Digital Fourier Transform(DFT) Fourier Series 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 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(𝝎) Each terms contains frequency as a integral multiple of original complex wave frequency. 𝜶𝟑 𝜶𝟏 𝜶𝟐 𝜶𝟒 𝝎 Fourier Series ∞ 𝒇 𝒙 = 𝒂𝟎 + 𝜶𝒏 𝒄𝒐𝒔 𝒏𝝎𝒕 + 𝜷𝒏 𝒔𝒊𝒏 𝒏𝝎𝒕 𝒏=𝟏 ∞ 𝒇 𝒙 = 𝒂𝟎 + 𝒏=𝟏 𝟐 𝜷𝒏 = 𝑻 𝟐 𝜶𝒏 = 𝑻 𝒂/𝟐 n=1,2,3,4……………… 𝟐𝒏𝝅𝒙 𝟐𝒏𝝅𝒙 𝜶𝒏 𝐜𝐨𝐬 + 𝜷𝒏 𝐬𝐢𝐧 𝒂 𝒂 n=1,2,3,4……………… 𝟐𝒏𝝅𝒙 𝒇 𝒙 𝐬𝐢𝐧( )𝒅𝒙 𝒂 −𝒂/𝟐 𝒂/𝟐 𝟐𝒏𝝅𝒙 𝒇 𝒙 𝐬𝐢𝐧 𝒅𝒙 𝒂 −𝒂/𝟐 Fourier Series Using the Euler's relation, e=cosθ+isinθ ∞ we can modify the Fourier series as 𝒊𝟐𝝅𝒏𝒙/𝒂 𝒇 𝒙 = 𝒆 𝒇𝒏 𝒏=−∞ The above summation of exponential summation series extends from negative infinity and the 𝑓𝑛 is the Fourier coefficient .If we can found out 𝑓𝑛 then the series is determined. we use the kronecker delta function to solve. 𝒂/𝟐 𝒅𝒙 𝒏=−𝒂/𝟐 a𝜹𝒏𝒎 = Due to this property the exponential function with integer value n is orthogonal −𝒊𝟐𝝅𝒏𝒙/𝒂 𝒊𝟐𝝅𝒏𝒙/𝒂 𝒆 𝒆 𝜹𝒏𝒎 𝟏, 𝒊𝒇 𝒎 = 𝒏 = 𝟎, 𝒊𝒇 𝒎 ≠ 𝒏 𝒂/𝟐 . We will use this to determine 𝒇𝒏 . −𝒂/𝟐 𝒂/𝟐 in nature 𝒅𝒙 𝒊𝟐𝝅𝒏𝒙/𝒂 𝒆 𝒏=−∞ 𝒅𝒙 𝒆 𝒏=−∞ −𝒂/𝟐 ∞ 𝒊𝟐𝝅𝒎𝒙 − 𝒆 𝒂 𝒇(𝒙) 𝒂/𝟐 ∞−𝒂/𝟐 = 𝒅𝒙 𝒊𝟐𝝅𝒎𝒙 − 𝒆 𝒂 𝒇(𝒙)= 𝒊𝟐𝝅𝒎𝒙/𝒂 𝒊𝟐𝝅𝒏𝒙/𝒂 𝒆 𝒇𝒏 ∞ 𝒇𝒏 = 𝒂𝜹𝒏𝒎 𝒇𝒏 𝒏=−∞ = 𝒂𝒇𝒏 ∞ 𝒇𝒏 = 𝒆 𝒊𝒌𝒏 𝒙 𝒇𝒏 𝒏=−∞ 𝟏 𝒂 𝒂/𝟐 𝒅𝒙 𝒆 −𝒊𝒌𝒏 𝒙 𝒇(𝒙) = 𝒇𝒏 −𝒂/𝟐 Where 𝒌𝒏 = 𝟐𝝅𝒏/𝒂 is the wave number Fourier 𝒇 𝒙 = 𝐥𝐢𝐦 Transform 𝒂→∞ ∞ 𝒆𝒊𝟐𝝅𝒏𝒙/𝒂 𝒇𝒏 𝒏=−∞ ∞ 𝒆𝒊𝒌𝒏𝒙 𝒇𝒏 𝒇 𝒙 = 𝐥𝐢𝐦 𝒂→∞ 𝒏=−∞ ∞ 𝒇 𝒙 = 𝐥𝐢𝐦 𝒂→∞ 𝒏=−∞ ∞ Δk 𝒊𝒌 𝒙 𝒆 𝒏 𝑭(𝒌𝒏 ) 𝑭(𝒌𝒏 )= 𝟐𝝅 𝒇 𝒙 = 𝐥𝐢𝐦 𝒂→∞ 𝒌𝒏 =nΔk, Δk= 𝟐𝝅/𝒂 𝒏=−∞ Δk 𝒊𝒌 𝒙 𝟐𝝅𝒇𝒏 𝒆 𝒏 𝟐𝝅 Δk dk 𝒊𝒌 𝒙 𝒇 𝒙 = 𝒆 𝒏 𝑭(𝒌) −∞ 𝟐𝝅 ∞ 𝟐𝝅𝒇𝒏 Δk 𝟐𝝅𝒇𝒏 𝑭(𝒌𝒏 ) = 𝐥𝐢𝐦 𝒂→∞ Δk 𝟐𝝅 𝟏 𝑭(𝒌𝒏 ) = 𝐥𝐢𝐦 𝒂→∞ 𝟐𝝅/𝒂 𝒂 𝒂/𝟐 𝒅𝒙 𝒆−𝒊𝒌𝒏 𝒙 𝒇(𝒙) −𝒂/𝟐 ∞ 𝑭(𝒌𝒏 ) = 𝒅𝒙 𝒆 −𝒊𝒌𝒙 𝒇(𝒙) −∞ ∞ dk 𝒊𝒌 𝒙 𝒇 𝒙 = 𝒆 𝒏 𝑭(𝒌) −∞ 𝟐𝝅 ∞ 𝑭(𝒌) = 𝒅𝒙 𝒆 −∞ −𝒊𝒌𝒙 𝒇(𝒙) 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] 𝑿[𝒏] = 𝒙[𝒌]𝒆 𝒋𝟐𝝅𝒌𝒏/𝑵 ,k=0,1,23………..(1) N(period) n data points n 2 1 t 𝒋𝟐𝝅𝒌𝒏/𝑵 𝒙[𝒌]𝒆 ,k=0,1,2…(1) X[n] 𝐗𝐧 = 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, 𝒆 𝒋𝟐𝝅𝒌𝒏/𝑵 =𝒆 𝒋𝟐𝝅(𝒌+𝑵)𝒏/𝑵 𝑵−𝟏 𝒙[𝒌]𝒆𝒋𝟐𝝅𝒌𝒏/𝑵 ………………..2 X[n]= 𝟎 To find out x[k], we multiply 𝒋𝟐𝝅𝒓𝒏/𝑵 equation 2 with 𝒆 on both sides and summing n=0 to N-1. 𝑵−𝟏 𝑵−𝟏 𝑵−𝟏 −𝒋𝟐𝝅𝒓𝒏/𝑵 𝒙[𝒏] 𝒆 = 𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵 𝒙[𝒌]𝒆 𝒏=𝟎 𝒌=𝟎 𝒏=𝟎 Recall the exponential summation of complex function as given in below 𝑵−𝟏 𝒋𝟐𝝅𝒌𝒏/𝑵 𝒆 𝒏=𝟎 𝑵 𝒊𝒇 𝒌 = 𝟎, ±𝑵, ±𝟐𝑵 = 𝟎 𝒊𝒇 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆 𝑵−𝟏 𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵 𝒆 𝑵 𝒊𝒇 𝒌 − 𝒓 = 𝟎, ±𝑵, ±𝟐𝑵 = 𝟎 𝒊𝒇 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆 𝒏=𝟎 𝑵−𝟏 𝒙[𝒏] 𝒆 −𝒋𝟐𝝅𝒓𝒏/𝑵 = 𝒏=𝟎 𝑵−𝟏 𝑵−𝟏 𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵 𝒙[𝒌]𝒆 𝒏=𝟎 𝒌=𝟎 If k-r=0,then k=r and the series is modified 𝟏 = 𝑿[𝒌] 𝑵 𝑵−𝟏 𝒙[𝒏] 𝒆 𝒏=𝟎 −𝒋𝟐𝝅𝒌𝒏/𝑵 k=0,1,2…N1…….(2) 𝑵−𝟏 𝒙[𝒏] 𝒏=𝟎 −𝒋𝟐𝝅𝒓𝒏/𝑵 𝒆 𝑵−𝟏 𝑵−𝟏 = 𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵 𝒙[𝒌]𝒆 𝒌=𝟎 𝒏=𝟎 𝑵−𝟏 𝑵−𝟏 𝒋𝟐𝝅(𝟏−𝒓)𝒏/𝑵 𝒙[𝟏]𝒆 𝒏=𝟎 𝑵−𝟏 + 𝒙[𝟐]𝒆 𝒏=𝟎 𝒙[𝑵 − 𝟏]𝒆 +…….. 𝒌=𝟎 𝒋𝟐𝝅(𝟐−𝒓)𝒏/𝑵 𝒋𝟐𝝅(𝑵−𝟏−𝒓)𝒏/𝑵 If k-r=0, only this term is ≠0 𝑵−𝟏 𝑵−𝟏 𝒙[𝒏] 𝒏=𝟎 𝑵−𝟏 𝒏=𝟎 𝑵−𝟏 𝒙[𝟐]𝒆 𝒋𝟐𝝅(𝟏−𝒓)𝒏/𝑵 + 𝒙[𝟏]𝒆 −𝒋𝟐𝝅𝒓𝒏/𝑵 = 𝒆 𝒋𝟐𝝅(𝟐−𝒓)𝒏/𝑵 +…. 𝒙[𝒌]𝒆𝒋𝟐𝝅(𝒌−𝒓)𝒏/𝑵 +…. 𝒌=𝟎 𝒏=𝟎 If k-r=0,then k=r and then 𝟏 𝑿[𝒌] = 𝑵 𝑵−𝟏 −𝒋𝟐𝝅𝒌𝒏/𝑵 𝒙[𝒏] 𝒆 𝒏=𝟎 k=0,1,2…N- X[k]N 𝟏 𝑿[𝒌] = 𝑵 𝑵−𝟏 −𝒋𝟐𝝅𝒌𝒏/𝑵 𝒙[𝒏] 𝒆 𝒏=𝟎 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? 𝑵−𝟏 𝟏 −𝒋𝟐𝝅(𝒌+𝑵)𝒏/𝑵 𝑿[𝒌 + 𝑵]= 𝒙[𝒏] 𝒆 𝑵 𝑵−𝟏 𝒏=𝟎 𝟏 −𝒋𝟐𝝅𝒌𝒏/𝑵 −𝒋𝟐𝝅𝑵𝒏/𝑵 =x[k] 𝒙[𝒏] 𝒆 𝒆 𝑵 𝒏=𝟎 −𝒋𝟐𝝅𝑵𝒏/𝑵 𝒆 = cos 𝟐𝝅𝒏 + 𝒋𝒔𝒊𝒏𝟐𝝅𝒏 = 𝟏 𝒇𝒐𝒓 𝒏 = 𝟎, 𝟏 … 𝑵 − 𝟏 Thus ,x[k] is discrete variable but with a period of N Discrete Time Fourier Series(DTFS) X[n] = 𝑥[𝑘]𝑒 𝑗2𝜋𝑘𝑛/𝑁 𝑋[𝑘] 1 = 𝑁 𝑁−1 𝑥[𝑛] 𝑛=0 −𝑗2𝜋𝑘𝑛/𝑁 𝑒 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]. ∞ 𝒙[𝒏] 𝒆−𝒋𝟐𝝅𝒌𝒏/𝑵 𝑿[𝒌] = 𝒏=−∞ As in the figure x[n] is periodic in nature, we can write X[n]=x[n+rN], where r is an integer