6.02 Fall 2012 Lecture #16 • DTFT vs DTFS • Modulation/Demodulation • Frequency Division Multiplexing (FDM) 6.02 Fall 2012 Lecture 16 Slide #1
Fast Fourier Transform (FFT) to compute samples of the DTFT for signals of finite dura tion For an x[n] that is zero outside of the interval [0,L-1], choose P ≥ L (with P preferably a power of 2; we’ll assume that it’s at least a multiple of 2, i.e., even): P − 1 ( P /2) − 1 1 X( Ω k ) = ∑ x [ m ] e − j Ω k m , ∑ X ( Ω k ) e j Ω k n x [ n ] = P m = 0 k =− P /2 where Ω k = k(2 π /P), and k ranges from –P/2 to (P/2)–1, or over any P successive integers. Simpler notation: X( Ω k ) = X k 6.02 Fall 2012 Lecture 16 Slide #2
Where do the Ω Ω k live? e.g., for P=6 (even ) Ω � 3 Ω � 2 Ω 0 Ω 1 Ω 2 Ω 3 Ω � 1 0 – � � exp ( j Ω 2 ) exp ( j Ω 1 ) j exp ( j Ω 3 ) exp ( j Ω 0 ) . –1 1 = exp ( j Ω � 3 ) –j exp ( j Ω � 1 ) exp ( j Ω � 2 ) 6.02 Fall 2012 Lecture 16 Slide #3
Fast Fourier Transform (FFT) to compute samples of the DTFT for signals of finite dura tion For an x[n] that is zero outside of the interval [0,L-1], choose P ≥ L (with P preferably a power of 2; we’ll assume that it’s at least a multiple of 2, i.e., even): P − 1 ( P /2) − 1 1 X( Ω k ) = ∑ x [ m ] e − j Ω k m , ∑ X ( Ω k ) e j Ω k n x [ n ] = P m = 0 k =− P /2 where Ω k = k(2 π /P), and k ranges from –P/2 to (P/2)–1, or over any P successive integers. Simpler notation: X( Ω k ) = X k Note that X( Ω ) =X( Ω -P/2 ). ) = X( � ) = X(– � P/2 The above formulas have essentially the same structure, and are both efficiently computed, with Plog(P) computations, by the FFT. 6.02 Fall 2012 Lecture 16 Slide #4
Some further details, assuming real x[n] P − 1 X( Ω ) = ∑ x [ m ] e − j Ω k m k m = 0 • X(0) = sum of x[m] = real • X( � ) = X(– � ) = alternating sum of x[m] = real • In general P-2 other complex values, but X(– Ω k ) = X*( Ω k ) • So: total of P numbers to be determined, given P values of x[m] ( P /2) − 1 1 ∑ X ( Ω k ) e j Ω k n x [ n ] = P k =− P /2 • Evaluating this eqn. for n in [0,P-1] recovers the original x[n] in this interval • Evaluating it for n outside this interval results in periodic replication of the values in [0,P-1], producing a periodic signal x [n] • So this eqn. is also called a DT Fourier Series (DTFS) for the periodic signal x [n]. Notation: A k =X( Ω k )/P=X k /P, Fourier c oefficient . 6.02 Fall 2012 Lecture 16 Slide #5
Why the periodicity of x x [n] is irrelevant in many applications x[n] y[n] h[.] Suppose x[n] is nonzero only over the time interval [0 , n x ], and h[n] is nonzero only over the time interval [0 , n h ] . In what time interval can the non-zero values of y[n] be guaranteed to lie? The interval [0 , n x + n h ] . Since all the action we are interested in is confined to this interval, choose P – 1 ≥ n x + n h . It’s now irrelevant what happens outside [0,P–1]. So we can use the FFT to go back and forth between samples of X( Ω), Η(Ω), Y( Ω) in the frequency domain and time-domain behavior in [0,P-1]. 6.02 Fall 2012 Lecture 16 Slide #6
Back to Modulation/Demodulation • You have: a signal x[n] at baseband (i.e., centered around 0 frequency) • You want: the same signal, but centered around some specific frequency Ω c • Modulation: convert from baseband up to Ω c , to get t[n] • Demodulation: convert from Ω c down to baseband Re(X k ) Re(T k ) modulation � Ω m + Ω m � Ω c + Ω c Im(X k ) Im(T k ) demodulation Signal centered at Ω c Signal centered at 0 6.02 Fall 2012 Lecture 16 Slide #7
Modulation by Heterodyning or Amplitude Modulation (AM) × x[n] t[n] Re(X ) k A cos( Ω c n) � Ω m + Ω m i.e., just replicate baseband Im(X k ) signal at ± Ω c , and scale B by ½. Re(T ) k A/2 � Ω c + Ω c Im(T k ) B/2 To get this nice picture, the baseband signal needs to be band-limited to some range of frequencies [– Ω m, Ω m ], where Ω m ≤ Ω c 6.02 Fall 2012 Lecture 16 Slide #8
–1000 Hz 1000 Hz 0 Hz Not great band-limiting, but maybe we can get away with it! 6.02 Fall 2012 Lecture 16 Slide #9
At the Receiver: Demodulation • In principle, this is (as easy as) modulation again: If the received signal is r[n] = x[n]cos( Ω c n) = t[n], (no distortion or noise) then simply compute d[n] = r[n]cos( Ω c n) = x[n]cos 2 ( Ω c n) = 0.5 {x[n] + x[n]cos(2 Ω c n)} If there is distortion (i.e., r[n] ≠ t[n]), then write y[n] instead of x[n] (and hope that in the noise-free case y[.] is related to x[.] by an approximately LTI relationship!) • What does the spectrum of d[n], i.e., D( Ω ), look like? • What constraint on the bandwidth of x[n] is needed for perfect recovery of x[n]? 6.02 Fall 2012 Lecture 16 Slide #10
Demodulation Frequency Diagra m Re(T ) k A/2 � Ω c + Ω c R( Ω )=T( Ω ) Im(T k ) B/2 Note combining of signals around 0 results in doubling of amplitude Re(D k ) A/2 � Ω m Ω m � 2 Ω c +2 Ω c D( Ω ) Im(D k ) B/2 What we want 6.02 Fall 2012 Lecture 16 Slide #11
Demodulation + LPF × r[n] d[n] x[n] LPF Cutoff @ ± Ω c Gain = 2 cos( Ω c n) 6.02 Fall 2012 Lecture 16 Slide #12
Phase Error In Demodulation When the receiver oscillator is out of phase with the transmitter: d [ n ] = r [ n ] ⋅ cos( Ω c n − ϕ ) = x [ n ] ⋅ cos( Ω c n ) ⋅ cos( Ω c n − ϕ ) But cos( Ω c n ).cos( Ω c n − ϕ ) = 0.5{cos( ϕ ) + cos(2 Ω c n − ϕ )} It follows that the demodulated output, after the LPF of gain 2, is y [ n ] = x [ n ].cos( ϕ ) So a phase error of φ results in amplitude scaling by cos( φ ). Note: in the extreme case where φ=̟/2 , we are demodulating by a sine rather than a cosine, and we get y[n]=0 . 6.02 Fall 2012 Lecture 16 Slide #13
Demodulation with sin( Ω Ω c n) Re(T k ) A/2 � Ω c + Ω c R( Ω ) Im(T k ) B/2 D( Ω ) +2 Ω c –2 Ω c + 6.02 Fall 2012 Lecture 16 Slide #14
… produces Note combining of signals around 0 results in cancellation! 6.02 Fall 2012 Lecture 16 Slide #15
Channel Delay Time delay of D samples d[n] t[n] × × x[n] y[n] D LPF Cutoff @ ±k in Gain = 2 cos( Ω c n) cos( Ω c n) Very similar math to the previous “phase error” case: d [ n ] = t [ n − D ] ⋅ cos( Ω c n ) = x [ n − D ] ⋅ cos[ Ω c ( n − D )] ⋅ cos( Ω c n ) Passing this through the LPF: Looks like a phase error of Ω c D y [ n ] = x [ n − D ] ⋅ cos( Ω c D ) D ) If Ω c D is an odd multiple of � /2, then y[n]=0 !! 6.02 Fall 2012 Lecture 16 Slide #16
Fixing Phase Problems in the Receiver So phase errors and channel delay both result in a scaling of the output amplitude, where the magnitude of the scaling can’t necessarily be determined at system design time: • channel delay varies on mobile devices • phase difference between transmitter and receiver is arbitrary One solution: quadrature demodulation × I[n] = x[n-D]·cos( θ ) LPF From θ = φ - Ω c D cos( Ω c n) channel × Q[n] = x[n-D]·sin( θ ) LPF sin( Ω c n) 6.02 Fall 2012 Lecture 16 Slide #17
Quadrature Demodulation jQ If we let x[n-D]sin( θ ) w [ n ] = I [ n ] + jQ [ n ] θ I then x[n-D]cos( θ ) I [ n ] 2 + Q [ n ] 2 = w [ n ] cos 2 θ + sin 2 θ = | x [ n − D ]| Constellation diagrams: = | x [ n − D ]| x[n] = { 0, 1 } Q Q OK for recovering x[n] if it I I never goes negative, as in on-off keying transmitter receiver 6.02 Fall 2012 Lecture 16 Slide #18
Dealing With Phase Ambiguity in Bipolar Modulation In bipolar modulation (x[n]=±1), also called Binary Q Phase Shift Keying (BPSK) since the modulated carrier changes phase by � /2 when x[n] switches I levels, the received constellation will be rotated with respect to the transmitter’s constellation. Which phase corresponds to which bit? Different fixes: 1. Send an agreed-on sign-definite preamble 2. Transmit differentially encoded bits, e.g., transmit a “1” by stepping the phase by � , transmit a “0” by not changing the phase 6.02 Fall 2012 Lecture 16 Slide #19
QPSK Modulation We can use the quadrature scheme at the transmitter too: × Samples from first bit stream Q I[n] (-1,1) (1,1) cos( Ω c n) + t[n] I sin( Ω c n) (-1,-1) (1,-1) Q[n] × Samples from second bit stream 6.02 Fall 2012 Lecture 16 Slide #20
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