University of Illinois Spring 2016ECE210 / ECE211 - Homework 08Solutions1. Consider the function f(t) = Re{2ejt+ 2e−j2t+ 2}. Determine its fundamental period T0, its fundamentalfrequency ω0, and plot it over at least two periods.Solution:f(t) = Re{2ejt+ 2e−j2t+ 2} = 2 cos(t) + 2 cos(2t) + 2. This function has frequency: ωo= 1 rad/s =2πT.Therefore, the period is T =2πωo= 2π s.2. For each one of the following functions of t, indicate whether they are periodic or not. If periodic, indicate itsperiod, and if not periodic, indicate why. Assume n is a positive integer.(a) sin(2√2πt) + cos√2π t(b) cos (2t) sin (2t)(c) ejπt+ e−j2πt(d) sin(πt) + sin(3π2t) + cos(2π5t)(e) cosn√2tSolution:(a) The function sin(2√2πt) + cos√2π tis periodic, its period T =√2 s.(b) Using a trigonometric identity: cos (2t) sin (2t) =12sin (4t). This function is periodic and has fundamentalfrequency ωo= 4 rad/s. Therefore the period isT =2πωo=π2s.(c) The function ejπt+e−j2πtis periodic, because all possible ratios of the individual frequencies are rational.The individual periods are T1= 2 s and T2= 1 s. Define LMM(a, b) =The least common multiple numberof a and b. Therefore the period isT = LMM(T1, T2) = 2 s.(d) The function sin (πt) + cos3π2t+ sin2π5tis periodic, because all possible ratios of the individual fre-quencies are rational. The individual periods are T1= 2 s, T2=43s, and T3= 5 s. Therefore the periodisT = LMM(T1, T2, T3) = 15 × T2= 4 × T3= 20 s.Page 1 of 4(e) The function cosn√2 tis periodic. In this case the fundamental frequency is ωo=√2n rad/s. The periodisT =2πωo=2π√2n=πn√2 s.3. The function f(t) is periodic with period T = 2 s. Between t = 0 and 2 s, the function is described byf(t) =(2, 0 < t < 1 s1, 1 < t < 2 s(a) Plot f(t) between t = −3s and t = 4s.Solution:The function is periodic, thus we can plot f(t) by periodic extension, f(t) = f(t + 2):(b) Determine the exponential Fourier coefficients Fnof f(t) for n = 0, n = ±1s and n = ±2s.Solution:Finding the Fourier coefficients using Fn=1TRTf(t)e−jnωotdt where T = 2 and ωo= π.F0=12[R102dt +R211dt] =12× [2 + 1] =32Fn=12[R102e−jnπtdt +R211e−jnπtdt] =12[jπn(e−jnπ+ e−j2πn− 2)]For n = ±1s and n = ±2s, we can calculateF−1=jπF+1= −jπF−2= 0F+2= 0(c) Using the results of part (b), determine the compact-form Fourier coefficients c0, c1and c2.Solution:By definition, c0= 2|F0| = 3, c1= 2|F1| =2πand c2= 0.4. For each one of the following functions, obtain their Fourier series in exponential, trigonometric, and compactforms.Page 2 of 4(a) f(t) = sin4(t)Solution:Since sin(t) = (ejt− e−jt)/(2j), we can calculate the exponential form:sin2(t) =e2jt+ e−2jt− 2−4sin4(t) = [e2jt+ e−2jt− 2−4]2=e4jt+ e−4jt+ 2 − 4e2jt− 4e−2jt+ 416Convert to trigonometric form.sin4(t) =38−12cos(2t) +18cos(4t)The compact form issin4(t) =38+12cos(2t + π) +18cos(4t)(b) f(t) = cos4(t)Solution:Note that cos4(t) = sin4(t +π2). Thus (b) is phase-shifted version of (a), and we can calculate the compactform:cos4(t) =38+12cos(2t) +18cos(4t)we have the trigonometric form is the same as compact form:cos4(t) =38+12cos(2t) +18cos(4t)Then, the exponential form iscos4(t) =38+12[(ej2t+ e−j2t)2] +18[(ej4t+ e−j4t)2] =38+14ej2t+14e−j2t+116ej4t+116e−j4t(c) The triangular function g(t) shown in the figure below. Make use of the results of Example 6.11 in textbook.Solution:Making use of the results of Example 6.11 in the textbook, we observe that g(t) is continuous and derivativeproperty can be applied. g(t) is the derivative of scaled version of f(t) in 6.11. Let h(t) =12f(t), by theresult of Example 6.11,h(t) =Xn=odd4nπcos(nπ6t −π2)|Hn| =4nπ2=2nπwhen n = oddangle(Hn) = −π2n > 0Page 3 of 4As real function, Hn= H∗−n, thus Hn= H−n= −2nπj. By derivative property, h(t) =dgdt, and ω0=2πT=π6.For n = odd,Gn=Hnjnω0= −12n2π2Consider the DC component, G0=1TRTg(t)dt = 3. Therefore, we have the exponential form:g(t) = 3 −Xn=odd12n2π2ejnπ6tSince g(t) is even function, Gn= G−n, bn= 0, an= Gn+ G−n= −24n2π2. Thus we can obtain thetrigonometric form:g(t) = 3 +Xn=odd−24n2π2cos(nπ6t)Therefore, the compact form is,g(t) = 3 +Xn=odd24n2π2cos(nπ6t + π)5. Let the function g(t) of problem 4(b) be the input to an LTI system with frequency response H(ω) = 2e−jωπ/2for ω ∈ [−2.5, 2.5] rad/s, and zero elsewhere. Obtain the steady state response yss(t) of the system to the inputg(t).Solution:The steady state response yss(t) has Fourier series coeffients Yn= H(nωo)Gn,Yo= H(0)Go=2e038=34,Y1= H(1)G1=2e−jπ(14) = −12, Y−1= H(−1)G−1=2ejπ(14) = −12,Therefore, in the time domain,yss(t) =34−12e−j2t−12ej2t.Page 4 of