Assignment Solution 01

North South University ETE 321 Spring 2010 Instructor Nahid Rahman concession 1 Total Marks 100 Worth 7. 5% 1. share the repulsivenessusoidally modulate DSB LC ratify shown be showtime. The common flattop DSB-LC frequence is ? c and the gist mark frequency is ? m. (a) look the modulation index m. tooth root Amax = 25 Amin = 5 ? 25 ? 5 = = 0. 67 + 25 + 5 (b) create verbally an expression for the modulate head ? (t). ancestor 1 1 ) = (25 ? 5) = 10 = ( ? 2 2 1 1 ) = (25 + 5) = 15 = ( + 2 2 = + romaine = romaine lettuce + ( ) romaine lettuce designation 1 colloidal suspension rascal 1 of 12 = 15 romaine lettuce + 10 romaineineine romaine lettuce lettuce (c) Derive time domain expressions for the upper and lower sidebands. dissolver = 15 co offensee + 10 romaineineine romaine lettuce = 15 cos + 5 cos( + ) + 5 cos( ? ) Upper sideband 5 cos( + ) ) Lower sideband 5 cos( ? (d) Determine the total mediocre ability of the modulated maneuver , the carrier power and the two sidebands. colloidal solutionution forcefulness of carrier bless = (15 cos )2 = + ? (15)2 2 cos cos = (cos( + ) + cos( ? )) 2 1 = 112. 5 W 2 (5)2 2 (5)2 Power of upper sideband = (5 cos( Power of lower sideband = (5 cos( ))2 = = 12. 5 W = 12. 5 W Power of modulated luff = 137. 5 W (e) Assuming that the heart and soul signal is a voltage signal, calculate the PEP (Peak Envelop Power) crossways a 100? load. colloidal suspensionution PEP = 2 ))2 = eed to obtain the RMS prize by dividing the billhook by v2. (f) Determine the modulation efficiency ?. dissolving agent 12. 5 + 12. 5 = = 18. 18% 137. 5 Amax is the peak value of the modulated signal. To calculate the DC power, we = ( )2 v2 = (25 )2 v2 100 = 3. 125 W Assignment 1 Sol Page 2 of 12 2. A DSB-SC modulated signal can be generated by multiplying the message signal with a hourly pulse generator and pas transgressg the resultant signal through a band-pass filter. = 2 cos two hundred + cos 600 ( )= 1 2 + 2 ? (? 1) ? 1 cos ( =1 2 ? 1 (2 ? 1)) (a) Find the DSB-SC signal component in V(t). solvent Input to the BPF = ? 1 1 2 ? (? ) = ( ) + cos ( (2 ? 1)) =1 2 ? 1 2 1 2 2 ? 1 2 1 = ( ) + cos + cos 3 + cos 5 + other terms 2 3 5 1 2 2 2 ( )+ ( ) cos ( ) cos 3 ( ) cos 5 = ? + + other terms 2 3 5 Output of the BPF 2 = cos 2 = 2 cos 200 + cos 600 cos (b) Specify the unwanted components in V(t) that need to be removed by a BPF of suitable design. Solution 1 2 2 ( ), ( ) cos 3 ( ) cos 5 , ,other terms 2 3 5 (c) necessitate the carrier frequency is 500 Hz. chalk out the spectral density of the resulting DSB-SC waveform. Solution = 2 cos 200 + cos 600 =2 ? 200 + 2 + 200 + ? 600 + + 600 Assignment 1 Sol Page 3 of 12 = 2 = 1 cos ? = 2 1 = 2 ? 00 rad = 1000 rad See piece below. (d) In the sketch for Part (c), specify lower and upper sidebands. + + 1 2 + 1 2 = 1 + 1 Assignment 1 Sol Page 4 of 12 3. Let f(t) be a real signal. The transmitter transmits the following modulated signal = cos + fault Wher e is the Hilbert transform of f(t). (a) Explain that the modulated signal is a lower sideband SSB signal u intrudeg an example of = cos . Solution Note that there was an break in the question. The frequency of f(t) should be ? m instead of ? c. Any students with a reasonable attempt to this question will be awarded full marks. However, the resultant role below refers to the corrected problem. cos = sin cos + sin sin = cos = cos ? Since, cos ? = cos cos + sin sin ? = + ? + ? + For ? > 0, the impulse perish is located to the left of the carrier frequency. For ? < 0, the impulse function is located to the right of ? c. Therefore, the modulating function produces lower sideband signals. (b) Determine the frequency of the modulated signal. Solution From the expression of , the frequency of the modulated signal is ? . Assignment 1 Sol Page 5 of 12 4. An SSB signal is generated by modulating an fc = 1 MHz carrier by the message signal = 2 cos 2000? t + cos 4000? t . The premium of the carrier signal is Ac = 1. a) Determine the Hilbert transform of f(t). Solution = 2 cos 2000? t + cos 4000? t ? ? = 2 cos 2000? t ? + cos 4000? t ? 2 2 = 2 sin 2000? t + sin 4000? t (b) Determine the time domain expression of the lower SSB and upper SSB signals. Solution ? t = cos ? sin ? 2 sin 2k? t + sin 4k? t sin = 2 cos 2k? t + cos 4k? t cos = 2 cos 2k? t + cos 4k? t cos ? 2 sin 2k? t + sin 4k? t sin = 2 cos 2k? t cos + cos 4k? t cos ? sin 4k? t sin ? 2 sin 2k? t sin = 2 cos 2k? t cos ? sin 2k? t sin + cos 4k? t cos ? sin 4k? t sin 2000 + cos 4000 = 2 cos (c) Sketch the magnitude spectrum of the lower SSB. Solution ? t = 2 cos ? 2000 + cos ? 000 ? ? = 2?? + ? 2000 + 2?? ? + 2000 + ?? + ? 4000 + ?? ? + 4000 A 2? ? -? c -? c+4000? -? c+2000? ?c-4000? ?c-2000? ?c ? (d) The coherent detection of the lower SSB signal consists of multiplying the true modulated signal by cos followed by a low pass filter. If the topical anaesthetic (receiver) oscillator generates a phase error ? (i. e. the message signal is now multiplied by cos + , write the expression at the production of the low-pass filter and discuss how the phase error will hit the demodulated signal. Solution Assignment 1 Sol Page 6 of 12 Input of the LPF = = = cos cos cos + + cos cos A cos B = sin A cos B = = os + sin + + cos sin + cos + cos A + B + cos A ? B sin A + B + sin A ? B cos 2 + + sin ? + sin 2 + sin ? A = ? sin A = cos + cos 2 + ? sin + sin 2 + Output of the LPF = = cos ? sin 2 cos 2000? t + cos 4000? t cos ? 1 2 sin 2000? t + sin 4000? t sin 2 = cos 2000? t cos + cos 4000? t cos 1 ? sin 2000? t sin ? sin 4000? t sin 2 1 cos 4000? t cos ? sin 4000? t sin 2 = cos 2000? t cos ? sin 2000? t sin + cos A cos B ? sin A sin B = cos A + B = cos 2000? t + ? + cos 4000? t + ? Assignment 1 Sol Page 7 of 12 5. A given DSB-LC transmitter develops an unmodulated power output of 1 KW across a 50-ohm resistive load.When a sine curveal test quality with a peak bounteousness of 5. 0 V is applied to t he gossip of the modulator, it is found that the spectral bloodline for each sideband in the magnitude spectrum for the output is 40% of the carrier line. Determine the following quantities in the output signal (a) The modulation index. Solution = + cos cos cos + cos cos = 1 1 = cos + cos ? + cos + 2 2 When a sinusoidal test tone with a peak amplitude of 5. 0 V is applied to the input of the modulator, it is found that the spectral line for each sideband in the magnitude spectrum for the output is 40% of the carrier line. 1 = 0. 40 2 = = 0. 0 (b) The peak amplitude of the lower sideband. Solution A given DSB-LC transmitter develops an unmodulated power output of 1 KW across a 50-ohm resistive load. = /v2 = 1000 Am is the amplitude (or peak) of the modulated signal. We need to use the rms value when calculating DC power. = 1000 2 = 10 = 316. 27 Peak amplitude of the sideband = = 158. 11 (c) The ratio of total sideband power to carrier power. Solution Total power of the sidebands = = Carrier Power = Ratio = = . cos = cos . + = . 0. 8 ? . = = 0. 32 + cos + (d) The total power of output. Solution Assignment 1 Sol Page 8 of 12 Total Power = + = 33kW e) The total honest power in the output if the peak amplitude of the modulation sinusoid is reduced to 4. 0 V. Solution Changing the modulation sinusoid peak amplitude will affect the modulation index. 4 = 5 4 = ? 0. 8 = 0. 64 5 Ratio of total sideband power to carrier power = . = . 0. 64 = 0. 2048 Total Power = + = 30. 12kW Assignment 1 Sol Page 9 of 12 6. estimate that a message signal f(x) has bandwidth B Hz. If f(x) is modulated by one of the modulation schemes DSB-SC or SSB or VSB, then for demodulation, the receiver must generate a (local) carrier in phase and frequency synchronous with the incoming carrier. This is referred to as synchronous or coherent demodulation. ) (a) Draw a block diagram for the demodulator. Solution (b) Assume that there is a frequency error in the local carrier (the phase is cor rectly estimated). Give the expression of the Fourier transform of the output of the demodulator for the case of DSB-SC modulation, sketch the spectrum of the output signal, and compare it with the spectrum of the original signal ( you may assume an arbitrary shape of F (w)). Solution = cos ? cos +? = = = = = 1 2 ? = 2 ? ? ? ? ? cos cos ? cos ? os ? +? cos +? +? cos 2 +? + cos 2 + ? + ? 2 ? +? Without frequency error Assignment 1 Sol Page 10 of 12 With frequency error (c) Repeat (b) for the case of SSB-SC modulation (you may do so by choosing either upper SSB or lower SSB). Solution + sin = cos Input to the LPF ? = cos = = = = ? ? ? ? +? + cos cos + ? cos cos cos cos ? 1 2 ? ? sin +? +? cos + + cos 2 ? +? sin sin +? +? cos cos +? +? + = 1 2 sin ? cos ? + +? + 1 2 ? ? sin 2 sin ? ? 2 ? = ? ? 2 ? ? ? 2 + ?? + ?? ? 2 ? +? Assignment 1 Sol Page 11 of 12 With frequency error d) Suppose that you are an engineer who responds to design a modulation system for a coarse environment in which it is difficult to generate a local carrier in frequency synchronous with the incoming carrier during roughly period of transmission. Which modulation system would you like to recommend, DSB-SC or SSB? Justify your answer. Solution For DSB-SC, we notice a distortion in frequency. For SSB, we only observe a frequency shift. Therefore, it would be better to use SSB for a coarse environment. Assignment 1 Sol Page 12 of 12