soln6 - EECS 216 SOLUTIONS TO PROBLEM SET#6 Winter 2008 1...

EECS 216 SOLUTIONS TO PROBLEM SET #6 Winter 2008
1. #4.5 Throughout F{ x ( t ) } = X ( ω ) = rect (( ω − 1) / 2) =
braceleftBig
1 for 0 ≤ ω ≤ 2
0 otherwise
.
Note: X ( ω ) negationslash = X ( − ω ) * so x ( t ) is NOT real–DON’T use conjugate symmetry!
1a. F{ x ( − t ) } = X ( − ω ) = rect (( ω + 1) / 2).
1b. F{ tx ( t ) } = j
dX
dω
= jδ ( ω ) − jδ ( ω − 2).
1c. F{ x ( t + 1) } = X ( ω ) e jω = e jω rect (( ω − 1) / 2).
1f. F{
dx
dt
} = jωX ( ω ) = jω for 0 ≤ ω ≤ 2 and 0 otherwise.
1g. F{ t
dx
dt
} = j
d
dω
[ jωX ( ω )] = 2 δ ( ω − 2) − rect (( ω − 1) / 2).
1i. F{ x ( t ) e
- j 2 t
} = X ( ω + 2) = rect (( ω + 1) / 2)=answer to (a).
2a. #4.14 X ( ω ) = F{ e - at u ( t ) } =
1
jω + a
,
H ( ω ) = F{
sin (2 t )
πt
} = rect ( ω/ 4)=0 outside | ω | < 2.
Y ( ω ) = H ( ω ) X ( ω ) =
rect ( ω/ 4)
jω + a
.
2b. Input energy=
integraltext
∞
-∞
| x ( t ) |
2
dt =
integraltext
∞
0
e
- 2 at
dt =
1
2 a
.
Output energy=
1
2 π
integraltext
2
- 2
|
1
jω + a
| 2 dω =
1
2 π
integraltext
2
- 2
1
ω 2 + a 2
dω =
1
aπ
tan - 1 2
a
.
So:
1
aπ
tan
- 1 2
a
=
1
2
1
2 a
→ tan
- 1 2
a
=
aπ
4 a
=
π
4
→ a = 2.
3. #4.24 X ( ω ) =
braceleftbigg
2 π for | ω | < ω
1
π for ω
1
< | ω | < ω
2
and 0 otherwise.
3a. Y ( ω ) = 2 π for | ω | < ω
f
→ y ( t ) = 2 π
sin( ω
f
t )
πt
= 2
sin( ω
f
t )
t
.
3b. Y ( ω ) =
braceleftbigg
2 π for | ω | < ω
1
π for ω
1
< | ω | < ω
f
→ y ( t ) =
sin( ω
1
t )
t
+
sin( ω
f
t )
t
3c. Since H ( ω ) passes all of x ( t ), y ( t ) = x ( t ) =
sin( ω
1
t )
t
+
sin( ω
2
t )
t
4. #4.19 Mult. by p ( t ) modulates x ( t ): shifts spectrum up and down by ω
0
and halves.
Filtering by H
1
( ω ) doesn’t affect this if ω
B
> ω
m
; clips this if ω
B
< ω
m
.