EE123 Digital Signal Processing

Transcription

1 EE123 Digital Signal Processing Lecture 2B D. T. Fourier Transform M. Lustig, EECS UC Berkeley

2 Something Fun gotenna Text messaging radio Bluetooth phone interface MURS VHF radio or 900MHz ISM 2W (VHF), 1W (900MHz) mile range encryption, mesh networking 2 for 150$ Lab 6 implements a similar approach -- (without the slick system integration) M. Lustig, EECS UC Berkeley

3 Discrete-Time LTI Systems The impulse response h[n] completely characterizes an LTI system DNA of LTI [n] LTI h[n] discrete convolution x[n] LTI y[n] =h[n] x[n] 1X y[n] = h[n m]x[m] N= 1 Sum of weighted, delayed impulse responses! M. Lustig, EECS UC Berkeley

4 BIBO Stability of LTI Systems An LTI system is BIBO stable iff h[n] is absolutely summable 1X k= 1 h[k] < 1 M. Lustig, EECS UC Berkeley

5 BIBO Stability of LTI Systems Proof: if y[n] = apple 1X h[k]x[n k] k= 1 1X k= 1 h[k] x[n k] apple B x apple B x 1 X k= 1 h[k] < 1 M. Lustig, EECS UC Berkeley

6 BIBO Stability of LTI Systems Proof: only if P 1 k= 1 h[k] = 1 suppose show that there exists bounded x[n] that gives unbounded y[n] Let: M. Lustig, EECS UC Berkeley

7 Discrete-Time Fourier Transform (DTFT) X(e j! )= x[n] = 1 Alternative 2 1X k= 1 Z X(f) = x[n] = x[k]e j!k X(e j! )e j!n d! 1X k= 1 Z x[k]e j2 fk X(f)e j2 fn df Why one is sum and the other integral? Why use one over the other?

8 Example 1: w[n] window -N N DTFT: W (e j! ) = NX e j!k k= N = e j!n 1+e j! + + e j!2n Recall: 1+p + p p M = 1 pm+1 1 p p = e j! M = 2N

9 Example 1 cont. DTFT:

10 Example 1 cont. DTFT: W (e j! ) = e j!n 1+e j! + + e j!2n = e jwn 1 ei!(2n+1) 1+e j! = e j!n e j!n e j! 1 e j! = e j!(n+ 1 2 ) e j!(n+ 1 2 ) e j! 2 e j! 2 = sin[(n )!] sin(! 2 ) j - periodic sinc e j! 2 e j! 2

11 Example 1 cont. W (e j! )= sin[(n )!] sin(! 2 ) also, Σx[n]! (2N + 1) as!! 0 from l Hôpital 2N +1 N =1, why?

12 Properties of the DTFT Periodicity: X(e j(!+2 ) )=X(e j! ) Conjugate Symmetry: X (e j! )=X(e j! ) if x[n] is real Re X(e j! ) = Re X(e j! ) Im X(e j! ) = Im X(e j! ) Big deal for: MRI, Communications, more...

13 Half Fourier Imaging in MR k-space (Raw Data) Image Complete based on conjugate symmetry Half the Scan time! Discrete Fourier transform

14 SSB Modulation Real Baseband signal has conjugate symmetric spectrum m[n] cos(! 0 n) AM modulation (DSB-SC)!! Single sideband (USB) half bandwidth!

15 SSB Amateur radio on shortwaves often use SSB modulation Example: Websdr

16 Properties of the DTFT cont. Time-Reversal x[n] $ X(e i! ) x[ n] $ X(e i! ) = X (e j! ) if x[n] 2Real If x[n] = x[-n] and x[n] is real, then: X(e j! ) = X (e j! )! X(e j! ) 2Real

17 Q: Suppose: x[n] $ X(e j! )? $ Re X(e j! ) A: Decompose x[n] to even and odd functions x[n] =x e [n]+x o [n] x e [n] := 1 2 x o [n] := 1 2 (x[n]+x[ n]) (x[n] x[ n]) x e [n]+x o [n]!re X(e j! ) + jim X(e j! )

18 Oops!

19 Properties of the DTFT cont. Time-Freq Shifting/modulation: x[n] $ X(e j! ) Good for MRI! Why x[n n d ] $ e j!n d X(e j! ) e j! 0n x[n] $ X(e j(!! o) )

20 Example 2 What is the DTFT of: 5 High Pass Filter 2N +1= e j n See 2.9 for more properties

21 Frequency Response of LTI Systems Check response to a pure frequency: e i!n 0 LTI y[n] y[n] = = 1X k= 1 0 1X k= 1 h[k]e j!(n k) h[k]e j!k C {z } 0 1 A ej!n 0 0 H e j!!=! 0

22 Frequency Response of LTI Systems Check response to a pure frequency: e i!n 0 LTI y[n] H(e j! )=DTFT{h[n]} y[n] =H e j!!=! 0 e j! 0n Output is the same pure frequency, scaled and phase-shifted! e j! 0n is an eigen function of LTI systems Recall eigen vectors satisfy: A =

23 Example 3 Frequency response of a causal moving average filter y[n] = x[n M]+ + x[n] M +1 Q: What type of filter is it? A: Low-Pass 1 M +1 0 M h[n] = 1 M +1 w[n n M 2 ]

24 Example 3 Cont. Frequency response of a causal moving average filter h[n] = 1 M +1 w[n M 2 ] Same as example 1, only: Shifted by N, divided by M+1, M=2N H(e j! )= e j! M 2 M +1 sin ( M )! sin(! 2 )

25 Example 3 Cont. Frequency response of a causal moving average filter H(e j! )= e j! M 2 M +1 sin ( M 2 + 1)! sin(! 2 ) Not a sinc! 1 H(e j! ) 3 \H(e j! )

26 Example 4: Impulse Response of an Ideal Low-Pass Filter H 1 LP (e j! )! c! c h LP [n] = 1 2 = 1 2 Z Z!c H LP (e jw )e j!n d!! c e j!n d!

27 Example 4 Impulse Response of an Ideal Low-Pass Filter Z h LP [n] = 1 2 = 1 2 Z!c H LP (e jw )e j!n d!! c e j!n d! = 1 2 jn ej!n! c! c =2j sin(w c n) = sin(w cn) n

28 Example 4 Impulse Response of an Ideal Low-Pass Filter h LP [n] = sin(w cn) n sampled sinc h LP [n] Non causal! Truncate and shift right to make causal

29 Example 4 Impulse Response of an Ideal Low-Pass Filter Non causal! Truncate and shift right to make causal How does it changes the frequency response? Truncation: h LP [n] =w N [n] h LP [n] property 2.9.7: H LP (e j! )= 1 2 Z H LP (e j )W (e j(! ) )d Periodic convolution

30 Example 4 We get smearing of the frequency response We get rippling ω real imag