University of Kentucky Department of Electrical and Computer Engineering. EE421G: Signals and Systems I Fall 2007

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1 University of Kentucky Department of Electrical an Computer Engineering EE4G: Signals an Systems I Fall 7 Issue: October 4, 7 Problem Set 6 Due: October, 7 (In class) Reaing Assignments: Rea Chapter 3.7, 3.8, 4., 4., 4.3 of Chen Computer Assignments: ) Simulating a iscrete-time LTI system in Matlab is easy. During lecture, you have learne that a causal ifference equation can be written as follows: a y[ n] + a y[ n ] am + y[ n M ] = bu[ n] + bu[ n ] bn + u[ n N] Given a finite uration signal u[n] (specifie as a vector in Matlab), you can compute the zero-state output of the system using the comman filter: >> y = filter(b, a, u) where a is the vector [a, a,, a M+ ] an b is the vector [b, b,, b N+ ]. a) Assuming zero initial conitions an input u[ n] = cos( ) 5 n q[ n], compute the first values of the output of the following ifferential equation: y [ n].43y[ n ] +.48y[ n ] =.675u[ n] +.349u[ n ] +.675u[ n ] MATLAB coe: b = [.675,.349,.675]; a = [, -.43,.48]; n = :; u = cos(pi/5*n); ya = filter(b, a, u); figure(); stem(ya); xlabel('n'); ylabel('y'); title('zero State Response'); set(gca, 'Xlim', [, ]);

2 .5 Zero State Response.5.5 y n b) Alternatively, you can first compute the impulse response of the LTI system using the following comman: >> [h, t] = impz(b, a, n) where n is the number of impulse response values to be compute. The vector h contains the values of the impulse response, an t contains the corresponing time inices. After computing h[n], you can then use the conv comman between h[n] an the input u[n] to compute the output. Repeat part a) using this metho an plot the outputs for both part a) an b) versus time using the comman stem. Explain any iscrepancies, if any. Remember, you can always use >> help <comman name> to fin out the proper syntax an usage of any comman MATLAB coe: [h, t] = impz(b, a, 3); yb = conv(h, u); figure(); stem(yb); xlabel('n'); ylabel('y');

3 title('zero State Response using Convolution'); set(gca, 'Xlim', [, ]);.5 Zero State Response using Convolution.5.5 y n ) Continuous-time LTI system is hanle a little ifferently in Matlab. Given the ifferential equation M N a y( am y( + am y( b u(... bn u( bn u( M + = N + you can specify the CT LTI system as a transfer function object as follows: >> sys = tf(b, a) where a is the vector [a, a,, a M+ ] an b is the vector [b, b,, b N+ ]. Matlab cannot irectly compute the output of a CT system because one cannot specify a continuous-time signal in a igital computer. All Matlab can o is to compute the output at specifie time instances using the comman lsim: >> y = lsim(sys,u, where sys is the transfer object function, t is a vector that contains the iscrete time instances, u is the vector of the input signal sample at those time instances an y is the output sample at the same time instances.

4 Using these comman, compute the zero-state output response of the following ifferential equation 3 3 ( y + 6 y( + y( + 6y( = u( + 3 u( + 4u( for time instances starting from. sec to sec at an internal of. sec. The input is u ( = log(cos(5. MATLAB coe: t =.:.:; u = log(*.*cos(5*; b = [, 3, 4]; a = [, 6,, 6]; sys = tf(b, a); y = lsim(sys, u, ; plot(t, y); xlabel('t'); ylabel('y'); title('zero State Response');.5 Zero State Response y t Paper an Pencil Assignments: The first three are the same as last homework. You o not nee to turn them in if you have one them previously. BUT DO NOT FORGET THE REMAINING THREE PROBLEMS!

5 ) Problem 3.4: Consier the positive feeback system show below. Fin its impulse response. r( + a u( Unit Time elay Element y( Solution: 3 ht = aδ t aδ t + aδ t 3 () ( ) ( ) ( ) k = ( ) k k a δ ( t k) = ) Problem 3.5: Compute the integral convolution of h i ( an u i (, shown below. h ( u ( h ( u ( Solution: Part :

6 Part :.5 t, t ( t 3) + 3, t 4 y () t =.5( t 6 ), 4 t 6, otherwise y =.5, y =, y 3 = 3 () ( ) ( ) y t 3) Problem 3.8: Fin a ifferential equation to escribe the following network. Ω + u - H F + y -

7 Solution: t u t y + y τ = ( τ) () y() t 4 y() t + y() t = u () t y () t ifferential equation: 4 y() t + y () t + y( = u ( 4) Problem 4.: What is the funamental frequency an funamental perio of the signal Express it in complex Fourier series an plot the magnitues an phases of its frequency components. x( = 3 + sin 6t cos6t + sin 9t cost Solution: The gc of {6, 9, } is 3. Thus the funamental frequency is ω = 3 ra / s an the funamental perio is P =. 3 j6t j6t j6t j6t j9t j9t jt jt e e e + e e e e + e xt () = 3+ + j j = 3 + j.5 e + + j.5 e +.5 j e +.5 je.5e.5e ( ) ( ) ( ) j Because + j.5 =.e =.e We have j6t j6t j9t j9t jt jt.68 j53.4 an j.5 =.e j53.4 j j53.4 j6t j53.4 j6t j9 j9t j9 j9t () = x t e e e e e e e e e j8 jt j8 jt +.5e e +.5e e

8 c m ω ω ) Problem 4.: Consier the full-wave rectifier in Figure.(a). What is its output y( if the input is u(=sint? What are the funamental perio an funamental frequency of y(? Express the output in Fourier series. Does the output contain frequency other than ra/s? Solution:

9 .5 y(.5 / 3/ t Funamental perio = Funamental frequency = = 4 ra / s = ω / j4mt cm = sin te ( 4m ) ( 4 sin cos ) 4+ ( j4m) j4mt e j m t t = j m e + 4 = = 4 6m 4 4 = ( m ) t= j 4mt y() t = e m= ( 4m ) It contains frequency 4m, for all integer m. 6) Problem 4.5: What is the total energy of the signal in Problem 4? What is its average power in one perio? How many percentage of the average power lies insie the frequency range [-7, 7]?

10 Solution: Its total energy is infinite. Its total average power is c = m ( ) ( ) ( ) ( ) ( ) is ( ) = = 6.9 The average power insie [ 7,7] =.5 It is.5 =.68 = 68% of the total average power. 6.9