This preview shows page 1. Sign up to view the full content.
Unformatted text preview: THE STATE UNIVERSITY OF NEW JERSEY RUTGERS
College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Problem Set 1 Spring 2003 1. Derive the convolution intergal from first principles (as outlined in class) given by y t d, where h t is the impulse response of a LTI system, x t the input and x h t y t the output. Convolution Integral: In time domain a linear system is described in terms of its impulse response which is defined as the response of the system (with zero initial conditions) to a unit impulse or delta funtion t applied to the input of a system. If the system is time invariant,then the shape of the impulse response is same no matter when the impulse is applied to the system. Let h(t) denote the impluse response of a LTI system. Let this system be subjected to an arbitrary excitation x(t). To determine the output y(t) we first approximate the input x(t) by staircase function composed of narrow rectangular pulses,each of duration . The approximation becomes better for smaller . As approaches zero, each pulse in the limit approaches a delta funtion weighed by a factor equal to the height of the pulse times . Consider a pulse which occurs at t n. By defenition,the response of the system to a unit impulse or a delta funtion t occuring at t 0 is h(t). Due to the time invariance property the response of the system to a delta funtion weighed by a factor x n occuring at t n,must be y t lim x n h t n Above is the response to one component of input occuring at t n. Since the system is linear we can apply superposition principle to find the response to the sum of the input components (which together constitute x(t)) occuring at different times as, The right hand side of the above equation by definition is the convolution integral that is, 2. For each of the systems described by the input output relationships below, determine which of the following properties apply to the system: Memoryless(M), causal(C), linear(L), timeinvariant(TI), stable(S). Justify your answers. (a) y t sin t 1xt 1 yt xht 0n d yt lim x n h t n n 0 Memoryless: y(t) depends on the input only at time t,so the system is memoryless. Time Invariant: We first compute output of the system and delay it by T, Now delay bt T,and compute the output of the system Linear:First check Homogenity, the response to ax t , Next check additivity. Let y1 t be the response to x1 t and y2 t be the response to x2 t is x2 t . Then the response to x1 t Thus the system is linear. Causal:The system is memoryless,so it is also causal. So for bounded inputs,the output is also bounded and the system is stable. (b) y n x 2 n 1 Memoryless: y[n] depends on the input at times other than n, so the system has memory. Time Invariant:First compute output of the system and then delay it by N, We now delay it by N, then compute output of the system. Homogenity is not satisfied and hence the system is not linear. 2 1 Linear:Let y n be the response to x n : y n the response to ax n , ax2 n Since w n z n , the system is time varying. x2 n 1. We first check the homogenity ayn wn x2 n N 1 zn yn N x2 n N 1 yt sin t 1xt sin t 1 xt xt Stable:Let the input x(t) be bounded for all t: x t K for all t. Then, K sin t 1 x1 t x2 t sin t 1 x1 t sin t 1 x2 t y1 t y2 t sin t 1 ax t asin t 1xt ay t Since z t w t ,the system is time varying. wt sin t 1xt T zt yt T sin t T
1xt T So a bounded input implies bounded output,so the system is stable. 3. Consider a continuous time system with the following input x t and impulse response h t , (a) Compute the output of the system y t xt ht Flipping and shifting h t gives FIGURE 3a. From FIGURE 3a, we find the following different expressions depending on the value of t: t 0. The nonzero portions of x and h t do not overlap,so y t 0. 0 t 2, 3 yt 2 0 1 exp 1 2t 2 exp t 4, 2 t d exp 4 1 1 exp 2 t exp 8 exp 4 4 2 4 d yt 2 0 1 exp 1 2 2 exp 2t d 1 exp 2t t 2 1 exp 1 2t 2 t 4, exp 1 2 1 yt t 01 exp 2 t d t 2t 0 exp 2 d exp 2t d and h t exp 2t u t xt 0 t 2 1 2 t 4 0 otherwise 1 yn x2 n 1 x2 n Stable:Let the input x n be bounded for all n: x n Causal:To get the output at n system is not causal. 0, the system has to look at the input at time 2. Thus,the K for all n. Then, 1 K 1 k otherwise We use the Parseval's relationship: k We can now find the average power: 2 3 1 k 4 5. Let x t exp at u t ,where u(t) is the unit step function. Find the Fourier transform of the following signals. 0 0 (b) y t xt 5 Using the time shifting property of Fourier transform, 4 Y j e j5 X j e j5 1 a j X j exp at exp jt dt exp (a) x t a ak 2 k k k 1 i 1 i 1 4 k 2 k 1 ak 2 When k 0, 1 2j k 2 1 4 1 4 i 1 k 1 4 k k 1 k 1 4 k k k ak 2 k 2 When k 0 1 2j 0 1 4 xt 1 T T 2 dt 1 T T 0 xt . Find the average power in the signal x(t), ak k 1 k 2j 0 1 2j 4. A signal x t is periodic with period T given by 10 3. The Fourier series coeffecients for x(t) are k 0 0 2 . Hint:Use Parseval's relationship! ak 2 k k j t dt 1 a j h t dt (b) Is the system stable? Is the system causal? The system is causal, since h t 0 for t 0. The system is stable, since 1 2t dt 2 0 exp (c) z t x t sin 240t Writing z(t) as complex exponentials gives Applying frequency shitfing property of Fourier transform to each term of z(t) gives, a 1 j2 fct e e j2 fct 2j Therefore,applying the frequency shifting property to the Fourier transform pair we find that the Fourier transform of the damped sinusiodal wave is fc 1 fc 1 7. Show that the overall system function H s for the feedback system in FIGURE 7 is given F s by H s 1 F sGs . + F(s) G(s) Rearranging the previous equation, 5 H s Y s X s 1 F s F s Gs System funtion F s X s Y s 1 F s Gs W s X s Gs Y s Y s F s W s F s X s Gs Y s G f 1 2j 1 1 j2 f 1 j2 f sin 2 fct 2 fc j2 f 2 2 fc 2 6. Evaluate the Fourier transform of the damped sinusoidal wave g t where u t is the unit step function. exp t sin 2 f ct u t , Z j X j 240t 1 X j 240t 2j 1 2j a 1 j 240 1 j 240 zt xt exp j240t 1 exp j240t 2j 1 x t exp j240t 2j 1 x t exp 2j j240t 8. A signal x t of finite energy is applied to a sqaurelaw device whose output is defined by yt x2 t . The spectrum of x t is limited to the frequency interval W f W . Hence show that the spectrum of y t is limited to 2W f 2W . Since multiplication in time domain corresponds to convolution in frequency domain, we may express the Fourier transform of y(t) as W In this integral we note that X f is limited to W f W . When W ,we find that 2W f 0. When W , 0 f 2W . Accordingly the Fourier transform Y(f) is limited to the frequency interval 2W f 2W . 6 Y f X X f W d where X(f) is the Fourier transform of x(t). However X(f) is zero for f Y f X X f d W . Hence, yt x2 t xt xt ...
View
Full
Document
This note was uploaded on 04/03/2008 for the course ECE ECE332 taught by Professor Rose during the Spring '08 term at Rutgers.
 Spring '08
 Rose

Click to edit the document details