Notice: Undefined offset: -1 in /var/www/html/wp-content/themes/theme/functions.php on line 810

Solution for Communication Systems Essay

Solutions Manual for: Communications Systems, 5th edition by Karl Wiklund, McMaster University, Hamilton, Canada Michael Moher, Space-Time DSP Ottawa, Canada and Simon Haykin, McMaster University, Hamilton, Canada Published by Wiley, 2009. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Chapter 2 2. 1 (a) g (t ) = A cos(2? f c t ) fc = 1 T ? ?T T ? t?? , ? ? 2 2? We can rewrite the half-cosine as: ? t ? A cos(2? f c t ) ? rect ? ? ? T ? Using the property of multiplication in the time-domain: G ( f ) = G1 ( f ) ? G2 ( f ) 1 sin(? fT ) [? ( f ? fc ) + ? ( f + fc )] ? AT ? fT 2 Writing out the convolution: ?

AT ? sin(?? T ) ? G( f ) = ? ? ? [? (? ? ( f + f c ) + ? (? ? ( f ? f c ) ] d ? 2 ? ?? T ? ?? = A ? sin(? ( f + f c )T ) sin(? ( f ? f c )T ) ? + ? ? f + fc f ? fc 2? ? ? ? ? A ? cos(? fT ) cos(? fT ) ? = ? ? 1 ? 2? ? f ? 1 ? f+ 2T 2T ? ? = (b)By using the time-shifting property: T exp(? j 2? ft0 ) g (t ? t0 ) t0 = 2 ? ? A ? cos(? fT ) cos(? fT ) ? G( f ) = ? ? exp(? j? fT ) ? 1 ? 2? ? f ? 1 ? f+ ? 2T 2T ? fc = 1 2T Copyright © 2009 John Wiley & Sons, Inc.

We will write a custom sample essay on
Solution for Communication Systems
specifically for you for only $13.9/page
Order now

All Rights Reserved. (c)The half-sine pulse is identical to the half-cosine pulse except for the centre frequency and time-shift. fc = 1 2Ta ? cos(? fTa ) cos(? fTa ) ? ? ? ? (cos(? fTa ) ? j sin(? fTa )) f + fc ? ? f ? fc A ? cos(2? fTa ) cos(2? fTa ) sin(2? fTa ) sin(2? fTa) ? = ? +j ? j ? ? 4? ? f ? f c f + fc f ? fc f + fc ? A 2? A ? exp(? j 2? fTa ) exp(? j 2? fTa ) ? ? ? ? 4? ? f ? fc f + fc ? G( f ) = = (d) The spectrum is the same as for (b) except shifted backwards in time and multiplied by -1. ? ? A ? cos(? fT ) cos(? fT ) ? ? ? exp( j? fT ) G( f ) = ? 1 ? 2? ? f ? 1 ? f+ 2T 2T ? ? ? ? A ? exp( j 2? fT ) exp( j 2? fT ) ? = ? ? ? 1 4? ? f ? 1 ? f+ 2T 2T ? ? (e) Because the Fourier transform is a linear operation, this is simply the summation of the results from (b) and (d) ? A ? exp( j 2? fT ) + exp(? j 2? fT ) exp( j 2? fT ) + (? j 2? fT ) ? ? G( f ) = ? ? 1 1 4? ? ? f? f+ 2T 2T ? ? ? ? A ? cos(2? fT ) cos(2? fT ) ? = ? ? 1 ? 2? ? f ? 1 ? f+ 2T 2T ? ? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 2 g (t ) = exp(? t ) sin(2? f c t )u(t ) = ( exp(? t )u(t ) )( sin(2? f c t ) ) ? 1 ? 1 ? ? (? ( f ? f c ) ? ? ( f + f c ) ) ? 1 + j 2? f ? 2 j ? ? 1 ? 1 1 = ? ? ? 2 j ? 1 + j 2? ( f ? f c ) 1 + j 2? ( f + f c ) ? ? G( f ) = 2. 3 (a) g (t ) = g e (t ) + g o (t ) 1 [ g (t ) + g (? t )] 2 ? t ? g e (t ) = Arect ? ? ? 2T ? g e (t ) = 1 [ g (t ) ? g (? t )] 2 ? ? 1 ? ? 1 ? t ? 2T ? ?t+ T ? rect ? 2 g o (t ) = A ? rect ? ? ? ? T ? ? T ? ? ? ? ? g o (t ) = ?? ?? ?? ?? ? ?? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. (b) By the time-scaling property g(-t) Ge ( f ) = G(-f) 1 [G ( f ) + G (? f ) ] 2 1 = [sinc( fT ) exp(? j 2? fT ) + sinc( fT ) exp( j 2? fT ) ] 2 = sinc( fT ) cos(? fT ) Go ( f ) = 1 [G ( f ) ? G (? f )] 2 1 = [sinc( fT ) exp(? j 2? fT ) ? sinc( fT ) exp( j 2? fT ) ] 2 = ? jsinc( fT ) sin(? fT ) Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 4. We need to find a function with the stated properties. We can verify that: G ( f ) = ? sgn( f ) + ju( f ? W ) ? ju(? f ? W ) meets the stated criteria. By duality g(f) G(-t) g (t ) = ?1 1 ? j ? ? (t ) ? ? exp(? j 2? Wt ) ? j 2? t ? ?2 1 sin(2? Wt ) = +j 2? t ? t 1 + ? t ? ? u2 ? exp ? ? 2 ? du ? ? ? ? ? t ? T 1 1 t +T ?1 1 ? j ? ? (t ) ? ? exp( j 2? Wt ) j 2? t ? ?2 2. 5 g (t ) = = ? t ? T ? 0 h(? )d? + 1 t +T ? ? h(? )d? 0 dg (t ) 1 1 = ? h(t ? T ) + h(t + T ) dt ? ? By the differentiation property: ? dg (t ) ? F? ? = j 2? fG ( f ) ? dt ? 1 = [ H ( f ) exp( j 2? f ? ) ? H ( f ) exp(? j 2? f ? )] ? = 2j ? H ( f ) sin(2? f ? ) But H ( f ) = ? exp(?? f 2? 2 ) 1 exp(?? f 2? 2 ) sin(2? fT ) ?

G( f ) = ? f sin(2? fT ) = exp(?? f 2? 2 ) ? f = 2T exp(?? f 2? 2 )sinc(2? fT ) lim G ( f ) = 2Tsinc(2? fT ) ? >0 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 6 (a) If g(t) is even and real then 1 [ g (t ) + g (? t )] 2 and g (t ) = g * (t ) ? G ( f ) = G* (? f ) g (t ) = 1 G* ( f ) = [G* ( f ) + G * (? f )] 2 1 * 1 G ( f ) = G* (? f ) 2 2 * G ( f ) = G( f ) ? G ( f ) is all real If g(t) is odd and real then 1 [ g (t ) ? g (? t )] 2 and g (t ) = g * (t ) ? G ( f ) = G* (? f ) g (t ) = 1 G ( f ) = [G ( f ) ? G (? f )] 2 1 1 G* ( f ) = G* ( f ) ? G* (? f ) 2 2 * * G ( f ) = ? G (? ) G* ( f ) = ? G ( f ) ? G ( f ) must be all imaginary (b) (? j 2? t )G (t ) t ? G (t ) d g (? f ) by duality df j d g (? f ) 2? df The previous step can be repeated n times so: dn g (? f ) (? j 2? ft ) n G (t ) df n But each factor (? j 2? ft ) represents another differentiation. t n ? G (t ) ? j ? (n) ? ? g (? f ) ? 2? ? Replacing g with h ? j ? (n) ? ? H (f) ? 2? ? n n t n h(t ) Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. (c) ? j ? (n) Let h(t ) = t g (t ) and H ( f ) = ? ? G (f) ? 2? ? n n ? j ? (n) ? h(t )dt = H (0) = ? 2? ? G (0) ? ? ?? ? n (d) g1 (t ) * g 2 (t ) G1 ( f ) G2 (? ) ? g1 (t ) g 2 (t ) * g1 (t ) g 2 (t ) ?? ? ? G (? )G ( f ? ? )d ? 1 2 ?? ? ? G (? )G (? ( f ? ? ))d ? 1 2 1 2 = (e) ?? ? G (? )G (? ? f )d ? ? * g1 (t ) g 2 (t ) ? ?? ? G (? )G (? ? f )d ? 1 2 ?? ? ? g (t ) g (t )dt 1 * 2 G (0) ? ?? ? ? * g1 (t ) g 2 (t )dt ?? ? * 2 ? G (? )G (? ? 0)d ? 1 2 ?? ? g (t ) g (t )dt 1 ?? ? G (? )G (? )d ? 1 2 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 7 (a) g (t ) ATsinc 2 ( fT ) ? ?? ? g (t ) dt = AT max G ( f ) = G (0) = ATsinc 2 (0) = AT ? The first bound holds true. (b) ? ?? ? dg (t ) dt = 2 A dt sin(? fT ) sin(? fT ) ? ? fT ? fT j 2? fG ( f ) = 2? ATsinc 2 ( fT ) = 2? fAT = 2A sin(? fT ) ? sin(? fT ) ? fT But, sin(? fT ) ? 1 ? f and sinc(? fT ) ? 1 ? f ? 2A sin(? fT ) ? sin(? fT ) ? 2 A ? fT ? j 2? fG ( f ) ? 2 A Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 7 c) ( j 2? f ) 2 G ( f ) = 4? 2 f 2G ( f ) sin 2 (? fT ) = 4? f AT (? fT ) 2 2 2 4A 2 sin (? fT ) T 4A ? T = The second derivative of the triangular pulse is plotted as: Integrating the absolute value of the delta functions gives: ? ?? ? d 2 g (t ) 4A dt = 2 dt T ? ? ( j 2? f ) 2 G ( f ) ? ?? ? d 2 g (t ) dt dt 2 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. . 8. (a) g1 (t ) ? g 2 (t ) G1 ( f )G2 ( f ) = G2 ( f )G1 ( f ) by the commutative property of multiplication b) g1 ( f ) ? [ g 2 ( f ) ? g 3 ( f ) ] G1 ( f ) [G2 ( f )G3 ( f ) ] Because multiplication is commutative, the order of the multiplication doesn’t matter. ? G1 ( f ) [G2 ( f )G3 ( f ) ] = [G1 ( f )G2 ( f ) ] G3 ( f ) ? G1 ( f ) [G2 ( f )G3 ( f ) ] [ g1 ( f ) ? g 2 ( f )] ? g3 ( f ) c) Taking the Fourier transform gives: G1 ( f ) [G2 ( f ) + G3 ( f )] Multiplication is distributive so: G1 ( f )G2 ( f ) + G2 ( f )G 3 ( f ) g1 (t ) g 2 (t ) + g1 (t ) g 2 (t ) Copyright © 2009 John Wiley , Inc.

All Rights Reserved. 2. 9 a) Let h(t ) = g1 (t ) ? g 2 (t ) dh(t ) j 2? fH ( f ) dt = j 2? fG1 ( f )G2 ( f ) = ( j 2? fG1 ( f ) ) G2 ( f ) ? dg1 (t ) ? ? dt ? ? g 2 (t ) ? ? d ? dg (t ) ? ? [ g1 (t ) ? g 2 (t )] = ? 1 ? ? g 2 (t ) dt ? dt ? ( j 2? fG1 ( f ) ) G2 ( f ) b) ?? ? g (t ) ? g (t )dt 1 2 t 1 G (0)G2 (0) ? ( f ) G1 ( f )G2 ( f ) + 1 j 2? f 2 ? 1 ? ? G (0) ? =? G1 ( f ) ? G2 ( f ) + ? 1 ? ( f ) ? G2 ( f ) ? 2 ? ? j 2? f ? ? 1 ? G (0) =? G1 ( f ) + 1 ? ( f ) ? G2 ( f ) 2 ? j 2? f ? t ? t ? ? ? g1 (t ) ? g 2 (t )dt = ? ? g1 (t ) ? ? g 2 (t ) ?? ? ?? ? Y( f ) = 2. 10. ?? ? X (? ) X ( f ?? )d? t X (? ? 0 if ? ? W X ( f ?? ) ? 0 if f ?? ? W ( f ?? ) ? W for f ? W +? when ? ? 0 and ? ? W ( f ?? ) ? ?W for f ? ?W +? when ? ? 0 and ? ? ? W ? ( f ?? ) ? W for 0 ? ? ? W when f ? 2W ( f ?? ) ? ?W for -W ? ? ? 0 when f ? ?2W ? Over the range of integration [ ? W ,W ] , the integral is non-zero if f ? 2W Copyright © 2009 John Wiley , Inc. All Rights Reserved. 2. 11 a) Given a rectangular function: g (t ) = always equal to 1, and the height is 1/T. 1 ? t ? rect ? ? , for which the area under g(t) is T ? T ? 1 ? t? rect ? ? T ? T ? sinc( fT ) Taking the limits: 1 ? t? lim rect ? ? = ? (t ) T >0 T ? T ? lim sinc( fT ) = 1 T >0 T b) g (t ) = 2Wsinc(2Wt ) 2Wsinc(2Wt ) ? f ? rect ? ? ? 2W ? W >? lim 2Wsinc(2Wt ) = ? (t ) ? 2 ? lim rect ? ? =1 ? 2W ? W >? 2. 12. G( f ) = 1 1 + sgn( f ) 2 2 By duality: 1 1 ? ( ? t ) ? j 2? t 2 j 1 ? g (t ) = ? (t ) + 2 2? t G( f ) Copyright © 2009 John Wiley , Inc. All Rights Reserved. 2. 13. a) By the differentiation property: 2 ( j 2? f ) G ( f ) = ? ki exp(? j 2? fti ) i ?G( f ) = ? 1 4? 2 f 2 ? k exp(? j 2? ft ) i i i b)the slope of each non-flat segment is: ± A tb ? t a ? 1 ?? A ? G( f ) = ? ? 2 2 ? ? ? [ exp( j 2? ftb ) ? exp( j 2? fta ) ? exp( j 2? fta ) + exp( j 2? tb ) ] ? 4? f ? ? tb ? ta ? A =? 2 2 [cos(2? ftb ) ? cos(2? fta )] 2? f ( tb ? ta ) 1 But: sin(? f (tb ? ta )) sin(? f (tb + ta )) = [ cos(2? fta ) ? cos(2? ftb ) ] by a trig identity. 2 A ? G( f ) = 2 2 [sin(? f (tb ? ta )) sin(? f (tb + ta ))] ? f (tb ? ta ) 2. 14 a) let g(t) be the half cosine pulse of Fig. P2. 1a, and let g(t-t0) be its time-shifted counterpart in Fig. 2. 1b ? = G ( f )G* ( f ) = G( f ) 2 ( G ( f ) exp(? j 2? ft0 ) ) ( G* ( f ) exp( j 2? ft0 ) ) = ( G ( f ) exp(? j 2? ft0 ) ) ( G* ( f ) exp( j 2? ft0 ) ) = G ( f ) exp(? j 2? ft0 ) exp( j 2? ft0 ) 2 G( f ) 2 Copyright © 2009 John Wiley , Inc.

All Rights Reserved. 2. 14 b)Given that the two energy densities are equal, we only need to prove the result for one. From before, it was shown that the Fourier transform of the half-cosine pulse was: AT 1 [sinc(( f + fc )T ) + sinc(( f ? f c )T )] for fc = 2 2T After squaring, this becomes: sin(? ( f + f c )T ) sin(? ( f ? f c )T ) ? A2T 2 ? sin 2 (? ( f + f c )T ) sin 2 (? ( f ? f c )T ) + +2 ? ? 4 ? (? ( f + f c )T ) 2 (? ( f ? f c )T ) 2 ? 2T 2 ( f + f c )( f ? f c ) ? The first term reduces to: ?? ? sin 2 ? ? fT + ? 2 cos 2 (? fT ) 2 ? cos (? fT ) ? = = 2 2 2 2 2 ?? ? ? ? T ( f + fc ) ? ? ? ? fT + ? ? ? fT + ? ? 2? ? ? The second term reduces to: ?? ? sin 2 ? ? fT ? ? cos 2 (? fT ) 2? ? = 2 2 2 2 ?? ? T ( f ? fc ) ? ? fT ? ? ? 2? ? The third term reduces to: sin(? ( f + f c )T ) sin(? ( f ? f c )T ) cos(? ) ? cos 2 (2? fT ) = 1 ? ? 2T 2 ( f + f c )( f ? f c ) ? ? 2T 2 ? f 2 ? 2 ? 4T ? ? ? 1 ? cos(2? fT ) = 1 ? ? ? 2T 2 ? f 2 ? 2 ? 4T ? ? 2 cos 2 (? fT ) =? 1 ? ? ? 2T 2 ? f 2 ? 2 ? 4T ? ? Summing these terms gives: ? ? ? 2 2 2 2 ? 2 cos (? fT ) A T ? cos (? fT ) cos (? fT ) ? + ? 2 2 2 1 ?? 1 ?? 4? 2T 2 ? ? ? 1 ? 1 ? ? ? f+ ?? f ? ? ?? f + ? ?f? ? 2T ? ? 2T ? ? ? 2T ? 2T ? ? ?? ? 2 Copyright © 2009 John Wiley , Inc.

All Rights Reserved. 2. 14 b)Cont’d By rearranging the previous expression, and summing over a common denominator, we get: ? ? ? 2 2 ? 2 A T ? cos (? fT ) ? 2 4? 2T 2 ? ? 2 1 ? ? ?? f ? 2 ? ? 4T ? ? ?? ? ? ? A2T 2 ? cos 2 (? fT ) = 2 4? ? 4? T ? 1 1 4T 2 f 2 ? 1 2 ? ( )? ? 16 T 4 ? ? A2T 2 cos 2 (? fT ) ? = 2 ? ? ? ( 4T 2 f 2 ? 1)2 ? ? ? Copyright © 2009 John Wiley , Inc. All Rights Reserved. 2. 15 a)The Fourier transform of Let g ‘(t ) = dg (t ) dt ? ?? dg (t ) dt j 2? fG ( f ) By Rayleigh’s theorem: ?W T 2 2 ? ? g (t ) dt = 2 2 ?? ? G( f ) 2 df ?t = ? t = 2 g (t ) dt ? ? f 2 G ( f ) df 2 (? 2 g (t ) dt 2 ) 2 2 (t ) dt ? ? g ‘(t ) g ‘* (t )dt 4? 2 ( ? g (t ) dt ) 2 2 ? t 2 g * (t ) g ‘(t ) ? tg (t )g ‘* (t )dt ? ? ? ?? 2 2 16? 2 ? g (t ) dt 2 ( ) d ? ? * ? ? t ? dt ( g (t ) g (t ) ) dt ? ? =? 2 16? 2 ? g (t ) g * (t )dt 2 ( ) Using integration by parts, we can show that: ? ? d 2 2 t ? g (t ) dt = ? g (t ) ? dt ?? ?? 1 16? 2 1 ? WT ? 4? ?W 2T 2 ? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 15 b) For g (t ) = exp(?? t 2 ) g (t ) exp(?? f 2 ) ? 2 2 2 2 ? t exp(? 2? t )dt ? ? f exp(? 2? f )df ? ?? ? ?W 2T 2 = ?? ?? ? exp(? 2? t ? 0 2 )dt Using a table of integrals: ?x 2 exp(? ax 2 )dx = 1 ? 4a a for a > 0 ? t 2 exp(? 2? t 2 )dt = ?? ? ? 1 1 4? 2 1 4? 1 2 ?? ? ? f 2 exp(? 2? t 2 )df = 2 ?? ? exp(? 2? t )= 1 2 1? ? 2? 2 ? 1 ? 4? 2 2 ? T W = ? 1 2 2 ? 1 ? =? ? ? 4? ? 1 ? TW = 4? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 2. 16. ? Given: ?? ? ? ? x(t ) dt < ? and 2 ?? ? ? h(t ) dt < ? , which implies that ? ?? ? h(t ) dt < ? . 4 However, if ?? ? x(t ) dt < ? then 2 ?? ? ? X ( f ) df < ? and 2 ?? ? X ( f ) df < ? . This result also applies to h(t). Y( f ) = H( f )X ( f ) ? ?? ? ? Y ( f ) df = 2 ?? ? ? X ( f )H ( f ) ? X ? 2 * ( f ) H * ( f )df = ? 2 X ( f ) H ( f ) df ? 2 ?? ?? ? Y ( f ) df ? ?? ? ? X ( f ) df 4 ?? ? H ( f ) df 4 0, the frequencies are shifted inwards by ? f. ?Vo ( f ) contains {99. 98,199. 98,399. 98} Hz (b) When the lower side-band is transmitted, and ? f>0, then the baseband frequencies are shifted outwards by ? f. ?Vo ( f ) contains {100. 02, 200. 02, 400. 02} Hz Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 3. 22. f1 = f c ? ?f ? W f 2 = f c + ? f v1 (t )v2 (t ) = A1 A2 cos(2? f1t + ? 1 ) cos(2? f 2t + ? 2 ) = A1 A2 [cos(2? ( f1 ? f 2 )t + ? ? ?2 ) + cos(2? ( f1 + f 2 )t + ? 1 + ? 2 )] 2 The low-pass filter will only pass the first term. 1 ? LFP(v1 (t )v2 (t )) = A1 A2 [cos(? 2? (W + 2? f )t + ? 1 ? ?2 )] 2 Let v0(t) be the final output, before band-pass filtering. vo (t ) = = ? W + 2? f ? 1 ? 1 ? ?2 A1 A2 [cos(? 2? ? ) ? A2 cos(2? f 2t + ? 2 )] ? t + 2 ? W / ? f + 2 ? W / ? f + 2 1 ? ?? ? ?? 2 A1 A2 [cos(? 2?? ft + 1 2 ? ?2 ) ? cos(2? f 2t + 1 2 + ? 2 )] 2 n+2 n+2 ? ?? ? ?? 1 2 = A1 A2 [cos(? 2? ( f c + 2? f ) + 1 2 ? ?2 ) + cos(? 2? f c t + 1 2 + ? 2 )] 4 n+2 n+2 After band-pass filtering, retain only the second term. ? vo (t ) = 1 ? ?? 2 A1 A2 [cos(? 2? c t + 1 2 + ? 2 ) 4 n+2 + ? 2 = 0 n+2 n+2 rearranging and solving for ? 2 : ? ?1 ?2 ?2 = ? ?1 n +1 (b) At the second multiplier, replace v2(t) with v1(t). expression for the phase: This results in the following ?1 n+2 ? ?2 n+2 + ? 1 = 0 ?1 = ?2 n+3 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 3. 23. Assume that the mixer performs a multiplication of the two signals. y1 (t ) ? {1, 2,3, 4,5, 6, 7,8,9} MHz y2 (t ) ? {100, 200,300, 400,500, 600, 700,800,900} kHz This system essentially produces a DSB-SC signal centred around the frequency of y1(t). The lowest frequencies that can be produced are: 1 yo (t ) = [cos(2? f1 ? f 2 )t ) + cos(2? ( f1 + f 2 )t )] 2 f1 = 1 MHz f1 ? f 2 = 0. 9 MHz f 2 = 100 kHz f1 + f 2 = 1. 1 MHz The highest frequencies that can be produced are: f1 = 9 MHz f 2 = 900 kHz f1 ? f 2 = 8. 1 MHz f1 + f 2 = 9. 9 MHz The resolution of the system is the bandwidth of the output signal. Assuming that no branch can be zeroed, the narrowest resolution occurs with a modulation frequency of 100 kHz. The widest bandwidth occurs when there is a modulation frequency of 900 kHz. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 3. 24 Given the presence of the filters, only the baseband signals need to be considered.

All of the other product components can be discarded. (a) Given the sum of the modulated carrier waves, the individual message signals are extracted by multiplying the signal with the required carrier. For m1(t), this results in the conditions: cos(? 1 ) + cos( ? 1 ) = 0 cos(? 2 ) + cos( ? 2 ) = 0 cos(? 3 ) + cos( ? 3 ) = 0 ?? i = ? i ± ? For the other signals: m2 (t ) : cos(?? 1 ) + cos(? ?1 ) = 0 cos(? 2 ? ?1 ) + cos( ? 2 ? ?1 ) = 0 cos(? 3 ? ?1 ) + cos( ? 3 ? ?1 ) = 0 Similarly: m3 (t ) : (? 1 ? ? 2 ) = ( ? 1 ? ? 2 ) ± ? (? 3 ? ? 2 ) = ( ? 3 ? ? 2 ) ± ? m4 (t ) : (? 1 ? ? 3 ) = ( ? 1 ? ?3 ) ± ? (? 2 ? ? 3 ) = ( ? ? ? 3 ) ± ? ?1 = ? 1 ± ? (? 2 ? ?1 ) = ( ? 2 ? ?1 ) ± ? (? 3 ? ?1 ) = ( ? 3 ? ?1 ) ± ? (b) Given that the maximum bandwidth of mi(t) is W, then the separation between fa and fb must be | fa- fb|>2W in order to account for the modulated components corresponding to fa- fb. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 3. 25 b) The charging time constant is (rf + Rs )C = 1? s The period of the carrier wave is 1/fc = 50 ? s. The period of the modulating wave is 1/fm = 0. 025 s. ?The time constant is much shorter than the modulating wave and therefore should track the message signal very well.

The discharge time constant is: Rl C = 100? s . This is twice the period of the carrier wave, and should provide some smoothing capability. From a maximum voltage of V0, the voltage Vc across the capacitor after time t = Ts is: T Vc = V0 exp(? s ) Rl C Using a Taylor series expansion and retaining only the linear terms, will result in the T linear approximation of VC = V0 (1 ? s ) . Using this approximation, the voltage will Rl C decay by a factor of 0. 94 from its initial value after a period of Ts seconds. From the code, it can be seen that the voltage decay is close to this figure.

However, it is somewhat slower than what was calculated using the linear approximation. In a real circuit, it would also be expected that the decay would be slower, as the voltage does not simply turn off, but rather decreases over time. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 3. 25 c) The output of a high-pass RC circuit can be described according to: V0 (t ) = I (t ) R Qc (t ) = C (Vin (t ) ? V0 (t )) dQc dt ? dV (t ) dV (t ) ? V0 (t ) = RC ? in ? 0 ? dt ? ? dt Using first order differences to approximate the derivatives results in the following difference equation: RC RC V0 (t ) = V0 (t ? ) + (Vin (t ) ? Vin (t ? 1)) RC + Ts RC + Ts I (t ) = The high-pass filter applied to the envelope detector eliminates the DC component. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 3. 25. MATLAB code function [y,t,Vc,Vo]=AM_wave(fc,fm,mi) %Problem 3. 25 %Inputs: fc % fm % mi Carrier Frequency Modulation Frequency modulation index %Problem 3. 25 (a) fs=160000; %sampling rate deltaT=1/fs; %sampling period t=linspace(0,. 1,. 1/deltaT); %Create the list of time periods y=(1+mi*cos(2*pi*fm*t)). *cos(2*pi*fc*t); %Create the AM wave %Problem 3. 5 (b) %%%%Create the envelope detector%%%% Vc=zeros(1,length(y)); Vc(1)=0; %inital voltage for k=2:length(y) if (y(k)>(Vc(k-1))) Vc(k)=y(k); else Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Vc(k)=Vc(k-1)-0. 023*Vc(k-1); end end %Problem 3. 25 (c) %%%Implement the high pass filter%%% %%This implements bias removal Vo=zeros(1,length(y)); Vo(1)=0; RC=. 001; beta=RC/(RC+deltaT); for k=2:length(y) Vo(k)=beta*Vo(k-1)+beta*(Vc(k)-Vc(k-1)); end Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Chapter 4 Problems Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 4. 7. s (t ) = Ac cos(? (t )) ? (t ) = 2? f c t + k p m(t ) Let ? = 0. 3 for m(t) = cos(2? fmt). ? s (t ) = A c cos(2? f c t + ? m(t )) = Ac [cos(2? f c t ) cos( ? cos(2? f mt )) ? sin(2? c t ) sin( ? cos(2? f mt ))] for small ? : cos( ? cos(2? f mt )) 1 sin( ? sin(2? f mt )) ? cos(2? f mt ) ? s (t ) = Ac cos(2? f ct ) ? ? Ac sin(2? f c t ) cos(2? fmt ) = Ac cos(2? f c t ) ? ? Ac [sin(2? ( f c + f m )t ) + sin(2? ( f c + f m )t ) 2 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 4. 14. 2 = av12 s (t ) = Ac cos(2? f c t + ? sin(2? f mt )) = Ac cos(2? f c t + ? m(t )) v2 = a ? s 2 (t ) = a ? cos 2 (2? f c t + ? m(t )) = a ? cos(4? f c t + 2 ? m(t )) 2 The square-law device produces a new FM signal centred at 2fc and with a frequency deviation of 2?. This doubles the frequency deviation. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 4. 17. Consider the slope circuit response: The response of |X1(f)| after the resonant peak is the same as for a single pole low-pass filter.

From a table of Bode plots, the following gain response can be obtained: | X 1 ( f ) |= 1 ? f ? fB ? 1+ ? ? ? B ? 2 Where fB is the frequency of the resonant peak, and B is the bandwidth. For the slope circuit, B is the filter’s bandwidth or cutoff frequency. For convenience, we can shift the filter to the origin (with X 1 ( f ) as the shifted version). | X 1 ( f ) |= 1 ? f ? 1+ ? ? ? B? =? f = kB 2 d | X1 ( f ) | df k B(1 + k ) 3 2 2 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Because the filters are symmetric about the central frequency, the contribution of the second filter is identical.

Adding the filter responses results in the slope at the central frequency being: d | X(f )| df =? f = kB 2k B(1 + k ) 3 2 2 In the original definition of the slope filter, the responses are multiplied by -1, so do this here. This results in a total slope of: 2k B(1 + k ) 3 2 2 As can be seen from the following plot, the linear approximation is very accurate between the two resonant peaks. For this plot B = 500, f1=-750, and f2=750. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 4. 23 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 4. 24 The amplitude spectrum corresponding to the Gaussian pulse p(t ) = c exp ? ?? c 2t 2 ? * rect[t / T ] ? ? is given by the magnitude of its Fourier transform. ? ? P ( f ) = F ? c exp ?? c 2t 2 ? F ? rect ( t / T ) ? ? ? = c exp ? ?? f 2 c 2 ? Tsinc [ fT ] ? ? where we have used the convolution theorem ) Problem 4. 25 The Carson rule bandwidth for GSM is BT = 2 ( ? f + W ) where the peak deviation is given by k c 1 ? f = f = B 2? / log(2) = 0. 75 B 2? 4 With BT = 0. 3 and T = 3. 77 microseconds, the peak deviation is 59. 7 kHz From Figure 4. 22, the one-sided 3-dB bandwidth of the modulating signal is approximately 50 kHz. Combining these two results, the Carson rule bandwidth is BT = 2 ( 59. 7 + 50 ) = 219. 4 kHz The 1-percent FM bandwidth is given by Figure 4. 9 with ? = vertical axis we find that ?f 59. 7 = = 1. 19 . From the W 50 BT = 6 , which implies BT = 6(59. 7) = 358. 2 kHz. ?f Copyright © 2009 John Wiley & Sons, Inc.

All Rights Reserved. Problem 4. 26. a) Beta 1 2 5 10 # of side frequencies 1 2 8 14 b)By experimentation, a modulation index of 2. 408, will force the amplitude of the carrier to be about zero. This corresponds to the first root of J0(? ), as predicted by the theory. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 4. 27. a)Using the original MATLAB script, the rms phase error is 6. 15 % b)Using the plot provided, the rms phase error is 19. 83% Problem 4. 28 a)The output of the detected signal is multiplied by -1. This results from the fact that m(t)=cos(t) is integrated twice.

Once to form the transmitted signal and once by the envelope detector. In addition, the signal also has a DC offset, which results from the action of the envelope detector. The change in amplitude is the result of the modulation process and filters used in detection. f ? ? b)If s (t ) = sin(2? f mt ) + 0. 5cos ? 2? m t ? , then some form of clipping is observed. 3 ? ? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. The above signal has been multiplied by a constant gain factor in order to highlight the differences with the original message signal. c)The earliest signs of distortion start to appear above about fm =4. kHz. As the message frequency may no longer lie wholly within the bandwidth of either the differentiator or the low-pass filter. This results in the potential loss of high-frequency message components. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 4. 29. By tracing the individual steps of the MATLAB algorithm, it can be seen that the resulting sequence is the same as for the 2nd order PLL. e(t ) is the phase error ? e (t ) in the theoretical model. The theoretical model of the VCO is: ?2 (t ) = 2? kv ? v(t )dt 0 t and the discrete-time model is: VCOState = VCOState + 2? kv (t ? )Ts which approximates the integrator of the theoretical model. The loop filter is a PI-controller, and has the transfer function: a H ( f ) = 1+ jf This is simply a combination of a sum plus an integrator, which is also present in the MATLAB code: Filterstate = Filterstate + e(t ) Integrator v(t ) = Filterstate + e(t ) Integrator +input b)For smaller kv, the lock-in time is longer, but the output amplitude is greater. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. c)The phase error increases, and tracks the message signal. d)For a single sinusoid, the track is lost if f m ?

K 0 where K 0 = k f kv Ac Av For this question, K0=100 kHz, but tracking degrades noticeably around 60-70 kHz. e)No useful signal can be extracted. By multiplying s(t) and r(t), we get: Ac Av ? sin(k f ? ? VCOState) + sin(4? f c t + k f ? + VCOState) ? ? 2 ? This is substantially different from the original error signal, and cannot be seen as an adequate approximation. Of particular interest is the fact that this equation is substantially more sensitive to changes in ? than the previous one owing to the presence of the gain factor kv Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc.

All Rights Reserved. Chapter 5 Problems 5. 1. ( x ? ? x )2 exp(? ) (a) Given f ( x) = 2 2 2? x 2?? x 1 exp(?? f 2 ) , then by applying the time-shifting and scaling properties: 2 2 2?? x exp(?? ( 2?? x ) 2 ? f 2 ) exp( j 2? f ? x ) and exp(?? t 2 ) F( f ) = 1 2?? 2 x 2 = exp(?? 2 2? x f 2 + j ? x 2? f ) 1 2 = exp( j?? x ? ? 2? x ) 2 and let ? = 2? f (b)The value of ? x does not affect the moment, as its influence is removed. Use the Taylor series approximation of ? x(x), given ? x = 0. ?x (? ) = exp(? ? 2? x2 ) x2 exp( x) = ? n =0 n ! E[ X n ] = d n? x (? ) d? n v =0 k ? 1 2 2k 2k ? ? 1? ? ? ? ? x (? ) = ? ? ? ? x k! 2? k =0 ?

For any odd value of n, taking d n? x (? ) leaves the lowest non-zero derivative as ? 2k-n. d? n When this derivative is evaluated for v=0, then E[ X n ] =0. For even values of n, only the terms in the resulting derivative that correspond to ? 2k-n = ? 0 are non-zero. In other words, only the even terms in the sum that correspond to k = n/2 are retained. ? E[ X n ] = n! ? x2 (n / 2)! Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 5. 2. (a) All the inputs for x ? 0 are mapped to y = 0. However, the probability that x > 0 is unchanged. Therefore the probability density of x ? 0 must be concentrated at y=0. ? b) Recall that ?? ? f x x)dx = 1 where f x ( x) is an even function. Because fy(y) is a probability distribution, its integral must also equal 1. ? ? ? 0 ? f x ( x)dx = 0. 5 and ? f y ( y )dy = 0. 5 0 + Therefore, the integral over the delta function must be 0. 5. This means that the factor k must also be 0. 5. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 5. 3 (a) p y ( y ) = p y ( y | x0 ) P( x0 ) + p y ( y | x1 ) P( x1 ) Assume: P( x0 ) = P( x1 ) = 0. 5 1 ? p y ( y ) = [ p y ( y | x0 ) + p y ( y | y1 ) 2 1 ( y + 1) 2 ( y ? 1) 2 py ( y) = [exp(? ) + exp(? )] 2? 2 2? 2 2 2?? 2 (b) P ( y ? ? ) = ? p y ( y )dy ? Use the cumulative Gaussian distribution, ? ? ,? 2 ( y ) = ?? ? y ( y ? ? )2 exp(? )dy 2? 2 2?? 2 1 1 ? P( y ? ? ) = [? ?1,? 2 (?? ) + ? 1,? 2 (?? )] 2 1 y?? )] But, ? ? ,? 2 ( y ) = [1 + erf ( 2 ? 2 1 ? ?? + 1 ? ? ?? ? 1 ? ? P( y ? ? ) = [2 + erf ? ? + erf ? ?] 2 ?? 2 ? ?? 2 ? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 4 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 5 If, for a complex random process Z(t) RZ (? ) = E [ Z *(t ) Z (t + ? ) ] then (i) The mean square of a complex process is given by RZ (0) = E [ Z *(t ) Z (t ) ] 2 = E ? Z (t ) ? ? ? (ii) We show RZ (? has conjugate symmetry by the following RZ (?? ) = E [ Z *(t ) Z (t ? ? ) ] = E [ Z *( s + ? ) Z ( s ) ] = E [ Z ( s ) Z ( s + ? )] * * = RZ (? ) where we have used the change of variable s = t – ?. (iii) Taking an approach similar to that of Eq. (5. 67) 2 0 ? E ? ( Z (t ) ± Z (t + ? ) ) ? ? ? ? ? = E ? ( Z (t ) ± Z (t + ? ) )( Z *(t ) ± Z *(t + ? ) ) ? ? ? = E [ Z (t ) Z *(t ) ± Z (t ) Z *(t + ? ) ± Z *(t ) Z (t + ? ) + Z (t + ? ) Z *(t + ? ) ] 2 2 = E ? Z (t ) ? ± E [ Z (t ) Z *(t + ? )] ± E [ Z *(t ) Z (t + ? ) ] + E ? Z (t + ? ) ? ? ? ? ? 2 = 2E ? Z (t ) ? ± 2 Re {E [ Z *(t ) Z (t + ? ) ]} ? ? = 2 RZ (0) ± 2 Re { RZ (? } Thus Re { RZ (? )} ? RZ (0) . Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 6 (a) E[ Z (t1 ) Z * (t2 )] = E[( A cos(2? f1t1 + ? 1 ) + jA cos(2? f 2t1 + ? 2 )) ? ( A cos(2? f1t2 + ? 1 ) + jA cos(2? f 2t2 + ? 2 ))] Let ? 1=2? f1 ? 2=2? f2 After distributing the terms, consider the first term: A2 E[cos(? 1t1 + ? 1 ) cos(? 1t2 + ? 1 )] = A2 E[cos(? 1 (t1 ? t2 )) + cos(? 1 (t1 + t2 ) + 2? 1 )] 2 The expectation over ? 1 goes to zero, because ? 1 is distributed uniformly over [-? ,? ]. This result also applies to the term A2 [cos(? 2t1 + ? 2 ) cos(? 2t2 + ? 2 )] . Both cross-terms go to zero. R(t1 , t2 ) = A2 [cos(? 1 (t 1 ? t2 )) + cos(? 2 (t1 ? t2 ))] 2 (b) If f1 = f2, only the cross terms may be different: E[ jA2 (cos(? 1t1 + ? 2 ) cos(? 1t2 + ? 1 ) + cos(? 1t1 + ? 2 ) cos(? 1t2 + ? 1 )] But, unless ? 1=? 2, the cross-terms will also go to zero. ? R(t1 , t2 ) = A2 cos(? 1 (t1 ? t2 )) (c) If ? 1=? 2, then the cross-terms become: ? jA2 E[cos((? 1t1 ? ?2t2 )) + cos((? 1t1 + ? 2t2 ) + 2? 1 ) + jA2 E[cos((? 2t1 ? ?1t2 )) + cos((? 1t1 + ? 2t2 ) + 2? 1 )] After computing the expectations, the cross-terms simplify to: jA2 [cos(? 2t1 ? ?1t2 ) ? cos(? 1t1 ? ?2t2 )] 2 ? RZ (t1 , t2 ) = A2 [cos(? 1 (t1 ? t2 )) + cos(? (t1 ? t2 )) + j cos(? 2t1 ? ?1t2 ) ? j cos(? 1t1 ? ?2t2 )] 2 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 7 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 8 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 9 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 10 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 11 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 12 Copyright © 2009 John Wiley & Sons, Inc.

All Rights Reserved. Problem 5. 13 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 14 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 15 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 16 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 17 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 18 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 19 Problem 5. 20 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

Problem 5. 21 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 22 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 23 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 24 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 25 Problem 5. 26 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 27 Copyright © 2009 John Wiley & Sons, Inc.

All Rights Reserved. Problem 5. 28 c)For a given filter, H ( f ) , let ? = ln H ( f ) ? and the Paley-Wiener criterion for causality is: ?? ? 1 + (2? f ) df < ? 2 ?( f ) For the filter of part (b) 1 ? ( f ) = [ ln(2) + ln( S x ( f ) ? ln( N 0 )] 2 The first and the last terms have no impact on the absolute integrability of the previous expression, and so do not matter as far as evaluating the above criterion. This leaves the only condition: ? ln S x ( f ) ? 1 + (2? f )2 df < ? ?? Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 29 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 30

Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 31 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 32 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 33 (a) The receiver position is given by x(t) = x0+vt Thus the signal observed by the receiver is ? ? x ?? r (t , x) = A( x) cos ? 2? f c ? t ? ? ? ? c ?? ? ? ? x + vt ? ? = A( x) cos t ? 2? f c ? t ? 0 c ?? ? ?? ? ? ? f v? x ? = A( x) cos ? 2? ? f c ? c ? t ? f c 0 ? c ? c? ? ? The Doppler shift of the frequency observed at the receiver is f D = fc v . (b) The expectation is given by E ? exp ( j 2? f n? ) ? = ? ? = 1 2? 1 2? ? ?? ? exp ( j 2? f ? cos? ) d? D n n ? ? = J 0 ( 2? f D? ) where the second line comes from the symmetry of cos and sin under a -? /2 translation. Eq. (5. 174) follows directly from this upon noting that, since the expectation result is real-valued, the right-hand side of Eq. (5. 173) is equal to its conjugate. ?? exp ( j 2? f ? sin? ) d? D n n Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 5. 34 The histogram has been plotted for 100 bins. Larger numbers of bins result in larger errors, as the effects of averaging are reduced.

Distance 0? 1? 2? 3? 4? Relative Error 0. 94% 2. 6 % 4. 8 % 47. 4% 60. 7% The error increases further out from the centre. It is also important to note that the random numbers generated by this MATLAB procedure can never be greater than 5. This is very different from the Gaussian distribution, for which there is a non-zero probability for any real number. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 5. 34 Code Listing %Problem 5. 34 %Set the number of samples to be 20,000 N=20000 M=100; Z=zeros(1,20000); for i=1:N for j=1:5 Z(i)=Z(i)+2*(rand(1)-0. ); end end sigma=sqrt(var(Z-mean(Z))); %Calculate a histogram of Z [X,C]=hist(Z,M); l=linspace(C(1),C(M),M); %Create a gaussian function with the same variance as Z G=1/(sqrt(2*pi*sigma^2))*exp(-(l. ^2)/(2*sigma^2)); delta2=abs(l(1)-l(2)); X=X/(20000*delta2); Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 5. 35 (a) For the generated sequence: ? ? y = ? 0. 0343 + j 0. 0493 ? 2 ? y = 5. 597 The theoretical values are: ? y = 0 (by inspection). 2 The theoretical value of ? y =5. 56. See 5. 35 (c) for the calculation. 5. 35 (b) From the plots, it can be seen that both the real and imaginary components are approximately Gaussian.

In addition, from statistics, the sum of tow zero-mean Gaussian signals is also Gaussian distributed. As a result, the filter output must also be Gaussian. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. 5. 35 (c) y (n) = ay (n ? 1) + w(n) Y ( z ) = aY ( z ) z ? 1 ? H ( z) = 1 1 ? az ? 1 -1 h( n) = a u ( n) n Rh(z) = H(z)H(z ) = 1 (1 ? az )(1 ? az ) ?1 = But, Ry(z) = Rh(z)Rw(z) Taking the inverse z-transform: ry (n) = 2 ? w a z ? 1 1 1 + 2 ? 1 2 1 ? a 1 ? az 1 ? a 1 ? az 1 ? a2 an ?? < n < ? From the plots, the measured and observed autocorrelations are almost identical. Copyright © 2009 John Wiley & Sons, Inc.

All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Chapter 6 Solutions Problem 6. 1 The transfer function of the filter can be readily found to be 1 H( f ) = 1 + j 2? fRC Likewise, the power spectral density of the filtered noise is N0 / 2 SN ( f ) = 1 + (2? fRC ) 2 Let a=1/RC, SN ( f ) = aN 0 / 2 a + (2? f ) 2 2 Therefore, noting that 2a exp(? a ? ) 2 a + (2? f ) 2 then the autocorrelation function of the output noise is RN (? ) = ? N0 exp(? ) 4 RC RC N0 4 RC For a zero-mean noise signal, the output power is simply RN(0), which is The output of the filtered sinusoid is: S( f ) = A 1 [? f ? f c ) + ? ( f + f c )] ? 2 1 + j 2? fRC And the resulting output power is A2 1 2 1 + ( 2? fRC )2 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Therefore the SNR is: 4 RCA2 1 10 ? log10 dB N 0 1 + (2? fRC ) 2 Problem 6. 2 The transfer function of the circuit can be found to be R H( f ) = 1 R + j 2? fL + j 2? fC where f c = 1 2? LC and Q = 1 L R C ?H( f ) = For Q 1 1 + jQ[( f / f c ) ? ( f c / f )] 1 , the transfer function may be approximated as follows: f >0 f 0 f >0 which is a symmetric function about the f = 0 axis. However, S NI (t ) = S NQ (t ) = S N ( f ? c ) + S N ( f + f c ) Around f = fc, this allows us to approximate the PSDs of the in-phase and quadrature components as follows Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. S NI ( f ) = S NQ ( f ) N0 1 + (2Qf / f c ) 2 The variance of the in-phase and quadrature components of n(t) however, are the same as the variance of n(t) itself. Therefore, by taking the inverse Fourier transform of the above PSD and setting ? =0, we obtain ?f 1 ? fc exp(? c ? ) 2 Q Q 1 ? fc RN (0) = 2 Q which is the approximate variance (power) of the narrow band noise. RN (? ) = Therefore, the SNR is, A2 Q 10 ? og10 dB ? fc Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 3 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 4 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 5 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 6 Problem 6. 7 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 8 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. Problem 6. 10 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 11 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 12 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 13 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 24 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 15 Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 16

The following Matlab script simulates the generation and detection of an AM-modulated signal in noise. %—————————————-% Matlab code for Problem 6. 16 %—————————————function Prob6_16() Fs = 143; % sample rate (kHz) t = [0: 1/Fs : 100]; % observation period (ms) Fc = 20; % carrier frequency (kHz) Fm = 0. 1; % modulation frequency (kHz) Ka = 0. 5; % modulation index SNRc = 25; % Channel SNR (dB) Ac = 1; tau = 0. 25/4; %————————————% Modulated signal %————————————m = cos(2*pi*fm*t); C = Ac*cos(2*pi*fc*t); s = (1 + ka*m). c; subplot(4,1,1), plot(t,s), grid on P = std(m)^2; %—————————————————% Add narrowband noise % Create bandpass noise by low-pass % filtering AWGN noise and converting to % bandpass %———————————————————–P_AM = Ac^2*(1+ka^2*P)/2; N = P_AM/10. ^(SNRc/10); sigma = sqrt(N); %— Create bandpass noise by low-pass filtering complex noise –noise = randn(size(s)) + j*randn(size(s)); LPFnoise = LPF(Fs, noise, tau); BPnoise = real(LPFnoise . * exp(j*2*pi*fc/Fs*[1:length(s)])); scale = 2*sigma / std(BPnoise); s_n = s scale * BPnoise; subplot(4,1,2), plot(t,s_n), grid on %— Envelope detection of both noisy and noise-free signals –ED = EnvDetector(t,s); ED_n = EnvDetector(t,s_n); %— Remove transient and dc –ED = ED(400:end); ED_n = ED_n(400:end); t = t(400:end); ED = ED – mean(ED); ED_n = ED_n – mean(ED_n); %— Low pass filter —BBsig = LPF(Fs,ED,tau); BBsig_n = LPF(Fs,ED_n,tau); Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. %— plot results —–subplot(4,1,3), plot(t,BBsig); subplot(4,1,4), plot(t,BBsig_n) %————————————–% Envelope Detector from Problem 3. 5 %————————————–function Vc = EnvDetector(t,s); Vc(1) = 0; % initial capacitor voltage for i = 2:length(s) if s(i) > Vc(i-1) % diode on Vc(i) = s(i); else % diode off Vc(i) = Vc(i-1) – 0. 023*Vc(i-1); end end % plot(t, Vc), grid on return; %————————————–% Low pass filter %————————————–function y = LPF(Fs, x, tau); % tau = 1; % time constant of RC filter (ms) t1 = [0: 1/Fs : 5*tau]; h = exp(-t1/tau) * 1/Fs; y = filter(h, 1, x); return; The Matlab script produces the following plot: Copyright © 2009 John Wiley & Sons, Inc.

All Rights Reserved. 2 0 -2 2 0 -2 0. 05 0 -0. 05 0. 05 0 -0. 05 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 Time (ms) 70 80 90 100 Figure 6. 16 Plot from Matlab script (a) AM modulated carrier (b) AM modulated carrier plus noise (c) AM demodulated signal in absence of noise (d) AM demodulated signal in noise Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. Problem 6. 17 The following Matlab script simulates the generation and detection of an FM modulated signal in noise. —————————————-% Problem 6. 17 %—————————————-function b = Prob6_17; %— Parameters ——————fc = 100; % Carrier frequency (kHz) Fs = 1024; % Sampling rate (kHz) fm = 0. 5; % Modulating frequency (kHz) Ts = 1/Fs; % Sample period (ms) t = [0:Ts:10]; % Observation period (ms) C_N = 20 % channel SNR (dB) Ac = 1; Bt = 20 % (kHz) W = 5; % (kHz) SNRc = C_N+10*log10(Bt/W); %— Message signal ————–m = cos(2*pi*fm*t); % modulating signal kf = 2. ; % modulator sensitivity index (~Bt/2) (kHz/V) %— FM modulate —————FMsig = FMmod(fc,t,kf,m,Ts); %— Add narrowband noise ——–%— Create bandpass noise by low-pass filtering complex noise –P = 0. 5; N = P/10. ^(SNRc/10); sigma = sqrt(N); noise = randn(size(FMsig)) + j*randn(size(FMsig)); LPFnoise = LPF(Fs, noise, 0. 05); % 0. 01 => Bt ` 50 kHz eq. Noise BW BPnoise = real(LPFnoise . * exp(j*2*pi*fc/Fs*[1:length(FMsig)])); scale = sigma / std(BPnoise); FMsign = FMsig + scale * BPnoise; subplot(4,1,1), plot(t,FMsig), grid on subplot(4,1,2), plot(t,FMsign), grid on — FM receiver —Rx_c = FMdiscriminator(fc,FMsig,Ts); Rx_n = FMdiscriminator(fc,FMsign,Ts); t = t(round(1/Ts):end); % remove transient subplot(4,1,3), plot(t,Rx_c); grid on subplot(4,1,4), plot(t,Rx_n); grid on %— Plot result ——–% FFTsize = 4096; %S = spectrum(FMsig,FFTsize); % % Freq = [0:Fs/FFTsize:Fs/2]; Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved. % subplot(2,1,1), plot(t,s), xlabel(‘Time (ms)’), ylabel(‘Amplitude’); % axis([0 0. 5 -1. 5 1. 5]), grid on % subplot(2,1,2), stem(Freq,sqrt(S/682)), xlabel(‘Frequency (kHz)’), ylabel(‘Amplitude Spectrum’); % axis([95 105 0 1]), grid on ———————————————–% FM modulator %———————————————–function s = FMmod(fc,t,kf,m,Ts); theta = 2*pi*fc*t+ 2*pi*kf * cumsum(m)*Ts; % integrate signal s = cos(theta); %———————————————–% FM discriminator %———————————————–function D3 = FMdiscriminator(fc,S, Ts) t = [0:Ts:10*Ts]; % for filter %— FIR differentiator (Fs = 1024 kHz, BT/2 = 10 kKhz) –FIRdiff = [ 1. 0385 0. 0 0. 0 0. 0 -0. 0 0. 0 0. 0 -0. 0 -0. 0 -0. 0 -1. 60385]; BP_diff = real(FIRdiff . * exp(j*2*pi*fc*t)); %— Lowpass filter – Fs = 1024 kHz, f3dB = 5 kHz —-LPF_B = 1E-4 *[ 0. 0706 0. 2117 0. 2117 0. 0706]; LPF_A = [1. 0000 -2. 9223 2. 8476 -0. 252]; D1 = filter(BP_diff, 1, S); % Bandpass discriminator D2 = EnvDetect(D1); % Envelope detection D2 = D2 – mean(D2); % remove dc D3 = filter(LPF_B,LPF_A, D2); % Low-pass filtering D3 = D3(round(1/Ts):end); % remove transient (approx 1s) %————————————–% Envelope Detector %————————————–function Vc = EnvDetect(s); Vc(1) = 0; % initial capacitor voltage for i = 2:length(s) if s(i) > Vc(i-1) % diode on Vc(i) = s(i); else % diode off Vc(i) = Vc(i-1) – 0. 05*Vc(i-1); end end return; %————————————–% Low pass filter %————————————–function y = LPF(Fs, x, tau); % tau = 1; % time constant of RC filter (ms) t1 = [0: 1/Fs : 5*tau]; h = exp(-t1/tau) * 1/Fs; y = filter(h, 1, x); return; Copyright © 2009 John Wiley & Sons, Inc. All Rights Reserved.

Haven’t Found A Paper?

Let us create the best one for you! What is your topic?

By clicking "SEND", you agree to our terms of service and privacy policy. We'll occasionally send you account related and promo emails.

Eric from Graduateway Hi there, would you like to get an essay? What is your topic? Let me help you

logo

Haven't found the Essay You Want?

Get your custom essay sample

For Only $13.90/page