# GEORGE MASON UNIVERSITY Solutions to Problem Set #6

## Transcription

GEORGE MASON UNIVERSITY Solutions to Problem Set #6

GEORGE MASON UNIVERSITY Signals and Systems I Spring 2015 Solutions to Problem Set #6 Regular Problems Problem 6.1F Consider the following function that is periodic with a period of 2π, x(t) = et + e−t ; −π < t < π (a) Find the complex Fourier series xn of x(t). (b) Express the Fourier series expansion of x(t) in terms of sines and cosines. (c) Use the derivative property to find the Fourier series coefficients of y(t) = et − e−t ; −π < t < π Solution (a) Since x(t) is a sum of two signals, let’s begin by finding the Fourier series coefficients of v(t) = et . With ω0 = 1, the Fourier coefficients are Z π π 1 1 1 vn = et(1−jn) et e−jnt dt = 2π −π 2π 1 − jn −π i 1 1 h π(1−jn) 1 1 −π(1−jn) = e −e = (−1)n eπ − e−π 2π 1 − jn 2π 1 − jn Now note that x(t) = v(t) + v(−t). We have already found the coefficients for v(t), and we may use the following property to obtain the coefficients for v(−t) (or you may find them easily by modifying the derivation above for vn , FS v(−t) ⇐⇒ v−n So, for the Fourier series coefficients of x(t) we have 2 1 1 1 1 xn = vn + v−n = (−1)n eπ − e−π + = (−1)n eπ − e−π 2π 1 − jn 1 + jn 2π 1 + n2 (b) Since the Fourier series coefficients are even, xn = x−n , then the Fourier series expansion of x(t) in terms of sines and cosines is ∞ ∞ X X x(t) = x0 + xn ejnt + e−jnt = x0 + 2xn cos(nt) n=1 n=1 Therefore, using the expression for xn in part (a) we have eπ − e−π x(t) = π ∞ X 2(−1)n 1+ cos(nt) 1 + n2 n=1 (c) Since y(t) = d x(t) dt and ω0 = 1, then yn = jnxn ! Problem 6.2F Find the Fourier series coefficients for the following signal that is periodic with a period T = 6 (only one period of x(t) is shown. x(t) 1 t -3 -2 -1 1 2 3 Solution There are several ways to do this problem. The first, which is perhaps the most tedious, is to find the FS coefficients directly by integration. This will involve find the integral over three separate intervals. This could be simplified a bit by noting that the signal is an even function. So one could find the Fourier series coefficients of the signal v(t) = x(t)u(t) i.e., only the positive time part of x(t). Then noting that x(t) = v(t) + v(−t) the Fourier series coefficients for x(t) may be easily found from the coefficients for v(t). Another approach would be to differentiate x(t) and use the derivative property. Since x0 (t) is a sum of two pulses, then finding the FS coefficients of the derivative is quite straightforward. Taking this one step further, let’s differentiate x(t) twice. This results in a signal that has four impulses over each period as shown in the figure below. y(t) = x00 (t) 1 -1 1 -2 t 2 -1 The Fourier series coefficients of y(t) = x00 (t) are given by Z T /2 yn = y(t)e−jnω0 t dt −T /2 where T = 6 and ω0 = 2π/6 = π/3. Since y(t) is a sum of four impulses, using the sifting property of the impulse, Z ∞ δ(t − t0 )f (t)dt = f (0) −∞ we have o 1 1 1 n jn2π/3 e − ejnπ/3 − e−jnπ/3 + e−jn2π/3 = cos(2nπ/3) − cos(nπ/3) yn = 6 3 3 Finally, since yn are the FS coefficients of the second derivative of x(t), then applying the derivative theorem twice, we divide yn by (jnω0 )2 to get xn , xn = 3 {cos(nπ/3) − cos(2nπ/3)} n2 π 2 Problem 6.3F The one-dimensional heat equation is ut (x, t) = c2 uxx (x, t) where u(x, t) is the temperature at time t and position x. When considering heat flow around a circular wire of length L, the general solution is: u(x, t) = ∞ X 2 fn e−(cnπ/L) t ej2πnx/L n=−∞ where fn are the Fourier series coefficients that describe the initial temperature distribution at time t = 0. In this problem, you are asked to solve the problem of heat flow along a finite length bar as shown below. Here, we assume that both ends of the bar are kept at a constant temperature of zero degrees (for any other temperature, all you would need to do is add a constant to the solution that is equal to the temperature at the end of the bar). Although the solution is not a periodic function of x as we had for the circular wire, we may consider this bar of length L to be one period of a infinitely long bar with a temperature distribution that is periodic in x. The solution to the heat equation then becomes u(x, t) = ∞ X 2 fn e−(cnπ/L) t sin n=−∞ πnx L where, as we see, the condition that the temperature at the ends is always equal to zero. Note that ∞ X u(x, 0) = fn sin n=−∞ πnx L so fn are the Fourier series coefficients of the sine terms in the Fourier series expansion of the initial temperature distribution, f (x) = u(x, 0), and these coefficients are given by fn = 2 L Z L f (x) sin 0 πnx dx L ; n = 1, 2, . . . (a) Find the temperature u(x, t) in a copper bar of length 80 centimeters if the initial temperature is f1 (x) = 100 sin(πx/80)◦ C and the ends are kept at 0◦ C. (b) How long will it take for the maximum temperature in the bar to drop to 50◦ C? The constant c2 in the heat equation is K c2 = ρσ where K is the thermal conductivity, ρ is the density of the material, and σ is the specific ◦ 2 heat. For copper, K = 0.95 cal/(cm sec C), ρ = 8.92 g/cm , and σ = 0.092 cal/(g◦ C). (c) Repeat parts (a) and (b) when the initial temperature distribution is f2 (x) = 100 sin(3πx/80)◦ C. (d) Does the maximum temperature drop to 50◦ C faster for the initial distribution f1 (x) = 100 sin(πx/80)◦ C or f2 (x) = 100 sin(3πx/80)◦ C, or is it the same for both? Can you give a reason why? Solution (a) We want to find the temperature u(x, t) in a copper bar of length 80 centimeters if the initial temperature is f1 (x) = 100 sin(πx/80)◦ C and the ends are kept at 0◦ C. This problem may seem hard, but if we sort out what in known, the solution is actually very simple. First note that we know that the solution of the heat equation is given by u(x, t) = ∞ X 2 πnx L 2 πnx 80 fn e−(cnπ/L) t sin n=1 and with L = 80 this becomes u(x, t) = ∞ X fn e−(cnπ/80) t sin n=1 At time t = 0, the solution has the form u(x, 0) = ∞ X fn sin n=1 πnx 80 and this corresponds to the initial temperature distribution. Since we are told that πx f (t) = u(x, 0) = 100 sin 80 then it is clear that f1 = 100 and all the rest of the coefficients are equal to zero. Therefore, the solution to the heat equation that satisfies the initial temperature distribution is 2 πnx u(x, t) = 100e−(cnπ/80) t sin 80 (b) To find out how long will it take for the maximum temperature in the bar to drop to 50◦ C, with c2 = K ρσ and with K = 0.95 cal/(cm sec ◦ C), ρ = 8.92 g/cm2 , and σ = 0.092 cal/(g◦ C), we find that (cnπ/80)2 = 1.785 · 10−3 sec−1 Therefore, u(x, t) = 100e−0.001785t sin πnx 80 and the maximum temperature is equal to 50◦ C when e−0.001785t = 0.5 =⇒ t= ln(0.5) = 388 sec ≈ 6.1 min −0.0001785 (c) When the initial temperature distribution is f2 (x) = 100 sin(3πx/80)◦ C, then f3 = 100 instead of f1 and the solution becomes 2 3πx u(x, t) = 100e−(3cπ/80) t sin 80 Now, the exponential term decays at a rate nine times faster than before, and so the maximum temperature reaches 50◦ C in approximately 43 sec. (d) As we see by comparing the results in parts (b) and (c), the maximum temperature drops to 50◦ C faster for the initial distribution f2 (x) = 100 sin(3πx/80)◦ C.