Communication Systems 2010 Spring – Midterm Thursday, April 22, 2010. 1. Closed book. 2. 13:10~16:00 (170 minutes). 1. [13] The Fourier transform of a signal g(t) is denote by G( f ). Prove the following properties: i. [3] If a real signal g(t) is an even function of time t, the Fourier transform G( f ) is purely real. If a real signal g(t) is an odd function of time t, the Fourier transform G( f ) is purely imaginary. n j (n) (n) G f , where G f is the nth derivative of G( f ) with respect to f. 2π n ii. [3] t g (t ) iii. [3] g1 (t ) g2* (t ) G1 ( )G2* ( f ) d n j iv. [2] t g (t ) dt G ( n ) 0 2π v. [2] n g1 (t ) g2* (t ) dt G1 ( )G2* ( ) d 2. [7] Please show that the Fourier transform of a periodic signal gT0 t of period T0 is given by: gT0 (t ) m g (t mT0 ) g (t ), g (t ) T0 0, Hint: gT0 (t ) 1 T0 G(nf ) ( f nf ) n 0 0 where G( f ) is the Fourier transform of T0 T t 0 2 2. elsewhere c n n exp j 2 nf 0t , where cn 1 T0 T0 2 T0 2 gT0 t exp j 2 nf 0t dt . 3. [5] Properties of Filter i. [1] For low-pass and band-pass filters, what’s the relationship between the number of zeros and the number of poles? ii. [1] In the s-plane, where are the poles of the transfer function for causal systems? iii. [1] Where are the zeros of the transfer function for the Butterworth filter? iv. [1] What kind of filter is said to have a maximally flat passband response? v. [1] Is the ideal low-pass filter causal or noncausal? Please explain your answer. 4. [10] A tapped delay-line filter consists of N weights, where N is odd. It is symmetric with respective to the center tap, that is, the weights satisfy the condition wn wN 1n 0 n N 1 i. [6] Find the amplitude response of the filter. Hint: H ( f ) h k k exp j 2 fk ii. [4] Show that this filter has a linear phase response. 5. [5] The analysis of a band-pass system can be replaced by an equivalent but much simpler low-pass analysis that completely retains the essence of the filtering process. Please illustrate the procedure. 6. [15] Please show that the generation of an AM wave may be accomplished by using the following switching modulator: 7. [10] Consider the models shown in the following figure for generating the VSB signal and for coherently detecting it. Please derive the condition of the filter H( f ) to permit accurate recovery of the message signal. 8. [10] Consider a square-law detector, using a nonlinear device whose transfer characteristic is defined by 2 t a1v1 t a2v12 t , where a1 and a2 are constants, 1 t is the input, and 2 t is the output. The input consists of the AM wave 1 t Ac 1 ka m t cos 2πf ct . i. [5] Evaluate the output 2 t . ii. [5] Find the conditions for which the message signal m(t) may be recovered from 2 t . 9. [10] Consider a message signal m(t) containing frequency components at 100, 200, and 400 Hz. This signal is applied to an SSB modulator together with a carrier at 100 kHz. In the coherent detector used to recover m(t), the local oscillator supplies a sine wave of frequency 100.02 kHz. The signal is transmitted by SSB modulator as follow: s t Ac m t cos 2πf ct mˆ t sin 2πf ct 2 i. [5] Assuming that only the upper sideband is transmitted. Determine the frequency components of the detector output. ii. [5] Repeat your analysis, assuming that only the lower sideband is transmitted. 10. [10] Consider the following figure where (a) is the carrier wave and (b) is the sinusoidal modulating signal. Please answer the following questions and justify your answers. i. [4] Which is the phase modulated signal? ii. [3] Which is the frequency modulated signal? iii. [3] Which is the amplitude modulated signal? 11. [5] In the case of sinusoidal modulation, the FM signal is given by s t Ac cos 2 f ct sin 2 f mt where is the modulation index, fc is the carrier frequency, and fm is the frequency of the sinusoidal modulating signal. Please show that the expression of a narrow-band FM signal is similar to that of an AM signal given by: 1 s AM t Ac cos 2 f ct Ac cos 2 f c f m t cos 2 f c f m t . 2 2sin cos =sin( )+sin( ) 2cos sin =sin( )-sin( ) sin sin 2sin cos 2 2 sin sin 2cos sin 2 2 sin A B sinAcosB cosAsinB 2sin sin =cos( )-cos( ) 2cos cos =cos( )+cos( ) cos cos 2cos cos 2 2 cos cos 2sin sin 2 2 cos A B cosAcosB sinAsinB sin A B sinAcosB cosAsinB cos A B cosAcosB sinAsinB 2 2 A 1 cos A cos 2 2 A 1 cos A sin 2 2 n n Fourier Series: gT0 t a0 an cos 2π t bn sin 2π t n 1 T0 T0 a0 1 T0 T0 2 T0 2 gT0 t dt ; an 2 T0 n 2 gT0 t cos 2 t dt ; bn 2 T0 T0 T0 2 T0 n gT0 t sin 2 t dt 2 T0 T0 2 T0
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