EECS 20. Midterm No. 2 Practice Problems Solution, November 10, 2004.

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1 EECS. Midterm No. Practice Problems Solution, November, 4.. When the inputs to a time-invariant system are: n, x (n) = δ(n ) x (n) = δ(n +), where δ is the Kronecker delta the corresponding outputs are y (n) = δ(n ) + δ(n 3) y (n) = δ(n +)+δ(n), respectively. Is this system is linear? Give a proof or a counter-example. Answer to The system is not linear. From time-invariance we see that for the second pair of input and output, x (n 3) = δ(n ) y (n 3) = δ(n ) + δ(n 3) So we can rewrite the first pair of input and output as x (n 3) = δ(n ) = x (n 3) y (n 3) = δ(n ) + δ(n 3) y (n 3) Therefore, the system is not linear.. Consider discrete-time systems with input and output signals x, y [Integers Reals]. Each of the following relations defines such a system. For each, indicate whether it is linear(l), time-invariant (TI), both(lti), or neither (N). Give a proof or counter-example. (a) y(n) =g(n)x(n) (b) y(n) =e x(n) Answer to (a) The system is linear: ˆx(n) = ax (n)+bx (n) ŷ(n) = g(n)(ax (n)+bx (n)) = ay (n)+by (n)

2 Also the system is time-varying if g is not constant (so there exist n, n so that g(n) g(n n )): ˆx(n) = x(n n ) ŷ(n) = g(n)ˆx(n) = g(n)x(n n ) y(n n ) = g(n n )x(n n ) (b) The system is non-linear: ˆx(n) = ax (n)+bx (n) ŷ(n) = eˆx(n) = e ax (n)+bx (n) = (y (n)) a (y (n)) b ay (n)+by (n) But the system is time-invariant: ˆx(n) = x(n n ) ŷ(n) = eˆx(n) = e x(n n ) = y(n n ) 3. (a) An LTI system with input signal x and output signal y is described by the differential equation ÿ(t)+ẏ(t)+y(t) =x(t). Suppose the input signal is t, x(t) =e iωt, where ω is fixed. What is its output signal y? (b) Another LTI system is subject to the differential equation ÿ(t)+y(t) =ẋ(t)+x(t) i. What is the frequency response? ii. What is the magnitude and phase of the frequncy response for ω =? Answer to 3 (a) The output signal is t, y(t) =H(ω)e iωt. It follows that ω H(ω)e iωt +iωh(ω)e iωt +H(ω)e iωt = e iωt, thus H(ω) = ω +iω+, Hence t, y(t) = ω +iω + eiωt

3 (b) (i) The frequency response is H(ω) = iω+ ω +. (ii) Hence H() = i 3 = 5 3, H() = π 6 4. For this problem, assume discrete time everywhere. Given two LTI systems S and T suppose signal f is input into S and g into T. The input and output signals are displayed in figure. Are the two systems identical, that is, S = T? f(n) (S(f))(n) g(n) (T(g))(n) Figure : Signals for problem 4 Answer to 4 No. S T Argue by contradiction. Assume S = T = R, say. Observe that f(n) is (g f)(n +). The figure below plots R(g f)(n) =T (g)(n) S(f)(n) and R((g f))(n +)=R(f)(n) =S(f)(n). But the second plot is not the first plot delayed by. 5. A system is described by the difference equation y(n) =x(n)+bx(n ) + ay(n ), () wherein a, b are constants. (a) Obtain the [A, b, c T,d] representation of this system by: i. choosing the state, ii. calculating A, b, c T,dfor your choice of state. (b) If x(n ) =,y(n ) =, calculate the zero-input (i.e. x(n) =,n ) state response. 3

4 (g-f)(n ) (T(g)-S(f))(n) ) (g-f)(n+) = f(n) S(f)(n) (c) Calculate the frequency response of this system. Answer to 5 (a) (i) Take the state as s(n) =[x(n ),y(n )] T. (ii) Writing s(n +) =As(n)+bx(n) in expanded form gives s(n +) = = [ ] [ x(n) x(n) = y(n) x(n)+bx(n ) + ay(n ) [ ][ ] [ ] x(n ) + x(n), b a y(n ) ] from which [ A = b a ], b = [ ] () and, since y =[b a] [ x(n ) y(n ) ] + x(n), so c T =[b a],d=. (b) The zero-input state response is s(n) =A n s(),n. So we need to calculate A n, with A givenin(). By induction, A n = [ a n b a n ] 4

5 and since s() = [ ] T, s(n) =[ a n ]. (c) To obtain the frequency response, substitute x(n) =e iomegan,y(n) =H(ω)e iωn in )() and simplify to get ω, H(ω) = +be iω ae iω. 6. For the linear difference equation y(n) =y(n ) + x(n), (a) Taking the state at time n to be s(n) =y(n ), write down the zero-input response, the zero-state impulse response h : Ints Reals, the zero-state response, and the (full) response. (b) Show that the zero-input response y zi is a linear function of the initial state, i.e. it is of the form n, y zi (n) =a(n)s(), for some constant coefficients a(n). Then show that lim y zi(n) = n (c) Suppose s is the initial state and the input is a unit step, i.e. x(n) =,n ; =,n<. Determine the response y(n),n, and calculate the steady state response y ss = lim y(n). n (d) Plot the input, output and the steady state value in the previous part. (e) Calculate the frequency response H : Reals Complex and plot the magnitude and phase response. (f) Suppose x(n) =, <n<. What is the output y(n), <n< and compare it with y ss. Answer to 6 (a) The a, b, c, d representation is (with s(n) =y(n )) s(n +)=s(n)+x(n), y(n) =s(n)+x(n). The zero-input response (x(n) =,n ) is s zi (n) = n s(), y zi (n) = n+ s() = n+ y( ). (3) The zero-state impulse response is n, h(n) = { d =, n = ca n b = n, n =n. 5

6 H(ω) H(ω) π/ π ω π/ π ω Figure : Plots for problem 6 So the full response is n y(n) = n+ y( ) + n m x(m),n. (4) m= (b) From (3) y zi is a linear (time-varying) function of the initial state with a(n) = n+. Clearly, y zi (n) as n. (c)in (4) take x(m) =,m to get y(n) = n+ s + m= n n m = n+ s + k y ss =as n = n k = n+ s = n+ (d) The plots are straightforward. (e) The frequency response is ω, H(ω) = the magnitude response is ω, H(ω) = the phase response is e iω, [.5 cos(ω)] /, ω, H(ω) =tan sin(ω) cos(ω). The plots in figure are for ω π: (f) In this case x(n) e in,soy(n) H()e in ==y ss. 6

7 7. Suppose x is a continuous-time periodic signal, with period p and exponential FS representation, t, x(t) = in which ω =π/p. k= X k exp(ikω t), (a) Write down the formula for X k in terms of x. (b) Consider the signal y, t, y(t) =x(αt), in which α>is some positive constant. i. Show that y is periodic and find its period q. ii. Suppose y has FS representation t, y(t) = Y k exp(kω t), k= What is ω? Determine the Y k in terms of the X k. Answer to 7 (a) The formula is X k = p p x(t)e ikω t dt. (5) (b) We want y(t) =x(αt) =y(t + q) =x(α(t + q)) = x(t + p),soαq = p or q = p/α. So the FS of y is y(t) = k = k Y k e ikω t X k e ikαω t from which ω = αω and Y k = X k. 8. Give an example of a nonlinear, time-invariant system S that is not memoryless. Time is discrete. (a) Show that S is nonlinear, time-invariant, and not memoryless. (b) Suppose x : Ints Reals is periodic with period p. Let y = S(x). Isy periodic? (c) Suppose Q is another discrete-time, time-invariant system. Is the cascade composition S Q time-invariant? Give a proof or a counterexample. (d) Define the system R by reversing time: x, n, R(x)(n) = S(x)( n). IsR timeinvariant? Why? If x is periodic as above and w = R(x), isw periodic? Why. 7

8 Answer to 8 One possible system is x, n, S(x)(n) =[x(n )]. (a) S is clearly nonlinear since, if x(n ), S(x)(n) =4[x(n )] [x(n )]. S is time-invariant, since for any integer T, S D T (x)(n) =[x(n T )] = D T S(x)(n). S is not memoryless, because if it is there is f : Reals Reals with S(x)(n) =f(x(n)). But this will not hold if we choose x, n, n so that x(n) =and [x(n )] f(). (b) Yes it is periodic, since S(x)(n + p) =D p Sx(n) =S D p (x)(n) =S(x)(n), since D p x = x because x is periodic with period p. (c) The composiiton of any two time-invariant systems is periodic, since D T (Q S) =Q D T S =(Q S) D T. (d) R is not time-invariant, because D T R(x)(n) = R(x)(n T )=S(x)( n + T )=[x( n + T )] R D T (x)(n) = S D T (x)( n) =[D T (x)( n )] =[x( n T )]. These two quantities are not equal for particular choices of x, n, T. w is periodic with the same period p, because by part (b) S(x) is periodic with period p, so w(n + p) =S(x)( n p) =S(x)( n) =R(x)(n) =w(n). 9. You are given three kinds of building blocks for discrete-time systems: one-unit delay; gains; and adders. (a) Use these building blocks to implement the system: y(n) =y(n ) + x(n)+x(n ). (6) (b) Take the outputs of the delay elements as the state and give a [A, b, c T,d] representation of this system. (c) You are allowed to set the output of the delay elements to any value at time n =. Select these values so that the output of your implementation is the solution y(n),n for any input x(n),n and initial conditions: y( ) =,y( ) =.8,x( ) =. Now suppose x() = x() = x() =. Calculate y(),y(),y(). 8

9 x(n) y(n) x(n-) D D D y(n-) y(n-) Figure 3: Implementation for problem 9 Answer to 9 (a) Figure 3 is one implementation. (b) Taking s(n) =[x(n ) y(n ) y(n )] T and using (6) weget x(n) s(n +) = y(n) = s(n)+ y(n ) y(n) = []s(n)+ x(n) from which we can read off A, b, c, d. (c) We take the initial state as s() = [x( ) y( ) y( )] T =[.8] T. Then y() = c T s()=[]s() =.4 y() = c T As() = = y() = c T A s() =.7 One can also get these directly from (6).. An integrator can be used as a building block: For any input x : Reals + Reals, its output is: t t, y(t) =y + x(s)ds. The initial condition y() can be set. Use integrators, gains and adders to implement the system: d y (t) y(t) =x(t), (7) dt with iniital condition y() =, ẏ() =.4. Hint First convert a differential equation into an integral equation and then implement. Answer to Figure 4 shows the implementation 9

10 x(t) y(t) t y(t) t y(t) y() y() Figure 4: Implementation for problem. A periodic signal x : Reals Reals is given by t, x(t) =[+cos(π t)] cos(π 4t). (a) What are the fundamental frequency ω and period T of x? Calculate the Fourier Series of x in the forms: t, x(t) = A + A k cos(kω t + φ k ) = k= k= X k e ikω t Is X k = X k? (b) Suppose the LTI system S has frequency response {, if π 395 ω π 45 ω, H(ω) =, otherwise Plot the magnitude and phase response of H. Repeat part a for y. Answer to Using gives cos(x)cos(y) = cos(x + y)+ cos(x y), x(t) =cos(π 4t)+ cos(π 39t)+ cos(π 4t), from which (a) ω =π rad/sec and t =.sec. Also A 39 =, A 4 =, A k =, else; kφ k =

11 and X k = A k e φksgn(k) in which sgn(k) =,k ; =,k <. So X 39 = X 39 = X 4 = X 4 =; X 4 = X 4 =; X k =, else. (b) This system is a bandpass filter, in which only sinusoids with frequencies within specified range go through unchanged and the others become. Thus So, t, y(t) =cos(π 4t); ω =π 4 rad/sec; T = 4 sec. A =; A k =,k ; φ k =, k, X = X =; X k =else.. Give the ABCD state space representation of a discrete-time system with frequency response H(ω), where: H(ω) = +e jω 3e 3jω Hint: First find a difference equation which has the given frequency response. Then find the state space representation. Answer to From H(ω)[ 3e 3jω ]=+e jω we see that H is the frequency response of the difference equation So we select y(n) 3y(n 3) = x(n)+x(n ). s(n) = A = x(n ) y(n ) y(n ) y(n 3) 3 B = [ C T = 3 ] D =

12 3. You are given the signal tx(t) =cos(πt) + sin(5πt) to use as input to a system with frequency response H(ω) = ω. Answer the following questions based on this setup. (a) Indicate the Fourier series expansion (in cosine format) of x by writing the nonzero values of A, A k, and φ k in the expansion x(t) =A + k= A k cos(kω t + φ k ). (b) Indicate the Fourier series expansion (in complex exponential format) of x(t) by writing the nonzero values of the complex coefficients X k in the expansion x(t) = k= X k e jkωt. (c) Give y, the output of the system with input x. Answer to 3 (a) First rewrite x(t) =cos(πt)+ sin(5πt) in terms of cosines: x(t) =+cos(πt)+cos(5πt + π ) Next find the fundamental frequency. The largest frequency that evenly divides both π and 5π is ω =5π. We rewrite x(t) in terms of nonzero coefficients: x(t) = + cos(4(5π)t + ) + cos(5(5π)t + π ) = A + A 4 cos(4ω t + φ 4 )+A 5 cos(5ω t + φ 5 ) We see from above that A =, A 4 =, φ 4 =, A 5 =, φ 5 = π, and all other A k and φ k are zero. (b) We can calculate the X k s directly, but since we ve already calculated the A k s, let s use them to derive the X k s. (See also page 3 in the text.) Note in particular that with complex exponentials, we have negative frequency and complex coefficients instead of phases, meaning that the X k s are complex and k can be negative. Recalling that cos(t) = ejt + e jt, we can say that, for positive k: A k cos(ω kt + φ k ) = A ke jφ k e jωkt + A ke jφ k = X k e jωkt + X k e jω ( k)t e jω kt In our case, we have three nonzero A k. We start with A. Since cos() = e j =,we conclude that X = A. For A 4, we relate the frequency components at ω = ±4ω : cos(4ω t)= e4jω t + e 4jω t and conclude that X 4 =/ and X 4 =/.

13 And finally, for A 5 and φ 5, we relate the frequency components at ω = ±5ω. cos(5ω t) = e jπ/ e 5jω t + e jπ/ e 5jω t = ie 5jω t ie 5jω t and conclude that X 5 = i and X 5 = i. (c) We can either apply the frequency response to the eigenfunctions or we can look at x(t) directly and see how it behaves when sent through the system. Let s start with the latter approach. Looking at x(t) =cos(πt) + sin(5πt), we see it has components at ω =, ω = π, and ω =5π. The frequency response is simple enough that we can see that the DC component (i.e. the component at ω =) gets completely attenuated (i.e. multiplied by ). The other two components are scaled by the absolute value of their frequency, leading to: y(t) = () + (π)cos(πt) (5π) sin(5πt) = π cos(πt) 5π sin(5πt) If the frequency response had been more complicated, we may have preferred another approach: We already have the complex exponential breakdown of the input signal, meaning that we know the input signal in terms of scaled eigenfunctions. We can therefore apply the frequency response: y(t) = H()X +X 4 H(4ω )e 4jωt + X 4 H( 4ω )e 4jω t +X 5 H(5ω )e j5ω t + X 5 H( 5ω )e 5jω t = + π eπt + π e πt + 5π ie 5πt + 5π ( i)e 5πt = π eπt + e πt = π eπt + e πt +5π(i ) e5πt e 5πt i 5π e5πt e 5πt i = π cos(πt) 5π sin(5πt) which is the same result as with the other method. 3