DISCRETE - TIME ROURIER TRANSFORM (DTFT)1.32007/12/06 00:58:33 US/Central2008/07/07 02:28:47.271 GMT-5PhuongHuuNguyennhphuong@hcmuns.edu.vnPhuongHuuNguyennhphuong@hcmuns.edu.vnWe 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 asx(n)↔DTFTX(ω)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqipDI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqaaeaadaqaaqaaaOqaaiaadIhacaGGOaGaamOBaiaacMcadaGd0aWcbaGaamiraiaadsfacaWGgbGaamivaaqabOGaayjLHaGaamiwaiaacIcacqaHjpWDcaGGPaaaaa@4240@where the transform and the inverse transform are given respectively asX(ω)=∑n=−∞∞x(n)e−jωn(DTFT)(analysisequation) size 12{X \( ω \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) e rSup { size 8{ - jωn} } } matrix {
{} # {}
} \( ital "DTFT" \) matrix {
{} # {}
} \( ital "analysis" matrix {
{}
} ital "equation" \) } {}x(n)=12π∫−ππX(ω)ejωndω(DTFT)(synthesisequation) size 12{x \( n \) = { {1} over {2π} } Int rSub { size 8{ - π} } rSup { size 8{π} } {X \( ω \) } e rSup { size 8{jωn} } dω matrix {
{} # {}
} \( ital "DTFT" \) matrix {
{} # {}
} \( ital "synthesis" matrix {
{}
} ital "equation" \) } {}ω size 12{ω} {} is the digital angular frequency of unit radians/sample ( section ). Its range
[−π,π] size 12{ \[ - π,π \] } {} corresponds to the Nyquist interval
[−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
fs 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:12π∫−ππX(ω)ejωndω=12π∫−ππ∑m=−∞∞x(m)e−jωnejωndω 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:12π∫−ππX(ω)ejωndω=12π∑m=−∞∞x(m)∫−ππe−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≠m size 12{n <> m} {} and equals 2π
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWbaa@37A2@
for
n=m size 12{n=m} {}. As a result the RHS reduces to just
x(n) size 12{x \( n \) } {} as expected.The DTFT has an important characterstic. i.e. the DTFT is periodic with period of
2π size 12{2π} {} radians, while the CTFT is not periodic at all. We need to showX(ω)=X(ω+2π),allω
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfacaGGOaGaeqyYdCNaaiykaiabg2da9iaadIfacaGGOaGaeqyYdCNaey4kaSIaaGOmaiabec8aWjaacMcacaaMf8UaaiilaiaaywW7caWGHbGaamiBaiaadYgacaaMf8UaeqyYdChaaa@4C3B@To this end, we replace
ω size 12{ω} {} in the analysis equation by
ω+2π size 12{ω+2π} {}:X(ω+2π)=∑n=−∞∞x(n)e−j(ω+2π)n=∑n=−∞∞x(n)e−jωne−j2πnalignl { 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=1 size 12{e rSup { size 8{ - j2πn} } =1} {}, the RHS is
X(ω) size 12{X \( ω \) } {} as expected.The DTFT
X(ω) size 12{X \( ω \) } {} of a sequence
x(n) size 12{x \( n \) } {} exists if
x(n) size 12{x \( n \) } {} converges, i.e∑n=−∞∞∣x(n)∣<∞ size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { lline x \( n \) rline } < infinity } {}For demonstration we take the alsolute value of both sides of the analysis quation:∣X(ω)∣=∣∑n=−∞∞x(n)e−jnω∣≤∑n=−∞∞∣x(n)∣∣e−jnω∣=∑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) 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:X(ω)=XR(ω)+jX1(ω)=∣X(ω)∣ejΦ(ω) size 12{X \( ω \) =X rSub { size 8{R} } \( ω \) + ital "jX" rSub { size 8{1} } \( ω \) = lline X \( ω \) rline e rSup { size 8{jΦ \( ω \) } } } {}where|X(ω)|=XR2(ω)+XI2(ω)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamiwaiaacIcacqaHjpWDcaGGPaaacaGLhWUaayjcSdGaeyypa0ZaaOaaaeaacaWGybWaa0baaSqaaiaadkfaaeaacaaIYaaaaOGaaiikaiabeM8a3jaacMcacqGHRaWkcaWGybWaa0baaSqaaiaadMeaaeaacaaIYaaaaOGaaiikaiabeM8a3jaacMcaaSqabaaaaa@4A9E@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} } \( ω \) } } } {} For real-valued signal x(n) the magnitude spectrum
∣X(ω)∣ size 12{ lline X \( ω \) rline } {} is symmetric, and the phase spectrum
∣Φ(ω)∣ size 12{ lline Φ \( ω \) rline } {} is antisymmetric:∣X(−ω)∣=∣X(ω)∣andΦ(−ω)=−Φ(ω) size 12{ lline X \( - ω \) rline = lline X \( ω \) rline matrix {
{} # {}
} ital "and" matrix {
{} # {}
} Φ \( - ω \) = - Φ \( ω \) } {}The notation
∣X(ω)∣ size 12{ lline X \( ω \) rline } {} is used for magnitude spectrum (only absolute value) and
X(ω) size 12{X \( ω \) } {} for amplitude spectrum (can be positive or negative) Because the spectrum is
2π size 12{2π} {}-periodic we need to compute
X(ω) size 12{X \( ω \) } {} for a range of
2π size 12{2π} {}, usually from
−π size 12{ - π} {} to
π size 12{π} {}, sometimes from 0 to
2π size 12{"2π"} {}. Furthermore, because the mentioned symmetry we need to compute
X(ω) size 12{X \( ω \) } {} only for
ω size 12{ω} {} from 0 to
π size 12{π} {} then take the even symmetry (mirror image) for the magnitude spectrum and take the odd-symmetry for the phase spectrum.Find the spectrum of a digital rectangular pulse having 2N+1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeqabiGaciaacaqabeaadaqaaqaaaOqaaiaaikdacaWGobGaey4kaSIaaGymaaaa@3912@
samples from
n=−N size 12{n= - N} {} to
n=N size 12{n=N} {} and amplitude A (Figure a). Plot the amplitude and phase spectrum when
A=1 size 12{A=1} {} and
N=2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeqabiGaciaacaqabeaadaqaaqaaaOqaaiaad6eacqGH9aqpcaaIYaaaaa@387B@. Inchapter 5 we will call M for 2N , hence the length of the pulse is
2N+1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeqabiGaciaacaqabeaadaqaaqaaaOqaaiaaikdacaWGobGaey4kaSIaaGymaaaa@3912@
or M+1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeqabiGaciaacaqabeaadaqaaqaaaOqaaiaad2eacqGHRaWkcaaIXaaaaa@3855@ Solution Apply the analysis equation and arrange as follows. X(ω)=x(0)e−j0ω+[x(1)e−j1ω+x(−1)ej1ω]+[x(2)e−j2ω+x(−2)ej2ω]+...+[x(N)e−jNω+x(−N)ejNω] 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 relatione−jωn+ejωn=2cosωn size 12{e rSup { size 8{ - jωn} } +e rSup { size 8{jωn} } =2"cos"ωn} {}Thus X(ω)=A+2A(cosω+cos2ω)+...+cosNω size 12{X \( ω \) =A+2A \( "cos"ω+"cos"2ω \) + "." "." "." +"cos"Nω} {}Or in compact formX(ω)=A(1+2∑n=1Ncosnω)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfacaGGOaGaeqyYdCNaaiykaiabg2da9iaadgeacaGGOaGaaGymaiabgUcaRiaaikdadaaeWbqaaiGacogacaGGVbGaai4Caiaad6gacqaHjpWDaSqaaiaad6gacqGH9aqpcaaIXaaabaGaamOtaaqdcqGHris5aOGaaiykaaaa@4ACC@On the other hand, in keeping the original form of the analysis equation, we writeX(ω)=∑n=−∞∞x(n)e−jωn=∑n=−NNAe−jωn size 12{X \( ω \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) e rSup { size 8{ - jωn} } = Sum cSub { size 8{n= - N} } cSup { size 8{N} } { ital "Ae" rSup { size 8{ - jωn} } } } } {}Applying the formula of middle geometric series∑n=N1N2an=aN1−aN2+11−a,N2≥N1
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqabeaabiGaciaacaqaaeaadaqaaqaaaOqaamaaqahabaGaamyyamaaCaaaleqabaGaamOBaaaaaeaacaWGUbGaeyypa0JaamOtamaaBaaameaacaaIXaaabeaaaSqaaiaad6eadaWgaaadbaGaaGOmaaqabaaaniabggHiLdGccqGH9aqpdaWcaaqaaiaadggadaahaaWcbeqaaiaad6eadaWgaaadbaGaaGymaaqabaaaaOGaeyOeI0IaamyyamaaCaaaleqabaGaamOtamaaBaaameaacaaIYaaabeaaliabgUcaRiaaigdaaaaakeaacaaIXaGaeyOeI0IaamyyaaaacaaMf8UaaiilaiaaywW7caaMf8UaamOtamaaBaaaleaacaaIYaaabeaakiabgwMiZkaad6eadaWgaaWcbaGaaGymaaqabaaaaa@562D@we will getX(ω)=Aejωn−e−jω(N+1)1−e−jω=Asin(N+12)ωsinω2,ω≠0A(2N+1),ω=0alignl { stack {
size 12{X \( ω \) =A { {e rSup { size 8{jωn} } - e rSup { size 8{ - jω \( N+1 \) } } } over {1 - e rSup { size 8{ - jω} } } } =A { {"sin" \( N+ { {1} over {2} } \) ω} over {"sin" { {ω} over {2} } } } matrix {
} , matrix {
{} # {} # {}
} ω <> 0} {} #
matrix {
matrix {
matrix {
{} # {} # {} # {}
} {} # {} # {}
} {} # {} # {}
} A \( 2N+1 \) matrix {
{} # {} # {}
} , matrix {
{} # {} # {}
} ω=0 {}
} } {}
(rectangular pulse of A = 1 , N = 2 , and spectra)
Comparing the two results, we have this trigonometric relation1+2∑n=1Ncosnω=sin(N+12)ωsinω2 size 12{1+2 Sum cSub { size 8{n=1} } cSup { size 8{N} } {"cos"nω= { {"sin" \( N+ { {1} over {2} } \) ω} over {"sin" { {ω} over {2} } } } } } {}By putting
X(ω)=0 size 12{X \( ω \) =0} {} we can solve for the zero-crossing points which are:
ω=±2π/5 size 12{ω= +- 2π/5} {},
±4π/5 size 12{ +- 4π/5} {}... Because
X(ω) size 12{X \( ω \) } {} is an even-symmetric function we need to evaluate it for
0≤ω≤π size 12{0 <= ω <= π} {} then take the mirror image to get the spectrum for
−π≤ω≤0 size 12{ - π <= ω <= 0} {}. Also notice the
2π size 12{2π} {}-periodicity.b is the amplitude spectrum, for the magnitude spectrum we let the positive value part as is and invert the negative value to be positive. The part of the spectrum from -2π
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWbaa@37A2@
/5 to 2π
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWbaa@37A2@
/5 is called the mainlobe, the rest is the sidelobes . Even if the amplitude spectrum is real, it still has a phase spectrum defined on its sign:Φ(ω)=0,X(ω)>0±π,X(ω)<0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeuOPdyKaaiikaiabeM8a3jaacMcacqGH9aqpcaaIWaGaaGzbVlaacYcacaaMf8UaaGzbVlaadIfacaGGOaGaeqyYdCNaaiykaiabg6da+iaaicdaaeaaaeaacaaMf8UaaGzbVlaaywW7cqGHXcqScqaHapaCcaaMe8UaaiilaiaaywW7caaMf8UaamiwaiaacIcacqaHjpWDcaGGPaGaeyipaWJaaGimaaaaaa@5ADB@Also, we interpret the phase spectrum so that it has an odd symmetry (c).Find the magnitude spectrum of the digital exponential in a.
Soluton Examining the decay pattern of the given causal signal we can check its mathematical expression asx(n)=0.5nu(n) size 12{x \( n \) =0 "." 5 rSup { size 8{n} } u \( n \) } {}Its DTFT isX(ω)=∑n=−∞∞x(n)e−jωn=∑n=∞(0.5e−jω)n size 12{X \( ω \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) e rSup { size 8{ - jωn} } = Sum cSub { size 8{n={}} } cSup { size 8{ infinity } } { \( 0 "." 5e rSup { size 8{ - jω} } \) rSup { size 8{n} } } } } {}Applying the formula of infinite geometric series ( Equation ), we will obtain X(ω)=11−0.5e−jω size 12{X \( ω \) = { {1} over {1 - 0 "." 5e rSup { size 8{ - jω} } } } } {}Notice that the convergence condition
∣0.5e−jω∣<1 size 12{ lline 0 "." 5e rSup { size 8{ - jω} } rline <1} {} is satisfied.
(decaying exponential and magnitude spectrum)
In order to compute the spectrum manually we expand the complex exponential in terms of trigonometric functions:X(ω)=11−0.5(cosω−jsinω)=1(1−0.5cosω)+j0.5sinω
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIfacaGGOaGaeqyYdCNaaiykaiabg2da9maalaaabaGaaGymaaqaaiaaigdacqGHsislcaaIWaGaaiOlaiaaiwdacaGGOaGaci4yaiaac+gacaGGZbGaeqyYdCNaeyOeI0IaamOAaiGacohacaGGPbGaaiOBaiabeM8a3jaacMcaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaaiikaiaaigdacqGHsislcaaIWaGaaiOlaiaaiwdaciGGJbGaai4BaiaacohacqaHjpWDcaGGPaGaey4kaSIaamOAaiaaicdacaGGUaGaaGynaiGacohacaGGPbGaaiOBaiabeM8a3baaaaa@6044@Thus the magnitude spectrum is |X(ω)|=1[(1−0.5cosω)2+(0.5sinω)2]1/2=1[1.25−cosω]1/2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@6540@The we proceed as follows: At
ω=0 size 12{ω=0} {},
∣X(ω)∣=2 size 12{ lline X \( ω \) rline =2} {} and is maximum; at
ω=π/4 size 12{ω=π/4} {},
∣X(ω)∣=1.36 size 12{ lline X \( ω \) rline =1 "." "36"} {} ; at
ω=π/2 size 12{ω=π/2} {},
∣X(ω)∣=0.89 size 12{ lline X \( ω \) rline =0 "." "89"} {}, and at
ω=π size 12{ω=π} {},
∣X(ω)∣=0.67 size 12{ lline X \( ω \) rline =0 "." "67"} {} and is minimum. b is the sketch of the magnitude response.