Homework 5 EE235, Summer 2013 Solution

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1 Homework 5 EE235, Summer 23 Solution. Fourier Series. Determine w and the non-zero Fourier series coefficients for the following functions: (a f(t 2 cos(3πt + sin(πt + π 3 w π f(t e j3πt + e j3πt + j2 [ej(πt+ π 3 e j(πt+ π 3 ] e j3πt + e j3πt + 2 e j π 2 [e j(πt+ π 3 e j(πt+ π 3 ] e j3πt + e j3πt + 2 e j π 2 [e jπt e j π 3 e jπt e π 3 ] e j3πt + e j3πt + 2 e j π 6 e jπt 2 e j 5π 6 e jπt e j3wt + e jwt + 2 e j π 6 e jw t 2 e j 5π 6 e jw t d 3, d 3, d 2 e j π 6, d 2 e j 5π 6 (b f(t [sin(4πt] 2 Hint: You don t need to take a dot product. he period of sin(4πt is /2, but when you square sine, the negative part becomes positive, so the period of f(t [sin(4πt] 2 is /4. Mathematically, using the trig identity sin 2 (x.5.5 cos(2x, f(t [sin(4πt] cos(8πt herefore, w 2π 8π Use the complex exponential representation of sin(t, f(t [ ] 2 j2 (ej4πt e j4πt 4 (ej8πt 2 + e j8πt his looks like a Fourier series now. Since we know that w 8π, f(t 4 ejwt + 2 ejwt 4 ej( wt So, the coefficients are d 4, d 2, d 4. (c f(t e jπk δ(t 4k 2 k Find a formula for d n and be sure to specify which d n are zero here.

2 We can simplify f(t: f(t ( k δ(t 4k 2 k here is a delta every 4 time units, alternating between positive and negative, so the period is 8 and the fundamental frequency is w 2π 8 π 4. Calculate the coefficients d n f(te jnwt Choose limits which cover one whole period (you could use (,8 too k ( k δ(t 4k 2e jnwt Replace the infinite sum with just the deltas within our limits of integration, (since all the other delta functions are zero between -4 and 4 So for even n, d n [δ(t 2 δ(t + 2]e jnwt 8 [ ej2nw + e j2nwt ] 2j 8 e j2nw e j2nwt 2j j 4 sin(2nw j 4 sin(nπ/2 (d f(t 3 cos(4t + 4 sin(πt Hint: this is a trick problem. his is not periodic, so it doesn t have a Fourier series. 2. Fourier Series Manipulation. Consider a real-valued signal f(t with the Fourier series representation f(t n d n e jnwt. Some transformations of f(t can be represented by simply modifying the Fourier series coefficients. (a Let g(t be the time-reversed version of f(t, g(t f( t. Let g(t have Fourier series coefficients, c n. Derive the relationship between the Fourier series coefficients, c n and d n. Since g(t is the time-reversed f(t, both signals have the same fundamental frequency, w. heir Fourier series expansion is f(t g(t n n d n e jnwt c n e jnwt 2

3 ime-reversing f(t g(t f( t d n e jnwt n Perform a change of variable, k n d k e jkwt k Compare to the first equation for g(t and match coefficients, so c n d n, (b Let y(t be the time-shifted version of f(t represented by y(t f(t t for real constant t. Let y(t have Fourier series coefficients denoted by b n. Derive the relationship between the Fourier series coefficients, b n and d n. Write the Fourier series for f(t and y(t (note y(t will have the same frequency and period as f(t f(t d n e jnwt y(t Write y(t as the time-shifted version of x, y(t x(t t n n n n b n e jnwt n d n e jnwt e jnwt d n e jnwt e jnwt d n e jnw(t t Matching terms, b n e jnwt d n. 3. Fourier series and compression. Many applications in electrical engineering, such as image compression, video streaming, etc., require a signal to be compressed using a lossy compression scheme. In this problem, we explore how much a signal is corrupted by discarding some of its Fourier series coefficients. If we can discard some coefficients without changing the signal drastically, we can store (or transmit fewer coefficients to represent the signal. Suppose we are given the Fourier coefficients of a signal that we wish to compress: C k a k, where a > and k...,,,,... Assume that the signal has period 2π. he signal can be expressed as the Fourier series x(t C k e jkt. k 3

4 (a Compute the energy average power in one period of this signal. Hint: you can use Parseval s theorem: C k 2 x(t 2 k Also recall the following geometric series properties: n r k rn+ and if < r <, r k k r k r. If you have calculated the Energy of the signal, and multiplied the formula with, that s fine, too, as long as you do it to the other series as well. Average Power x(t 2 k k C k 2 a k 2 since a is positive, a k will always be positive k since C is counted twice with the new limits since a 2 k a 2 k, since < a 2 <, a 2 k a 2 + ( k a 2 k + a 2 k k ( + a 2 k + a 2 k k k k a 2 k a 2k ( a 2 k k + 2 a 2 a2 a 2 + 2a2 a 2 + a2 a 2 k 4

5 (b We want to store the signal using just the Fourier series coefficients. Since there are an infinite number of coefficients, we cannot store them all. Instead, we keep only the coefficients C k where N k N, and discard all the coefficients for k > N (we assume they are zero. How much energy is in one period of this compressed signal? Let ˆx be the approximation of x(t using the coefficients where N k N. since a 2 k a 2 k, since < a 2 <, Average Power N k N N k N N k N C k 2 a k 2 a 2 k a 2 + ( k N ( N a 2 k + a 2 k k ( N ( N + a 2 k + a 2 k k N k N k a 2 k a 2k N ( a 2 k k k + 2 ( a 2 (N+ a a2 a 2N a 2 + a2 2a 2N a 2 (c If a.2 and we only keep the 2 coefficients for k, what percentage of the energy of the original signal is captured in the compressed signal? + a 2 2a 2N a % 4. Fourier series and real signals (a Prove that if the Fourier series coefficients of a signal f(t come in complex conjugate pairs d n d n (where d n r n e jθn is the complex conjugate of d n r n e jθn, that the signal f(t 5

6 must be real. Note that d d d. f(t n d n e jnwt Separate the positive and negative n terms d + d n e jnwt + n ( n d n e jnwt Let n k in the second sum ( d + d n e jnwt + d k e jkwt n k d + d n e jnwt + d n e jnwt n d + d n e jnwt + d ne jnwt n d + r n e jθn e jnwt + r n e jθn e jnwt n d + r n e j(nwt+θn + r n e j(nwt+θn n d + 2r n cos(nw t + θ n n Since d d, d must be real. herefore, f(t is a sum of real signals, so f(t is real. (b State the converse of part (a. (Which is also true. You do not have to prove this, but you should know this property. If a signal is real then the Fourier series coefficients must come in complex conjugate pairs d n d n (or equivalently d n d n. 5. Given the signal f(t 52 + cos(t + 2sin(3t + cos(7t. State the w and the coefficient values d n such that f(t is expressed as a weighted sum of complex exponentials: Rewriting the signal f(t yeilds, f(t n d n e jwnt. f(t 5e 7jt + je 3jt + 2 e jt ejt je 3jt + 5e 7jt. Since,3 and 7 are all prime a likely candidate for ω is. herefore d n is nonzero for the values n ±7, ±3, ±, and where, d 7 d 7 5, d d 2, d 3 j, d 3 j, d 52. 6

7 6. Let x(t n series expansion of x(t. u(t n u(t n 4 for some constant. Find the coefficients d n in the Fourier Period. One integration interval is from to 4 Find X (k : X Find X k (k : 4 t 4 4 X k 4 e jkωt (time in which pulse is active. [e jkωt 4 jkω (e jkω 4 j kω 2 e jkω 8 (e jkω 8 2j kω e jkω 8 2 e jkω 8 sin(kω kω 8 e jkω 8 sinc(kω Given the signal x(t {, < t <, t >, state the Fourier ransform, X(jω. For a-f state the Fourier ransform given the time domain property. For e and f, use a general form for y(t, i.e. y(t Y (jω. he Fourier ransform of a rect function is, X(jω 2sin(ω 2 sinc(ω. ω (a Scaling, ax(t? ax(t a2 sinc(ω (b Frequency Shift, e jγt x(t? e jγt x(t 2 sinc((ω γ (c ime Shift, x(t t? x(t t e jωt 2 sinc(ω d (d Differentiation, x(t? d x(t jω2 sinc(ω (e Convolution, x(t y(t? x(t y(t 2 sinc(ω Y (jω (f Multiplication, x(ty(t? x(ty(t 2π 2 sinc(ω Y (jω 8. Calculate the Fourier ransforms of the following using the integral definition: (a δ(t + δ(t 2 7

8 F (jω f(te jωt (δ(t + δ(t 2e jωt δ(te jωt + δ(t 2e jωt e jωt t + e jωt t2 + e j2ω (b e 2t u(t 3 F (jω 3 f(te jωt e 2t u(t 3e jωt e (2+jωt e (2+jωt (2 + jω 3 (2 + jω (e (2+jω e (2+jω3 e 6 e 3jω 2 + jω 8

9 (c e 4 t F (jω 2 2 f(te jωt e 4 t e jωt (cos(ωt jsin(ωte j4t + (cos(ωt + jsin(ωte j4t + cos(ωte j4t e ωt + e ωt e j4t 2 e (jω j4t + e (jω+j4t e(jω j4t (jω j4 + e (jω+j4t (jω + j4 (jω j4 + (jω + j4 jω + j4 jω + j4 j(ω 4(jω + j4 8 (ω ω 2 (cos(ωt jsin(ωte j4t (cos(ωt jsin(ωte j4t 9. Find the Fourier transform of the following signals using tables: (a te t u(t h(t H(jω te t u(t ( + jω 2 (b sin(2πte t u(t h(t sin(2πte t u(t ( h(t ( ej2πt e j2πt e t u(t 2j h(t 2j ej2πt e t u(t 2j e j2πt e t u(t e t u(t + jω e jαt s(t S(j(ω α H(jω 2j ( + j(ω 2π + j(ω + 2π 9

10 Or using your tables h(t sin(2πte t u(t (2 H(jω 2π (2π 2 + ( + jω 2 (c d (te 2t sin(tu(t d (te 2t sin(tu(t d (te 2t u(t ejt e jt 2j d 2j (te 2t u(te jt te 2t u(te jt x(t d 2j (g(t X(jω 2j jω(g(jω X(jω ω 2 G(jω g(t te 2t u(te jt te 2t u(te jt g(t h(te jt h(te jt G(jω H(j(ω H(j(ω + h(t te 2t u(t H(jω (2 + jω 2 G(jω (2 + j(ω 2 (2 + j(ω + 2 X(jω ω 2 ( (2 + j(ω 2 (2 + j(ω + 2. Find the inverse Fourier transform of the following signals (Use of tables is acceptable. Requires partial fractions: (a X(jω jω 2 ω 2 + 5jω + 4 X(jω jω 2 ω 2 + 5jω + 4 A jω B jω + Solving for A and B yeilds the result X(jω 2 jω + 4 jω + x(t (2e 4t e t u(t

11 6jω + 6 (b X(jω (jω 2 + 5jω + 6 Solving for A and B yeilds the result X(jω X(jω 6jω + 6 (jω 2 + 5jω + 6 A jω B jω jω jω + 2 x(t (2e 3t + 4e 2t u(t. State in words what the LI systems do with these transfer functions: (a H(jω 5 Amplifies the input by 5. (b H(jω Inverts the input. (c H(jω e 3jω Delays the input signal by 3. (d H(jω e jω Amplifies the input by and delays it by. 2. For an LI system, y(t h(t x(t. (a Express the system s transfer function H(jω in terms of the input signal Fourier transform X(jω and output signal Fourier transform Y (jω. Y (jω H(jωX(jω H(jω Y (jω X(jω (b Find H(jω for the LI system: y(t x(t 8x(t 3 y(t x(t 8x(t 3 Y (jω X(jω 8e 3jω X(jω Y (jω X(jω( 8e 3jω H(jω 8e 3jω (c Find H(jω for the LI system: y(t y(t 3 5x(t y(t y(t 3 5x(t Y (jω e 3jω Y (jω 5X(jω Y (jω( e 3jω 5X(jω H(jω 5 e 3jω

12 3. A causal LI system is described by the following ordinary differential equation d 2 y(t + 5 dy(t (a Find the frequency response H(jω for this system. + 6y(t dx(t. d 2 y(t + 5 dy(t + 6y(t dx(t (jω 2 Y (jω + 5jωY (jω + 6Y (jω jωx(jω H(jω Y (jω X(jω H(jω jω 6 + 5jω + (jω jω jω (b Also find its impulse response h(t. h(t ( 3e 3t + 2e 2t u(t 2