Problems on Modulation techniques Quote of the day “Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.” ― Albert Einstein Important formulae to solve AM problems e(t ) Ec cos(c t ) where c 2f c Em m Ec em (t ) Em cosm t Emax Ec Em Bandwidth 2 f m Pt (total ) E 2c 2 Emin Ec Em Emax Emin m Emax Emin E 2c 2 2 E E Pc 2 c Pc c R 2R 2 m2 m2 1 1 Pc 2 2 m 2 PLSB PUSB for R 1 m 2 E 2c m 2 Pc 8 4 / 2 Pc m2 transmission efficiency 2 1 m 2 Pc 2 m 2 For an AM, amplitude of modulating signal is 0.5 V and carrier amplitude is 1V.Find Modulation Index. Solution: Ec = 1 V and Em = 0.5 V We know that Em m Ec 0.5 m 1 m 0.5 When the modulation percentage is 75%, an AM transmitter radiates 10KW Power. How much of this is carrier Power? Solution: Pt = 10 KW and m=0.75 We know that m 1 2 2 Pt (total ) Pc 10 103 Pc 0.752 1 2 Pc Pt (total ) m2 1 2 Pc 7.8 103W Pc 7.8KW An AM transmitter radiates 20KW. If the modulation Index is 0.7. Find the carrier Power.(June-July2009(2002 scheme)) Solution: Pt = 20 KW and m=0.7 We know that m 1 2 2 Pt (total ) Pc 20 103 Pc 0.7 2 1 2 Pc Pt (total ) m2 1 2 Pc 16.064103W Pc 16.064KW The total Power content of an AM signal is 1000W. Determine the power being transmitted at carrier frequency and at each side bands when modulation percentage is 100%. .(Dec 2014-Jan 2015) Solution: Pt = 1000 W and m=1 We know that Pt (total ) m2 Pc 1 2 1103 Pc 12 1 2 PUSB PLSB m2 Pc 4 Pc Pt (total ) m2 1 2 Pc 666.67W 1 PUSB PLSB 666.67 4 Pc 166.67W A500W, 100KHz carrier is modulated to a depth of 60% by modulating frequency of 1KHz. Calculate the total power transmitted. What are the sideband components of AM Wave?(Dec –Jan 2010(2006 scheme)) Solution: Pc = 500 W, fc =100 KHz, m = 60% = .6 and fm =1 KHz We know that Pt (total ) m2 Pc 1 2 Pt (total ) 0.6 2 500 1 2 Pt (total ) 590 W We know that fUSB f c f m f LSB f c f m fUSB 101 KHz f LSB 99 KHz A400W, 1MHz carrier is amplitude-modulated with a sinusoidal signal 0f 2500Hz. The depth of modulation is 75%. Calculate the sideband frequencies, bandwidth, and power in sidebands and the total power in modulated wave. (June-July2008(2006 scheme)) Solution: Pc = 400 W, fc =1 MHz, m = 75% = .75 and fm =2.5 KHz fUSB We know that fc fm f LSB fc fm BW 2 f m fUSB 1002.5 KHz f LSB 997.5 KHz BW 2 2.5KHz 5KHz We know that Pt (total ) m2 Pc 1 2 Pt (total ) 0.752 400 1 2 Pt (total ) 512.5 W PUSB PLSB m2 Pc 4 PUSB PLSB 0.752 400 56.25 W 4 A Carrier of 750 W,1MHz is amplitude modulated by sinusoidal signal of 2 KHz to a depth of 50%. Calculate Bandwidth, Power in side band and total power transmitted. Solution: Pc = 750 W, fc =1 MHz, m = 50% = .5 and fm =2 KHz We know that fUSB f c f m f LSB f c f m fUSB 1002 KHz f LSB 998 KHz BW BW fUSB 2f LSB f m 44 KHz KHz We know that Pt (total ) m2 Pc 1 2 Pt (total ) 0.52 750 1 2 Pt (total ) 843.75 W PUSB PLSB m2 Pc 4 PUSB PLSB 0.52 750 46.875 W 4 Calculate the percentage power saving when one side band and carrier is suppressed in an AM signal with modulation index equal to 1.(VTU Model QP-2014) Solution: m = 1 We know that Pt (total ) Psuppressed m2 3 Pc Pc 1 2 2 m2 Pc 4 Psuppressed Pc PLSB 5 1 Pc Psuppressed Pc Pc Pc 4 4 Amount of power saved 5 4 3 2 Psuppressed Ptotal Pc 10 Amount of power saved 0.833 83.3% Pc 12 Calculate the percentage power saving when one side band and carrier is suppressed in an AM signal if percentage of modulation is 50%. Solution: m = 0.5 We know that Pt (total ) Psuppressed m2 1 2 9 Pc Pc 8 m2 Pc 4 Psuppressed Pc PLSB 17 1 P Pc Pc Pc suppressed Pc 16 16 Amount of power saved Amount of power saved 9 8 17 16 Psuppressed Ptotal Pc 9 16 0.944 94.4% Pc 8 17 A Sinusoidal carrier frequency of 1.2MHz is amplitude modulated by a sinusoidal voltage of frequency 20KHz resulting in maximum and minimum modulated carrier amplitude of 110V & 90V respectively. Calculate I. frequency of lower and upper side bands II. unmodulated carrier amplitude III. Modulation index IV. Amplitude of each side band. Solution: fc =1.2 MHz, Emax = 110V, Emin = 90V and fm =20 KHz We know that fUSB f c f m fUSB 1220 KHz We also know that f LSB f c f m f LSB 1180 KHz E max E min Emax Emin Emax Emin also m & Em Ec E max E min 2 2 110 90 110 90 110 .90 m 0 1 E 10V E EE m m m c c 100V 2 2 110 90 An audio frequency signal 10 sin(2π×500t) is used to amplitude modulate a carrier of 50 sin(2π×105t).Calculate. (JAN2015) I.frequency of side bands IV. Amplitude of each side band. II. Bandwidth V. Transmission efficiency III. Modulation index VI. Total power delivered to a load of 600Ω. Solution: fm =500 Hz , Em = 10V, fc =100 KHz and Ec = 50V. We know that fUSB f c f m f LSB f c f m BW 2 f m f LSB 99.5 KHz BW 2 500Hz 1KHz E 10 m 0.2 50% We also know that m m m 50 Ec Transmission mEc 0.2 50 5V Efficiency Amplitude of side band 2 2 m2 eff. 2 2 Ec 2 m 2500 also carrier power Pc 2.08 Pc 2 0 . 2 2 R 2 600 eff. 2 2 0.2 2 m Pc Pt 2.125 and total power Pt 1 eff. 0.196 1.96% fUSB 100.5 KHz 2