EEO 401 Digital Signal Processing Prof. Mark Fowler
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1 EEO 401 Digital Signal Processing Pro. Mark Fowler Note Set #10 Fourier Analysis or DT Signals eading Assignment: Sect. 4.2 & 4.4 o Proakis & Manolakis Much o Ch. 4 should be review so you are expected to read it to reresh your memory. We ll ocus on a ew topics you might not have seen beore 1/18
2 Fourier Series or DT Periodic Signals (DTFS) The idea here is the same as FS or CT periodic signals just a ew details are dierent. Let x[n] be a periodic signal with period o N: x[n+n] = x[n] We deine the undamental requency as 2π/N in units o rad/sample n exactly the same way as or CT FS we can decompose the periodic signal x[n] into a sum o complex sinusoids with requencies that are integer multiples o the undamental The key dierence here is that or DT requencies we can limit our selves to the range 0 to 2π. So the requencies o interest are: 2 k / N, k 0,1,2,, N 1 FS or DT Signal N 1 [ ] k j2 kn/ N, 0,1,2,, 1 k 0 xn ce n N Though extendable to other n via periodicity 2/18
3 Similar to how we ind the FS coeicients or CT the DTFS coeicients are N 1 1 j2 kn/ N ck x[ n] e, k 0,1,2,, N 1 N n0 t is easy to veriy that c k+n = c k. n other words, the DTFS coeicients themselves have a periodic nature in requency. DTFS vs DTFT xn [ ] N 1 ce k k0 j2 kn/ N 1 jn xn [ ] ( ) e d 2 2 c k 1 N N 1 n0 x[ n] e j2 kn/ N j n ( ) x[ n] e n ninite summation so issues o convergence & truncation 3/18
4 CT Signals DT Signals Aperiodic Signals Periodic Signals Periodicity w/ period α in one domain automatically implies discretization with spacing o 1/α in the other domain and vice versa! 4/18
5 Convergence o DTFT The equation or inding the DTFT o a signal is j n ( ) x[ n] e n t involves an ininite sum so there are issues with its convergence. ecall our study o OC or ZT and the link between ZT & DTFT! Deine: j n ( ) x[ n] e N Mathematicians have many dierent ways to characterize convergence the one we consider irst is called Uniorm Convergence. We say that Like we saw in our studies o ZT a suicient condition is absolute summability o x[n]. The DTFT converges uniormly i n N nn N ( ) converges uniormly to xn [ ] ( ) when limsup ( ) N ( ) 0 N A more general concept o max This roughly says that the largest dierence between the two keeps getting smaller However in DSP the class o signals o interest are energy signals ( square summable ) and not all o those are absolutely summable 5/18
6 To allow us to deal with energy signals we have to relax our expectation on the type o convergence we ll accept. We consider Mean Square Convergence deined as 2 lim ( ) N ( ) d 0 N This says that were are looking or the area o squared error to go to zero. However, the area can go to zero even though the largest dierence between the two does not go to zero. Example DTFT o sinc unction Peak o ripple stays same Area o squared error goes to zero 6/18
7 elationship o DTFT to ZT We ve already discussed this but we ll see it again here. z n ( z) x[ n] z, OC: z n 2 1 the OC contains the UC then we can replace z by its values on the UC: z j e z n z j jn jn ( ) [ ] ( ) [ ] ( ) [ ] z xnz e xne xne n n n Viewed as a unction o ω yields the DTFT So the OC contains the UC then we get the DTFT by evaluating the ZT on the UC But there are some signals w/ DTFT that do not come rom a ZT. One o those is the sinc unction t has a DTFT but does not have a ZT! 7/18
8 8/18
9 DTFT o Signals w/ Poles on UC Even though these don t satisy the UC in OC criteria i we allow the DTFT to contain delta unctions we can get DTFT results or such signals. z 1 Consider the signal x[n] = u[n] whose ZT is ( z) 1 1 z t has a pole on the UC so clearly the OC does not contain the UC. But in Signals & Systems we said that its DTFT is ( ) ( 2 k) 1 1 j e k Evaluate ZT on UC except at poles Accounts or pole at z = 1, which is at ω = 2πk 9/18
10 Symmetry Properties o the DTFT Due to similarities between all the Fourier tools these hold or the other Fourier tools as well n this section we will allow the signal to be complex valued in general (Obviously, the DTFT is also in general complex valued) xn [ ] x[ n] jx[ n] ( ) ( ) j ( ) jn ( ) x[ n] e x [ n] jx [ n] e n n jn n n x [ n] jx [ n] cos( n) jsin( n) x [ n]cos( n) x [ n]sin( n) j x [ n]sin( n) x [ n]cos( n) 10/18
11 Thus. ( ) ( ) j ( ) with. ( ) x [ n ]cos( n ) x [ n ]sin( n ) n ( ) x [ n ]sin( n ) x [ n ]cos( n ) n Similarly. xn [ ] x[ n] jx[ n] with. 1 [ ] ( )cos( ) ( )sin( ) 2 x n n n d 1 [ ] ( )sin( ) ( )cos( ) 2 x n n n d We can now use these general results to explore several special cases! 11/18
12 eal Signals x[ n] 0 x[ n] x[ n] ( ) x [ n ]cos( n ) x [ n ]sin( n ) n ( ) x [ n ]sin( n ) x [ n ]cos( n ) n ( ) x[ n]cos( n) n ( ) x[ n]sin( n) n Even Function because cos is even ( ) ( ) Odd Function because sine is odd ( ) ( ) ( ) ( ) * ( ) ( ) ( ) ( ) Hermitian Symmetry Magn is Even Phase is Odd Proo: ( ) ( ) j ( ) ( ) j( ) * ( ) 12/18
13 eal Signals (cont.) 1 [ ] [ ] ( )cos( ) ( )sin( ) 2 xn x n n n d Even = Even x Even Even = Odd x Odd So only need to integrate over hal the range then x2 1 [ ] ( )cos( ) ( )sin( ) 0 xn n n d 13/18
14 eal & Even Signals From the act that x[n] is real we recall that ( ) x[ n]cos( n) n ( ) x[ n]sin( n) n Even = Even x Even Odd = Even x Odd ( ) x[0] 2 x[ n]cos( n) n1 ( ) 0 x x n n n1 ( ) ( ) [0] 2 [ ]cos( ) Even & eal From the act that x[n] is real we recall that 1 [ ] ( )cos( ) ( )sin( ) 0 xn n n d 1 [ ] ( )cos( ) 0 xn nd 14/18
15 eal & Odd Signals Using a similar argument ( ) 0 ( ) 2 x[ n]sin( n) n1 j j x n n n1 ( ) ( ) 2 [ ]sin( ) Odd & maginary From the act that x[n] is real we recall that 1 [ ] ( )cos( ) ( )sin( ) 0 xn n n d 1 [ ] ( )sin( ) 0 xn nd 15/18
16 Purely maginary Signals ( ) x [ n ]cos( n ) x [ n ]sin( n ) n ( ) x [ n ]sin( n ) x [ n ]cos( n ) n ( ) x[ n]sin( n) n x [ n] 0 x [ n] x[ n] ( ) x[ n]cos( n) n Odd Function because sin is odd ( ) ( ) Even Function because cos is even ( ) ( ) 1 [ ] ( )sin( ) ( )cos( ) 0 xn n n d 16/18
17 maginary & Even Signals ( ) 0 ( ) x[0] 2 x[ n]cos( n) n1 ( ) ( ) [0] 2 [ ]cos( ) j jx x n n n1 Even & maginary 1 [ ] ( )cos( ) 0 xn nd maginary & Odd Signals ( ) 0 xn n ( ) 2 [ ]sin( ) n1 x n n n1 ( ) ( ) 2 [ ]sin( ) Odd & eal 1 [ ] ( )sin( ) 0 xn nd 17/18
18 Also see Table 4.4 in the Textbook 18/18
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