ECE 314 { Signals and Systems Fall/2012

1 ECE 314 Signals and SystemsFall/2012 Solutions to Homework 3 Problem clarity, let us rename the signal described in the text-book fromxtox . Now differentiatexand observe that the derivative iszero outside the interval ( /2, /2), and it is 1over this , lim 0x (t) = 0 for anyt6= 0. At the same time , the integral ofx (t) over the interval ( , ) is always unity. These two properties arethose that define a delta Systems that follow have inputx(t) orx[n] and outputy(t) ory[n]. For each system , determine whether it is (i) memoryless, (ii)stable, (iii) causal, and (v) time -invariant.(a)y(t) = cos(x(t))Solution:(i) Is the system memoryless?Yes, sincey(t) only depends on the presentvalue ofx(t).(ii) Is the system stable?Yes, since|y(t)| 1 (property of the cosine func-tion).

2 (iii) Is the system causal?Yes, since it is memoryless, it only depends onthe present input (For a system to be causal, its present output must notdepend on future values of the input).(v) Is the system time -invariant?Yes, sincey(t+ ) = cos(x(t+ )), for anytand .(b)y[n] = 2x[n]u[n]Solution:(i) Is the system memoryless?Yes, sincey[n] only depends on the presentvalue ofx[n].(ii) Is the system stable?Yes, since|y(t)|= 2|x[n]|u[n] 2|x[n]|. Hence, ifx[n] is bounded byM(|x[n]| M), theny[n] is bounded by 2M.(iii) Is the system causal?Yes, since it is memoryless, it only depends onthe present input.(v) Is the system time -invariant?No. The time -invariance condition doesnot hold, because the signal that is being multiplied byx[n] varies with (d)y(t) = t/2 x( )d Solution:(i) Is the system memoryless?

3 No, since the integral is evaluated on theinput over all the time from tot/2.(ii) Is the system stable?No. A simple counter example is when the inputsignal isx(t) 1, which is obviously bounded, while the outputy(t) is notfinite, since the integral of 1 from tot/2 is not finite.(iii) Is the system causal?No. For negativet, the output depends on allvalues ofx(t), from tot/2, which is greater thant. Hence, the outputdepends on future values ofx(t).(v) Is the system time -invariant?No. The outputyd(t) for a time -shiftedversion of the inputx(t d) isyd(t) = t/2 x( d)d = t/2 d x(s)ds= (t 2d)/2 x(s)ds=y(t 2d).Therefore, it does not obey the time -invariance condition.(f)y(t) =ddtx(t)Solution:(i) Is the system memoryless?No, since the derivative of a function at aspecific pointtocannot be determined just from the knowledge of the valueof the function onto.

4 (ex. you cannot determine the derivative ofx(t) att= 2, if you only know thatx(2) = 10).(ii) Is the system stable?No. A counter example is whenx(t) = 1 t2, 1<t <1, and it is zero (t) is bounded, but its derivative, which isgiven bydx(t)dt= t 1 t2,goes to , whentapproaches (iii) Is the system causal?Yes, since the derivative can be determined fromthe expressiondx(t)dt= limh 0+x(t) x(t h)h,which only depends on past values ofx(t).(v) Is the system time -invariant?Yes. The derivative of a time -shifted signalisyd(t) =ddt[x(t d)] =dxdt(t d)ddt(t d) =dxdt(t d) =y(t d).(i)y(t) =x(2 t)(i) Is the system memoryless?No, since the output depends on the value ofthe input at a time -instant other thent.(ii) Is the system stable?Yes, since|y(t)| M, if|x(t)| M.(iii) Is the system causal?No.

5 For negativet, 2 tis positive, therefore theoutput depends on the future.(v) Is the system time -invariant? (t) =x(2 t d)) =x(2 (t+d)) =y(t+d).Problem (a) Yes. Consider the system defined by the ruleO(i)(t) = t 1M(s)f(s)ds,whereM(t) = 50e (t). Show that this system is linear but it is timevariant. Can you give an example of a physical system that can be modeledby the above system ?(b) The equation for this circuit is:i(t)R(t) +v2(t) =vi(t). Assume thatv2( ) = 0. Sincei(t) =cv 2(t), we can rewrite the circuit equation asv 2(t) +1R(t)Cv2(t) =1R(t)Cvi(t). Following class notes on linearity of ODE s,show that a system represent by a differential equation with time varyingcoefficients is still linear systemHhas the input-ouput pairs depicted inFig. (a) (in the book).

6 Answer the following questions, and explain youranswers:3(a)Could this system be causal?Solution:No. The system is linear, therefore, for an inputx(t) 0, the out-put should bey(t) 0. This is true because, by the homogeneity property,whenx(t) = 0 f(t), the output must bey(t) = 0 H(f)(t) the system is causal, it doesn t know anything about the future. There-fore, if for some inputx( ) = 0, for < t, then the output must bey( ) = 0,for < t. Because, as for what the system knows,x(t) could be 0 for the figure, we notice thaty2(t) = 1 fort (0,1), whilex2(t) = 0 fort ( ,1). This contradicts the conclusions discussed before. Therefore,the system cannot be discrete- time system is both linear and the output due to an inputx[n] = [n] is given in Fig. (a) (inthe book).(a)Find the output due to an inputx[n] = [n 1]Solution:Let s call the signal in Fig.

7 (a)h[n]. Since the system istime-invariant, forx[n] = [n 1],y[n] =h[n 1]. n 2 113456 0 11232y[n](b)Find the output due to an inputx[n] = 2 [n] [n 2].Solution:Now we use the linearity property, as well as time -invariance:x[n] = 2 [n] [n 2] y[n] = 2h[n] h[n 2].4 n 2 12356 0 1123y[n]4 2 314(c)Find the output due to the input depicted in Fig. (b).Solution:Now, we can see thatx[n] = [n+ 1] [n] + 2 [n 1].Therefore,y[n] =h[n+ 1] h[n] + 2h[n 1]. n 2 113456 0 1123 2 3 4456y[n]2 Problem (a)The solution of a linear differential equation is givenbyx(t) = 10e t 5e MATLAB, plotx(t) versustfort = 0 :A possible MATLAB code to do that is:t = 0 : : 5 ;x = 10 * exp(-t) - 5 * exp( * t);plot(t,x); 10123456