CHAPTER 5. Additional Problems (a) The AM signal is defined by st () = A c. k a A c 1

Transcription

1 CHAPTER 5 Additional Problems 5.7 (a) The AM signal is defined by st () A c ( + k a mt ()) cos( ω c k a A c t cos( ω c To obtain 5% modulation, we choose k a, which results in the modulated waveform of Fig. for s( with A c volt: (+k a )A c Figure (b) The DSB-SC modulated signal is defined by st () A c mt () cos( ω c A c t cos( ω c + which has the modulated waveform of Fig. for A c volt: A c Figure Typically, a DSB-SC signal exhibits phase reversals. No such reversals are exhibited in Fig. because the modulating signal m( is nonnegative for all time t.

2 5.8 The AM signal is defined by st () A c ( + k a mt ()) cos( ω c where A c cos( ω c that is, khz. is the carrier and k a is a constant. We are given ω c π x 5 rad/sec; (a) mt () A cos( π 3 ω π 3 rad/sec khz The frequency components of s( for positive frequencies are: f khz + f + khz f 99 khz (b) mt () A cos( π 3 + A cos( 4π 3 which consists of two sinusoidal components with frequencies f khz and f khz. Hence, the frequency components of s( for positive frequencies are: khz + f + khz f 99 khz + f + khz f 98 khz (c) First, we note that cos( X) sin( Y ) -- sin( Y + X) + -- sin( Y X) Hence, mt () A cos( π 3 sin( 4π 3 A A sin( 6π sin( π 3 which consists of two sinusoidal components with frequencies f 3 khz and f khz. Correspondingly, the frequency components of s( for positive frequencies are: khz + f khz f 3 97 khz + f + khz

3 f 99 khz (d) Here we use the formula: cos θ -- [ + cos( θ) ] Hence, mt () A cos ( π 3 A [ + cos( 4π 3 ] which consists of dc component and sinusoidal component of frequency f khz. The frequency components of s( for positive frequencies are therefore: khz + f + khz f 98 khz (e) For this part of the problem, we find use of the formulas: cos θ -- [ + cos( θ) ] sin θ -- [ cos( θ) ] Hence, mt () -- 4π 3 [ + cos( ] + -- [ cos( 8π 3 ] -- 4π 3 + cos( -- cos( 8π 3 which consists of dc component, and two sinusoidal components with f khz and f 4 khz. Correspondingly, the frequency components of s( for positive frequencies are: khz + f + khz f 98 khz + f khz f 4 96 khz (f) We first use the formula 3 θ cos cosθ -- [ + cos( θ) ] 3

4 -- cosθ + -- cosθcos( θ) -- θ -- cos + -- cos( θ + θ) + -- cos( θ θ) 3 -- cos( 3θ) + -- cosθ 4 4 Hence, mt () A cos ( π 3 A π 3 3A cos( cos( π which consists of two sinusoidal components with frequencies f khz and f 3 khz. The frequency components of s( are therefore: khz + f + khz f 99 khz + f khz f 3 97 khz Note: For negative frequencies, the frequency components of s( are the negative of those for positive frequencies. 5.9 (a) For a square wave with equal mark-to-space ratio and frequency f 5 Hz, alternating between and, has the frequency expansion (see Example 3.3 of the tex: mt () mk [ ]e jkω t π, ω T k- sin( kω where mk [ ] T ) π π , ω Tkω T T T π -- kπ kπ sin That is, 4

5 mt () -- π -- k- kπ -- sin k e jkω t + -- cos( ω π cos3( ω 3π cos5( ω 5π + With ω π f π 5 rad/sec, m( consists of a dc component, and sinusoidal components of frequencies.5 khz,.5 khz,.5 khz (i.e., odd harmonics). Hence, the following components of the AM signal s( for positive frequencies (with progressively decreasing amplitude) are as follows: f khz f khz f khz f khz f khz f khz f khz and so on. (b) When the square modulating wave m( alternates between - and +, the dc component is zero. Nevertheless, the frequency components of the AM signal m( remain the same as in part (a); the only difference is in the carrier amplitude. Note: The frequency components for negative frequencies are the negative of those for positive frequencies. 5. (a) For positive frequencies, the spectral content of the AM signal consists of the following: Carrier: khz Upper sideband, occupying the band from.3 to 3. khz Lower sideband, occupying the band from 99.7 to 96.9 khz (b) For positive frequencies, the spectral content of the AM signal consists of the following: Carrier: khz Upper sideband, occupying the band from.5 to 5 khz Lower sideband, occupying the band from to 85 khz 5

6 Note: For negative frequencies, the spectral contents of the AM signal s( are the negative of those for positive frequencies. 5. The percentage modulation is defined by A max A min + A max A min % 5. Building on the solution to Problem 5.7, the frequency components of the DSB-Sc signal for positive frequencies are as follows: (a) (b) (c) (d) (e) (f) + f khz f 99 khz + f khz f 99 khz + f khz f 98 khz + f 3 khz f 97 khz + f khz f 99 khz + f khz f 98 khz + f khz f 98 khz + f 4 khz f 96 khz + f khz 6

7 f 99 khz + f 3 khz f 97 khz Note: For negative frequencies, the frequency components of the DSB-SC signal are the negative of those for positive frequencies. 5.3 Building on the solution for Problem 5.9, we find that the DSB-SC signal has the same frequency content for both forms of the square wave described in parts (a) and (b), as shown by: For positive frequencies, we have the frequency components.5, 99.5,.5, 98.5,.5, 97.5 khz, and so on, with progressively decreasing amplitude. For negative frequencies, the frequency components are the negative of those for positive frequencies. Note: By definition, the carrier is suppressed from the modulated signal. 5.4 (a) For positive frequencies, we have Upper sideband, extending from.3 to 3. khz Lower sideband, extending from 99.7 to 96.9 khz (b) For positive frequencies, we have Upper sideband, extending from.5 to 5 khz Lower sideband, extending from to 85 khz Note: For negative frequencies, the spectral contents of the DSB-SC signal are the negatives of those for positive frequencies. 5.5 Building on the solution to Problem 5.8, we may describe the frequency components of SSB modulator for positive frequencies as follows: (i) Upper sideband transmission: (a) + f khz (b) + f khz f khz 7

8 (c) + f 3 khz f khz (d) + f khz (e) + f khz f 4 khz (f) + f khz + f 3 khz (ii) Lower sideband transmission: (a) f 99 khz (b) f 99 khz f 98 khz (c) f 97 khz f 99 khz (d) f 98 khz (e) f 98 khz f 96 khz (f) f 99 khz f 97 khz Note: For both (i) and (ii), the frequency contents of SSB modulated signal for negative frequencies are the negative of the frequency contents for positive frequencies. 5.6 With the carrier suppressed, the frequency contents of the SSB signal are the same for both square waves described under (a) and (b), as shown by (for positive frequencies): (i) Upper sideband transmission:.5,.5,.5 khz, and so on. (ii) Lower sideband transmission: 99.5, 98.5, 97.5 khz, and so on. 8

9 Note: For negative frequencies, the frequency components of the SSB signal are the negative of those for positive frequencies. 5.7 (i) Upper sideband transmission: (a) Upper sideband, occupying the band from.3 to 3. khz (b) Upper sideband, occupying the band from.5 to 5 khz (ii) Lower sideband transmission: (a) Lower sideband, occupying the band from 99.7 to 96.9 khz (b) Lower sideband, occupying the band from 99.5 to 85 khz Note: For negative frequencies, the spectrum contents of the SSB signal are the negative of those for positive frequencies. 5.8 The spectra of the pertinent signals are as follows: Magnitude Carrier - Magnitude f ω/(π), khz Modulating signal -4-4 f ω/(π), khz Modulating signal Upper side frequency for negative frequencies Lower side frequency for positive frequencies Carrier Magnitude Lower side frequency for negative frequencies Carrier Upper side frequency for positive frequencies f (ω/π), khz We clearly see that this spectrum suffers from frequency overlap, with the lower side frequency for negative frequencies coinciding with the carrier for positive frequencies; similarly, for negative frequencies. This spectrum is therefore radically different from the spectrum of a regular AM signal; hence, it is not possible to recover the modulating signal using an envelope detector. 9

10 5.9 (a) Modulating signal Upper side frequency for negative frequencies Lower side frequency for positive frequencies Magnitude Lower side frequency for negative frequencies Upper side frequency for positive frequencies The modulated signal therefore consists of two sinusoidal components, one at frequency khz and the other at frequency 6 khz. (b) A conventional coherent detector consists of a product modulator followed by a lowpass filter. The product modulator is supplied with a carrier of frequency khz. The spectrum of this modulator consists of the following components: dc component (at zero frequency) sinusoidal component at 4 khz another sinusoidal component at 8 khz f (ω/π), khz To extract the sinusoidal modulating signal of 4 khz, the low-pass filter has a cutoff frequency slightly in excess of 4 khz. The resulting output therefore consists of the desired sinusoidal modulating signal, plus a dc component that is undesired. To suppress the dc component, we have to modify the coherent detector by passing the detector output through a capacitor. 5.3 For a message bandwidth ω m.5π x 3 rad/s and carrier frequency ω c π x 3 rad/s, the spectra of the message signal and double sideband-suppressed carrier (DSB-SC) modulated signal may be depicted as follows: M(jω) M() -π π ω( 3 rad/s) S(jω) -4.5π -.5π -.5π.5π.5π 4.5π ω( 3 rad/s)

11 The important point to note from the picture depicted here is that there is a clear separation between the sidebands lying in the negative frequency region and those in the positive region. Consequently, when the DSB-SC modulated signal is applied to a coherent detector, the resulting demodulated signal is a replica of the original message signal except for a scale change. When, however, the carrier frequency is reduced to ω c.5π x 3 rad/s, the lower sideband for negative frequencies overlaps the lower sideband for positive frequencies, and the situation changes dramatically as depicted below: M(jω) M() -π -π π π ω( 3 rad/s) S(jω) -3.5π -.5π.5π 3.5π -.5π.5π ω( 3 rad/s) V(jω) -π π ω( 3 rad/s) The spectrum labeled V(jω) refers to the demodulated signal appearing at the output of the coherent detector. Comparing this spectrum with the original message spectrum M(jω), we now see that the modulation-demodulation process results in a message distortion. The conclusions to be drawn from the results presented above are:. To avoid message distortion on demodulation due to sideband overlap, we must choose the carrier frequency in accordance with the condition ω c ω m. The minimum acceptable value of ω c is therefore ω m.. For ω c < ω m, we have sideband overlap and therefore message distortion. 5.3 The two AM modulator outputs are s () t A c ( + k a mt ()) cos( ω c

12 s () t A c ( k a mt ()) cos( ω c Here it is assumed that the two modulators are identical, that is, they have the same amplitude modulation sensitivity k a and the same carrier A c cos(ω c. Subtracting s ( from s (: st () s () t s () t A c ( + k a mt ()) cos( ω c A c ( k a mt ()) cos( ω c k a A c mt () cos( ω c which is the formula for a DSB-SC modulated signal. For this method of generating a DSB-SC modulated signal to work satisfactorily, the two AM modulators must be carefully matched. 5.3 From Eq. (5.3) of the text, we have S( jω) ---- M( j( ω kω T s ))H( jω) s k- () where H( jω) T c ( ωt ( π) )e jωt sin We are given T µs π ω π 5 rad/s M( jω) A m A m [ δ( jω jω m ) + δ( jω + jω m )] amplitude ω m π 3 rad/s π ω s T s With ω ω m, it follows that the effect of flat-top sampling is small enough to approximate Eq. () as follows: S( jω) ---- M( j( ω kω T s )) s k-

13 With ω s frequencies) > ω m, the side frequencies of the modulated signal are as follows (for positive k : k : k : k 3: and so on. ω m ω s - ω m, ω s + ω m ω s - ω m, ω s + ω m 3ω s - ω m, 3ω s + ω m For negative frequencies, the side frequencies are the negative of those for positive frequencies (a) The radio frequency (RF) pulse is defined by st () A c cos( ω c, T T -- t --, otherwise The modulated signal s( is obtained by multiplying the carrier A c cos(ω c bya rectangular pulse of unit amplitude and duration T (centered about the origin). The Fourier transform of A c cos(ω c is A c A c -----δ( ω ω. c ) δ( ω+ ω c ) The Fourier transform of the rectangular pulse is equal to the sinc function Tsinc(ωT). Since multiplication in the time domain is transformed into convolution in the frequency domain, we may express the Fourier transform of s( as A c T S( jω) [ sinc( T ( ω ω () c )) + sinc( T ( ω+ ω c ))] When ω c T is much greater than π, the overlap between the two sinc functions sinc(t(ω - ω c )) and sinc(t(ω + ω c )) is correspondingly small. We may then approximate Eq. () as follows: S( jω) A c T sinc( T ( ω ω c )), ω >, ω A c T sinc( T ( ω+ ω c )), ω < (b) For ω c T π, use of the approximate formula of Eq. () is justified. The width of the main lobe of the sinc function is 4π/T ω c /5. The width of each sidelobe is ω c /. We may thus plot the magnitude spectrum of S(jω) as shown on the next page. () 3

14 5.34 (a) The spectrum of each transmitted radar pulse is closely defined by Eq. () given in the solution to Problem The periodic transmission of each such pulse has the equivalent effect of sampling this spectrum at a rate equal to the pulse repetition frequency of the radar. Accordingly, we may express the spectrum of the transmitted radar signal s( as follows: S( jω) a n δω ( ω c nω ) for ω > n for ω a n δω ( + ω + c nω ) for ω < n where ω fundamental frequency π T and the coefficient a n is defined by T A a n c c nt sin T T where T is the pulse duration. (b) The spectrum S(jω) defined in Eq. () is discrete in nature, consisting of a set of impulse functions located at ω +(ω c + nω ), where n,+, +,... The envelope of the magnitude spectrum S(jω) is therefore as shown on the next page. () 4

15 T A c T The mainlobe of the spectrum has a width of 4π π 6 rad/s T and the sidelobes have a width of π x 6 rad/s. The impulse functions are separated by ω π 3 rad/s and the carrier frequency ω c π x 9 rad/s. Thus, there are impulse functions enveloped by the mainlobe and impulse functions enveloped by each sidelobe The DSB-SC modulated signal is defined by st () A c mt () cos( ω c Let the local oscillator output in the coherent detector be denoted by cos(ω c t + ω, where ω is the frequency error. Multiplying s( by the local carrier yields s () t st () cos( ω c t + ω A c mt () cos( ω c cos( ω c t + ω A c -----m() t[ cos( ω + cos( ω c t + ω ] Low-pass filtering s ( results in the output A c s () t -----m() tcos( ω In words, the effect of frequency error ω in the local oscillator is to produce a new DSB- SC modulated signal with an effective carrier frequency of ω. It is only when ω that the coherent detector works properly. 5

16 5.36 The mixer produces an output signal of frequency equal to the sum or the difference between its input signal frequencies. The range of sum- or difference-frequencies is from khz (representing the difference between input frequencies MHz and 9 khz) to 9.9 MHz (representing the sum of input frequencies 9 MHz and 9 khz). The frequency resolution is khz The basic similarity between full AM and PAM is that in both cases the envelope of the modulated signal faithfully follows the original message (modulating) signal. They differ from each other in the following respects:. In AM, the carrier is a sinusoidal signal; whereas in PAM, the carrier is a periodic sequence of rectangular pulses.. The spectrum of an AM signal consists of a carrier plus an upper sideband and a lower sideband. The spectrum of a PAM signal consists of a carrier plus an upper sideband and a lower sideband, which repeat periodically at a rate equal to the sampling rate (a) (b) (c) gt () sinc( This sinc pulse corresponds to a bandwidth of Hz. Hence the Nyquist rate is Hz, and the Nyquist interval is (/) seconds. gt () sinc ( This signal may be viewed as the product of the sinc signal sinc( with itself. Since multiplication in the time domain corresponds to convolution in the frequency domain, we find that the signal g( has a bandwidth twice that of the sinc pulse sinc(, that is, Hz and the Nyquist interval is /4 seconds. gt () sinc( + sinc ( The bandwidth of g( is determined by its highest frequency component. With sinc( having a bandwidth of Hz and sinc ( having a bandwidth of Hz, it follows that the bandwidth of g( is Hz. Correspondingly, the Nyquist rate of g( is 4 Hz and its Nyquist interval is /4 seconds (a) With a sampling rate of 8 khz, the sampling interval is T s µs There are 4 voice channels and synchronizing pulse, so the time allotted to each channel is τ T channel µs 5 6

17 (b) If each voice signal is sampled at the Nyquist rate, the sampling rate would be twice the highest frequency component 3.4 khz, that is, 6.8 khz. The sampling interval is then T s µs Hence, 47 T channel µs (a) The bandwidth required for each single sideband modulated channel is khz. The total bandwidth for such channels is x khz. (b) The Nyquist rate for each channel is xkhz. For TDM signals, the total data rate is x 4 khz. By using a sinc pulse whose amplitude varies in accordance with the modulating signal, and with zero crossings at multiples of (/4)ms, we would need a minimum bandwidth of khz (a) The Nyquist rate for s ( and s ( is 6 Hz. Therefore, must be greater than 6 Hz. Hence, the maximum value of R is 3. (b) With R 3, we may use the signal format shown in Fig. to multiplex the signals s ( and s ( into a new signal, and then multiplex s 3 (, s 4 ( and s 5 ( markers for synchronization: Marker 3 s Marker R 4 s..... Time s 3 s 4 s s 3 s 4 s s 3 s 4 s 3 s 4 s 3 s 4 s 3 s 4 s 3 s 4 s 3 s 4 s 3 s 4 s s 3 s 4 s Figure (/7) s. zero samples 7

18 Based on this signal format, we may develop the multiplexing system shown in Fig.. 4 Hz clock.. 4 s. _ s delay delay. 7 s delay Marker generator. 7 s delay Sampler s ( Sampler M U X s 3 ( s ( s 5 ( Sampler Sampler M U X Figure Multiplexed signal s 4 ( Sampler 5.4 The Fourier transform provides a tool for displaying the spectral content of a continuoustime signal. A continuous-wave (CW) modulated signal s( involves the multiplication of a message signal m( by a sinusoidal carrier c( A c cos(ω c in one form or another. The CW modulated signal s( may be viewed as a mixture signal that involves, in one form or another, the multiplication of m( by c(. Multiplication in the time-domain is transformed into the convolution of two spectra, namely, the Fourier transform M(jω) and the Fourier transform C(jω). The Fourier transform C(jω) consists of a pair of impulses at +ω c. Hence, the Fourier transform of the modulated signal, denoted by S(jω), contains a component M(jω - jω c ) for positive frequencies. For negative frequencies, we have the image of this spectrum with respect to the vertical axis. The picture so portrayed teaches us that the carrier frequency ω c must be greater than the highest frequency component of the message spectrum M(jω). Moreover, given that this condition is satisfied, Fourier analysis teaches us that recovery of the original message signal m( from the modulated signal s( is indeed a practical reality. For example, we may use a product modulator consisting of a product modulator followed by a low-pass filter. The product modulator, supplied by a local carrier of frequency ω c, produces a replica of the original message signal m( plus a new modulated signal with carrier frequency ω c. By designing the lowpass filter to have a cutoff frequency slightly higher than the highest frequency of m(, recovery of m( except for a scaling factor is realized. Consider next the case of pulse-amplitude modulation (PAM). The simplest form of PAM consists of multiplying the message signal m( by a periodic train of uniformly spaced impulse functions, with adjacent impulses being spaced T s seconds apart. What we have just described is the instantaneous form of sampling. Thus, the sampling process represents another form of a mixture signal. The Fourier transform of the periodic pulse train just described consists of a new uniformly spaced periodic train of impulses in the frequency domain, with each pair of impulses spaced apart by /T s hertz. As already mentioned, multiplication of two time functions is transformed into the convolution of 8

19 their spectra in the frequency domain. Accordingly, Fourier analysis teaches us that there will be no spectral overlap provided that the sampling frequency /T s is not less than twice the highest frequency component of the original message signal m(. Hence, provided that this condition is satisfied, the recovery of a replica of m( is indeed possible at the receiver. This recovery may, for example, take the form of a low-pass interpolation filter with a cutoff frequency just slightly greater than the highest frequency component of m(. Advanced Problems 5.43 The nonlinear device is defined by i () t a v i () t + a v i () t () The input v i ( is given by v i () t A c cos( ω c + A m cos( ω m () where A c cos(ω c is the carrier wave and A m cos(ω m is the modulating wave. Substituting Eq. () into (): i () t a ( A c cos( ω c + A m cos( ω m ) + a ( A c cos( ω c + A m cos( ω m ) a A c cos( ω c + a A m cos( ω m + a A c cos + a A m cos ( ω m ( ω c + a A c A m cos( ω c cos( ω m (3) Using the trigonometric identity cos θ -- ( cos( θ) + ) we may rewrite Eq. (3) in the equivalent form (after a rearrangement of terms): i () t a A c ( ω c + a A c A m cos( ω c cos( ω m + a A m cos( ω m + --a A m cos( ω m + --a A c cos( ω c + --a A c + --a A m (4) We may now recognize the following components in the output i (:. An amplitude modulated component (AM) represented by a A m st () a A c cos( ω m cos( ω c a. A set of undesirable components represented by the remaining components of Eq. (4). 9

20 (a) In amplitude modulation, the carrier frequency ω c is typically much larger than the modulation frequency ω m. Hence, the frequency context of i ( may be depicted as follows (showing only the positive frequency components so as to simplify the presentation): ω m ω m ω c -ω m ω c ω c +ω m ω c ω c (b) From this figure, we see that the desired AM component occupies a frequency band extending from the lower side-frequency ω c - ω m to the upper side frequency ω c + ω m. To extract this component, we need to pass the output of the nonlinear device, i (, through a band-pass filter. The frequency specifications of this filter are as follows: Passband of a width slightly larger than ω m, centered on the carrier frequency ω c. Lower stopband, lying below the lower side-frequency ω c - ω m and thereby suppressing the dc component as well as the frequency components ω m and ω m. Upper stopband, lying above the upper side-frequency ω c + ω m and thereby suppressing the frequency component ω c (a) For the special case of an infinite unipolar sequence (binary sequence consisting of a square wave of equal mark-to-space ratio), we have mt () cos( ω π cos( 3ω 3π cos( 5ω 5π π k cos( ( k+)ω k+ where ω πf π(/t )π/t. (See the solution to Problem 5.9). The resulting K signal is given by (assuming a sinusoidal carrier of unit amplitude and frequency ω c ) st () mt () cos( ω c -- cos( ω c + -- π cos( ( k+)ω k+ cos( ω c k ()

21 m( T -3T -T T 3T 5T Using the formula cos( A) cos( B) -- [ cos( A+ B) + cos( A B) ] we may rewrite Eq. () in the equivalent form st () -- cos( ω () c + -- π [ cos( (( k+)ω k+ +ω c )t + cos( ( k+)ω -ω c )] k The spectrum of s( defined in Eq. () is depicted in Fig. (for positive frequencies): Figure (b) For the BPSK signal, the binary sequence is represented in its polar nonreturn-to-zero sequence with the following waveform: m( + -3T -T -T T T 3T 4T t - Correspondingly, m( has the Fourier series representation:

22 mt () 4 -- π cos( ( k + )ω k + k where ω π/t. The binary sequence of Eq. (3) differs from that of Eq. () in two respects: It has no dc component It is scaled by a factor of. The resulting BPSK signal is defined by (assuming a carrier of unit amplitude and frequency ω c ) st () mt () cos( ω c 4 -- π cos( ( k + )ω k + cos( ω c k -- π [ cos( (( k + )ω k + +ω c ) + cos( (( k + )ω -ω c )] k The spectrum of s( defined in Eq. (4) is depicted in Fig. : (3) (4) Figure (c) The energy of the unipolar signal defined in Eq. () is given by T --- T st () dt Sk ( ) k- The left-hand side of Eq. (5) yields T E -- st () T T dt st () T t s() t T t s() t dt T d d + T T T T T ( ) dt T -- The right-hand side of Eq. (5) yields, in light of Eq. (), the energy (5) (6)

23 E -- ( π) ( ) π π ---- π which checks the result of Eq. (). For the polar nonreturn-to-zero signal of Eq. (3), the energy is T E -- st () dt T T T st () T t s() t T dt + s() t dt T d + T T T T ( ) T t ( ) T dt + ( ) dt T d + T T T T T T T Application of Parseval s theorem yields E ( 4 π) ( ) ( ) π π π (d) To raise the energy of the unipolar signal to equal that of the polar nonreturn-to-zero signal, the amplitude of the former signal has to be increased by the scale factor. (e) Examining the OOK signal of Eq. (), we see that it contains a carrier component in addition to side frequencies, hence the similarity to an AM signal. On the other hand, examining Eq. (4) for the BPSK signal, we see that it lacks a carrier component, hence the similarity to a DSB-SC signal. 3

24 5.45 The multiplexed signal is st () A c m () t cos( ω c + A c m () t sin( ω c where m ( and m ( are the incoming message signals. Taking the Fourier transform of s(: S( jω) A c [ M ( jω jω c ) + M ( jω + jω c )] A c [ M j ( jω jω c ) + M ( jω + jω c )] FT FT where m () t M ( jω) and m () t M ( jω). With H(jω) denoting the transfer function of the channel, the Fourier transform of the received signal is R( jω) H( jω)s( jω) A c -----H( jω) [ M ( jω jω c ) + M ( jω + jω c )] A c H( jω) [ M ( jω jω c ) + M ( jω + jω c )] FT where rt () R( jω). To recover m (, we multiply the received signal r( by cos(ω c and then pass the resulting output through a low-pass filter with cutoff frequency equal to the message bandwidth. The result of this processing is a signal with the spectrum S ( jω) -- ( R( jω jω c ) + R( jω + jω c )) A c -----H( jω jω 4 c )[M ( jω jω c ) + M ( jω) + --M j ( jω jω c ) --M j ( jω)] A c H( jω jω 4 c )[M ( jω) + M ( jω + jω c ) + --M () j ( jω) --M j ( jω + jω c )] For a real channel, the condition H( jω c + jω) H * ( jω c jω) is equivalent to H( jω + jω c ) H( jω jω c ) This equivalence follows from the fact that for a channel with real-value impulse response h(, we have H( jω) H * ( jω). Hence, substituting this condition into Eq. (): S ( jω) A c -----H( jω jω c )M ( jω) A c H( jω jω 4 c )[M ( jω jω c ) 4

25 + M ( jω + jω c ) + --M j ( jω jω c ) --M j ( jω + jω c ) Passing the signal defined by S (jω) through a low-pass filter outoff frequency equal to the message bandwidth, ω m, we get an output whose Fourier transform is equal to A c -----H( jω jω for c )M ( jω) ω m ω ω m Recognizing the band-pass nature of H(jω), we immediately see that the output so obtained is indeed a replica of the original message signal m (. Similarly, to recover m (, we multiply r( by sin(ω c and then pass the resulting signal through a low-pass filter. In this case, we get an output with a spectrum equal to A c -----H( jω jω c )M ( jω) 5.46 The block diagram of the scrambler is as follows: m( Product v ( High-pass v ( Product v 3 ( Low-pass s( modulator filter modulator filter cos(ω c cos(ω c t + ω b Figure (a) The first product modulator output is v () t mt () cos( ω c When v ( is processed by the high-pass filter, we get v (. The second product modulator output is v 3 () t v () t cos( ω c t + ω b The magnitude spectra of m(, v (, v (, v 3 ( and s( are illustrated in Fig.. From this figure, we observe that v ( is a DSB-SC modulated signal, v ( is a SSBmodulated signal, v 3 ( is a DSB-SC modulated signal whose lowest frequency is zero, and s( is a low-pass signal whose spectrum is uniquely defined by m(. M(jω) -ω b -ω a ω a ω b ω 5

26 V (jω) ω c - ω b ω c + ω b ω c ω c - ω a ω c ω c + ω a ω Figure V (jω) ω ω c - ω b ω c - ω a ω c ω c ω c + ω a ω c + ω b V 3 (jω) ω b + ω a ω b - ω a ω c + ω b ω S(jω) ω b + ω a ω b - ω a To find an expression for the scrambled voice signal s(, we need to invoke the Hilbert transform. Specifically, the Hilbert transform of m( is defined by m( τ) mˆ () t -- π d t τ τ The mˆ () t may be viewed as the convolution of m( with /(π. The Fourier transform of /(π is equal to -jsgn(ω), where sgn( ω) for ω > for ω for ω < Hence, Mˆ ( jω) jsgn( ω)m( jω) ω 6

27 Using this relation, it is a straightforward matter to show that the spectrum of s( defined previously is the same as that of the following expression: st () --m() tcos( ω 4 b + --mˆ () t sin( ω 4 b (b) With s( as the input to the scrambler, the output of the first product modulator is v () t st () cos( ω c --m() tcos( ω 4 b cos( ω c --mˆ () t sin( ω 4 b cos( ω c --m() t[ cos( ω 8 c t + ω b + cos( ω c t ω b ] + --mˆ () t [ sin( ω 8 c t + ω b sin( ω c t ω b ] The high-pass filter output is therefore v () t --m() tcos( ( ω 8 c + ω b ) + --mˆ () t sin( ( ω 8 c + ω b ) Correspondingly, the output of the second product modulator is v 3 () t v () t cos( ( ω c + ω b ) + --m() tcos (( ω 8 c + ω b ) + --mˆ () t sin( ( ω 8 c + ω b ) cos( ( ω c + ω b ) -----m() t m() tcos( ( ( ω 6 c + ω b )) mˆ () t sin( ( ω 6 c + ω b ) The scrambler output is therefore v () t -----m() t 6 which is a scaled version of the original message signal. 7

28 5.47 Consider a message signal m( whose frequency content lies inside the band ω l ω ω m. Assuming that the carrier frequency ω c > ω m, we may describe the spectrum of the SSB modulated signal s( (with only its upper sideband retained) as follows: M(jω) -ω m -ω l ω l ω m ω -ω c - ω m -ω c ω c ω c + ω m -ω c - ω l ω c + ω l To demodulate s(, we apply it to the coherent detector of Fig. P5.47. Let v( denote the output of the product modulator in this figure, and let v ( denote the output of the lowpass filter. We may then describe the corresponding spectra of v( and v ( in graphical forms as shown in Fig.. ω Figure S(jω) -ω c - ω m -ω c ω c ω c + ω m ω - ω c - ω l ω c + ω l V(jω) -ω c - ω m -ω c -ω m -ω l ω l ω m ω c ω c + ω l ω -ω c - ω l ω c + ω m Figure From this figure we see that the product modulator output v( consists of two components: A scaled version of the original message signal m( A new SSB modulated signal with carrier frequency ω c 8

29 The latter component is suppressed by the low-pass filter, leaving the original message signal as the detector output as shown by the spectrum of Fig. 3. V (jω) - ω m - ω l ω l ω m ω Figure The multiplexed signal is 4 st () [ cos( ω a t + α k- ) + cos( ω b t + β k- )]m k () t k where α β. The corresponding output of the product modulator in the coherent detector of the receiver is v i () t st ()[ cos( ω a t + α i- ) + cos( ω b t + β i- )] () where i,,3,4. There using Eq. () in (): 4 v i () t m k () t [ cos( ω a t + α k- ) + cos( ω b t + β k- )] k Expanding terms: 4 [ cos( ω a t + α i- ) + cos( ω a t + β i- )] v i () t m k () t [ cos( ω a t + α k- ) cos( ω a t + α i- )] k 4 k + cos( ω a t + α k- ) cos( ω b t + β i- ) + cos( ω b t + β k- ) cos( ω a t + α i- ) [ + cos( ω b t + β k- ) cos( ω b t + β i- )] -- m k () t [ cos( α k- α i- ) + cos( β k- β i- )] + cos( ω a t + α k- + α i- ) + cos( ω b t + β k- + β i- ) + cos( ( ω a + ω b )t + α k- + β i- ) + cos( ( ω a ω b )t + α k- β i- ) + cos( ( ω a + ω b )t + α i- + β k- ) + cos( ( ω a ω b )t + α i- β k- )] The low-pass filter in the coherent detector removes the six high-frequency components of v i (, leaving the output () 9

30 v i () t 4 -- m k () t [ cos( α k- α i- ) + cos( β k- β i- )] k The requirement on α k and β k is therefore cos( α k- α i- ) + cos( β k- β i- ) where (i,k),,3,4., i k, i k 5.49 Consider an incoming AM signal of bandwidth khz and carrier frequency ω c that lies in the range ( ) MHz. It is required to frequency translate this modulated signal to a fixed band centered at.455 MHz. The problem is to determine the range of tuning that must be provided in the local oscillator. Let ω local denote the local oscillator frequency, which is required to satisfy the condition ω c ω local π rad/s or f local.455 MHz where both and f local are expressed in MHz. That is, f local.455 When.535 MHz, f local MHz. When.65 MHz, f local MHz. Thus the required range of tuning of the local oscillator is MHz, independent of the AM signal s bandwidth. 5.5 Consider a periodic train of rectangular pulses, each of duration T. Assuming that a pulse of the train is centered on the origin, we may expand it as a Fourier series: ct () f s c( nf s T )e jπnf st sin n- where f s is the pulse repetition frequency and the amplitude of each rectangular pulse is /T (i.e., each pulse has unit area). The assumption that f s T >> means that the impulse functions in the spectrum of the periodic pulse train c( are well separated from each other. Multiplying a message signal g( by c( yields the PAM signal st () ct ()gt 3

31 n- f s sinc( nf s T )gt ()e jπnf st Hence the Fourier transform of s( is given by S( jω) f s sinc( nf s T )G( jω jnω s ) n- where ω s πf s. Thus the spectrum of a naturally sampled signal consists of frequencyshifted replicas of the message spectrum G(jω), with the nth replica being scaled in amplitude by the factor f s sinc(nf s T ), which decreases with increasing n. 5.5 (a) The spectrum of the carrier cn [ ] cos( Ω c n), n ±, ±, is defined by Ce jω ( ) -- [ δ( jω jω, c n) + δ( jω + jω c n) ] n ±, ±, where Ω c > Ω m, with Ω m denoting the highest frequency component of m[n] inside the range Ω π. The spectrum of the modulated signal sn [ ] cn [ ]mn [ ], namely, S(e jω ), is obtained by convolving Ce ( jω ) with M( e jω ). (i) n : S(e jω ) π + Ω c Ω c π + Ω c Ω -π+ω c -Ω m -π+ω c +Ω m Ω c -Ω m Ω c +Ω m (ii) n -: S(e jω ) π - Ω c -Ω c π - Ω c Ω -π-ω c -Ω m -π-ω c +Ω m Ω c -Ω m - Ω c +Ω m π-ω c -Ω m π-ω c +Ω m 3

32 (iii) n S(e jω ) Ω c -π Ω m Ω c -π+ω m Ω c -π Ω c Ω c +π Ω (iv) n - Ω c -Ω m Ω c +Ω m Ω c +π-ω m Ω c +π+ω m S(e jω ) π-ω c +Ω m -Ω c +Ω m Ω -π - Ω c -Ω c π - Ω c -π-ω c -Ω m -Ω c -Ω m π-ω c -Ω m π-ω c +Ω m and so on for n +3,... The transmitter includes a band-pass digital filter designed to pass the spectral bands centered on +Ω c and reject all other bands. (b) Following the coherent detector described in Section 5.5 for demodulation of a DSB- SC modulated signal of the continuous-time variety, we may postulate the scheme of Fig. for demodulation of the discrete-time DSB-SC modulated signal: s[n] Low-pass digital filter m[n] cos(ω c n) Figure Here it is assumed that () the local oscillator supplying cos(ωn) is synchronous with the carrier generator used in the transmitter, and () the low-pass digital filter is designed with a cutoff frequency slightly greater than Ω m. 3

33 Computer Experiments %Problem 5.5 clear fm e3; fc e4; mu.75; t e-6:e-6:.; y (+mu*cos(*pi*fm*).*cos(*pi*fc*; figure() plot(t,y); xlim([.]) xlabel( Time (s) ) ylabel( Amplitude ) title( AM Wave: fc e4 Hz, fm e3 Hz, \mu.75 ) L length(y); Y fft(y,l); Y fftshift(y); Ys Y.*conj(Y); f e6/l*(-(l/):l/-); figure() plot(f,ys) xlim([-.5e5.5e5]) xlabel( Frequency (Hz) ) ylabel( Power ) title( Spectrum AM Wave: fc e4 Hz, fm e3 Hz, \mu.75 ) AM Wave: fc e4 Hz, fm e3 Hz, µ Amplitude.5.5 Figure : Modulated Signal Time (s) x 3 33

34 .5 x 9 Spectrum AM Wave: fc e4 Hz, fm e3 Hz, µ.75.5 Power Frequency (Hz) x 4 Figure : Spectrum of modulated signal %Problem 5.53 fo ; t :.:5; m sawtooth(*pi*fo*t,.5); figure() plot(t,m) xlabel( Time (s) ) ylabel( Amplitude ) title( Triangular Wave Hz ) mu.8; y (+mu*m).*cos(*pi*5*; figure(); plot(t,y); xlim([ ]) xlabel( Time (s) ) ylabel( Amplitude ) title( Modulated Wave ) figure(3) L length(y); Y fft(y,l); Y fftshift(y); Ys abs(y); f e4/l*(-(l/):l/-); plot(f,ys) 34

35 xlim([-5 5]) xlabel( Frequency (Hz) ) ylabel( Magnitude ) title( Spectrum AM Wave: fc 5 Hz, fm Hz, \mu.8 ) Triangular Wave Hz Amplitude Time (s) (a) Figure : Waveform of modulating signal Modulated Wave.5.5 Amplitude.5.5 (b) Figure : Modulated signal Time (s) 35

36 .5 x 4 Spectrum AM Wave: fc 5 Hz, fm Hz, µ.8.5 Magnitude.5 Figure Frequency (Hz) x 4 Spectrum AM Wave: fc Hz, fm Hz, µ.8.5 Carrier Magnitude.5 Message Overlap Frequency (Hz) Figure 3 x 4 Spectrum AM Wave: fc Hz, fm Hz, µ.8.5 Carrier Magnitude.5 Message Frequency (Hz) Figure 36

37 %Problem 5.55 fo ; t :.:5; m sawtooth(*pi*fo*t,.5); figure() plot(t,m) xlabel( Time (s) ) ylabel( Amplitude ) title( Triangular Wave Hz ) mu.8; y m.*cos(*pi*5*; figure(); plot(t,y); xlim([ ]) xlabel( Time (s) ) ylabel( Amplitude ) title( DSSC Modulated Wave ) figure(3) L length(y); Y fft(y,l); Y fftshift(y); Ys abs(y); f e4/l*(-(l/):l/-); plot(f,ys) xlim([-5 5]) xlabel( Frequency (Hz) ) ylabel( Magnitude ) title( Spectrum of DSSC AM Wave: fc 5 Hz, fm Hz, \mu.8 ) 37

38 DSSC Modulated Wave Amplitude (a) Figure : Modulated signal Time (s) Spectrum of DSSC AM Wave: fc 5 Hz, fm Hz, µ.8 8 Magnitude Frequency (Hz) (b) Figure : Spectrum of modulated signal 38

39 %Problem 5.56 clear;clc; wc.5*pi; res.; Fs ; sam floor(/(fs*res)); t :res:-res; m sin(wc*; f /res * ([:/(length(-):] -.5); T.5; %Pick pulse duration value (.5,.,.,.3,.4,.5) durinsam floor(t/res); %number of samples in pulse sm zeros(size(); %Reserve vector for output waveform for i:floor(length(/sam)-; %loop over number of samples sm((i*sam)+:(i*sam)+durinsam) m(i*sam+); end y fftshift(abs(fft(sm))); figure() subplot(,,) plot(t,sm); xlabel( Time (s) ) ylabel( Amplitude ) title( PAM Wave: T_.5 s ) subplot(,,) plot(f,y) xlabel( Frequency (Hz) ) ylabel( Amplitude ) xlim([- ]) 39

40 PAM Wave: T.5 s.5 Amplitude Time (s) 5 Amplitude Frequency (Hz) Figure : T.5s PAM Wave: T. s.5 Amplitude Time (s) Amplitude Frequency (Hz) Figure : T.s 4

41 PAM Wave: T. s.5 Amplitude Time (s) Amplitude Frequency (Hz) Figure 3: T.s PAM Wave: T.3 s.5 Amplitude Time (s) 8 Amplitude Frequency (Hz) Figure 4: T.3s 4

42 PAM Wave: T.4 s.5 Amplitude Time (s) 4 Amplitude Frequency (Hz) Figure 5: T.4s PAM Wave: T.5 s.5 Amplitude Time (s) 5 Amplitude Frequency (Hz) Figure 6: T.5s 4

43 %Problem 5.57 clear;clc; res e-6; t :res:.-res; modf; m sin(modf**pi*; %Samples per sampling period freqt e4; periodt /freqt; pulsedur e-5; samplesperpulse floor(pulsedur/res); samplespert floor(periodt/res); %Number oomplete sampling cycles we can get i on the signal m num floor(length(m) / samplespert); r[]; for i:num, r [r ones(,samplesperpulse) zeros(,samplespert-samplesperpulse)]; end figure() subplot(,,) yr.*m; plot(t,y) title( Naturally Sampled Waveform ) xlabel( Time (s) ) ylabel( Amplitude ) subplot(,,) plot(t,y) title( Naturally Sampled Waveform ) xlabel( Time (s) ) ylabel( Amplitude ) set(gca, xlim, [.]) figure() Fy fftshift(abs(fft(y))); f /res * ([:/(length(-):]-.5); subplot(,,) plot(f,fy) title( Spectrum of the Modulated Waveform ) xlabel( Frequency (Hz) ) ylabel( Magnitude ) 43

44 subplot(,,) plot(f,fy) title( Spectrum of the Modulated Waveform ) xlabel( Frequency (Hz) ) ylabel( Magnitude ) set(gca, xlim, [-e5 e5]) Naturally Sampled Waveform.5 Amplitude Time (s) Naturally Sampled Waveform.5 Amplitude Time (s) x 3 (a) Figure 5 Spectrum of the Modulated Waveform 4 Magnitude Frequency (Hz) x 5 Spectrum of the Modulated Waveform 5 4 Magnitude Frequency (Hz) x 5 (b) Figure 44