MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6 003 Signals and Systems Spring 2004 Tutorial 9 Monday April 12 and Tuesday April 13 2004 Announcements There is no problem set due this week Quiz 2 will be held on Thursday April 15 7 30 9 30 p m in Walker Memorial The quiz will cover material in Chapters 1 7 of O W through Section 7 4 Lectures and Recitations through April 2 Problem Sets 1 6 and that part of Problem Set 7 involving problems from Chapter 7 The TAs will jointly hold o ce hours from 2 8 p m on Wednesday April 14 and again from 10 a m 3 p m on Thursday April 15 A schedule is posted on the 6 003 web site A quiz review package is available on the 6 003 web site TAs will hold two identical optional quiz review sessions on Monday April 12 and Tuesday April 13 7 30 9 30 p m in 34 101 Because of the Patriot s Day holiday next week there will be no tutorials next Monday and Tuesday and no lecture on Tuesday Today s Agenda Fourier Transform Pitfalls 2 factors Sampling Pitfalls Impulses in the frequency domain Do we really have to sample at the Nyquist rate 191 1 Fourier Transform Pitfalls 1 1 2 factors We ve seen 2 and 1 2 factors appear all over the place in Fourier transform formulae and got headaches trying to remember them They all stem from the fact that we use angular frequency instead of cyclic frequency f where 2 f We view angular frequency as being more natural but many practical problems use cyclic frequency so we need to remember when to add in factors of 2 With this convention we saw that the synthesis and analysis equations for the CT and DT Fourier transforms become 1 X j ej t d CT synthesis inverse CTFT 2 x t e j t dt CT analysis CTFT 1 X ej ej n d DT synthesis inverse DTFT 2 2 x n e j n DT analysis DTFT x t X j x n X ej n The 2 factor is manifested in the following FT pairs and properties Value of a signal at zero time x 0 x 0 1 X j d 2 1 X ej d 2 2 Constant signal F x t 1 F x n 1 X j 2 X ej 2 2 l l Complex exponentials x t ej 0 t x n ej 0 n F X j 2 0 F j X e 2 0 2 l l Multiplication property F r t s t p t r n s n p n F 1 1 S j P j S j P j d 2 2 1 1 S ej P ej R j S ej P ej d 2 2 2 R j 192 2 Sampling Pitfalls 2 1 Impulses in the frequency domain We need to be careful when there are impulses in the frequency domain Let s consider the simplest case where Xc j A so xc t A 2 and we sample xc t with period T Recall that we label impulses by the area obtained when they are integrated I shall follow the convention of surrounding that value with a pair of parentheses to remind us of this To aid us in the process let s consider the impulse A to be the limit as 0 of a rectangular pulse with width and height A For our purposes here it does not matter how this pulse is centered Xc j lim 0 Xc j A 2 0 A 2 0 Thus Xp j has the height A T retains the width and is replicated the area is now A T A T Xp j A T 2 T 2 0 2 2 T lim 0 Xp j A T 2 T 0 2 T Finally Xd ej retains the height A T but its width is T the area is now A T T A 193 Xd ej A T 2 T2 0 T 2 2 lim 0 Xd ej A 2 0 2 But wait a minute didn t we say earlier that C D conversion scaled down the FT by T The impulse apparently breaks our rules But it actually follows them because we label a impulse by its area NOT by its height which is in nite Since C D conversion preserves the Fourier area Area Invariance of Impulses in the Frequency Domain The labeling of an impulse in the frequency domain does not change when a CT signal is converted into a DT sample sequence 194 2 2 Do we really have to sample at the Nyquist rate It was said earlier that we must sample at strictly greater than twice the highest frequency present in a signal to recover it However there exist signals where sampling at exactly twice that frequency is su cient Let s look at two related examples which will not only illustrate this point but will also highlight the fact that we cannot ignore phase when we analyze signals First let s try to sample xc t sin 0 t at s 2 0 xp t xc t xr t 1 2T T 0 T t 2T T 2T Xc j j 0 0 0 j jT jT t T 0 Xp j 2T T 2T jT jT t 2T Xr j 3 0 T 0 jT 0 jT 0 0 jT 3 0 jT 0 0 0 As we can see in both the time and frequency domains the interpolation is zero the two impulses in the sine have opposing signs and exactly cancel each other out by aliasing Since we did not satisfy Nyquist we should not be surprised that this happened Now let s consider xc t cos 0 t xp t 1 xc t 1 1 2T T 0 T t 2T T 2T Xc j 0 2 0 0 T 3 0 0 2T T 2T 0 T t 2T 1 2 Xp j 2 T 1 t T 0 1 xr t 1 0 195 Xr j 2 T T 0 3 0 0 0 0 Something subtle happened here As usual we performed bandlimited interpolation by lowpass ltering with cuto frequency s 2 which is 0 in this case But what happens exactly at that frequency do we lter out the whole thing or do with keep the whole impulse area 2 Our gures for lters were always ambiguous as to what the value of the lter was at the cuto frequency We never really cared about this sort of boundary case before but here it is important The time domain picture resolves this issue it shows that we end up with the cosine again Since the frequency domain and time domain pictures must agree and produce the same result this means that we compromise and keep half of each impulse on either side This example con rms that The Value of Signals at Discontinuities Whenever Fourier analysis is used we …
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