EEE 303 HW#2 Solutions 1.43 a.

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1 EEE 33 HW # SOLUTIONS Problem.5 Since x(t) =fort<3, we have that x(t) =U(t 3)x(t). a. x( t) =x( t)u( t 3) = x( t)u( t ) which is zero for t>. b. x( t)+x( t) =x( t)u( t ) + x( t)u( t ) which is zero for t>. c. x( t)x( t) =x( t)u( t )x( t)u( t ) = x( t)x( t)u( t ) which is zero for t>. d. x(3t) =x(3t)u(3t 3) which is zero for t<. e. x(t=3) = x(t=3)u(t=3 3) which is zero for t<9.

2 EEE 33 HW# Solutions.43 a. Let H denote a TI system such that y = H[x] and suppose that x is periodic. That is, x(t+t)=x(t), for all t. Alternatively, in terms of the shift operator T, T -T [x] = x. Now, for y to be periodic, we should also have T -T [y] = y. Substituting y = H[x], we get T -T [y] = T -T H[x] = HT -T [x] (TI system, commutes with shifts) = H[x] (x is periodic) = y (The derivations hold in discrete time as well.) b. The zero system (y = ) is an easy but trivial example. A more interesting case is a low-pass filter eliminating the high frequency component of, say, x(t) = sin t + sin πt. (More details later.)

3 EEE 33 HW#3 Solutions. a. ( ) cos( ) ( )cos( ) ( )cos() u t t dt δ t t dt = 5 sin( = δ t dt = b. π ) ( 3), because δ(t+3) = for t in [,5]. 5 t δ t + dt = d[cos(πt)] = t dt u ( t c. u τ )cos(πτ ) dτ π sin(π ), since is in [-5,5]. 5 ( = = = Alternatively, using integration by parts and Then, 5 5 u ( τ )cos(πτ ) dτ = ) d =, we have that ( τ ) = δ ( ) τ dδ ( t) dt dδ ( τ ) cos(πτ ) dτ = dτ 5 5 d cos πτ [ δ ( t)cos(πt) ] + δ ( τ ) dτ = sin(π) = where we used the fact that δ(-5)= δ(+5) =. dτ u. τ d.47 Let y = h *x. Then, a. [ h ( t)]*[x ( t)] = ( h * x ) = y, (linearity of convolution) b. [ h ( t)]*[ x ( t) x ( t )] = ( h * x )( t) [ h ( t)]*[ x ( t )] = y ( t) y ( t ), where the first equality follows from linearity and the second from time-invariance. c. Using time-invariance, [ h ( t + )]*[ x ( t )] = ([ h ( t)]*[ x ( t )])( t + ). But, [ h ( t)]*[ x ( t )] = ([ h ( t)]*[ x ( t)])( t ) = y ( t ). Hence, [ h ( t + )]*[ x ( t )] = y ( t + ) = y ( t ). Alternatively, using the definition of the shift operator T and its commutativity with convolution, [ h = y ( t + )]*[ x ( t )] = ( T h ) *( T x ) = T T ( h * x ) T. d. There is not enough information to determine the output in this case. To prove this statement we need to find two pairs (h,x ) that produce the same response y and such that their corresponding transformations produce different responses. For this, consider h (t) = U(t) (a unit step) and x (t) = {a pulse in [,] with amplitude ½}. As a second pair, consider h (t) = y (t), x (t) = δ(t). For the first pair, it follows that [h (t)]*[x (-t)] = y (t+). But for the second pair, [h (t)]*[x (-t)] = y (t). e. Reflections are not time-invariant operations so we need to work with the convolution integral itself. Using a transformation of the integration variable σ = τ, we have that y (-t) can be expressed as: ( t y t) = h ( t τ ) x ( τ ) dτ = h ( t + σ ) x ( σ ) dσ = h( t σ ) x( σ ) dσ = ( h * x)( ) Notice that this is not a simple transformation of the independent variable. That is, we may not simply state that since y (t) = h (t)*x (t), then y (-t) = h (-t)*x (-t). Even though the final relationship is true, this statement is wrong as an implication! The error is caused by the bad notation (h(t)*x(t) is nonsense). This error becomes apparent when we consider scaling of the independent variable. That is, using the previous derivation it follows that y (at) = a [h (at)]*[x (at)] (and not h (at)*x (at), as one might expect!) f. Using the property of the unit doublet u, that x = u *x, we have that y = h * x = h' = y * x' = ( u * h ) *( u * x ) = u * u *( h * x ) From the graph of y, it follows that y(t) = ½ δ(t) ½ δ(t-). ''

4 EEE 33 HW # 4 SOLUTIONS Problem.3 Using the linearity and time invariance of convolution, we have that (T denotes the shift) Ã! X h x = h T kt ± k = X h (T kt ±) k = X k = X k T kt (h ±) T kt h So, nally, y(t) = X k h(t kt) T = 4 T = T =.5 T =

5 EEE 33 HW#4 Solutions 3.5 The Fourier series coefficients of the output y(t), say b k are related to the Fourier series coefficients of the input (a k ) by b k = H(jkw ) a k where w = π / T =. Since H(jw) = for w >, it follows that b k = for k >, or k > /, or k 9. Since it is given that x(t) = y(t), then a k = b k (uniqueness of the FS expansion), implying that a k must be zero for k 9.

6 EEE 33 HW Solutions 3..a.a Consider the derivative of the given function: dx ( t) = δ ( t n ) dt n= The train of deltas is periodic with period. Its Fourier Series expansion can be found either directly (integrals of delta-times-a-function) or by using the tables and applying the shift property. We have, jkw jkwt e ak = FS δ ( t n ) = δ ( t ) e dt = n where w = π/ = π. The FS coefficients of the function are if k = and otherwise. Thus, the coefficients of the derivative are jkπ e if k = b = FS{ dx k t } dt ( ) = jkπ e otherwise Next, the FS coefficients of x(t), say c k, are related to b k by b k = jkw c k. This can be solved for c k except for k =. This can be found by applying the FS definition directly to x(t). Thus, we find if k = ck = FS{ x( t) } = jkπ e /( jkπ ) otherwise Notice that the direct computation of c here is essential. Both x(t) and x(t)+const. have the same derivative and their FS cannot be distinguished after differentiation. REMARK: Applying Parseval s theorem on x(t) we find: ck = = = = t dt k k π k = k π 3 This yields the well-known expression π = k 6 k = 3..a.b Using the same technique as in a.a, let x be the square wave t <.5 x ( t) =, x ( t + 6) = x ( t) otherwise whose Fourier Series coefficients are given in Table 4.. Differentiating the given signal, we find dx ( t) = x ( t +.5) x ( t.5) dt Hence, applying the linearity and time-shift properties of the Fourier Series, we get sin( kπ / 6) π / π / sin( kπ / 6) FS{ ( )} ( ) dx jk jk dt t = e e j sin( kπ / ) kπ = kπ Next, using the differentiation property, we find dx sin( kπ / 6) ak = FS{ x( t)} = FS{ ( )} dt t /( jkw ) = 6 sin( kπ / ), k ( kπ ) while for k =, the FS coefficient is found by direct integration as a = (+/+/)/6 = ½.

7 EEE 33 HW # 9 SOLUTIONS Problem 3.6b x [n] is composed of two periodic functions, so we can apply superposition: y s [n] = H(e j ) + jh(e j3¼=8 )j sin(n3¼=8+¼=4+argfh(e j3¼=8 )g) But H(e j )=andh(e j3¼=8 )=,so y s [n] = sin(n3¼=8+¼=4) Alternatively, one could write the Fourier series expansion for x and use the ltering property. Problem 3.3 The set of equations for a k is fá nk g n;k fa k g k = fx[n]g n where f:g k denotes the vector or matrix indexed by k and Á = e j¼=4. Expanding the matrices, we get Á Á Á Á a x() 6 Á Á Á Á a 7 4 Á Á Á 4 Á a 5 = 6 x() 7 4 x() 5 = Á Á 3 Á 6 {z Á 9 } a 3 x(3) The matrix has a very special structure. Its columns are of the form [; ½ i ;½ i ;½3 i ]>. Such matrices are called Vandermonde matrices and they are nonsingular if the ½ i 's are distinct. In our case ½ = Á =,½ = Á = j, ½ 3 = Á =, ½ 4 = Á 3 = j, which are indeed distinct. In general, Vandermonde matrices are numerically ill-conditioned except when the ½ i 's are the n-th roots of unity, which is our case. When this happens, the columns of the matrix are orthogonal to each other and the matrix is easy to invert. (Observe that P k (Ákn ) Á kn = P k Ák(n n ) which is if n 6= n. This property is precisely the reason why we can write a simple expression for the Fourier series coe±cients a k.) Regardless of any simpli cations, we can use Matlab (or any other numerical package) to solve the above system of equations. Substituting the numerical values we have 6 4 j j j j a a a a = and Matlab produces the solution a ==, a =( j)=4, a =,a 3 =( +j)=4. Needless to say, we arrive at the same result with the Fourier Series equation

8 EEE 33 HW # 5 SOLUTIONS Problem 4.34 a. From the de nition of the transfer function as the Fourier transform of the impulse response, we have Y (jw) X(jw) = (jw)+4 (jw) +5(jw)+6 ) [(jw) +5(jw)+6]Y (jw)=[(jw)+4]x(jw) Taking the inverse Fourier transform of both sides, we recover the di erential equation that describes the input-output relationship for the system d dt y(t)+5d dt y(t)+6y(t) = d dt x(t)+4x(t) b. We have that h(t) =F fh(jw)g; using a partial fraction expansion, the roots of the denominator are ; 3 and H(jw)= A jw + + B jw +3 where A =[(jw +)H(jw)]j jw= =andb =[(jw +3)H(jw)]j jw= 3 =. From the tables, ½ ¾ ½ ¾ F fh(jw)g = F + F =e t U(t) e 3t U(t) jw + jw +3 c. The response of the system to the input x(t) =e 4t U(t) te 4t U(t) can be found by either a direct evaluation of the convolution of the impulse response with the input or by Fourier transform methods. Using the latter approach, we nd the Fourier transform of x(t) as Then, Y (jw) =H(jw)X(jw) so, Y (jw) = = = = X(jw)= jw +4 (jw +4) µ jw +4 (jw +)(jw +3) jw +4 (jw +4) (jw +)(jw +3) (jw +)(jw +3)(jw +4) (jw +) + (jw +3) = (jw +) (jw +3) = (jw +4) = (jw +) = (jw +4) It is now straightforward to evaluate the time response: y(t) =F fy (jw)g = e t U(t) e 4t U(t) Notice the cancellation of the pole at 3 by the zero of X(jw) =(jw + 3)=(jw + 4). Also, the transfer function zero at 4 cancelled one of the poles of X(jw) and we did not need to evaluate the partial fraction expansion for multiple poles. Problem 4.35 a. For the transfer function H(jw)= a jw a + jw we have that jh(jw)j = p a +( w) = p a + w =. Such transfer functions are called \all-pass." Further, 6H(jw)=tan ( w=a) tan (w=a) = tan (w=a).

9 b. The impulse response of the system is the inverse Fourier transform of its transfer function: ½ ¾ ½ ¾ a + a a jw a h(t) =F fh(jw)g = F = F a + jw a + jw =ae at U(t) ±(t) c. The response of the system to the input x(t) =cos(t= p 3) + cos(t)+cos( p 3t) can be evaluated through Fourier transforms. X(jw) will contain 6 impulses and so will the product X(jw)H(jw). Then the inverse transform will contain 6 complex exponentials, but the simpli cation of that formula is quite tedious. Alternatively, using the sinusoid response formula and for a =, y(t) = cos(t= p 3 tan (= p 3)) + cos(t tan ()) + cos( p 3t tan ( p 3)) = cos(t= p 3 ¼=3) + cos(t ¼=) + cos( p 3t ¼=3) The following gure shows the response of the all-pass lter to the three cosines. For comparison, the response of the lter to the input is x(t)u(t) is also included. In the latter case, notice that the appearance of the ±(t) in the impulse response causes the initial output to be y(+) = x(+) = 3. 3 output response to 3 cosines 3 Transient vs. steady state response detail Figure : Left: Output response to the three cosines. Right: Detail showing the di erence between the steadystate response and the transient response. (Steady-state response: solid line.) TheresponseofH(jw) tocos(w t)isjh(jw )j cos(w t + 6 H(jw )).

10 EEE 33 HW # 9 SOLUTIONS Problem 5. Let A = P n and de ne the sequence x[n] = n u[n]. Then X(e jw )= n e jw =. The transform of the sequence nx[n] isj dx(ejw ) dw = e jw = ( e jw =). Then, A = Ffnx[n]gj e j = = ( =) = Problem 5.9 By inspection, the transfer function is H(z) = 6 z 6 z where z = e jw. The poles (roots of denominator) are z = = and =3. Taking the Partial Fraction Expansion we have Using the tables to nd the inverse Fourier, we get h[n] = 3 µ 5 3 H(z) = ( z )( + 3 z ) = z n u[n]+ 5 µ 3 n u[n] + 3 z Note: The impulse response can also be computed directly from the recursion. First, since the system is linear, its response to the zero input sequence is zero. Furthermore since it is causal, then its response to ±[n] must be identical to the zero-input response up to the index n =. So,y[n] =,forn<. Then, y[] = ++x[] = y[] = =6 + = =6 y[] = (=6)(=6) + =6 + = 7=36. which is the same as the previous analytical computations.

11 EEE 33 HW # 6 SOLUTIONS Problem 6.9 The lterisstablesothesteady-stateresponseiswell-de ned. Thesteady-statepartofthestepisthe constant function. The lter transfer function is H(jw) = jw+5 which at w = is /5. Therefore, the steady-state part of the step response is the constant function (/5)() = /5, which is also the nal value of y s (t); t!. The step response is Y s (s) = s+5 s. Taking the inverse Laplace via partial fraction expansion we nd: ½ ¾ ½ y s (t) =L =5 = L s +5s s + =5 ¾ = s +5 5 u(t) 5 e 5t u(t) = 5 ( e 5t )u(t) The same result is obtained by using the Fourier transform, but notice the di erence in the transform of the step function: µ Y s (jw) = jw +5 jw + ¼±(w) = 5 ¼±(w)+ jw +5jw ½ ¾ ½ y s (t) = F fy s (jw)g = F 5 ¼±(w)+ = F jw +5jw 5 ¼±(w)+=5 jw + =5 ¾ jw +5 ½ µ = F 5 jw + ¼±(w) + =5 ¾ = jw +5 5 u(t) 5 e 5t u(t) As t!, this expression yields y s () = =5, as found in the rst part. Then, the time t at which y s (t )=y s ()( e )ist ==5. Problem 6. The transfer function of the cascade connection is H(jw) =H (jw)h (jw). Hence, H (jw) = H(jw) H (jw). Taking the magnitude and log of both sides, log jh (jw)j =log jh(jw)j log jh (jw)j )jh (jw)j db = jh(jw)j db jh (jw)j db Thus the straight-line approximation of jh (jw)j db is described as follows: for w<: constant with magnitude (-)-(6) = -6 db; for < w < 8: straight line with slope ()-() = - db/dec; for 8 <w<4: straight line with slope (-4)-() = -4 db/dec; for 4 <w: straight line with slope (-4)-(-) = - db/dec;

12 EEE 33 HW # 6 SOLUTIONS Problem 6.7 (a.) Using the di erentiation property of the Fourier transform (or Laplace) the transfer function of the system is: H(jw)= jw + or, H(s) ==(s + ). The Laplace version is more advantageous since in the subsequent derivations it is convenient to keep jw as one argument. With s = jw, the frequency response of this transfer function has the following magnitude and phase: jh(jw)j = p w + 6H(jw) = tan (w=) These are easy to plot; in MATLAB, the relevant command is bode([],[ ]); (see also help bode). (b.) For the group delay, di erentiate tan (w=) with respect to w. (c. & d.) When x(t) =e t U(t), Ffxg ==(s + ), with s = jw. Using the convolution property, Y (jw)= (s +)(s +) s=jw For its partial fraction expansion (PFE), we need to nd A; B such that Y (s) = A (s +) + B (s +) The values of A; B can be computed using the formulas in the Appendix, or simply by converting the partial fractions into one fraction and equating the coe±cients of equal powers of s in the numerator. (This is e ective when only a few terms are involved.) A = ; B = (s +) s= (s +) Now, the time response can be computed easily by using the tables: (e-i.) Here X(jw)= +jw +jw,so Converting the PFE back to one fraction we have that y(t) =()e t U(t)+( )e t U(t) s= Y (s) = s + (s +) = A (s +) + A (s +) A s +A + A = s +) A =; A = Hence, y(t) =()e t U(t)+( )te t U(t). Alternatively, since the fraction involves only one pole, we have that y(t) =y (t)+ d dt y (t), where y (t) = F =(jw +) ª = te t U(t). (Verify this!) s+ (e-ii.) Here Y (s) = (s+)(s+) = (s+) and, clearly, y(t) =e t U(t). (e-iii.) Here Y (s) = (s +) (s +) = A (s +) + B (s +) + B (s +) Using the PFE formulas, we nd A =,B =, B =. Hence, y(t) =()e t U(t)+( )e t U(t)+( )te t U(t).

13 EEE 33 HW # 6 SOLUTIONS Problem 7.4 The Nyquist rate of x(t) isw,sox(t) is bandlimited in ( w =; w =). In other words, the maximum frequency where x(t) has energy is w M = w =.. Ffx(t )g = e jw Ffx(t)g is also bandlimited in ( w M ;w M ). Hence, Ffx(t ) + x(t)g is bandlimited in ( w M ;w M ), and the Nyquist rate of x(t ) + x(t) isw m = w.. F dx dt ª = jwffx(t)g which is again bandlimited in ( wm ;w M ) and, hence, the Nyquist rate of dx=dt is w. Notice that this result hinges on the assumption that X(jw) is zero for jwj >w M. This can be deceptive in a practical situation where the Nyquist rate is de ned in terms of the signal \bandwidth," i.e., the frequency beyond which the energy content of the signal is small. In such a case, di erentiation of the signal will increase the magnitude (and energy) of the signal at high frequencies and will change the e ective Nyquist rate. 3. Let y(t) =x (t). Then Ffyg = ¼ Ffxg Ffxg. Since Ffxg is bandlimited in ( w M;w M ), we have that Ffyg is bandlimited in ( w M ; w M ); hence the Nyquist rate of y(t) isw. 4. Let y(t) =x(t)cos(w t). Then Ffyg = ¼ FfxgFfcos w tg = [X(j(w w )) + X(j(w + w ))]. Hence, Ffyg is bandlimited in ( w w M ;w + w M )=( 3w M ; 3w M )andthenyquistrateofy(t) is6w M =3w. Problem 7. Let x (t) be bandlimited in ( ¼; ¼) andx (t) be bandlimited in ( ¼; ¼). a. Let y(t) =(x x )(t). Then Ffyg = Ffx gffx g which is bandlimited in the intersection of the two frequency intervals, that is ( ¼; ¼). Hence the Nyquist rate of y(t) is¼ and the maximum sampling time to allow perfect reconstruction is T = ¼=(¼) = =. b. Let y(t) =¼x (t)x (t). Then Ffyg = Ffx g Ffx g which is bandlimited in the frequency interval ( ¼ ¼; ¼ +¼). Hence the Nyquist rate of y(t) is6¼ and the maximum sampling time to allow perfect reconstruction is T =¼=(6¼) ==3.

14 EEE 3 3 HW # 7 SOLUT I ONS Problem 8.4 Since v(t) = g(t)sin4¼t is passed through an ideal lowpass lter and this operation is best described in the frequency domain, we need to compute V (jw). This can be achieved by either trigonometric identities (for this speci c problem) or by performing the various frequency-domain convolutions. Let w =¼. Then, X(jw) = ¼ j ±(w w ) ¼ j ±(w + w )+ ¼ j ±(w w ) ¼ j ±(w +w ) G(jw) = ¼ ¼ X(jw) j ±(w w ) ¼ j ±(w +w ) = j X(j(w w )) j X(j(w +w )) V (jw) = ¼ ¼ G(jw) j ±(w w ) ¼ j ±(w +w ) = j G(j(w w )) j G(j(w +w )) = 4 X(j(w 4w )) 4 X(jw) 4 X(jw)+ 4 X(j(w +4w )) = 4 X(j(w 4w )) + X(jw) 4 X(j(w +4w )) = ¼ 4 j ±(w 5w ) ¼ j ±(w 3w )+ ¼ j ±(w 6w ) ¼ j ±(w w ) + ¼ j ±(w w ) ¼ j ±(w + w )+ ¼ j ±(w w ) ¼ j ±(w +w ) ¼ 4 j ±(w +3w ) ¼ j ±(w +5w )+ ¼ j ±(w +w ) ¼ j ±(w +6w ) = ¼ j ±(w 6w ) ¼ 4j ±(w 5w )+ ¼ 4j ±(w 3w )+ 3¼ j ±(w w )+ ¼ j ±(w w ) ¼ j ±(w + w ) 3¼ j ±(w +w ) ¼ 4j ±(w +3w )+ ¼ 4j ±(w +5w )+ ¼ j ±(w +6w ) Filtering v(t) with a lowpass lter with cuto frequency w and passband gain of we get an output, say y(t), that has frequency content in jwj w. This operation is ambiguous in that there is an impulse exactly at the cuto frequency. Formally, the ideal lowpass lter will only allow half of that impulse to go through. Y (jw)=h(jw)v (jw)=+ 3¼ j ±(w w )+ ¼ j ±(w w ) ¼ j ±(w + w ) 3¼ j ±(w +w ) Hence, converting back to time domain y(t) =F fy (jw)g =sin¼t + 3 sin 4¼t Problem 8.34 Let us rst de ne the intermediate signals v(t) =x(t)+cosw c t, z(t) =v (t). Then V (jw) = X(jw)+¼±(w w c )+¼±(w + w c ) Z(jw) = V (jw) V (jw) ¼

15 = X(j(w w c)) + ¼ ±(w w c)+ ¼ ±(w) + X(j(w + w c)) + ¼ ±(w)+ ¼ ±(w +w c) + X(j(w w c)) + X(j(w + w c)) + X(jw) X(jw) ¼ = X(j(w w c )) + X(j(w + w c )) + ¼ ±(w w c)+ ¼ ±(w +w c)+¼±(w)+ X(jw) X(jw) ¼ The exact expression of X(jw) X(jw) is irrelevant here; we only need the fact that, since X(jw) is bandlimited in ( w M ;w M ), then X(jw) X(jw) is bandlimited in ( w M ; w M ). The same result is obtained by squaring v(t) in time-domain and then taking its Fourier transform z(t) = x (t)+cos w c t +x(t)cosw c t = x (t)+ + cos w ct +x(t)cosw c t Z(jw) = ¼ X(jw) X(jw)+¼±(w)¼ ±(w w c)+ ¼ ±(w +w c)+x(j(w w c )) + X(j(w + w c )) For y(t) =x(t)cosw c t, we need to have that Y (jw) =H(jw)Z(jw)= [X(j(w w c)) + X(j(w + w c ))] Hence, H should be a bandpass lter with a gain A == and w M <w l <w c W M, w c + w M <w h < w c. The feasibility constraints for this solution are w M <w c W M,or w c > 3w M. (This also imples that w c + w M < w c.)

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20 EEE 3 3 HW # 8 SOLUT I ONS Problem.54 a. Let S N = P N n= an. Then, as N = P N n= an.therefore,( a)s N = a a N,fromwhichweget S N = N X n= a n = an a provided that a 6=. Ifa =, then simply S N = P N n= n = N. b. Let S = P n= an = lim N! S N, provided that the limit exists. For this, the limit of a N should exist as N!which occurs if jaj <. In the complex case, a N =(jaje jµ ) N = jaj N e jµn. So, if jaj <, lim N! a N =, implying X S = a n = a c. Observe that for jaj <, a ds da = a X da n da = X n= n= n= na n = a d µ = da a ( a) Notice that the exchange of summation and di erentiation makes sense since the series are absolutely convergent. d. Again for jaj <, X X k X a n = a n a n = a ak a = ak a n=k n= Problem. a. Let x(n) =a n U(n), h(n) =b n U(n). Then n= y(n) = (h x)(n) = = nx b n k a k k= X k= b n k U(n k)a k U(k) X n ³ a = b n k ; (assuming b 6= a) b k= = b n (a=b)n+ (a=b) = an+ b n+ a b Obviously, if a = b, theny(n) =(n +)b n. b. Let x(n) =U(n) U(n 5), h(n) =h (n ) + h (n ), where h (n) =U(n) U(n 6). Then, x(n) h(n) =y (n ) + y (n ), where y (n) =x(n) h (n). The last sequence is easy to compute: n y (n) From this, we get n y (n ) y (n ) x(n) h(n) Problem.3

21 Consider the di erence equation y(n)+y(n ) = x(n)+x(n ) with x(n) =f:::; ; ; 3; ; ; ; ;:::g for n = f:::; 3; ; ; ; ; ; 3; 4;:::g, respectively. Assuming that y(n) is\atrest"at, the above recursion (y(n) = y(n ) + x(n)+x(n )) yields:. (y(n) = ; for n< 3) y( 3) = y( 4) + x( 3) + x( 5) = y( ) = y( 3) + x( ) + x( 4) = y( ) = y( ) + x( ) + x( 3) = y() = y( ) + x() + x( ) = 5 y() = y() + x() + x( ) = 4 y() = y() + x() + x() = 6 y(3) = y() + x(3) + x() = 7 y(4) = y(3) + x(4) + x() = 58 y(5) = y(4) + x(5) + x(3) = 4 y(6) = y(5) + x(6) + x(4) = 4 y(7) = y(6) + x(7) + x(5) = 4 4 y(8) = y(7) + x(8) + x(6) = 8 4. (y(n) = 4( ) n 5 ; for n 5) Problem.55 a (extra) y[] = x[] =. h[n] satis es the equation h[n] = h[n ]; n with axuliary condition h[] =. The characteristic polynomial is given by: z = then the solution for the homogeneous di erence equation is h[n] = C µ n U[n] h[] = so C =then h[n] = µ n U[n]

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