ECE 301 Fall 2011 Division 1. Homework 1 Solutions.

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1 ECE 3 Fall 2 Division. Homework Solutions. Reading: Course information handout on the course website; textbook sections.,.,.2,.3,.4; online review notes on complex numbers. Problem. For each discrete-time (DT) signal below, do the following. (i) Sketch signal x as a function of n by hand, i.e. do not use Matlab or any other software. Carefully label the plots. (ii) Calculate its total energy, i.e. n (iii) Calculate its average power, i.e. lim N 2N + software. x[n] 2. Do not use Matlab or any other software. n N x[n] 2. Do not use Matlab or any other (iv) Find the smallest number L which satisfies the following inequality for every integer n: x[n] L. a. x[n] { if n, 3 n if n >. b. x[n] cos(πn/2). Please note that both signals are defined for all integer n x 3 n u[n ].2.5 x cos(π n/2) n (a) 5 5 n (b) Figure : The graphs of (a) x[n] 3 n u[n ] and (b) x[n] cos(πn/2). Solution.

2 a. x[n] 3 n u[n ] (i) The plot of this signal is in Fig. (a). (ii) E {x} n n x[n] 2 3 2n 3 2n n n (/9) n Since /9 <, we can now use the formula for summing an infinite geometric series, nm q n qm q, with q /9. In our formula for E {x}, the summation has been conveniently manipulated to have m, and therefore E {x} / (iii) Since the total energy is finite, the average power is zero. (iv) As evident from the plot, the global maximum of x occurs at n, and is: x[] /3. b. x[n] cos(πn/2). (i) The plot of this signal is in Fig. (b). (ii) Note that x[n] for all even values of n and x[n] for all odd values of n. Therefore, to compute the energy we would have to sum an infinite number of ones, which means that the energy is infinite. (iii) P {x} lim N 2N + n N x[n] 2 lim N 2N + n N cos(πn/2) 2. If N is even, the summation has N + ones and N zeros and is therefore equal to N +. If N is odd, the summation has N ones and N + zeros, and is therefore equal to N. The values of the fraction are thus (N + )/(2N + ) and N/(2N + ) for even and odd N, respectively. The limit of both as N is /2, and therefore the overall limit is /2 as well. Answer: P {x} /2. (iv) Since x[n] for all n and x[n] for some n, we have that L. Problem 2. For each continuous-time (CT) signal below, do the following. (i) Sketch signal x as a function of t by hand, i.e. do not use Matlab or any other software. Carefully label the plots. (ii) Calculate its total energy, i.e. + x(t) 2 dt. Do not use Matlab or any other software. 2

3 (iii) Calculate its average power, i.e. lim 2T x(t) 2 dt. Do not use Matlab or any other software. (iv) Find the smallest number L which satisfies the following inequality for every real number t: x(t) L. a. x(t) { if t, 3 t if t >. b. x(t) cos(πt/2). Please note that both signals are defined for all real t. x 3 t u(t) t (a) x cos(π t/2) t (b) Figure 2: The graphs of (a) x(t) 3 t u(t) and (b) x(t) cos(πt/2). Solution. a. x(t) 3 t u(t) (i) The plot of this signal is in Fig. 2(a). (ii) E {x} x(t) 2 dt ( ) t dt 9 e ln( 9)t dt 3 2t dt 9)t ln(/9) eln( ( ) ln(/9) ln 9 2ln

4 (iii) Since the total energy is finite, the average power is zero. (iv) Since x is a continuous function for t >, the values x(t) get arbitrarily close to as t +. Also, x(t) < for all t. Therefore, L. b. x(t) cos(πt/2). (i) The plot of this signal is in Fig. 2(b). (ii) E {x} (iii) x(t) 2 dt P {x} lim 2T lim lim 2T 4T cos(πt/2) 2 dt (iv) Since max x(t), we have that L. x(t) 2 dt lim 2T cos(πt/2) 2 dt ( + cos(πt))/2dt lim [t + π ] T 4T sin(πt) [2T + π sin(πt) π ] sin( πt) /2. Problem 3. For each of the following signals, determine whether or not it is periodic, and if it is, find its fundamental period, its fundamental frequency in radians per second, and its fundamental frequency in Hertz. (Note that the signals in Parts (a,b,c) are DT signals, and those in Parts (d,e,f) are CT signals.) (a) x[n] sin(πn/3) + cos(πn/2). Solution. Both components of this signal are periodic: the first component with fundamental period N 6, and the second component with fundamental period N 2 4. Therefore, their sum is also periodic, with fundamental period which is the smallest common multiple of N and N 2, i.e., N 2. In other words, the values of the first component repeat every six samples, and the values of the second component repeat every four samples. Therefore, their sum repeats every twelve samples. The fundamental frequency is therefore /2 Hertz or π/6 radians per second. (b) x[n] n. Solution. This signal never repeats itself, i.e., there are no values of n and N such that x[n] x[n + N]. Therefore, this signal is not periodic. (c) x[n]. Solution. This signal is periodic with fundamental period N, and fundamental frequency Hertz or 2π radians per second. Saying that the fundamental frequency is zero is also correct since this signal can be written as cos( n). (Recall that in discrete time, two frequencies are the same if they differ by an integer multiple of 2π.) 4

5 (d) x(t) sin(2t). Solution. This is a CT sinusoid and is therefore periodic. The fundamental period is π. The fundamental frequency is 2 radians per second or /π Hz. (e) x(t) e πjt. Solution. This is a CT complex exponential with magnitude and is therefore periodic. The fundamental period is 2. The fundamental frequency is /2 Hz or π radians per second. (f) x(t) e (3+πj)t. Solution. This is a CT complex exponential whose magnitude e t is an increasing function of time. Therefore, it is not periodic. 5

ECE 301 Fall 2010 Division 2 Homework 10 Solutions. { 1, if 2n t < 2n + 1, for any integer n, x(t) = 0, if 2n 1 t < 2n, for any integer n.

ECE 301 Fall 2010 Division 2 Homework 10 Solutions. { 1, if 2n t < 2n + 1, for any integer n, x(t) = 0, if 2n 1 t < 2n, for any integer n. ECE 3 Fall Division Homework Solutions Problem. Reconstruction of a continuous-time signal from its samples. Consider the following periodic signal, depicted below: {, if n t < n +, for any integer n,