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EE160: Digital and Analog Communication SystemsSan Jose State UniversitySpring 2003Lecture Notes # 4February 10, 20031Real signals and the trigonometric Fourier seriesAs discussed in class:x(t)=a02+∞n=1ancos 2πntT0+ bnsin 2πntT0. (1)wherean=2T0α+T0αx(t)cos 2πntT0dt,bn=2T0α+T0αx(t)sin 2πntT0dt, (2)Since xn= x −n, it follows thatxnej2πntT0+ x−ne−j2πntT0=2|xn| cos 2πntT0+ ∠xn,and the following alternative Fourier trigonometric seriesexpansion of a real and periodic signal x(t) is obtainedx(t)=x0+2∞n=1|xn| cos 2πntT0+ ∠xn. (3)22.2 Fourier transfomThe Fourier transform is an extension of the Fourier seriesto other signals — periodic and nonperiodic.If x(t) is “well behaved” (meeting the Dirichlet condi-tions), then the Fourier transformof x(t), deﬁned asX(f)=∞−∞x(t)e−j2πftdt, (4)exists.Notation: Capital letters denote signals in the frequency domain,lower-case letters denote signals in the time-domain.The original signal x(t) can be obtained from its Fouriertransform, via the inverse Fourier transform,x(t)=∞−∞X(f)ej2πftdf , (5)Notation• Operational: X(f)=F{x(t)},x(t)=F−1{X(f)}• Short-hand: x(t) ⇐⇒ X(f)3The Fourier transform of an impulseWrite a signal x(t) as the inverse Fourier transform of itsspectrumX(f), use the deﬁnition of X(f)intermsofx(t),and rearrange integrals:x(t)=∞−∞X(f)ej2πftdf=∞−∞ ∞−∞x(τ)e−j2πfτdτej2πftdf=∞−∞ ∞−∞ej2πf(t−τ)dfx(τ) dτ (6)From the deﬁnition of the unit impulse δ(t)asameasurefunction, a signal x(t) can be written asx(t)=∞−∞δ(t − τ)x(τ) dτ. (7)Then, combining Eqs. (6) and (7), we obtainδ(t − τ )=∞−∞ej2πf(t−τ)df , (8)or (redeﬁning t as t − τ),δ(t)=∞−∞ej2πftdf . (9)This shows thatδ(t) ⇐⇒ 14Example: (Rectangular pulse)Find the Fourier transform of Π(t).Solution:F{Π(t)} =1/2−1/21 · e−j2πftdt=1−j2πfe−j2πft1/2−1/2=j2πf−ejπf+ e−jπf=−2j22πfsin(πf) = sinc(f).Π(t)tfsinc(f)11/2-1/211-123-3-252.1. Fourier transforms of real symmetric signalsx(t) X(f)real, even real, evenreal, odd imaginary, oddTo see this, write the Fourier transform using Euler’s for-mula for the complex exponential:X(f)=∞−∞x(t)e−j2πftdt=∞−∞x(t)cos(2πft) dt − j∞−∞x(t)sin(2πft) dt= XR(f) − jXI(f)If the signal x(t)isreal and even symmetric about zero,then x(t)sin(2πft) is real and odd symmetric and conse-quently, XI(f)=0.Similarly, if signal x(t)isreal and odd symmetric aboutzero,thenXR(f)=0.62.2.3 Fourier transform of periodic signalsLet x(t) be a periodic signal with period T0.Thenx(t)=∞n=−∞xnej2πnT0t⇐⇒ X(f)=∞n=−∞xnδ f −nT0,(10)where the following Fourier transform pair has been used(prove this as an exercise):ej2πnT0t⇐⇒ δ f −nT0.This shows that the Fourier transform of a periodic signalconsists of a sequence of impulses weighted by the Fourierseries coeﬃcients xn.This suggests the following scheme for computing theFourier series of a periodic signal, based on properties ofthe Fourier transform:Use a “truncated signal” xT0(t) deﬁned over a period ofx(t):xT0(t)=x(t), |t|≤T0/2;0, otherwise,(11)So that x(t) can be expressed asx(t)=∞n=−∞xT0(t − nT0) (12)7Observe that∞n=−∞xT0(t − nT0)=∞n=−∞xT0(t) δ(t − nT0)= xT0(t) ∞n=−∞δ(t − nT0) (13)Using the convolution theorem property and the Fouriertransform of a train of impulses:X(f)=XT0(f)1T0∞n=−∞δ f −nT0=1T0∞n=−∞XT0 nT0δ f −nT0(14)Comparing Eqs. (10) with (14), it is concluded thatxn=1T0XT0 nT0(15)This nice result gives a shortcut for computing the Fourierseries coeﬃcients, based on the shape of a periodic signalover an interval (α, α + T0], ∀α.8Example: (Pulse train)xT0(t)=Π(t)tfsinc(f)11/2-1/211-123-3-2x(t)tnτ/T0sinc(nτ/T0)11/2-1/21/55-51015-15-10……Make periodicSam ple atMultiples of 1/T0(τ/T0= 1/5)T0-T09READING ASSIGNMENT• Review properties of the Fourier transform. Section2.2.2 of textbook.• Example 2.2.3 on page 40 of textbook.HOMEWORK # 1• To be assigned next class, 2/13/03• Due one week later,

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