University of Illinois Spring 2016ECE210 / ECE211 - Homework 10Solutions1. Let f(t)=f1(t)+f2(t) such that f1(t) $ F1(!) and f2(t) $ F2(!).(a) Show thatf(t) $ F1(!)+F2(!).(b) Now let f(t)=f1(t)+f2(t) be the input signal of an LTI system with a frequency resp on se H(!)=|H(!)|ej(!),where(!)=\H(!). The functions F1(!), F2(!), H(!) and (!) are given graphically asfollows:4F1(!)F2(!)|H(!)|(!)10⇡ rad/s10⇡ rad/s2⇡ rad10⇡ rad/s10⇡ rad/s10⇡ rad/s10⇡ rad/s10⇡ rad/s10⇡ rad/s!!!!Express the output y(t) of the system as a superposition of scaled and/or shifted versions of f1(t) and f2(t).Hint: y(t)=y1(t)+y2(t), with Y1(!)=H(!)F1(!).Solution:(a) Proving the addition property of the Fourier transform:F (!)=Z11(f1(t)+f2(t)) ej!tdt=Z11f1(t)ej!tdt +Z11f2(t)ej!tdt= F1(!)+F2(!).(b) We know that in an LTI system, the input and output in the Fourier domain are related asY (!)=H(!)F (!).With F (!)=F1(!)+F2(!), we haveY (!)=H(!)[F1(!)+F2(!)]= H(!)F1(!)| {z }Y1(!)+ H(!)F2(!)| {z }Y2(!).Now, for the region where F1(!) 6= 0, we haveH(!)=2.Page 1 of 5Therefore,Y1(!)=2F1(!) ! y1(t)=2f1(t).Also, for the region where F2(!) 6= 0, we notice a phase that is changing linearly with slope1/10. Hence,H(!)=4ej110!.Consequently,Y2(!)=4ej110!F2(!) ! y2(t)=4f2✓t 110◆.Finally, addin g the two results, we obtainy(t)=2f1(t)+4f2✓t 110◆.2. Determine the response y(t) of the circuit shown below with an arbitrary input f(t) in the form of an inverseFourier transform and then evaluate y(t) for the case f (t)=et6u(t) V.+-y(t)f(t)2⌦+-3FSolution:The equivalent circuit in the Fourier domain is+-Y (!)F (!)2⌦+-1j!3⌦Using voltage division, we haveY (!)=F (!)1j!32+1j!3= F (!)1616+ j!.Applying the inverse Fourier transform,y(t)=12⇡Z11Y (!)ej!td! =12⇡Z1116F (!)16+ j!ej!td!.For the input f (t)=et/6u(t) the Fourier transform paireatu(t) $1a + j!,a>0,yieldsF (!)=116+ j!.Substituting this F (!) into the inverse Fourier transform, we obtainy(t)=12⇡Z111616+ j!2ej!td!.Finally using the Fourier tr an sf orm pairteatu(t) $1(a + j!)2,a>0,gives,y(t)=16te16tu(t).Page 2 of 53. Given that f(t)e±j!0t$ F (! ⌥ !0), determine the Fourie r trans for m of f (t)sin(!0t).Solution:Based on Euler’s Formula, we havesin(!0t)=12j[e+j!0t ej!0t]Therefore, our target becomesf(t)sin(!0t)=f(t)12j[e+j!0t ej!0t]=f(t)12je+j!0t f(t)12jej!0tMaking use of the gi ven tr ans for m f (t) e±j!0t$ F (! ⌥ !0), we get the Fourier transform of f (t)sin(!0t) asf(t)sin(!0t) $12j[F (!  !0)  F (! + !0)]4. A signal f(t) is bandlimited to the interval ! 2 [⌦, ⌦] and modulated by a cosine carrier of frequency !c>⌦ . The resulting modulated signal r(t) is t h en coherently demodulated with a mismatched carrier signalcos(!ct + ✓), and filtered with an ideal low pass filter HLPF(!)=rect!2⌦as sh own in the figure below.f(t)cos(!ct) cos(!ct + ✓)HLPF(!)y(t)r(t)=f(t) cos(!ct) m(t)(a) Find an expression for y(t) in terms of f(t) and ✓.(b) For what values of ✓ is the ampli t u de of y(t) smallest and largest?(c) Consider what would happen when ✓ is slowly time varying (e.g. ✓ =!t,where0< ! ⌧ ⌦). If youwere to play y(t) on a loudspeaker, what qualitative e↵ect with this have on the signal you hear?Solution:(a) Using the identity cos(↵) cos()=12(cos(↵ + ) + cos(↵  )) yieldsm(t)=f(t) cos(!ct) cos(!ct + ✓)=f(t)2[cos(2!ct + ✓) + cos(✓)]=f(t)2cos✓2!c✓t +✓2!c◆◆+ cos(✓)Fourier transform i ng this expr ession yieldsM(!)=14F (!  2!c)ej✓+ F (! +2!c)ej✓+cos(✓)2F (!)The first term comes from the fact thatF {(f( t) c os( !ct + ✓)} =12Fnf(t)⇣ej(!ct+✓)+ ej(!ct+✓)⌘o=12⇥Ff(t)ej!ct ej✓+ Ff(t)ej!ct ej✓⇤=12⇥F (!  !c)ej✓+ F (! + !c)ej✓⇤.Since the ideal LPF HLPF(!) removes frequencies outside the range (⌦, ⌦), the modulate d componentsare removed. Thus we haveY (!)=cos(✓)2F (!)) y(t)=cos(✓)2f(t).Page 3 of 5(b) Since the amplitude of the output is determined by cos(✓), all we have to do is find the values of ✓ suchthat | cos(✓)| is maximized and m in i mi ze d. Thus the output is maximized at ✓ = n⇡, n 2 Z, and minimize dwhen ✓ =2n+12⇡, n 2 Z ( odd half integers of ⇡).(c) If ✓ =!t,thenwehaveY (!)=14(F (!  !)+F (! +!))) y(t)=f(t)2cos(!t).This means that that if you played the signal on a louds peaker, the signal would be constantly cutting inand out every time | cos(!t)| was max i mi ze d or minimized. This example illustrates the importance ofhaving accurate phase in coherent demodulation.5. A li n ear sy st em with freq u en cy re sponse H(!) is excited with an inputf(t) $ F (!).H(!) and F (!) are plotte d below:1! rad/s! rad/sF (!)H(!)4⇡4⇡8⇡8⇡2(a) Sketch the Fourie r trans for m Y (!) of the system output y(t) and calculat e the energy Wyof y(t).(b) It is observed that output q(t) of the following system equals y(t) determined in part (a).⇥cos(!ot)p(t)q(t)Sketch P (!) and determine !o.Solution(a) The output of the system isY (!)=H(!)F (!),but since F (!) = 1, for all H(!) 6= 0, we can easily sketchY (!)=H(!).! rad/sY (!)4⇡ 8⇡224⇡8⇡Using Parseval’s theorem, the energy of the signal isWy=12⇡Z11|Y (!)|2d! =1⇡Z8⇡4⇡22d!=4⇡(8⇡  4⇡) = 16 J.(b) Using Y (!)=Q(!) and the modulation property, we haveY (!)=12P (!  !0)+P (! + !0).So P (!) can be sketched asPage 4 of 5! rad/sY (!)4⇡ 8⇡224⇡8⇡! rad/sP (!)2⇡42⇡!0!0Clearly, from the graph we have!0=6⇡rads.Page 5 of