Tutorial 3 answers

TUTORIAL 3 (answer)

1.

Discuss the advantages of Frequency Modulation (FM) over Amplitude Modulation (AM) Solution:  Less sensitivity to noise. Noise is amplitude variations that are eliminated in the receiver.  Capture effect. Only the strongest signal on the receive frequency gets through.  Transmission efficiency. More efficient transmitter amplifiers can be used with FM.  The SNR can be increased without increasing transmitted power about 25dB higher than in AM  Certain forms of interference at the receiver are more easily to suppressed, as FM receiver has a limiter which eliminates the amplitude variations and fluctuations.  The modulation process can take place at a low level power stage in the transmitter, thus a low modulating power is needed.  Power content is constant and fixed, and there is no waste of power transmitted  There are guard bands in FM systems allocated by the standardization body, which can reduce interference between the adjacent channels.

2. Describe why FM is more resistant to noise than AM. Noise & interference usually corrupt the signals by changing their voltage, by adding spikes or some other ways by changing the shape of the signal; however it cannot change its frequency. Since FM signals only change their frequency, not their amplitude, it is possible to design a receiver so it ignores amplitude changes. (For instance, an FM radio often contains a limiter circuit which clips the tops off the signal to make the signal the same height no matter what level it comes in). This won’t work on very weak signals, since then there’s nothing to clip, but for strong signal, FM can be remarkably free of noise.

3. Give three comparisons between the narrowband FM (NBFM) and the wideband FM (WBFM).

4. Relation between PM and FM

 (t ) Phase in PM is directly proportional to m(t) Phase in PM is directly proportional to integral of m(t)

 k p m(t ) t

 (t )  2 k f  m( )d 

t

2 k f  m f ( )d  k p m p (t ) 

mp (t ) 

2 k f

m f (t ) 

kp



t



m f ( )d

k p dmp (t ) 2 k f

dt

5.

A carrier of frequency 100 MHz is frequency modulated by a signal x(t)=20sin (200πx103t ). What is the bandwidth of the FM signal if the frequency sensitivity of the modulator is 25 kHz per volt?

Solution: Fc=100Mhz; kf =25kHz/V 3 Vm(t) = 20 sin( 200x10 t )

2πfm=200π x 103t 200  10 3 2 Therefore, fm= =100000Hz

kf (Vm)  25000(20) / 100000  5 fm m=

From Bessel Table, m=5; n=8 Hence, BW= 2 (n x fm)= 2 (8 x 100000)= 1.6Mhz 6. What is the bandwidth required for an FM wave in which the modulating frequency signal is 2 KHz and the maximum frequency deviation is 12 KHz? Solution: Fm=2kHz; ∆f=12kHz; f 12k M=  6 fm 2k From Bessel Table, m=6; n=9. Hence, BW= 2 (n x fm)= 2 (9 x 2000)= 36kHz

9.

A message signal vm (t )  2 sin( 10000t ) volt is used to frequency modulate the carrier vc (t )  10 sin( 180 x10 6 t ) volt. Given the percent modulation is 70%. Determine: i) The frequency sensitivity of the modulator circuit. ii) The frequency deviation of the modulated carrier. Solution: i) k1=m(fm) / vm = 0.7(5000)/2 =1.75kHz/V ii) ∆f = vmk1 =m(fm) = 0.7(5000) = 3500kHz

10.

A message signal vm (t )  2 sin( 30000t ) volt is used to frequency modulate the carrier vc (t )  10 sin( 200 x10 6 t ) volt. The frequency sensitivity of the modulating circuit is k1 = 25 kHz/V. Determine the modulation index. Solution: vmk1 2(25 kHz/V) m = ------- = -------------------= 3.33 fm 15000

11.

Input to an FM modulator with a modulation index m=1.5, is a modulating signal Vm(t)=Vm sin (2π1000t), and an unmodulated carrier Vc(t)=20 sin(2π500kt).Using the Bessel function table, determine i) ii) iii) iv) v)

number of sets of significant side frequencies Their amplitudes Draw the frequency spectrum showing their relative amplitudes The bandwidth The bandwidth if the amplitude of the modulating signal increases by a factor 2.65 Assume load resistance RL =50Ω; determine the unmodulated carrier power The total power in the angle modulated wave.

vi) vii)

Solution: i)

From Bessel table/graph, m=1.5 yields a reduced carrier component and 4 sets of significant side frequencies.

ii)

Relative amplitudes: J0 = 20(0.51) = 10.2V J1 = 20(0.56) =11.2 V J2 = 20(0.23) = 4.6 V J3 = 20(0.06) = 1.2 V J4 = 20(0.01) = 0.2 V

iii) 11.2VVV 11.2VVV 10.2V 4.6V

4.6V

1.2V

1.2V

0.2V

0.2V f(kHz)

496

497

498

499 500

501

502

503

504

iv)

Bandwidth = 2 (n x fm) = 2x4x1000 = 8000 Hz

v)

Bandwidth will be different for the same deviation sensitivities, kf k1 = m*fm/ Vm = 1500/Vm m = k1Vm/fm = 1500/Vm*2.65Vm / 1kHz = 3.975 ~ 4 for m = 4, no of sidebands = 7 BW = 2*n*fm = 2*7*1kHz = 14 kHz

vi) Pc

(Vc)2 (20)2 = ------ = -------- = 4W 2R 2(50)

vii) 2 2 2 2 V 2 ( V ) ( V ) 2 ( V ) c 2 n 1 2 P  P  P  P  ..  P     ..  t 0 1 2 n 2 R2 R 2 R 2 R

2 2 2 2 2 ( 10 . 2 ) 2 ( 11 . 2 ) 2 ( 4 . 6 ) 2 ( 1 . 2 ) 2 ( 0 . 2 ) P      4 . 002 W t 2 ( 50 ) 2 ( 50 ) 2 ( 50 ) 2 ( 50 ) 2 ( 50 )