?x4?n?x4?n?2?,x4?n?1??0 ?y4?n???x4?n?1?otherwise?0,??x4?n?x4?n?2?,x4?n?1??0 ?ay5?n????ay4?n?x4?n?1?otherwise?0,?Therefore, the system is homogenous.
(a) Invertible. Inverse system y(t)=x(t+4)
(b)Non invertible. The signals x(t) and x1(t)=x(t)+2?give the same output (c) ?[n] and 2?[n] give the same output d) Invertible. Inverse system; y(t)=dx(t)/dt
(e) Invertible. Inverse system y(n)=x(n+1) for n?0 and y[n]=x[n] for n<0 (f) Non invertible. x(n) and –x(n) give the same result (g)Invertible. Inverse system y(n)=x(1-n) (h) Invertible. Inverse system y(t)=dx(t)/dt (i) Invertible. Inverse system y(n) = x(n)-(1/2)x[n-1] (j) Non invertible. If x(t) is any constant, then y(t)=0
?[n] and 2?[n] result in y[n]=0 (k)
(l) Invertible. Inverse system: y(t)=x(t/2)
(m) Non invertible x1 [n]= ?[n]+ ?[n-1]and x2 [n]= ?[n] give y[n]= ?[n] (n) Invertible. Inverse system: y[n]=x[2n]
(a) Note that x2[t]= x1 [t]- x1 [t-2]. Therefore, using linearity we get y2 (t)= y1 (t)- y1 (t-2).this is shown in Figure
(b)Note that x3 (t)= x1 [t]+ x1 [t+1]. .Therefore, using linearity we get Y3 (t)= y1 (t)+ y1 (t+2). this is shown in Figure y2 y3
2 t 4 t
0 2 -1 0 2
Figure
-2 All statements are true (1) x(t) periodic with period T; y1 (t) periodic, period T/2 (2) y1 (t) periodic, period T; bx(t) periodic, period2T (3) x(t) periodic, period T; y2 (t) periodic, period2T (4) y2(t) periodic, period T; x(t) periodic, period T/2; (1) True x[n]=x[n+N]; y1 (n)= y1 (n+ N0). periodic with N0=n/2
if N is even and with period N0=n if N is odd.
(2)False. y1 [n] periodic does no imply x[n] is periodic . Let x[n] = g[n]+h[n] where
neven?0,?1,neven g[n]??andh[n]??n?0,nodd?(1/2),noddThen y1 [n] = x [2n] is periodic but x[n] is clearly not periodic. (3)True. x [n+N] =x[n]; y2 [n+N0] =y2 [n] where N0=2N
(4) True. y2 [n+N] =y2 [n]; y2 [n+N0 ]=y2 [n] where N0=N/2 . (a) Consider
n????x[n]?x[0]???x[n]?x[?n]?n?1??If x[n] is odd, x[n] +x [-n] =0. Therefore, the given summation evaluates to zero. (b) Let y[n] =x1[n]x2[n] .Then
y [-n] =x1[-n] x2[-n] =-x1[n]x2[n] =-y[n]. This implies that y[n] is odd.
???(c)Consider 222[n]??[n]?[n] eon???n??? n ???
?x?x??x?
Using the result of part (b), we know that xe[n]xo[n] is an odd signal .Therefore, using the result of part (a) we may conclude that
n????x[n]?e2n????x?2o[n]2?Therefore,
??n???x[n]x[n]?0eo?2?2eon????x[n]???x[n]??x[n].n???n??????2(d)Consider
?x???2(t)dt??xe(t)dt?????e2???xo(t)dt?2?02???x(t)x(t)dt.e0Again, since xe (t) xo (t) is odd,
?x(t)x(t)dt?0.Therefore,
?x???2(t)dt?????xe(t)dt??2???x(t)dt.o2. We want to find the smallest N0 such that m(2π /N) N0 =2πk or N0 =kN/m,
where k is an integer, then N must be a multiple of m/k and m/k must be an integer .this implies that m/k is a divisor of both m and N .Also, if we want the smallest possible N0, then m/k should be the GCD of m and N. Therefore, N0=N/gcd(m,N).
j?0(n?N)T,where.?o?2?/T0. This implies that ．(a)If x[n] is periodic e2?TkNT?2?k???a rational number . T0ToN (b)T/T0 =p/q then x[n] = efrequency is
j2?n(p/q),The fundamental period is q/gcd(p,q) and the fundmental
??T2?2?pgcd(p,q)?gcd(p,q)?0gcd(p,q)?0gcd(p,q).qpqpp (c) p/gcd(p,q) periods of x(t) are needed .
.(a) From the definition of ?xy(t).We have
?xy(t)??x(t??)y(?)d??????y(?t??)x(?)d??????yx(?t). (b) Note from part(a) that ?xx(t)??xx(?t).This implies that ?xy(t)is even .Therefore,
the odd part of ?xx(t).is zero. (c) Here, ?xy(t)??xx(t?T).and?yy(t)??xx(t). .(a) We know that 2?V(2t)??V/2(t).Therefore This implies that
1lim?V(2t)?lim?V/2(t).V??V??21?(2t)??(t).
2(b)The plot are as shown in Figure . We have
limuV(t)?(t)?limuV(0)?(t)?0.
V?0V?0 Also,
1limuV(t)?V(t)??(t).V?02
-Δ/
-Δ
-Δ We have ???uΔ（t） 1 1/2 0 1 1/2 0
Δ
'
1 uΔ
（t）
'
uΔ（t） Δ
'
t
t
0 uΔ'Δ（ t） 21 1/2
t
t
t
t
-Δ
0
Δ
t t
uΔ（t） 1 1 1/2et -t/Δ'uΔ（t） 1/2 0 Δ '1-1/2e-t/Δ 0 1 ?Δ -Δ t Figure g(t)??u(?)?(t??)d???u(?)?(t??)d?
0
Therefore,
t?0?0,Q?(t??)?0? g(t)??1,t?0Qu(?)?(t??)??(t??)?undefinedfort?0?.(a) If a system is additive ,then also, if a system is homogeneous,then
0?x(t)?x(t)?y(t)?y(t)?00?0.x(t)?y(t).0?0(b) y(t)=x(t) is such a systerm . (c) example,consider y(t) y(t)?2
?t??x(?)d? with x(t)?u(t)?u(t?1).Then x(t)=0
for t>1,but y(t)=1 for t>1.
. (a) y[n]=2x[n].Therefore, the system is time invariant.
(b) y[n]=(2n-1)x[n].This is not time-invariant because y[n- N0]≠(2n-1)2x[n- N0].
nn-1
(c) y[n]=x[n]{1+(-1)+1+(-1)}=2x[n].Therefore, the system is time invariant .
.(a) Consider two system S1 and S2 connected in series .Assume that if x1(t) and x2(t) are
the inputs to S1..then y1(t) and y2(t) are the .Also,assume that if y1(t) and y2(t) are the input to S2 ,then z1(t) and z2(t) are the outputs, respectively . Since S1 is linear ,we may write
ax1?t??bx2?t??ay1?t??by2?t?,
where a and b are constants. Since S2 is also linear ,we may write
s1ay1?t??by2?t??az1?t??bz2?t?,
We may therefore conclude that
1s2ax1(t)?bx2(t)?s??az1(t)?bz2(t)
s2Therefore ,the series combination of S1 and S2 is linear.
Since S1 is time invariant, we may write
x1?t?T0??y1?t?T0?
s1and
y1?t?T0??z1?t?T0?
Therefore,
s2x1?t?T0??z1?t?T0?
Therefore, the series combination of S1 and S2 is time invariant. (b) False, Let y(t)=x(t)+1 and z(t)=y(t) corresponds to two nonlinear systems. If these systems are connected in series ,then z(t)=x(t) which is a linear system.
(c) Let us name the output of system 1 as w[n] and the output of system 2 as z[n] .Then
s1s211y[n]?z[2n]?w[2n]?w[2n?1]?w[2n?2]
2411?x?n??x?n?1??x?n?2?
24The overall system is linear and time-invariant.
. (a) We have
sx?t????y(t)
Since S is time-invariant.
sx?t?T????y(t?T)
Now if x (t) is periodic with period T. x{t}=x(t-T). Therefore, we may conclude that
y(t)=y(t-T).This implies that y(t) is also periodic with T .A similar argument may be made in discrete time . (b)
(a) Assumption : If x(t)=0 for t same as x1(t)for t< t0. But for t> t0 , x2(t) ≠x1(t),Since the system is linear, x1?t??x2?t??y1?t??y2?t?, Since x1?t??x2?t??0for t< t0 ,by our assumption =y1?t??y2?t??0for t< t0 .This implies that y1?t??y2?t?for t< t0 . In other words, t he output is not affected by input values for t?t0. Therefore, the system is causal . Assumption: the system is causal . To prove that :If x(t)=0 for t< t0 .then y(t)=0 for t< t0 . Let us assume that the signal x(t)=0 for t< t0 .Then we may express x(t) as x(t)?x1?t??x2?t?, Where x1?t??x2?t? for t< t0 . the system is linear .the output to x(t) will be y(t)?y1?t??y2?t?.Now ,since the system is causal . y1?t??y2?t? for t< t0 .implies that y1?t??y2?t? for t< t0 .Therefore y(t)=0 for t< t0 . (b) Consider y(t)=x(t)x(t+1) .Now , x(t)=0 for t< t0 implies that y(t)=0 for t< t0 .Note that the system is nonlinear and non-causal . (c) Consider y(t)=x(t)+1. the system is nonlinear and causal .This does not satisfy the condition of part(a). (d) Assumption: the system is invertible. To prove that :y[n]=0 for all n only if x[n]=0 for all n . Consider x[n]?0?y[n]. Since the system is linear : 2x[n]?0?2y[n]. Since the input has not changed in the two above equations ,we require that y[n]= 2y[n].This implies that y[n]=0. Since we have assumed that the system is invertible , only one input could have led to this particular output .That input must be x[n]=0 . Assumption: y[n]=0 for all n if x[n]=0 for all n . To prove that : The system is invertible . Suppose that x1[n]?y1[n] and x2[n]?y1[n] Since the system is linear , x1[n]?x2[n]??y1[n]?y2[n]?0 By the original assumption ,we must conclude that x1[n]?x2[n].That is ,any particular y1[n] can be produced that by only one distinct input x1[n] .Therefore , the system is invertible. 2 (e) y[n]=x[n]. . (a) Consider , x1?t??y1(t)??hx1(t) and x2?t??y2(t)??hx2(t). ss
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