MASSACHUSETTS INSTITUTE OF TECHNOLOGYDepartment of Electrical Engineering and Computer Science6.003: Signals and Systems — Spring 2004Tutorial 9Monday, April 12 and Tuesday, April 13, 2004Announcements• 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 willcover material in Chapters 1–7 of O&W through Section 7.4, Lectures and Recitations through April2, Problem Sets #1–6, and that part of Problem Set #7 involving problems from Chapter 7.• The TAs will jointly hold office hours from 2–8 p.m. on Wednesday, April 14 and again from 10 a.m.–3p.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 quizreview 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– 2π factors• Sampling Pitfalls– Impulses in the frequency domain– Do we really have to sample at the Nyquist rate?1911 Fourier Transform Pitfalls1.1 2π factorsWe’ve seen 2π and 1/2π factors appear all over the place in Fourier transform formulae and got headachestrying to remember them. They all stem from the fact that we use angular frequency ω instead of cyclicfrequency, f, where ω =2πf. We view angular frequency as being more “natural,” but many practicalproblems use cyclic frequency, so we need to remember when to add in factors of 2π. With this convention,we saw that the synthesis and analysis equations for the CT and DT Fourier transforms become:x(t)=12π+∞−∞X(jω)ejωtdω (CT synthesis, inverse CTFT)X(jω)=+∞−∞x(t)e−jωtdt (CT analysis, CTFT)x[n]=12π2πX(ejω)ejωndω (DT synthesis, inverse DTFT)X(ejω)=+∞n=−∞x[n]e−jωn(DT analysis, DTFT)The 2π factor is manifested in the following FT pairs and properties:• Value of a signal at zero timex(0) =12π+∞−∞X(jω)dωx[0] =12π2πX(ejω)dω• Constant signalx(t)=1F←→ X(jω)=2πδ(ω)x[n]=1F←→ X(ejω)=2π+∞l=−∞δ(ω − 2πl)• Complex exponentialsx(t)=ejω0tF←→ X(jω)=2πδ(ω − ω0)x[n]=ejω0nF←→ X(ejω)=2π+∞l=−∞δ(ω − ω0− 2πl)• Multiplication propertyr(t)=s(t)p(t)F←→ R(jω)=12π+∞−∞S(jθ)P (j(ω − θ)) dθ =12π{S(jθ) ∗ P (jθ)}r[n]=s[n]p[n]F←→ R(jω)=12π2πS(ejθ)P (ej(ω−θ))dθ =12π S(ejθ) P (ejθ)1922 Sampling Pitfalls2.1 Impulses in the frequency domainWe need to be careful when there are impulses in the frequency domain. Let’s consider the simplest casewhere Xc(jω)=Aδ(ω)(soxc(t)=A/(2π)) and we sample xc(t) with period T . Recall that we label impulsesby the area obtained when they are integrated. I shall follow the convention of surrounding that value witha pair of parentheses to remind us of this. To aid us in the process, let’s consider the impulse Aδ(ω)tobethe limit as ∆ → 0 of a rectangular pulse with width ∆ and height A/∆. For our purposes here, it does notmatter how this pulse is centered:Xc(jω)ω∆2−∆2A∆0lim∆→0Xc(jω)ω(A)0Thus, Xp(jω) has the height A/(T ∆), retains the width ∆, and is replicated; the area is now A/(T ∆)·∆=A/T :Xp(jω)ω∆2−∆2AT ∆02πT−2πT......lim∆→0Xp(jω)ωAT02πT−2πT......Finally, Xd(ejΩ) retains the height A/(T ∆), but its width is T ∆; the area is now A/(T ∆) · T ∆=A:193XdejΩΩT ∆2−T ∆2AT ∆02π−2π......lim∆→0XdejΩΩ(A)02π−2π......But wait a minute... didn’t we say earlier that C/D conversion scaled down the FT by T ? The impulseapparently breaks our rules. But it actually follows them, because we label a impulse by its area, NOT byits height, which is infinite. 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 aCT signal is converted into a DT sample sequence.1942.2 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 asignal to recover it. However, there exist signals where sampling at exactly twice that frequency is sufficient.Let’s look at two related examples, which will not only illustrate this point, but will also highlight the factthat we cannot ignore phase when we analyze signals. First, let’s try to sample xc(t)=sin(ω0t)atωs=2ω0.xc(t)t0T 2T−T−2T1xp(t)t0T 2T−T−2Txr(t)t0T 2T−T−2TXc(jω)ωπj−πjω0−ω00Xp(jω)ωπjTπjTπjTπjT−πjT−πjT−πjT−πjTω03ω0−ω0−3ω0 0......Xr(jω)ωω0−ω0 0As we can see in both the time and frequency domains, the interpolation is zero; the two impulses in thesine have opposing signs and exactly cancel each other out by aliasing. Since we did not satisfy Nyquist, weshould not be surprised that this happened.Now let’s consider xc(t)=cos(ω0t):xc(t)t0T 2T−T−2T1xp(t)t0T 2T−T−2T(1)(−1)(1)(−1)(1)xr(t)t0T 2T−T−2T1Xc(jω)ω(π)(π)ω0−ω0 0Xp(jω)ω2πT2πT2πT2πTω03ω0−ω0−3ω0 0......Xr(jω)ω(π)(π)ω0−ω0 0195Something subtle happened here. As usual, we performed bandlimited interpolation by lowpass filteringwith cutoff frequency ωs/2, which is ω0in this case. But what happens exactly at that frequency: do wefilter out the whole thing, or do with keep the whole impulse (area 2π)? Our figures for filters were alwaysambiguous as to what the value of the filter was at the cutoff frequency. We never really cared about thissort of boundary case before, but here it is important. The time domain picture resolves this issue: it showsthat we end up with the cosine again. Since the frequency domain and time domain pictures must agree andproduce the same result, this means that we compromise and keep half of each impulse on either side! Thisexample confirms that:The Value of Signals at Discontinuities:Whenever Fourier analysis is used, we should consider the value of a signalat discontinuities (in both time and frequency) to be the mean of its limitsfrom both sides. However, we can usually ignore this fact.196Problem 9.1(From the 6.003 Fall 2002 Quiz 2)Answer each of
View Full Document
Unlocking...