Signals and Systems Fall 2003 Lecture #10 7 October 2003 Examples of the dt fourier Transform 2. Properties of the dt Fourier transform 3. The Convolution Property and its Implications and Uses
Signals and Systems Fall 2003 Lecture #10 7 October 2003 1. Examples of the DT Fourier Transform 2. Properties of the DT Fourier Transform 3. The Convolution Property and its Implications and Uses
DT Fourier transform pair x]→X(1) X(-)=∑mln Analysis equation FT 2丌 X( jEjUn Synthesis equation Inverse ft
DT Fourier Trans form Pair – Analysis Equation – F T – Synthesis Equation – I n v e r s e F T
Convergence Issues Synthesis Equation: None, since integrating over a finite interval Analysis equation: Need conditions analogous to CTFT, e.g ∑ Finite energy or ∑aml<∞- Absolutely summable
Convergence Issues Synthesis Equation: None, since integrating over a finite interval Analysis Equation: Need conditions analogous to CTFT, e.g. — Absolutely summable — Finite energy
Examples Parallel with the ct examples in lecture #8 X()=∑6mle-ln 2)rn=dn-no -shifted unit sample X()=∑6n 0e4 Same amplitude(1) as above, but with a linear phase -wno
Examples Parallel with the CT examples in Lecture #8
More examples 3)m=anan,a< 1- Exponentially decaying function ∑(a Infinite sum formula <1 ae (1-a w)+jasin e X(eu) 2a cos+a =0:X( a<0 2a+ X(eu) √1+2a+a 0
More Examples Infinite sum formula
Still more 4)DT Rectangular pulse (Drawn for N=2) ●●●●b●●●●●●鲁●●●●●● ●鲁●●●●●●●●●●●●●●●鲁 N10N x(1)=∑c=∑(c)n=-m(M+2)=x(2(2) sin(w 2 N 52N+1
Still More 4) DT Rectangular pulse (Drawn for N1 = 2)
X(el 2 wo w 2 sinw a00| 2丌 W 2丌 x[n] T ●。。·°d , ●QQ●● ● 0
5)
DTFTS of Sums of Complex exponentials Recall ct result: r(t)=c1u←→X()=26(u-) What about dt won X(e1)=? a) We expect an impulse(of area 2T ato=o b) But X(el@) must be periodic with period 2T X(0)=2x∑6(a-o-27 m Note: The integration in the synthesis equation is over 2T period only need X(el) in one 2T period. Thus / 2丌 2丌 ∑6(a-40-27m)lnod m=-0 X(eju)
DTFTs of Sums of Complex Exponentials Recall CT result: What about DT: Note: The integration in the synthesis equation is over 2π period, only need X(ejω) in one 2π period. Thus, a) We expect an impulse (of area 2π) at ω = ωo b) But X(ejω) must be periodic with period 2π In fact
dtFt of periodic Signals an= n+N Kwon 2丌 DTFS 0 k= synthesis eq From the last page: e Kwon→2x∑6(-ko-2xm) Linearity X()=∑ ak|27 ∑ S(-kwo-2Tm) of DTFT k= m=- 2x∑ak6(-k)=2x∑a 2丌k k=-0o k=-00
DTFT of Periodic Signals Linearity of DTFT DTFS synthesis eq
Example #1: DT Sine function n= sinon=oe Jc0_二p-Jc x)=∑0-0-2xm)-x∑0(+-=2xm) X(e0) 2-0 2丌 2+O0 21
Example #1: DT sine function
