We now discuss the Discrete-time Fourier transform (DTFT) which is the counterpart of the continuous-time Fourier transform (CTFT) for analog signals and systems.

We can evolve the DTFS having line spectrum to the DTFT having continuous spectrum in the same way we did the CTFS to the CTFT (section ). The transform pair is denoted as

x(n) ↔ DTFT X(ω) x(n) ↔ DTFT X(ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqipDI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqaaeaadaqaaqaaaOqaaiaadIhacaGGOaGaamOBaiaacMcadaGd0aWcbaGaamiraiaadsfacaWGgbGaamivaaqabOGaayjLHaGaamiwaiaacIcacqaHjpWDcaGGPaaaaa@4240@

where the transform and the inverse transform are given respectively as

ωω size 12{ω} {} is the digital angular frequency of unit radians/sample (section). Its range [−π,π][−π,π] size 12{ \[ - π,π \] } {} corresponds to the Nyquist interval [−fs/2,fs/2][−fs/2,fs/2] size 12{ \[ - f rSub { size 8{s} } /2, {f rSub { size 8{s} } } slash {2} \] } {} with respect to the analog frequency , where fsfs size 12{f rSub { size 8{s} } } {} is the sampling frequency (sampling rate). Readers might need to look back the Equations , Equations and Equations for the relations between digital frequency and analog frequency, involving the sampling frequency.

Above we took the transform pair for granted, now we check that the two equations above are indeed a transform pair. For this, we show for a given analysis equation, the synthesis equation will hold, or vice versa. Let’s replace the RHS of the analysis equation into the RHS of the analysis equation:

1 2π ∫ − π π X ( ω ) e jωn dω = 1 2π ∫ − π π ∑ m = − ∞ ∞ x ( m ) e − jωn e jωn dω 1 2π ∫ − π π X ( ω ) e jωn dω = 1 2π ∫ − π π ∑ m = − ∞ ∞ x ( m ) e − jωn e jωn dω size 12{ { {1} over {"2π"} } Int rSub { size 8{ - π} } rSup { size 8{π} } {X \( ω \) } e rSup { size 8{jωn} } dω= { {1} over {"2π"} } Int rSub { size 8{ - π} } rSup { size 8{π} } { Sum cSub { size 8{m= - infinity } } cSup { size 8{ infinity } } {x \( m \) e rSup { size 8{ - jωn} } } e rSup { size 8{jωn} } dω} } {}

We then change the order of integral and summation:

1 2π ∫ − π π X ( ω ) e jωn dω = 1 2π ∑ m = − ∞ ∞ x ( m ) ∫ − π π e − jω ( n − m ) dω 1 2π ∫ − π π X ( ω ) e jωn dω = 1 2π ∑ m = − ∞ ∞ x ( m ) ∫ − π π e − jω ( n − m ) dω size 12{ { {1} over {2π} } Int rSub { size 8{ - π} } rSup { size 8{π} } {X \( ω \) e rSup { size 8{jωn} } dω={}} { {1} over {2π} } Sum cSub { size 8{m= - infinity } } cSup { size 8{ infinity } } {x \( m \) Int rSub { size 8{ - π} } rSup { size 8{π} } {e rSup { size 8{ - jω \( n - m \) } } } } dω} {}

By the property of orthogonality of exponentials, the integral on the RHS is zero for n≠mn≠m size 12{n <> m} {} and equals 2 π π MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWbaa@37A2@ for n=mn=m size 12{n=m} {}. As a result the RHS reduces to just x(n)x(n) size 12{x \( n \) } {} as expected.

The DTFT has an important characterstic. i.e. the DTFT is periodic with period of 2π2π size 12{2π} {} radians, while the CTFT is not periodic at all. We need to show

To this end, we replace ωω size 12{ω} {} in the analysis equation by ω+2πω+2π size 12{ω+2π} {}:

X ( ω + 2π ) = ∑ n = − ∞ ∞ x ( n ) e − j ( ω + 2π ) n = ∑ n = − ∞ ∞ x ( n ) e − jωn e − j2πn X ( ω + 2π ) = ∑ n = − ∞ ∞ x ( n ) e − j ( ω + 2π ) n = ∑ n = − ∞ ∞ x ( n ) e − jωn e − j2πn alignl { stack { size 12{X \( ω+2π \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) } e rSup { size 8{ - j \( ω+2π \) n} } } {} # matrix { {} # {} # {} # {} } = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) e rSup { size 8{ - jωn} } e rSup { size 8{ - j2πn} } } {} } } {} {}

Because e−j2πn=1e−j2πn=1 size 12{e rSup { size 8{ - j2πn} } =1} {}, the RHS is X(ω)X(ω) size 12{X \( ω \) } {} as expected.

The DTFT X(ω)X(ω) size 12{X \( ω \) } {} of a sequence x(n)x(n) size 12{x \( n \) } {} exists if x(n)x(n) size 12{x \( n \) } {} converges, i.e

For demonstration we take the alsolute value of both sides of the analysis quation:

∣ X ( ω ) ∣ = ∣ ∑ n = − ∞ ∞ x ( n ) e − jn ω ∣ ≤ ∑ n = − ∞ ∞ ∣ x ( n ) ∣ ∣ e − jn ω ∣ = ∑ n = − ∞ ∞ ∣ x ( n ) ∣ < ∞ ∣ X ( ω ) ∣ = ∣ ∑ n = − ∞ ∞ x ( n ) e − jn ω ∣ ≤ ∑ n = − ∞ ∞ ∣ x ( n ) ∣ ∣ e − jn ω ∣ = ∑ n = − ∞ ∞ ∣ x ( n ) ∣ < ∞ alignl { stack { size 12{ lline X \( ω \) rline = lline Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) e rSup { size 8{ - ital "jn"ω} } } rline <= Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { lline x \( n \) rline lline e rSup { size 8{ - ital "jn"ω} } rline } } {} # matrix { matrix { {} # {} # {} } {} # {} # {} # {} } = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { lline x \( n \) rline } < infinity {} } } {}

Usually the DTFT X(ω)X(ω) size 12{X \( ω \) } {} is a complex quantity and in computation it is decomposed into the real and imaginary parts and then the magnitude and phase spectra:

where

| X(ω) |= X R 2 (ω)+ X I 2 (ω) | X(ω) |= X R 2 (ω)+ X I 2 (ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamiwaiaacIcacqaHjpWDcaGGPaaacaGLhWUaayjcSdGaeyypa0ZaaOaaaeaacaWGybWaa0baaSqaaiaadkfaaeaacaaIYaaaaOGaaiikaiabeM8a3jaacMcacqGHRaWkcaWGybWaa0baaSqaaiaadMeaaeaacaaIYaaaaOGaaiikaiabeM8a3jaacMcaaSqabaaaaa@4A9E@

Φ(ω)=∠X(ω)=tan−1XI(ω)XR(ω)Φ(ω)=∠X(ω)=tan−1XI(ω)XR(ω) size 12{Φ \( ω \) "=∠"X \( ω \) ="tan" rSup { size 8{ - 1} } { {X rSub { size 8{I} } \( ω \) } over {X rSub { size 8{R} } \( ω \) } } } {}

The notation ∣X(ω)∣∣X(ω)∣ size 12{ lline X \( ω \) rline } {} is used for magnitude spectrum (only absolute value) and X(ω)X(ω) size 12{X \( ω \) } {} for amplitude spectrum (can be positive or negative)

Because the spectrum is 2π2π size 12{2π} {}-periodic we need to compute X(ω)X(ω) size 12{X \( ω \) } {} for a range of 2π2π size 12{2π} {}, usually from −π−π size 12{ - π} {} to ππ size 12{π} {} , sometimes from 0 to 2π2π size 12{"2π"} {}. Furthermore, because the mentioned symmetry we need to compute X(ω)X(ω) size 12{X \( ω \) } {} only for ωω size 12{ω} {} from 0 to ππ size 12{π} {} then take the even symmetry (mirror image) for the magnitude spectrum and take the odd-symmetry for the phase spectrum.

Apply the analysis equation and arrange as follows.

X ( ω ) = x ( 0 ) e − j0ω + [ x ( 1 ) e − j1ω + x ( − 1 ) e j1ω ] + [ x ( 2 ) e − j2ω + x ( − 2 ) e j2ω ] + . . . + [ x ( N ) e − jN ω + x ( − N ) e jN ω ] X ( ω ) = x ( 0 ) e − j0ω + [ x ( 1 ) e − j1ω + x ( − 1 ) e j1ω ] + [ x ( 2 ) e − j2ω + x ( − 2 ) e j2ω ] + . . . + [ x ( N ) e − jN ω + x ( − N ) e jN ω ] size 12{X \( ω \) =x \( 0 \) e rSup { size 8{ - j0ω} } + \[ x \( 1 \) e rSup { size 8{ - j1ω} } +x \( - 1 \) e rSup { size 8{j1ω} } \] + \[ x \( 2 \) e rSup { size 8{ - j2ω} } +x \( - 2 \) e rSup { size 8{j2ω} } \] + "." "." "." + \[ x \( N \) e rSup { size 8{ - ital "jN"ω} } +x \( - N \) e rSup { size 8{ ital "jN"ω} } \] } {}

The idea is to utilize the well known relation