectra. X[k] represents the phase and amplitude of corresponding frequency. k > N? π΅−π π −πππ (π+π΅)π/π΅ πΏ[π + π΅]= π[π] π π΅ π΅−π π=π π −πππ ππ/π΅ −πππ π΅π/π΅ =x[k] π[π] π π π΅ π=π −πππ π΅π/π΅ π = cos ππ π + ππππππ π = π πππ π = π, π … π΅ − π Thus ,x[k] is discrete variable but with a period of N Discrete Time Fourier Series(DTFS) X[n] = π₯[π]π π2πππ/π π[π] 1 = π π−1 π₯[π] π=0 −π2πππ/π π Frequency Analysis of Discrete aperiodic signals Since the time period for a aperiodic signal is infinite (N=∞), the time domain spectra contain the number of data points (-∞ to ∞) extends to infinite. we can add this to summation of exponential series to determine x[k]. ∞ π[π] π−πππ ππ/π΅ πΏ[π] = π=−∞ As in the figure x[n] is periodic in nature, we can write X[n]=x[n+rN], where r is an integer
