University of Illinois Spring 2016ECE210 - Homework 111. For each part of this problem, find and sketch y (t) = f(t) ∗ h(t).(a) f(t) = u(t),h(t) = u(t)(b) f(t) = tu(t),h(t) = u(t)(c) f(t) = δ(t) − 5δ(t − 2) + δ(t − 4),h(t) = rect(t/2)(d) f(t) = u(t),h(t) = e−tsin(t)u(t)Solution:(a) In the convolution formula we choose to flip h(t), writingy(t) = h(t) ∗ f(t) =Z∞−∞f(τ )h(t − τ)dτ =Z∞−∞f(τ )u(t − τ)dτ.In this convolution, we can identify 2 regions for calculating the area under the product curve: τ < t, andτ > t. By replacing u(t − τ) in the integrant by 1 and changing the upper integration limit from ∞to t, wehave:y(t) =tZ−∞f(τ )dτ =tZ−∞u(τ)dτ =(0, t < 0Rt0dτ, t > 0= tu(t)Sketching y(t):(b) In this convolution, we can identify 2 regions for calculating the area under the product curve: τ < t, andτ > t. By replacing u(t − τ) in the integrant by 1 and changing the upper integration limit from ∞to t, wehave:y(t) =tZ−∞f(τ )dτ =tZ−∞τu(τ)dτ =(0, t < 0Rt0τdτ, t > 0=(t22, t > 00, t < 0Sketching y(t):Page 1 of 6(c) From table 9.3 entry 1, we have δ(t) ∗ f(t) = f(t), δ(t − t0) ∗ f(t) = f (t − t0). Therefore, we will havey(t) = f(t) ∗ h(t) = (δ(t) − 5δ(t − 2) + δ(t − 4)) ∗ rect (t/2) = rect(t/2) − 5rect(t − 22) + rect(t − 42)Sketching y(t):(d) We utilize the commutative property of convolutiony(t) =tZ−∞h(τ)dτ =(0, t < 0−12(e−t(sin t + cos t) − 1), t > 0Sketching y(t):Page 2 of 6t-10 -8 -6 -4 -2 0 2 4 6 8 10y(t)00.10.20.30.40.50.62. Sketch the convolution of these two signals. You do not need to specify the functional form of y(t) (you don’tneed to give it in the form of an equation), but your sketch should specify the values of y(t) at each integer t,and should show roughly what the shape looks like in between these points.f(t) =1 − |t + 5|, −6 ≤ t ≤ −41 0 ≤ t ≤ 10 otherwiseh(t) =(1 0 ≤ t ≤ 10 otherwisePage 3 of 63. Postponed4. Use row 13 of table 7.1, together with the Fourier transform and convolution properties of the impulse, to proverow 8 of table 7.1.Solution:Entry 13:f(t) ∗ g(t) ←→ F (ω)G(ω)Entry 8:f(t − t0) ←→ F (ω)e−jωt0f(t − t0) = δ(t − t0) ∗ f(t) ←→ e−jωt0F (ω) W ith table7.2 entry16Page 4 of 65. Any real-valued signal x(t) can be written as the sum of a real-valued even part, xe(t), and a real-valued oddpart, xo(t), such thatxe(−t) = xe(t) (1)xo(−t) = −xo(t) (2)x(t) = xe(−t) + xo(t) (3)(a) Prove that Xe(ω), the Fourier transform of xe(t), is pure real. Hint: use the following equation, which istrue whenever its component integrals are well defined:Z∞−∞f(t)dt =Z∞−∞(f(t) + f (−t))dt(b) Prove that Xo(ω), the Fourier transform of xo(t), is pure imaginary.(c) The energy spectrum of a signal is defined asWx(ω) = |X(ω)|2Use Eq. 3 and the results of parts (a) and (b) to express Wx(ω) in terms of X2e(ω) and X2o(ω). Note thatthese are the squared Fourier transforms, not the squared magnitude Fourier transforms, therefore one ofthem is positive and one of them is negative (which one? why?).(d) The auto correlation of a signal is defined as the inverse FT of its energy spectrum, thusUse the results of parts (a) through (c) to express rx(t) in terms of xe(t) and xo(t).rx(t) =12πZ∞−∞Wx(ω)ejωtdω(e) Consider the following signalx(t) =3, −1 ≤ t ≤ 01, 0 ≤ t ≤ 10, otherwisei. Find xe(t) and xo(t).ii. Find rx(t) . Specify the values rx(−2) ,rx(−1) ,rx(0) ,rx(1) ,rx(2) .Solution:(a)Xe(ω) =Z∞−∞xe(t)e−jωtdt =Z∞0xe(t)e−jωt+ xe(t)ejωtdt =Z∞0xe(t)2 cos(ωt)dtBecause b oth xe(t) and cos(ωt) is purely real, then Xe(ω) is pure real.(b)Xo(ω) =Z∞−∞xo(t)e−jωtdt =Z∞0xo(t)e−jωt− xo(t)ejωtdt =Z∞0xo(t)(−2j sin(ωt))dtBecause b oth xo(t) and sin(ωt) is real and with a j, Xo(ω) is pure imaginary.(c)Wx(ω) = |X(ω)|2= ( Xe(ω))2− (Xo(ω))2(d)rx(t) =12πZ∞−∞Wx(ω)ejωtdω =12πZ∞−∞(X2e− X2o)ejωtdω = xe(t) ∗ xe(t) − xo(t) ∗ xo(t)Page 5 of 6(e) i.xe(t) =x(t) + x(−t)2=2, −1 ≤ t ≤ 02, 0 ≤ t ≤ 10, otherwisexo(t) =x(t) − x(−t)2=1, −1 ≤ t ≤ 0−1, 0 ≤ t ≤ 10, otherwiseii.rx(t) = xe(t) ∗ xe(t) − xo(t) ∗ xo(t)It’s easier to do the convolution graphically. Each division is 1 in the plot.rx(−2) = rx(2) = 0, rx(−1) = rx(1) = 1, and rx(0) = 6Page 6 of