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Unformatted text preview: E13279 ‘  Winter 20064007 
. Professor Cox '  Solution to Finai P1.(15 pts) For the Signal x(t) shown belOw, We use narrowband PM modulator and then  a frequency multiplier to modulate the signal. Wi'ite an expression for the transmitted
signal y(t) and Y(f). (hint: make appropriate approximation for xc (1‘)) 3:0) Ac simmer) xc(t) V  Frequengyl 3/3ult1pher ‘ ya) Solution: I(5 pts) xc (t) 2 Ac .005 wet — 7: / 100x(t) sin act 2 Ac cos(a)ct + it /100x(t))7
£6 pts) 510) = Ac cos(l 0(3th + 72x(t))‘= —c(1‘) cos(l 00606!) Where c(t) is shown below. inf)amag—1001;)+5(f+100;;))* icnaf—n/zgy
. (4pts)  m ' w “M”
=1/22cn6(f—n/1‘3—IOOjZ)+1/2.ché‘(f—n/1}, +100fc) n=—oo  n=—no
Where, 0, 74 even _ n_—1
c" " (1)91) 2, nadd
n71 . P2.(1‘5 pts) We want to transmit signal x (t) , which is a baseband signal with bandwidth I
W and power SI, using SSB modulation. ‘ ‘. vttl 3’0) 7
x60) 1: ——> —> HD(f) —* yo“)
GU)
2cos(a)ct) I
NO The receiver block diagram is shown above. GC f ) = elf—f“l f > 0 , H1; (f) is an 2 .
ideal (unity gain, rectangularshape) bandpass ﬁlter with bandwidth Br , and HD (f) is an
ideal (unity gaim rectangular shape) lowpass ﬁlter with bandwidth W. Find the ratio between (%)D for USSB and LSSB modulation. Solution:
(i) USSB
HR(f) 1
(2.5 pts)
fc
<————>
W Jgc (t) = % [x(t) cos (act —— ﬁt) sin act] 7' v0): Ac ' A  .
2 [x(t) cos met — x(t)su1a)ct]+ 111000st — n“T (I) sm wet ’
' A A =[% 170‘) coswct 4% ni(t)]cosa)ct —[ 26 x(t)+ nq (1)]sinwct yam = x0) + (0] A2 no _ WNO
(3.5 pts) SD =Tsx,ND =4£Gni(f)df=2 2 A3 .
H _ 73* _A38x 1
N 0m No(1—e'W) 4N0W1_—efW e‘fdf =No(l 49W) (ii) LSSB (2.5 pts) H
W i xc(t)= 2 [x(t) cosrwct + 32(1) sin war] v_(t) = 1—:9—[x(r) cos wct + 120‘) sin war] + n1.(t) cos coat — mg (I) simmer = x(t) (:03 war + ni (1‘)] cos (act — I} 3—4; 550‘) + nq (0] sin wet yDa) = [i x0) + m] 2
A2 on WN ‘
(3.513315!) = 4:5pr JGni(f)dfm2 e‘fdf'mNOG—e'w)
_ —a0 0 2
is [g] _ 4r._Afo 1.
N pm Noﬂ—e'w 4N0W1'—e‘W (3 pts) Therefore the ratio is. one.  _  ’ EE279 Final Exam Solutions Winter 2006 w 07
Prof. Donald Cox Solutions to problem 3 in EE279 ﬁnal 3. a) (5 pts.) The output at®is all Gaussian noise.
Explanation: Because of the scrambler, the lowest frequency in the bit stream is rb /16 = 62.5 kHz. Thus, the signal part at ® has no power in the range of
fc :t 62.5 kHz. The BPFH has a total bandwidth of 90 kHz or fa i45 kHz so the output of BPFH does not contain any signal part, only the AWGN ﬁltered to a
bandwidth of 90 kHz. 3. b) (5 pts.) The noise power N H at @is the total power that is given as 5 ><10’7 watts
where N H % N OBH with N 0 being the power spectral density of the noise. Thus, N0 = 5.56x10“12 watts/Hz
Then the noise power at®is N0 018 MHz =1><10C4 watts = N R The total power at® is 2x10"4 watts =SRRF +NRRF so SW =10"4 watts and SRRF =1:> OdB NRRF I 3. c) (6 pts.) The critical factor for part c) is to check whether the into the limiter is above threshold. Points were taken off for not checking for threshold. The signal
power out of BPEL is 10—4 watts. The noise power out of BPFL is N0 1 MHz = 5.56x10"2 .106 = 5.56x10‘6 watts Sm_ 10*t
NM 5.56x10—6 out of BPFL the =18:>12.5 dB Thus, SRFL >> N RFL at limiter input so the limiter is above threshold. (Carlson p. 420 says threshold is > 10 or 10 dB)
(Also, tan"i 0.24 :s 0.24) Phasor sketch from p. 413 of Carlson or from lecture notes: 3. 0) Then E (I) = A, (t) cos(a)ct + q}, The limiter output amplitude is Hemizl‘ Possible representations are: 1) A,(r)=i1 and a, (:)=¢,C (194% 0‘) m  (see p. 413 of Carlson for description of terms)
2) A, (t) =1 and a, (t) = ¢(t)+¢RC(t)+n‘1(t)\/ﬁ
R with Mr) :0 or a" and ﬁg (I) = quadrature noise SR : signal power Note 1: ¢RC (t) is an alternating i small phase that results from the pulse shaper output not going through 0 exactly half way between each bit period. am (1‘) will depend on the sequences of 1 and 0 in the bit stream, but it will have
essentially no power at frequencies below the minimum frequency in the bit stream, i.e. r, / 16. I did not expect you to notice this ¢RC (I) term and I did not take off points if it was not included since it was not discussed in class or in the
book. Note 2: Either A!7 (t) = i1 or (6(1) = 0 or 7r represent the modulation by the
pulse shaped bit stream. 3. d) (5 pts.) The squaring circuit will produce a signal component at DC and at 2 from cos2 y = i + 1cos 2y . The data modulation will be “removed”, i.e., either 2 2
(i1)2 or cos 221' = cosO , depending on representation of E (r) The DC term is rejected by BPFGi that has a bandwidth of 400 Hz. This ﬁlter obviously will also reject all ﬁequencies in ¢RC (t) above rb [16. Thus, since the modulation wasremoved by the squaring circuit, the only components remaining are
a “carrier” at 2 fc and the quadrature component of the noise (i.e., phase noise) ﬁltered by a 400 Hz ﬁlter. The noise effective deviation was multiplied by 2 in the
squaring circuit. The frequency divider (+2.) is way above threshold (400 Hz ﬁlter) so its output is a component at f6 and noise with its effective phase deviation divided by 2. The output of ili’PFG2 is then a carrier at fc and noise heavily ﬁltered by the 200 Hz
bandwidth BPFGZ. Note: the phase noise spectrum is the spectrum of mg (I) which is essentially ﬂat. See Carlson Figure 10.3 4 3 on p. 414. 3. e) (4 pts.) The signal at ©is a “recovered carrier” at fC with phase adjusted to be the same as the phase of the signal at The noise power at @is much less than the
noise power at® within the 1 MHz bandwidth of the signal because of the 200 Hz bandwidth BPFGZ. Thus, the very high % “coherent recovered carrier” at @ will contribute negligible noise power at the demodulator multiplier output so the % at the multiplier output can be taken to be the same as at®. I
orb 23..
Non . See lecture last week of class. 3D(5PtS) P..=Q[ —4
SR — 2w— : 35.97 m 36 Nari _ (5.56x10'12Waﬁ%{Z)(106Hz) $4.0 Pc = Q(6.0) =10"9 {from graph Table T.6 on page 791 of Carlson} 2 Here the N, has been assumed to be the same as at @because of the high % of the recovered coherent carrier at Perfect matched ﬁlter and synchronization
have been assumed as stated in the problem. 3. g) (10 pts.) With BPFL removed, the noise power at the input to the limiter is N RRF .
The signal has been decreased by 9.54 dB where 9.54 dB —> factor of 0.1 1 12 Therefore, the % at the limiter input for part g) is 0.1112 or 9.54 dB which is
considerably below threshold. The limiter output, L’ (t) , then is essentially all noise that has approximately a uniform distribution in phase from 0 to 221' . Therefore, the “recovered carrier” a©ias no “signal component” and is all phase
noise. The multiplier output then is all noise and ﬁe w 0.5. P4.(20 pts) a —> w l
Gn(f)_2 . x0 (0 = 35010“) 0 x0 takes gr; with equal probability 0 p(t) = rect(2t) , and P100 = rect(t — 1) _
i The Sample and Hold (S/II) device samples the signal at to .
o Decision—device is ‘ x0 = 51311020) 1,y0 2 0
—— Latherwise thl s1 0,0therwise r_ect(t) = { and sign(yo) ={ a) Find to such that we get the minimum probability of error. 2
b) What is the matched ﬁlter for 130‘) and what is y—g if we have the matched ﬁlter
. _ . 0 instead ofh(t)? _ Solutions: a) (5pts)ye = We”) = t; Lotto — r)h(r)dr p(t) * h(t) is shown below: PU) * W) —1/2 1/2  3/2 _ ' 5/2 2
(5 pts)T0 maximize y—oz which is equivalent to minimizing the probability of error, we
0' shoulcl choose to such that yg becomes maximum. As Shown in the ﬁgure y: is
maximumforI/ZStOS3/2. I b) The casual matched ﬁlter is (5 pts) h(t) = rect(l .— 2t) A2
2_
yo 4
2 WW 77
a=— t 4dr:
2£0011) 2 ' 2 2
M) A 5 ts "=— ( p )62 2” ' 5. _ (10 pts.)
The inphase If (I) and quadrature 1,1,0) modulating waveforms are as sketched below. The curves are segments of sinusoids. The waveforms modulate a canier
(ac = 27:}: so that the modulated signal can berepresented by I? ' Izc(é)=zi(?)COSwJzq(055110.21! 
éAé(t)cos[mctf¢(t)] . . ' I a) What is the modulation represented by the sketched waveforgms and
' equations above? ' An fw e. r . M If b) What is the amplitude A(r)? : [I c). What is the bit sequence represented by the sketched waveforms and 
' equations above? ' ' ...
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This note was uploaded on 05/23/2008 for the course EE 279 taught by Professor Hashemi,h during the Winter '08 term at Stanford.
 Winter '08
 Hashemi,H

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