EE160. Spring 2003. San Jos´e State UniversityPROBLEM SET # 3 April 12, 20031. Textbook, problem 3.3The following figure shows the modulated signals for A =1andf0= 10. As it isobserved both signals have the same envelope but there is a phase reversal at t =1forthe second signal Am2(t)cos(2πf0t) (right plot). This discontinuity is shown clearly inthe next figure where we plotted Am2(t)cos(2πf0t)withf0=3.-1-0.8-0.6-0.4-0.200.20.40.60.8100.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1-0.8-0.6-0.4-0.200.20.40.60.8100.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1-0.8-0.6-0.4-0.200.20.40.60.800.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 22. Textbook, problem 3.4y(t)=x(t)+12x2(t)= m(t)+cos(2πfct)+12m2(t)+cos2(2πfct)+2m(t)cos(2πfct)= m(t)+cos(2πfct)+12m2(t)+14+14cos(2π2fct)+m(t)cos(2πfct)Taking the Fourier transform of the previous, we obtainY (f)=M(f)+12M(f ) M(f)+12(M(f − fc)+M(f + fc))+14δ(f )+12(δ(f − fc)+δ(f + fc)) +18(δ(f − 2fc)+δ(f +2fc))The next figure depicts the spectrum Y (f)1/4-2fc -fc -2W 2W fc 2fc1/81/23. Textbook, problem 3.5u(t)=m(t) · c(t)= 100(2 cos(2π2000t)+5cos(2π3000t)) cos(2πfct)Thus,U(f )=1002 δ(f − 2000) + δ(f + 2000) +52(δ(f − 3000) + δ(f + 3000)) [δ(f − 50000) + δ(f + 50000)]=50 δ(f − 52000) + δ(f − 48000) +52δ(f − 53000) +52δ(f − 47000)+δ(f + 52000) + δ(f + 48000) +52δ(f + 53000) +52δ(f + 47000)A plot of the spectrum of the modulated signal is given in the next figure.................................................................66 66 6666125500-53 -52 -48 -47 47 48 52 53 KHz4. Textbook, problem 3.151) The modulated signal isu(t) = 100[1 + m(t)] cos(2π8 × 105t)= 100 cos(2π8 × 105t) + 100 sin(2π103t)cos(2π8 × 105t)+500 cos(2π2 × 103t)cos(2π8 × 105t)= 100 cos(2π8 × 105t) + 50[sin(2π(103+8× 105)t) − sin(2π(8 × 105− 103)t)]+250[cos(2π(2 × 103+8× 105)t)+cos(2π(8 × 105− 2 × 103)t)]Taking the Fourier transform of the previous expression, we obtainU(f ) = 50[δ(f − 8 × 105)+δ(f +8× 105)]+25 1jδ(f − 8 × 105− 103) −1jδ(f +8× 105+103)−25 1jδ(f − 8 × 105+103) −1jδ(f +8× 105− 103)+125δ(f − 8 × 105− 2 × 103)+δ(f +8× 105+2× 103)+125δ(f − 8 × 105− 2 × 103)+δ(f +8× 105+2× 103)= 50[δ(f − 8 × 105)+δ(f +8× 105)]+25δ(f − 8 × 105− 103)e−jπ2+ δ(f +8× 105+103)ejπ2+25δ(f − 8 × 105+103)ejπ2+ δ(f +8× 105− 103)e−jπ2+125δ(f − 8 × 105− 2 × 103)+δ(f +8× 105+2× 103)+125δ(f − 8 × 105− 2 × 103)+δ(f +8× 105+2× 103)rrrr...........................................................................................................................................................666 6 666666|U(f )|∠U(f )−π2π2fc+2×103fc−2×103fc−2×103fc−103fc+103−fcfcfc+2×10325501252) The average power in the carrier isPcarrier=A2c2=10022= 5000The power in the sidebands isPsidebands=5022+5022+25022+25022= 650003) The message signal can be written asm(t)=sin(2π103t)+5cos(2π2 × 103t)= −10 sin(2π103t)+sin(2π103t)+5As it is seen the minimum value of m(t)is−6 and is achieved for sin(2π103t)=−1ort =34×103+1103k,withk ∈ Z. Hence, the modulation index is α =6.4) The power delivered to the load isPload=|u(t)|250=1002(1 + m(t))2cos2(2πfct)50The maximum absolute value of 1 + m(t)is6.025 and is achieved for sin(2π103t)=120or t =arcsin(120)2π103+k103.Since2× 103 fcthe peak power delivered to the load isapproximately equal tomax(Pload)=(100 × 6.025)250=72.60125. Textbook, problem 3.18The signal x(t)ism(t)+cos(2πf0t). The spectrum of this signal is X(f)=M(f )+12(δ(f − f0)+δ(f + f0)) and its bandwidth equals to Wx= f0. The signal y1(t)aftertheSquareLawDeviceisy1(t)=x2(t)=(m(t)+cos(2πf0t))2= m2(t)+cos2(2πf0t)+2m(t)cos(2πf0t)= m2(t)+12+12cos(2π2f0t)+2m(t)cos(2πf0t)The spectrum of this signal is given byY1(f)=M(f ) M(f )+12δ(f )+14(δ(f − 2f0)+δ(f +2f0)) + M(f − f0)+M(f + f0)and its bandwidth is W1=2f0. The bandpass filter will cut-off the low-frequencycomponents M (f )M(f )+12δ(f ) and the terms with the double frequency components14(δ(f − 2f0)+δ(f +2f0)). Thus the spectrum Y2(f)isgivenbyY2(f)=M(f − f0)+M(f + f0)and the bandwidth of y2(t)isW2=2W . The signal y3(t)isy3(t)=2m(t)cos2(2πf0t)=m(t)+m(t)cos(2πf0t)with spectrumY3(t)=M(f)+12(M(f − f0)+M(f + f0))and bandwidth W3= f0+W . The lowpass filter will eliminate the spectral components12(M(f − f0)+M(f + f0)), so that y4(t)=m(t) with spectrum Y4= M(f)andbandwidth W4= W . The next figure depicts the spectra of the signals x(t), y1(t),y2(t), y3(t)andy4(t)....................SSSQQQQQQSSSSSSSSSSSSSSSHHHHH66 66 6%%%eee1412−WW−WW f0+Wf0−W−f0+W−f0−W−f0+W−f0−Wf0−Wf0+W−2f0−f0−W −f0+W −2W 2Wf0−Wf0+W 2f0−WW−f0f0Y4(f)Y3(f)Y2(f)Y1(f)X(f)6. Textbook, problem 3.281) If the output of the narrowband FM modulator is,u(t)=A cos(2πf0t + φ(t))then the output of the upper frequency multiplier (×n1)isu1(t)=A cos(2πn1f0t + n1φ(t))After mixing with the output of the second frequency multiplier u2(t)=A cos(2πn2f0t)we obtain the signaly(t)=A2cos(2πn1f0t + n1φ(t)) cos(2πn2f0t)=A22(cos(2π(n1+ n2)f0+ n1φ(t)) + cos(2π(n1− n2)f0+ n1φ(t)))The bandwidth of the signal is W = 15 KHz, so the maximum frequency deviation is∆f = βfW =0.1 × 15 = 1.5 KHz. In order to achieve a frequency deviation of f =75KHz at the output of the wideband modulator, the frequency multiplier n1should beequal ton1=f∆f=751.5=50Using an up-converter the frequency modulated signal is given byy(t)=A22cos(2π(n1+ n2)f0+ n1φ(t))Since the carrier frequency fc=(n1+ n2)f0is 104 MHz, n2should be such that(n1+ n2)100 = 104 × 103=⇒ n1+ n2= 1040 or n2= 9902) The maximum allowable drift (df) of the 100 kHz oscillator should be such that(n1+ n2)df=2=⇒ df=21040= .0019 Hz6. Textbook, problem 3.311) The modulation index isβ =kfmax[|m(t)|]fm=∆fmaxfm=20 × 103104=2The modulated signal u(t) has the formu(t)=∞n=−∞AcJn(β)cos(2π(fc+ nfm)t + φn)=∞n=−∞100Jn(2) cos(2π(108+ n104)t + φn)The power of the unmodulated carrier signal is P =10022= 5000. The power in thefrequency component f = fc+ k104isPfc+kfm=1002J2k(2)2The next table shows the values of Jk(2), the frequency fc+kfm, the amplitude 100Jk(2)and the power Pfc+kfmfor various values of k.Index k Jk(2) Frequency Hz Amplitude 100Jk(2) Power Pfc+kfm0 .2239