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CONTINUOUS - TIME FOURIER TRANSFORM (CTFT)

Module by: Nguyen Huu Phuong

The Fourier series expansion of a signal reveals its frequency structure we need. But unfortunately, the Fourier series expansion only applies to periodic signals while in reality signals are mostly aperiodic, and the Fourier transform was developed for the latter.

The Fourier transform pair

Figure 1: The evolution from Fourier series to Fourier transform
Figure 1 shows the evolution from Fourier series to Fourier transform.
In Equation we replace Ω0Ω0 size 12{ %OMEGA rSub { size 8{0} } } {} by 2πF02πF0 size 12{2πF rSub { size 8{0} } } {}, and write X(nF0)X(nF0) size 12{X \( ital "nF" rSub { size 8{0} } \) } {} for XnXn size 12{X rSub { size 8{n} } } {}, then
x(t)= n= X(n F 0 ) e j2π F 0 t dt x(t)= n= X(n F 0 ) e j2π F 0 t dt MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOaGaamiDaiaacMcacqGH9aqpdaaeWbqaaiaadIfacaGGOaGaamOBaiaadAeadaWgaaWcbaGaaGimaaqabaGccaGGPaGaamyzamaaCaaaleqabaGaamOAaiaaikdacqaHapaCcaWGgbWaaSbaaWqaaiaaicdaaeqaaSGaamiDaaaakiaadsgacaWG0baaleaacaWGUbGaeyypa0JaeyOeI0IaeyOhIukabaGaeyOhIukaniabggHiLdaaaa@5049@ (1)
The expansion coefficients are given by
X(nF0)=1T0T0/2T0/2x(t)ej2πnF0tdtX(nF0)=1T0T0/2T0/2x(t)ej2πnF0tdt size 12{X \( ital "nF" rSub { size 8{0} } \) = { {1} over {T rSub { size 8{0} } } } Int rSub { - T rSub { size 6{0} } /2} rSup {T rSub { size 6{0} } /2} {x \( t \) e rSup { size 8{ - j2π ital "nF" rSub { size 6{0} } t} } ital "dt"} } {} (2)
Periodic signals have line (discrete) spectrum.
Now we replace the analysis equation X(nF0)X(nF0) size 12{X \( ital "nF" rSub { size 8{0} } \) } {} for x(t) in the synthesis equation:
x ( t ) = n = 1 T 0 T 0 / 2 T 0 / 2 x ( t ) e j2π nF 0 t dt e j2π nF 0 t x ( t ) = n = 1 T 0 T 0 / 2 T 0 / 2 x ( t ) e j2π nF 0 t dt e j2π nF 0 t size 12{x \( t \) = Sum cSub {n= - infinity } cSup { infinity } { left [ { {1} over {T rSub { size 8{0} } } } Int rSub { - T rSub { size 6{0} } /2} rSup {T rSub { size 6{0} } /2} {x \( t \) e rSup { size 8{ - j2π ital "nF" rSub { size 6{0} } t} } ital "dt"} right ] size 12{e rSup {j2π ital "nF" rSub { size 6{0} } t} }} } {}
Let T0T0 size 12{T rSub { size 8{0} } rightarrow infinity } {} to push all periods on both sides of the central period of x(t) to infinity, thus transforming the periodic signal into an aperiodic one. On the other hand when the period T0T0 size 12{T rSub { size 8{0} } rightarrow infinity } {}, 1/T0dt1/T0dt size 12{1/T rSub { size 8{0} } rightarrow ital "dt"} {} (an infinitesimal quantity), nF0FnF0F size 12{ ital "nF" rSub { size 8{0} } rightarrow F} {}(continuous frquency) and the limits ±T0±±T0± size 12{ +- T rSub { size 8{0} } rightarrow +- infinity } {}, the discrete spectrum becomes continuous. Thus as T0T0 size 12{T rSub { size 8{0} } rightarrow infinity } {},
x ( t ) = n = 1 T 0 T 0 / 2 T 0 / 2 x ( t ) e j2π nf 0 dt x ( t ) = n = 1 T 0 T 0 / 2 T 0 / 2 x ( t ) e j2π nf 0 dt size 12{x \( t \) = Sum cSub {n= - infinity } cSup { infinity } { left [ { {1} over {T rSub { size 8{0} } } } Int rSub {T rSub { size 6{0} } /2} rSup {T rSub { size 6{0} } /2} {x \( t \) e rSup { size 8{ - j2π ital "nf" rSub { size 6{0} } ital "dt"} } } right ]} } {}
The term in brackets is, by definition, the Fourier transform (or the Fourier integral) X(F)X(F) size 12{X \( F \) } {} of x(t)x(t) size 12{x \( t \) } {}.
Thus
X(F)=x(t)ej2πFtdtX(F)=x(t)ej2πFtdt size 12{X \( F \) = Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } {x \( t \) e rSup { size 8{ - j2π ital "Ft"} } ital "dt"} } {} (3)
The inverse transform is then
x(t)=n=dFX(F)ej2πFtX(F)ej2πFtdFx(t)=n=dFX(F)ej2πFtX(F)ej2πFtdF size 12{x \( t \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { ital "dFX" \( F \) e rSup { size 8{j2π ital "Ft"} } } drarrow Int rSub { size 8{ infinity } } rSup { size 8{ infinity } } {X \( F \) e rSup { size 8{j2π ital "Ft"} } ital "dF"} } {} (4)
x(t)x(t) size 12{x \( t \) } {}and X(F)X(F) size 12{X \( F \) } {} form a continuous-time Fourier transform (CTFT) pair:
x ( t ) X ( F ) x ( t ) X ( F ) size 12{x \( t \) leftarrow rightarrow X \( F \) } {}

Magnitude and phase spectra

The Fourier transform X(F)X(F) size 12{X \( F \) } {} is, in general, complex and we write
X(F)=XR(F)+jXI(F)=X(F)e(F)X(F)=XR(F)+jXI(F)=X(F)e(F) size 12{X \( F \) =X rSub { size 8{R} } \( F \) + ital "jX" rSub { size 8{I} } \( F \) = lline X \( F \) rline e rSup { size 8{jΦ \( F \) } } } {} (5)
where X(f)X(f) size 12{ lline X \( f \) rline } {} is the magnitude spectrum, and Φ(F)Φ(F) size 12{Φ \( F \) } {} the phase spectrum:
X(F)=XR2(F)+XI2(F)X(F)=XR2(F)+XI2(F) size 12{ lline X \( F \) rline = sqrt {X rSub { size 8{R} } rSup { size 8{2} } \( F \) +X rSub { size 8{I} } rSup { size 8{2} } \( F \) } } {} (6)
Φ(F)=tan1XI(F)XR(F)Φ(F)=tan1XI(F)XR(F) size 12{Φ \( F \) ="tan" rSup { size 8{ - 1} } { {X rSub { size 8{I} } \( F \) } over {X rSub { size 8{R} } \( F \) } } } {} (7)
It can be shown that if the signal x(t)x(t) size 12{x \( t \) } {} is real, then XR(F)XR(F) size 12{X rSub { size 8{R} } \( F \) } {}is an even-symmetric (symmetric) and XI(F)XI(F) size 12{X rSub { size 8{I} } \( F \) } {} an old-synmetric (antisymmetric), thus the magnitude spectrum X(F)X(F) size 12{ lline X \( F \) rline } {} is an even-symmetric and the phase spectrum is an odd-symmetric, i.e.
| X(F) |=| X(F) |andΦ(F)=Φ(F) | X(F) |=| X(F) |andΦ(F)=Φ(F) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamiwaiaacIcacaWGgbGaaiykaaGaay5bSlaawIa7aiabg2da9maaemaabaGaamiwaiaacIcacqGHsislcaWGgbGaaiykaaGaay5bSlaawIa7aiaaywW7caWGHbGaamOBaiaadsgacaaMf8UaeuOPdyKaaiikaiaadAeacaGGPaGaeyypa0JaeyOeI0IaeuOPdyKaaiikaiabgkHiTiaadAeacaGGPaaaaa@5418@ (8)
Example 1 
(a) Find the Fourier transform, the magnitude and phase spectrum of an even-symmetric rectangular pulse of amplitude A and width τ τ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes8a0baa@37AA@ .
(b) Apply above result to find the transform of the unit impulse τ τ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes8a0baa@37AA@ (t) (delta Dirac function).
(c) Repeat the question when the above pulse is delayed by t 0 t 0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadshadaWgaaWcbaGaaGimaaqabaaaaa@37C4@ .
Solution
(a) The given pulse is plotted in Figure 2. It can be denoted as
x(t)=Ap[ τ 2 , τ 2 ] x(t)=Ap[ τ 2 , τ 2 ] MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOaGaamiDaiaacMcacqGH9aqpcaWGbbGaamiCamaadmaabaGaeyOeI0YaaSaaaeaacqaHepaDaeaacaaIYaaaaiaacYcadaWcaaqaaiabes8a0bqaaiaaikdaaaaacaGLBbGaayzxaaaaaa@44A6@
Figure 2: Example 1 (the pulse and its delay)
The CTFT Fourier transform is
X(F)= x(t) e j2πFt df = τ/2 τ/2 A e j2πFt df =A [ e j2πFt j2πFt ] τ/2 τ/2 =Aτ sinπFτ πFτ X(F)= x(t) e j2πFt df = τ/2 τ/2 A e j2πFt df =A [ e j2πFt j2πFt ] τ/2 τ/2 =Aτ sinπFτ πFτ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@97EC@
Notice that the result has the form of sinx/x function (section). The magnitude spectrum is
| X(F) |=| Aτ sinπFτ πFτ |=Aτ| sinπFτ πFτ | | X(F) |=| Aτ sinπFτ πFτ |=Aτ| sinπFτ πFτ | MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamiwaiaacIcacaWGgbGaaiykaaGaay5bSlaawIa7aiabg2da9maaemaabaGaamyqaiabes8a0naalaaabaGaci4CaiaacMgacaGGUbGaeqiWdaNaamOraiabes8a0bqaaiabec8aWjaadAeacqaHepaDaaaacaGLhWUaayjcSdGaeyypa0Jaamyqaiabes8a0naaemaabaWaaSaaaeaaciGGZbGaaiyAaiaac6gacqaHapaCcaWGgbGaeqiXdqhabaGaeqiWdaNaamOraiabes8a0baaaiaawEa7caGLiWoaaaa@6072@
For the amplitude spectrum we leave its negative part as is (i.e. we do not take the alsolute value) (Figure 3).
Figure 3: Example 1(magnitude and phase spectra)
As X(F) is a real function, its phase is zero at all frequency. But in Fourier analysis the phase spectrum is interpreted differently. That is for a real function, the phase spectrum is zero for positive value and is π π MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWbaa@37A2@ for negative value. Besides, to ensure the phase spectrum is an odd-symmetric function, the phase is understood to be + π π MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWbaa@37A2@ for positive frequency and - π π MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWbaa@37A2@ for negative frequency (Figure 3).
(b) In order to find the transform of δ(t) δ(t) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjaacIcacaWG0bGaaiykaaaa@39DC@ we consider the amplitude A as 1/ τ τ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes8a0baa@37AA@ , hence the spectrum of the rectangular pulse of amplitude 1/ τ τ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes8a0baa@37AA@ , width τ τ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes8a0baa@37AA@ , is
X(F)= 1 τ τ sinπFτ πFτ = sinπFτ πFτ X(F)= 1 τ τ sinπFτ πFτ = sinπFτ πFτ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfacaGGOaGaamOraiaacMcacqGH9aqpdaWcaaqaaiaaigdaaeaacqaHepaDaaGaeqiXdq3aaSaaaeaaciGGZbGaaiyAaiaac6gacqaHapaCcaWGgbGaeqiXdqhabaGaeqiWdaNaamOraiabes8a0baacqGH9aqpdaWcaaqaaiGacohacaGGPbGaaiOBaiabec8aWjaadAeacqaHepaDaeaacqaHapaCcaWGgbGaeqiXdqhaaaaa@564B@
Now let τ0τ0 size 12{τ rightarrow 0} {} then X(F)1X(F)1 size 12{X \( F \) rightarrow 1} {} which is the transform of δ(t)δ(t) size 12{δ \( t \) } {}:
δ ( t ) 1 δ ( t ) 1 size 12{δ \( t \) ↔1} {}
The amplitude spectrum is 1 at all frequencies and the phase spectrum is zero. From the amplitude spectrum of the rectangular pulse of Figure 3, as τ0 τ0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes8a0jabgkziUkaaicdaaaa@3A51@ the points 1/ τ τ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes8a0baa@37AA@ and -1/ τ τ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes8a0baa@37AA@ go to ± ± MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgglaXkabg6HiLcaa@3944@ and the central lobe extends to ± ± MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgglaXkabg6HiLcaa@3944@ the amplitude spectrum of δ(t) δ(t) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjaacIcacaWG0bGaaiykaaaa@39DC@ is 1 for all frequencies.
(c) The delayed pulse is denoted as
x(t t 0 )=Ap[ t 0 τ 2 , t 0 + τ 2 ] x(t t 0 )=Ap[ t 0 τ 2 , t 0 + τ 2 ] MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIhacaGGOaGaamiDaiabgkHiTiaadshadaWgaaWcbaGaaGimaaqabaGccaGGPaGaeyypa0JaamyqaiaadchadaWadaqaaiaadshadaWgaaWcbaGaaGimaaqabaGccqGHsisldaWcaaqaaiabes8a0bqaaiaaikdaaaGaaiilaiaaysW7caWG0bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaaSaaaeaacqaHepaDaeaacaaIYaaaaaGaay5waiaaw2faaaaa@4DBD@
Its CTFT is
X(F)= t 0 τ/2 t 0 +τ/2 A e j2πFt dt =A [ e j2πFt j2πFt ] t 0 τ/2 t 0 +τ/2 =Aτ sinπFτ πFτ e j2πF t 0 X(F)= t 0 τ/2 t 0 +τ/2 A e j2πFt dt =A [ e j2πFt j2πFt ] t 0 τ/2 t 0 +τ/2 =Aτ sinπFτ πFτ e j2πF t 0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@9630@
Notice the appearance of the phase factor ej2πFt0ej2πFt0 size 12{e rSup { size 8{ - j2π ital "Ft" rSub { size 6{0} } } } } {}. The magnitude response is exactly as in (a). However the appearance of the phase factor makes the phase spectrum quite different. It is understood as
Φ(F)=( sinπFτ πFτ 2πF t 0 ) Φ(F)=( sinπFτ πFτ 2πF t 0 ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfA6agjaacIcacaWGgbGaaiykaiabg2da9iabgcIiqlaacIcadaWcaaqaaiGacohacaGGPbGaaiOBaiabec8aWjaadAeacqaHepaDaeaacqaHapaCcaWGgbGaeqiXdqhaaiabgkHiTiaaikdacqaHapaCcaWGgbGaamiDamaaBaaaleaacaaIWaaabeaakiaacMcaaaa@4F1C@
Figure 4: Example 1 (phase spectrum of the delay pulse)
Where MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgcIiqdaa@3783@ denotes the phase (or argument). The phase spectrum is detailed as follows:
Φ(F)=2πF t 0 , sinπFτ πFτ >0andF>0orF<0 =2πF t 0 +π, sinπFτ πFτ <0andF>0 =2πF t 0 π, sinπFτ πFτ <0andF<0 Φ(F)=2πF t 0 , sinπFτ πFτ >0andF>0orF<0 =2πF t 0 +π, sinπFτ πFτ <0andF>0 =2πF t 0 π, sinπFτ πFτ <0andF<0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpW