signals and systems hw1 solution
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Signals and systems HW1 solution
1.
Probs. 1.6 (b)
The signal is not periodic.
A periodic continuous-time signal π₯(π‘) has the property that there is a
positive value of π» for which
π₯(π‘) = π₯(π‘ + π)
for all values of π‘.
For this problem,
π₯2(π‘) = π₯2(π‘ + π)
βΉ π(β1+π)π‘ = π(β1+π)(π‘+π)
βΉ π(β1+π)π‘ = π(β1+π)π‘+(β1+π)π
βΉ 1 = π(β1+π)π
βΉ 0 = (β1 + π)π
βΉ π = 0
Since π» is not a positive value, we conclude that the signal π₯2(π‘) is not
periodic.
Probs. 1.6 (d)
The signal is periodic.
π₯4[π] = 3ππ3π(π+12
) 5β= 3ππ
3π10 β ππ
3π5
π
The general discrete-time complex exponential signal can be expressed in the
form
π₯[π] = πΆπΌπ
where
πΆ = |πΆ|πππ
and
Ξ± = |πΌ|πππ0
Then
π₯4[π] = 3ππ3π10 β ππ
3π5
π = πΆπΌπ
where
πΆ = 3ππ3π10 = |πΆ|πππ
and
Ξ± = ππ3π5 = |πΌ|πππ0 βΉ π0 =
3π
5
Since |Ξ±| = 1, the fundamental period is given by
π = π (2π
π0) = π (
2π
3π 5β) = π (
10
3)
By choosing π = 3, we obtain the fundamental period to be 10.
Probs. 1.25 (b)
The signal is periodic.
π₯(π‘) = ππ(ππ‘β1) = πβπ β ππππ‘
The general continuous-time complex exponential signal can be expressed in
the form
π₯(π‘) = πΆππΌπ‘
where
πΆ = |πΆ|πππ
and
Ξ± = π + ππ0
Then,
π₯(π‘) = πβπ β ππππ‘ = πΆππΌπ‘
where
πΆ = πβπ = |πΆ|πππ
and
Ξ± = ππ = π + ππ0 βΉ π0 = π
Since Ξ± is purely imaginary, the fundamental period is given by
π =2π
π0=
2π
π= 2
We obtain the fundamental period to be 2.
Probs. 1.25 (c)
The signal is periodic.
π₯(π‘) = [cos (2π‘ βπ
3)]
2
=1 + cos (4π‘ β
2π3 )
2=
1
2+
1
2cos (4π‘ β
2π
3)
The general continuous-time sinusoidal signal can be expressed in the form
π₯(π‘) = π΄ cos(π0π‘ + π)
Then, the frequency for 1
2cos (4π‘ β
2π
3) is π0 = 4.
The fundamental period is given by
π =2π
π0=
2π
4=
π
2
We obtain the fundamental period to be π
π.
Probs. 1.25 (f)
The signal is not periodic.
π₯(π‘) = β πβ(2π‘βπ)π’(2π‘ β π)
β
π=ββ
π₯(π‘) = π₯(π‘ + π)
β β πβ(2π‘βπ)π’(2π‘ β π)
β
π=ββ
= β πβ(2(π‘+π)βπ)π’(2(π‘ + π) β π)
β
π=ββ
β β πβ(2π‘βπ)π’(2π‘ β π)
β
π=ββ
= πβ2π β πβ(2π‘βπ)π’(2(π‘ + π) β π)
β
π=ββ
β β πβ(2π‘βπ)[π’(2π‘ β π) β πβ2ππ’(2(π‘ + π) β π)]
β
π=ββ
= 0
β π’(2π‘ β π) β πβ2ππ’(2(π‘ + π) β π) = 0
β π = 0
Since π» is not a positive value, we conclude that the signal π₯(π‘) is not
periodic.
Probs. 1.26 (c)
The signal is periodic.
π₯[π] = cos (π
8π2)
π₯[π] = π₯[π + π]
βΉ cos (π
8π2) = cos (
π
8(π + π)2)
βΉ cos (π
8π2) = cos (
π
8π2 +
π
4ππ +
π
8π2)
βΉπ
4ππ +
π
8π2 = 2ππ, π = 0, Β±1, Β±2, β¦
βΉπ
4ππ = 2ππ1, π1 = 0, Β±1, Β±2, β¦ πππ
π
8π2 = 2ππ2, π2 = 0, Β±1, Β±2, β¦
For π
4ππ = 2ππ1, π1 = 0, Β±1, Β±2, β¦
ππ
8= π1 βΉ π = 8
For π
8π2 = 2ππ2, π2 = 0, Β±1, Β±2, β¦
π2
16= π2 βΉ π = 4
Thus, the fundamental period is the least common multiple of 8 and 4
which is
lcm(8,4) = 8
The answer is 8.
Probs. 1.26 (d)
The signal is periodic.
π₯[π] = cos (π
2π) cos (
π
4π)
= (1
2ππ
π2
π +1
2πβπ
π2
π) (1
2ππ
π4
π +1
2πβπ
π4
π)
=1
4ππ
3π4
π +1
4ππ
π4
π +1
4πβπ
π4
π +1
4πβπ
3π4
π
For 1
4ππ
3π
4π
, π = π2π
(3π 4β )= π
8
3= 8
For 1
4ππ
π
4π
, π = π2π
(π 4β )= 8
For 1
4πβπ
3π
4π =
1
4ππ(β
3π
4+2π)π =
1
4ππ
5π
4π
, π = π2π
(5π 4β )= π
8
5= 8
For 1
4πβπ
π
4π =
1
4ππ(β
π
4+2π)π =
1
4ππ
7π
4π
, π = π2π
(7π 4β )= π
8
7= 8
The answer is 8.
Probs. 1.26 (e)
The signal is periodic.
π₯[π] = 2 cos (π
4π) + sin (
π
8π) β 2 cos (
π
2π +
π
6)
2 cos (π
4π) has a fundamental period of π = π
2π
(π 4β )= 8.
sin (π
8π) has a fundamental period of π = π
2π
(π 8β )= 16.
β2 cos (π
2π +
π
6) has a fundamental period of π = π
2π
(π 2β )= 4.
Thus, the fundamental period is the least common multiple of 8, 16 and 4
which is
lcm(8,16,4) = 16
The answer is 16.
2.
Probs. 1.21 (c)
Probs. 1.21 (d)
Probs. 1.21 (e)
Probs. 1.21 (f)
Probs. 1.22 (d)
Probs. 1.22 (e)
The answer is the same as the original signal.
π₯[π]π’[3 β π] = π₯[π]
Probs. 1.22 (f)
Probs. 1.22 (g)
3.
Probs. 1.23 (a)
Probs. 1.23 (b)
Probs. 1.24 (b)
Even
Odd
Probs. 1.24 (c)
Even
Odd
4.
Probs. 1.27 (b)
y(π‘) = [cos(3π‘)]π₯(π‘)
(1)
The system is memoryless, as the value of y(π‘) at any particular time
depends only on the value of π₯(π‘) at that time, not π₯(π‘ + π) or π₯(π‘ β π).
(2)
y(π‘ β π‘0) = [cos(3(π‘ β π‘0))]π₯(π‘ β π‘0)
Let π₯2(π‘) = π₯(π‘ β π‘0)
π₯2(π‘) β π¦2(π‘) = [cos(3π‘)]π₯2(π‘) = [cos(3π‘)]π₯(π‘ β π‘0)
β [cos(3(π‘ β π‘0))]π₯(π‘ β π‘0) = y(π‘ β π‘0)
The system is not time invariant, as π₯(π‘ β π‘0) β y(π‘ β π‘0).
(3)
Assume
π₯1(π‘) β π¦1(π‘) = [cos(3π‘)]π₯1(π‘)
π₯2(π‘) β π¦2(π‘) = [cos(3π‘)]π₯2(π‘)
Let
π₯3(π‘) = Ξ±π₯1(π‘) + π½π₯2(π‘)
Then
π₯3(π‘) β π¦3(π‘) = [cos(3π‘)]π₯3(π‘) = [cos(3π‘)][Ξ±π₯1(π‘) + π½π₯2(π‘)]
= Ξ±[cos(3π‘)]π₯1(π‘) + π½[cos(3π‘)]π₯2(π‘) = Ξ±π¦1(π‘) + π½π¦2(π‘)
The system is linear, as Ξ±π₯1(π‘) + π½π₯2(π‘) β Ξ±π¦1(π‘) + π½π¦2(π‘).
(4)
The system is causal, as the value of y(π‘) at any particular time depends only
on the value of π₯(π‘) at present and past time, not π₯(π‘ + π) for π > 0.
(5)
Let
|π₯(π‘)| < π½
Since
|cos(3π‘)| < 1
Then
|cos(3π‘)||π₯(π‘)| < π½ β |π¦(π‘)| < π½
The system is stable.
Probs. 1.27 (e)
y(π‘) = {0, π₯(π‘) < 0π₯(π‘) + π₯(π‘ β 2), π₯(π‘) β₯ 0
(1)
The system is not memoryless, as the value of y(π‘) depends on the value of
π₯(π‘) at π₯(π‘ β 2).
(2)
y(π‘ β π‘0) = {0, π₯(π‘ β π‘0) < 0
π₯(π‘ β π‘0) + π₯(π‘ β π‘0 β 2), π₯(π‘ β π‘0) β₯ 0
Let π₯2(π‘) = π₯(π‘ β π‘0)
π₯2(π‘) β π¦2(π‘) = {0, π₯2(π‘) < 0
π₯2(π‘) + π₯2(π‘ β 2), π₯2(π‘) β₯ 0
= {0, π₯(π‘ β π‘0) < 0
π₯(π‘ β π‘0) + π₯(π‘ β π‘0 β 2), π₯(π‘ β π‘0) β₯ 0= y(π‘ β π‘0)
The system is time invariant, as π₯(π‘ β π‘0) β y(π‘ β π‘0).
(3)
Assume
π₯1(π‘) β π¦1(π‘) = {0, π₯1(π‘) < 0
π₯1(π‘) + π₯1(π‘ β 2), π₯1(π‘) β₯ 0
π₯2(π‘) β π¦2(π‘) = {0, π₯2(π‘) < 0
π₯2(π‘) + π₯2(π‘ β 2), π₯2(π‘) β₯ 0
Let
π₯3(π‘) = Ξ±π₯1(π‘) + π½π₯2(π‘)
Then
π₯3(π‘) β π¦3(π‘) = {0, π₯3(π‘) < 0
π₯3(π‘) + π₯3(π‘ β 2), π₯3(π‘) β₯ 0
= {0, Ξ±π₯1(π‘) + π½π₯2(π‘) < 0
Ξ±π₯1(π‘) + π½π₯2(π‘) + Ξ±π₯1(π‘ β 2) + π½π₯2(π‘ β 2), Ξ±π₯1(π‘) + π½π₯2(π‘) β₯ 0
= Ξ± {0, Ξ±π₯1(π‘) + π½π₯2(π‘) < 0
π₯1(π‘) + π₯1(π‘ β 2), Ξ±π₯1(π‘) + π½π₯2(π‘) β₯ 0
+ π½ {0, Ξ±π₯1(π‘) + π½π₯2(π‘) < 0
π₯2(π‘) + π₯2(π‘ β 2), Ξ±π₯1(π‘) + π½π₯2(π‘) β₯ 0= Ξ±π¦1(π‘) + π½π¦2(π‘)
The system is linear, as Ξ±π₯1(π‘) + π½π₯2(π‘) β Ξ±π¦1(π‘) + π½π¦2(π‘).
(4)
The system is causal, as the value of y(π‘) at any particular time depends only
on the value of π₯(π‘) at present and past time, not π₯(π‘ + π) for π > 0.
(5)
Let
|π₯(π‘)| < β πππ πππ π‘
Since
|π₯(π‘ β 2)| < β
Then
|y(π‘)| = {0, |π₯(π‘)| < 0|π₯(π‘)| + |π₯(π‘ β 2)|, |π₯(π‘)| β₯ 0
< β β |π¦(π‘)| < β
The system is stable.
Probs. 1.28 (b)
y[π] = π₯[π β 2] β 2π₯[π β 8]
(1)
The system is not memoryless, as the value of y[π] depends on the value of
π₯[π] at π₯[π β 2] and π₯[π β 8].
(2)
y[π β π0] = π₯[π β π0 β 2] β 2π₯[π β π0 β 8]
Let π₯2[π] = π₯[π β π0]
π₯2[π] β π¦2[π] = π₯2[π β 2] β 2π₯2[π β 8]
= π₯[π β 2 β π0] β 2π₯[π β 8 β π0]
= π₯[π β π0 β 2] β 2π₯[π β π0 β 8] = y[π β π0]
The system is time invariant, as π₯[π β π0] β y[π β π0].
(3)
Assume
π₯1[π] β π¦1[π] = π₯1[π β 2] β 2π₯1[π β 8]
π₯2[π] β π¦2[π] = π₯2[π β 2] β 2π₯2[π β 8]
Let
π₯3[π] = Ξ±π₯1[π] + π½π₯2[π]
Then
π₯3[π] β π¦3[π] = π₯3[π β 2] β 2π₯3[π β 8]
= Ξ±π₯1[π β 2] + π½π₯2[π β 2] β 2Ξ±π₯1[π β 8] β 2π½π₯2[π β 8]
= Ξ±(π₯1[π β 2] β 2π₯1[π β 8]) + π½(π₯2[π β 2] β 2π₯2[π β 8])
= Ξ±π¦1[π] + π½π¦2[π]
The system is linear, as Ξ±π₯1[π] + π½π₯2[π] β Ξ±π¦1[π] + π½π¦2[π].
(4)
The system is causal, as the value of y[π] at any particular time depends only
on the value of π₯[π] at present and past time, not π₯[π + π0] for π0 > 0.
(5)
Let
|π₯[π]| < β πππ πππ π‘
Since
|π₯[π β 2]| < β and |π₯[π β 8]| < β
Then
|π₯[π β 2]| β 2|π₯[π β 8]| < β β |π¦(π‘)| < β
The system is stable.
Probs. 1.28 (e)
y[π] = {π₯[π], π β₯ 10, π = 0
π₯[π + 1], π β€ β1
(1)
The system is not memoryless, as the value of y[π] depends on the value of
π₯[π] at π₯[π + 1].
(2)
y[π β π0] = {
π₯[π β π0], π β π0 β₯ 10, π β π0 = 0
π₯[π β π0 + 1], π β π0 β€ β1
Let π₯2[π] = π₯[π β π0]
π₯2[π] β π¦2[π] = {π₯2[π], π β₯ 10, π = 0
π₯2[π + 1], π β€ β1= {
π₯[π β π0], π β₯ 10, π = 0
π₯[π + 1 β π0], π β€ β1
β {
π₯[π β π0], π β π0 β₯ 10, π β π0 = 0
π₯[π β π0 + 1], π β π0 β€ β1= y[π β π0]
The system is not time invariant, as π₯[π β π0] β y[π β π0].
(3)
Assume
π₯1[π] β π¦1[π] = {π₯1[π], π β₯ 10, π = 0
π₯1[π + 1], π β€ β1
π₯2[π] β π¦2[π] = {π₯2[π], π β₯ 10, π = 0
π₯2[π + 1], π β€ β1
Let
π₯3[π] = Ξ±π₯1[π] + π½π₯2[π]
Then
π₯3[π] β π¦3[π] = {π₯3[π], π β₯ 10, π = 0
π₯3[π + 1], π β€ β1
= {Ξ±π₯1[π] + π½π₯2[π], π β₯ 1
0, π = 0Ξ±π₯1[π + 1] + π½π₯2[π + 1], π β€ β1
= Ξ± ({π₯1[π], π β₯ 10, π = 0
π₯1[π + 1], π β€ β1) + π½ ({
π₯2[π], π β₯ 10, π = 0
π₯2[π + 1], π β€ β1)
= Ξ±π¦1[π] + π½π¦2[π]
The system is linear, as Ξ±π₯1[π] + π½π₯2[π] β Ξ±π¦1[π] + π½π¦2[π].
(4)
The system is not causal, as the value of y[π] depends on the future value
π₯[π + 1].
(5)
Let
|π₯[π]| < β πππ πππ π‘
Since
|π₯[π + 1]| < β
Then
{|π₯[π]|, π β₯ 10, π = 0
|π₯[π + 1]|, π β€ β1< β β |π¦(π‘)| < β
The system is stable.
5.
Probs. 1.31
Probs. 1.42