ECE-350 Midterm: ECE 350 Boise State Exam1 sample Solutions f08
15 Feb 2019
School
Department
Course
Professor
EE 350 Signals and Systems
Fall 2008
Sample Exam #1 - Solutions
1) a. x(t) = u(t)+u(t-1)-2u(t-2) b. x[n] = ∑
−=
−
5
5
][
k
knk
δ
c.
2 1
2
3 t
1
x
(t) = u(t)+u(t-1)-2u(t-2)
x[n] = ∑
−=
−
5
5
][
k
knk
δ
-
5
1
5
5n
-5
-2x(2-4t)
-4
1
1/2
0t
2) a - Linearity Property
Additivity property:
add the inputs:
x3(t) = x1(t)+x2(t), so also x3(t+2) = x1(t+2)+x2(t+2)
y3(t) = (t-2)x3(t+2) <- from system definition
y3(t) = (t-2)[x1(t+2)+x2(t+2)]
add the outputs:
y1(t)+y2(t)=(t-2)x1(t+2) + (t-2)x2(t+2)
because y3(t)=y1(t)+y2(t), (response to sum of inputs
is sum of outputs) additivity is satisifed.
Scalability property:
scale the input:
x4(t) = ax1(t), so also x4(t+2) = ax1(t+2)
y4(t) = (t-2)x4(t+2) <- from system definition
y4(t) = (t-2)[ax1(t+2)]
scale the output:
ay1(t) = a[(t-2)x1(t+2)]
because y4(t) = ay1(t), (response to scaled input is
scaled output) scalability is satisfied
because additivity and scalability are both satisfied, then the system is linear. <- make sure to state this final answer
2) b - Time Invariance Property
y1(t) = (t-2)x1(t+2) (1) define system example 1
y
2(t) = y1(t-t0) (2) shift output
y
2(t) = ((t-t0)-2)x1(t-t0) (3) apply time shift from (2) to system in (1) (replace ALL
t’s by t-t0)
x
2(t) = x1(t-t0) (4) shift input
y
2(t) = (t-2)x2(t) (5) define system example 2
y
2(t) = (t-2)x1(t-t0) (6) apply input shift from (4) (inside parentheses of all x(*),
insert a “-t0”)
because y2(t) ≠ y1(t-t0) (steps 3 vs 6), (response to shifted input is NOT shifted output) time invariance is not satisifed.
Therefore the system is not time invariant, or in other words it is time variant.
This can be seen by inspection by observing the (t-1) outside the input signal.
2) c - Not Causal: y(0)=(-2).x(2) , y(0) depends on the input x(t) at time 2 which hasn't occurred yet.
2) d - Not memoryless: y(t) depends only on x(t+2) which isn't at the same time
also - memoryless Æ causal so not causal Æ not memoryless.
2) e - Not stable: as t Æ infinity, y(t) Æ infinity for most bounded inputs , from (t-2) multiplicative factor.
3) a) y(t) = h(t) + h(t-6)
t
2
-1 2
y(t)
6
b) y(t) = h(t) + h(t-2)
t
2
-1 2
y(t)
4
t
2
-1 2
y(t)
4
c) y(t) = h(t) + 2h(t-1)
t
2
-1 2
y(t)
4
4
t
2
-1 2
y(t)
4
4
5
4) x2(t) = x1(t-1) + 2x1(t-5) so y2(t) = y1(t-1) + 2y1(t-5)
y1(t)
t
2
0 3 1 7 5
5) h(t) = etu(-1-t)
a. Causality – not causal. Because the u(-1-t) is 1 for t<-1, the function has non zero values for t<-1<0.
b. Stability – stable
∞<=−===−− −∞−−
−
∞−
−
∞−
∞
∞−
∫∫ 11
1
1
)1( eeeedtedttue ttt
c. Memory – has memory because function is not a scaled impulse at t=0: )()( tkth
δ
≠
6) Determine y(t) if y(t) = x(t) * h(t) .
t
2
-2 2
x(t)
h(t)
t
1
-2 ,
τ
2
-2 2
x(
τ
)
τ
+2
−
τ+2 h
(
t-τ
)
τ
1
t t+2
Case I
τ
2
-2 2
x(τ)
τ+2 −τ+2
t t+2
t+2 < -2
t < 4
y(t) = 0
Case II
τ
2
-2 2
x(τ)
τ+2 −τ+2
t t+2
t < -2, t+2 < 0, t+2 > -2
so
-4 < t < -2
∫
+
−
+=
2
2
)2)(1()(
t
dty
ττ
y(t) = t2/2 + 4t + 8
Case III
τ
2
-2 2
x(τ)
τ+2 −τ+2
t t+2
t > -2, t < 0, and
t+2 > 0, t+2 < 2
so
-2 <t <0
∫∫
+
+−++=
2
0
0
)2)(1()2)(1()(
t
t
ddty
ττττ
y(t) = -t2 – 2t + 2
Case IV
Document Summary
Sample exam #1 - solutions: a. x(t) = u(t)+u(t-1)-2u(t-2) -2 so. )2 y(t) = t2/2 + 4t + 8 t > -2, t < 0, and t+2 > 0, t+2 < 2 so.
