ECE314: Signals and Systems
Fall 2009
Solution: Homework2
Professor: Balu Santhanam
Problem 1.64
by[n] = 2x[n]u[n]
i- the system is memoryless because the value of the output signal y[n] at time n0
depends only on the input x[n] at time n0 also.
ii- the system is BIBO because if:
|x[n]| < B
⇒ |2x[n]u[n]| < 2B
⇒ |y[n]| < 2B
iii- the system is causal same argument for i
iv- the system is linear because:
L[C1 x1 [n] + C2 x2 [n]] = 2(C1 x1 [n] + C2 x2 [n])u[n]
= C1 (2x1 [n]u[n]) + C2 (2x2 [n]u[n])
= C1 y1 [n] + C2 y2 [n]
v- the system is time variant because:
L[x[n − N ]] = 2x[n − N ]u[n]
y[n − N ] = 2x[n − N ]u[n − N ]
So,
y[n − N ] 6= L[x[n − N ]]
d-
Z
t/2
y(t) =
x(τ )dτ
−∞
i- the system has memory because at any time t, y(t) depends on all the values of x(t)
from −∞ up to t/2.
ii- the system is Not BIBO because if we take the bounded input x(t) = 1 for all t the
system output will be unbounded (equal to ∞)
HW2-1
iii- the system is noncausal because if we take for instance t = −2,y(−2) will be a
function of x(t) where t is between −∞ and −1.
iv- the system is linear because:
Z
L[C1 x1 (t) + C2 x2 (t)] =
t/2
−∞
(C1 x1 (τ ) + C2 x2 (τ ))dτ
Z
= C1
Z
t/2
−∞
x1 (τ )dτ + C2
t/2
−∞
x2 (τ )dτ
= C1 y1 (t) + C2 y2 (t)
v- the system is time variant because:
Z
t/2
L[x(t − T )] =
x(τ − T )dτ
−∞
Z t/2−T
=
x(τ 0 )dτ 0
−∞
Z
y(t − T ) =
t−T
2
x(τ )dτ
−∞
So,
y(t − T )y(t − T ) 6= L[x(t − T )]
hy(t) =
ª
d © −t
e x(t)
dt
i- the system has memory because at any time t, in order to find the derivative we
need the value at t− in order to find the slope at that point.
ii- the system is Not BIBO because if we take the bounded input x(t) = 1 for all t.
the derivative of e−t will go to infinity as t → − ∞
iii- the system is causal because in order to compute the derivative we only need the
value of x(t) at t and at t− .
iv- the system is linear because:
ª
d © −t
e (C1 x1 (t) + C2 x2 (t))
dt
ª
ª
d © −t
d © −t
e x1 (t) + C2
e x2 (t)
= C1
dt
dt
= C1 y1 (t) + C2 y2 (t)
L[C1 x1 (t) + C2 x2 (t)] =
HW2-2
v- the system is time variant because:
L[x(t − T )] =
y(t − T ) =
ª
d © −t
e x(t − T )
dt
ª
d © −t+T
e
x(t − T )
d(t)
So,
y(t − T ) 6= L[x(t − T )]
iy(t) = x(2 − t)
i- the system has memory because at any time t 6= 1, y(t) is a function of the future
or past value of the input function x(t).
ii- the system is BIBO because if:
|x(t)| < B
⇒ |x(2 − t)| < B
⇒ |y(t)| < B
iii- the system is noncausal because y(t) is a function of the future value of x(t) when
t < 1.
iv- the system is linear because:
L[C1 x1 (t) + C2 x2 (t)] = C1 x1 (2 − t) + C2 x2 (2 − t)
= C1 y1 (t) + C2 y2 (t)
v- the system is time invariant because:
L[x(t − T )] = x(2 − t + T )
y(t − T ) = x(2 − t + T )
So,
y(t − T ) = L[x(t − T )]
Problem 1.75
1.
• The system has memory: we can still see an output in all the signals when the
input x(t) goes to zero.
• The system maybe causal for all the examples that we have the output does not
start before the input.
HW2-3
• the system maybe Time Invariant we have x2 (t) = −x1 (t − 2) and the relation
between their outputs is the same (y2 (t) = −y1 (t − 2).
• The system is not Linear we have x3 (t) = x1 (t) − x2 (t) however this relation is
not the same after they pass through H (we have y3 (t) = y1 (t) + y2 (t)).
2.
• The system has memory: we can still see an output in all the signals when the
input x(t) goes to zero.
• The system is noncausal. For instance, x1 (t) starts at 0 however the system
response y1 (t) starts at −1.
• the system is Time Variant we have x4 (t) = x2 (t − 1). However the relation
between their outputs is y4 (t) 6= y4 (t − 1).
• The system maybe Linear we have x2 (t) = 2(x1 (t) + x3 (t)) and this relation is
exactly the same after they pass through H (we have y2 (t) = 2(y1 (t) + y3 (t))).
Problem 1.77
HW2-4
HW2-5
Problem 1.85
HW2-6