University of Illinois Spring 2016ECE210 - Homework 12Solutions (Wednesday, April 20 at 6:00 p.m.)1. Simplify the following expressions involving the impulse and/or shiftedimpulse and sketch the results:(a) g(t) = cos(πt)(dudt+ δ(t − 0.5)).(b) a(t) =´t−∞δ(τ − 1)dτ + rect(t3)δ(t − 2).(c) b(t) = δ(t − 4) ∗ u(t − 2).Solution:(a)g(t) = cos (π(0)) δ(t) + cos (π(0.5)) δ(t − 0.5)= δ(t).Sketch for (a):(b)a(t) =ˆt−∞δ(τ − 1)dτ + rect(t3)δ(t − 2)= u(t − 1) + 0.Sketch for (b):1(c)b(t) = δ(t − 4) ∗ u(t − 2)= u(t − 6).Sketch for (c):2. Consider an LTI system with impulse responseh(t) = e−4tu(t)2Find the Fourier series represention of the output y(t) for the followinginputs.(a) f(t) = cos(2πt)(b) f(t) =P∞n=−∞δ(t − n)Solution:(a) By convolution property, we knowY (ω) = H(ω)F (ω) =14 + jω×π[δ(ω−2π)+δ(ω+2π)] =π4 + j2πδ(ω−2π)+π4 − j2πδ(ω+2π)The inverse Fourier transform is y(t) =12π×[π4+j2πej2πt+π4−j2πe−j2πt].Thus the Fourier series coefficients areF1=π4 + j2π×12π=18 + j4πF−1=π4 − j2π×12π=18 − j4π.(b) For a impulse train, we know that the corresponding Fourier trans-form isF (ω) = 2π∞Xn=−∞δ(ω −2nπ)Therefore, we haveY (ω) = H(ω)F (ω) =14 + jω×2π∞Xn=−∞δ(ω−2nπ) = 2π∞Xn=−∞14 + j2nπδ(ω−2nπ)Similarly, we have the inverse Fourier transform y(t) =P∞n=−∞14+j2nπej2πnt.The Fourier series coefficients areFn=14 + j2nπ3. Determine the Fourier transform of the following signals — simplify the re-sults as much as you can. For parts (a), (b), and (c), sketch the magnitudeand phase of the result:(a) f(t) = 5 cos(5t) + 3 sin(15t).3(b) x(t) = cos2(6t).(c) y(t) = e−tu(t) ∗ cos(2t).(d) z(t) = (1 + cos(3t))e−tu(t).Solution:(a) Using Table 7.2, we obtainF (ω) = 5π [δ(ω − 5) + δ(ω + 5)] + j3π [δ(ω + 15) − δ(ω − 15)]Sketching |F(ω)| and ∠F (ω):|F (ω)| = 5π [δ(ω − 5) + δ(ω + 5)] + 3π [δ(ω + 15) + δ(ω − 15)]∠F (ω) =0, for ω = 5, and ω = −5 rad/sπ2for ω = −15 rad/s−π2for ω = 15 rad/s(b) Using trigonometric identities, we havex(t) =12+12cos(12t).Now, using Table 7.2, we obtainX(ω) = πδ(ω) +π2[δ(ω − 12) + δ(ω + 12)] .Sketching X(ω). Since the function is real and positive for all ω,then the phase will be zero for all frequencies. We only need tosketch X(ω).4(c) Using the time convolution property, and Table 7.2, we haveY (ω) =11 + jωπ [δ(ω − 2) + δ(ω + 2)]=π1 + j2δ(ω − 2) +π1 − j2δ(ω + 2).Sketching |Y (ω)| and ∠Y (ω):|Y (ω)| =π√5δ(ω − 2) +π√5δ(ω + 2)∠Y (ω) =(arctan(2) ≈ 1.107 rad, for ω = −2 rad/s−arctan(2) ≈ −1.107 rad, for ω = 2 rad/s|Y (ω)|ω (rad/s)6Y (ω) (rad)ω (rad/s)2π√5arctan(2)−22−2π√5−arctan(2)(d) Rearranging,z(t) = e−tu(t) + e−tcos(3t)u(t).Now, using Table 7.2, we obtainZ(ω) =11 + jω+1 + jω(1 + jω)2+ 9.4. The inverse of the sampling interval T –that is, T−1– is known as thesampling frequency and usually is specified in units of Hz. Determine theminimum sampling frequencies T−1needed to sample the following analogsignals without causing aliasing error.5(a) Arbitrary signal f(t) with bandwidth 20 kHz.(b) f(t) = sinc(4000πt).(c) g(t) = sinc(4000πt) cos(8000πt).Solution:(a) Using the Nyquist criterion, we haveT <12B=140 kHz.Therefore, the minimum sampling frequency isT−1min= 40 kHz.(b) Sincef(t) = sinc(4000πt) ←→14000rectω8000π,the bandwidth of f(t) is B = 4000π rad/s= 2000 Hz. So, using theNyquist criterion, we haveT <12B=14 kHz.Therefore, the minimum sampling frequency isT−1min= 4 kHz.Sampling at that frequency will lead to the following FT of the sam-pled signal∞Xn=−∞f(t)δ(t − nT ) ←→ FT(ω) =∞Xn=−∞1TF (ω −2πTn).Sketching FT(ω):(c) In this case, we have the signal from part (b) modulating a carrier sig-nal with carrier frequency ωc= 8000π rad/s . Using the modulationproperty, the FT of g(t) isG(ω) =12[F (ω − 8000π) + F (ω + 8000π)] =18000rectω − 8000π8000π+ rectω + 8000π8000π6Consequently, the bandwidth of the signal g(t) isB = 8000π + 4000 πrad/s = 12000π rad/s = 6 kHz.Hence,T−1min= 2B = 12 kHz.5. Determine whether of the following LTIC systems are BIBO stable andexplain why or why not:(a) h1(t) = 5δ(t) + 2e−2tu(t) + 3te−2tu(t)(b) h2(t) = δ(t) + u(t)(c) h4(t) = −2δ(t − 3) − te−5tu(t)Solution:(a) is BIBO stable. Because every term of h1(t) is absolutely integrable,h1(t) is then absolutely integrable.(b) is NOT BIBO stable. Because u(t) is NOT absolutely integrable.(c) is BIBO stable. Note that te−5tu(t) is absolutely integrable. Intu-itively, t increases much slower than the decay of e−5t.6. Consider the given zero-state input-output relations for a variety of sys-tems. In each case, determine whether the system is zero-state linear, timeinvariant, and causal(a) y(t) = f(t − 1) + f(t + 1)(b) y(t) = 5f(t) ∗ u(t)(c) y(t) = δ(t − 4) ∗ f(t) +´t−2−∞f2(τ)dτ(d) y(t) =´t+2−∞f(τ)dτ(e) y(t) =´t+2−∞f(τ2)dτ(f) y(t) = f3(t − 1)(g) y(t) = f((t − 1)2)Solution:(a) y(t) is linear, time-invariant, but NOT causal. Because f (t + 1) isthe future data.(b) y(t) is linear, time-invariant, and causal.(c) y(t) is time-invariant, causal but NOT linear. Because of the 2ndorder function term´t−2−∞f2(τ)dτ.(d) y(t) is linear, time-invariant, but NOT causal. Because´t+2−∞f(τ)dτhas future data involved.7(e) y(t) is linear, but NOT time-invariant. Also it is NOT causal. Be-cause´t+2−∞f(τ2)dτ has future data involved and has 2nd order vari-able term.(f) y(t) is time-invariant, causal but NOT linear. Because of the 3rdorder function term f3(t − 1).(g) y(t) is linear, but NOT time-invariant. Also it is NOT causal. Be-cause f ((t − 1)2) has 2nd order term and has future data involved(say input is y(3), the output is determined by f ((t − 1)2) = f (4)which is the future