Tutorial7_ans - EE 3008. AM Tut. Solutions 1.(i) fc = 200...

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EE 3008. AM Tut. Solutions 1.(i) f c = 200 Hz output g(t) = f(t)cos2 π f c t=cos100 π t cos400 π t=0.5 cos300 π t+0.5 cos500 π t. G(f)=(1/4)[ δ (f-150)+ δ (f+150)+ δ (f-250)+ δ (f+250)] (ii) g(t)=f(t)cos 2 π f c t, G(f)=(1/2)F(f+f c )+(1/2)F(f-f c ), f c =200 2. This question is to let you have some practices on trigonometric functions. (i) e i (t)=4cos800 π t+cos2000 π t e o (t)=a 1 e i (t)+a 2 [e i (t)] 2 =a 1 cos2000 π t+4a 1 cos800 π t+8.5a 2 +8a 2 cos1600 π t+ 0.5a 2 cos4000 π t+4a 2 cos1200 π t+4a 2 cos2800 π t At 1 kHz, output is 1V, therefore a 1 = 1V =1000mV At 600 Hz, output is 4a 2 and at 1kHz output is a 1 , therefore (4a 2 )/a 1 = 0.002, a 2 =5 × 10 -4 V = 0.5 mV (ii) At 800 Hz, V 1 = 8a 2 = 4 mV At 1400 Hz, V 2 =4a 2 = 2 mV -250 -150 150 250 G( f ) f 1/4 -100 100 F( f ) 1 G( f ) -300 -200 -100 100 200 300 1/2
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3. System 1: Then the modulated signal is φ 1 (t)=mcos2 π f m t cos2 π f c t + cos 2 π f c t = (1+mcos2 π f m t)cos2 π f c t System 2: The modulated signal is φ 2 (t)=(1+mcos2 π f m t)cos2 π f c t =(1+mcos2 π f m t)cos2 π f c t No, a dc level in the input becomes a portion of the carrier in the
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This note was uploaded on 01/11/2011 for the course EE 3008 taught by Professor Pingli during the Fall '08 term at City University of Hong Kong.

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Tutorial7_ans - EE 3008. AM Tut. Solutions 1.(i) fc = 200...

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