Signals & Systems
(Second Edition)
—Learning Instructions
(Exercises Answers)
Department of Computer Engineering
2005.12
Contents
Chapter 1 ······················································· 2
Chapter 2 ······················································· 17
Chapter 3 ······················································· 35
Chapter 4 ······················································· 62
Chapter 5 ······················································· 83
Chapter 6 ······················································· 109
Chapter 7 ······················································· 119
Chapter 8 ······················································· 132
Chapter 9 ······················································· 140
Chapter 10 ·······················································160
1
Chapter 1 Answers
1.1 Converting from polar to Cartesian coordinates:
1.2 converting from Cartesian to polar coordinates:
,
,
,
,
,
1.3. (a)
=
,
=0, because
,
.Therefore,
=
=
=
,
(b)
=
(c)
=cos(t). Therefore,
=
=
=
(d)
,
. Therefore,
=0,because
< .
=
=
,
(e)
(f)
=
=
=
=
,
=1. therefore,
=
=
,
. Therefore,
=
=
=
,
.
1.4. (a) The signal x[n] is shifted by 3 to the right. The shifted signal will be zero for n<1, And n>7.
(b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0.
(c) The signal x[n] is flipped signal will be zero for n<-1 and n>2.
(d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be
zero for n<-2 and n>4.
(e) The signal x[n] is flipped and the flipped and the flipped signal is shifted by 2 to the left.
1.5. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal by 1 to the right.
Therefore, x (1-t) will be zero for t>-2.
(b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1,
(c) x(3t) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be
This new signal will be zero for n<-6 and n>0.
Therefore, x (1-t) +x(2-t) will be zero for t>-2.
zero for t<1.
2
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(d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be
(b) x2[n]=1 for all n. Therefore, it is periodic with a fundamental period of 1.
(c) x3[n] is as shown in the Figure S1.6.
zero for t<9.
1.6 (a) x1(t) is not periodic because it is zero for t<0.
…
x3[n
…
…
-3
1
1
]
-4
…
…
1.7. (a)
1
4
5
…
…
…
n
0
…
…
1
…
-1
…
…
…
-1
Therefore, it is periodic with a fundamental period of 4.
-1
=
Therefore,
is zero for
>3.
(b) Since x1(t) is an odd signal,
is zero for all values of t.
(c)
Therefore,
is zero when
<3 and when
(d)
Therefore,
is zero only when
.
1.8. (a)
(b)
(c)
(d)
.
1.9. (a)
is a periodic complex exponential.
(b)
is a complex exponential multiplied by a decaying exponential. Therefore,
(c)
is not periodic.
is a periodic signal.
=
=
.
is a complex exponential with a fundamental period of
(d)
is a periodic signal. The fundamental period is given by N=m(
.
)
=
By choosing m=3. We obtain the fundamental period to be 10.
(e)
is not periodic.
is a complex exponential with
=3/5. We cannot find any integer m
such that m(
) is also an integer. Therefore,
is not periodic.
1.10.
x(t)=2cos(10t+1)-sin(4t-1)
Period of first term in the RHS =
Period of first term in the RHS =
.
.
Therefore, the overall signal is periodic with a period which the least common
multiple of the periods of the first and second terms. This is equal to
.
3
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1.11.
x[n] = 1+
−
Period of first term in the RHS =1.
Period of second term in the RHS =
=7 (when m=2)
Period of second term in the RHS =
=5 (when m=1)
Therefore, the overall signal x[n] is periodic with a period which is the least common
Multiple of the periods of the three terms inn x[n].This is equal to 35.
1.12. The signal x[n] is as shown in figure S1.12. x[n] can be obtained by flipping u[n] and then
Shifting the flipped signal by 3 to the right. Therefore, x[n]=u[-n+3]. This implies that
M=-1 and no=-3.
1.13
y(t)=
X[n]
-3
-2
-1
0
1
2
3
n
Figure S 1.12
=
=
Therefore
1
-1
0
1
2
t
0
x(t)
g(t)
1.14 The signal x(t) and its derivative g(t) are shown in Figure S1.14.
Therefore
Figure S 1.14
-2
-1
-3
2
t
1
-3
)
This implies that A =3, t =0, A =-3, and t =1.
1.15 (a) The signal x [n], which is the input to S , is the same as y [n].Therefore ,
y [n]= x [n-2]+
x [n-3]
= y [n-2]+
y [n-3]
=2x
[n-2] +4x
[n-3] + ( 2x [n-3]+ 4x [n-4])
=2x [n-2]+ 5x [n-3] + 2x [n-4]
The input-output relationship for S is
y[n]=2x[n-2]+ 5x [n-3] + 2x [n-4]
4
74jne25jne7/425/22tdtx)(dtt))2()2((2,022,12,0,ttt224dtEkkktkttg12(3)2(3)(1122221222121211112111111
(b) The input-output relationship does not change if the order in which S and S are connected series
reversed. . We can easily prove this assuming that S follows S . In this case , the signal x [n], which is the
input to S is the same as y [n].
Therefore y [n] =2x [n]+ 4x [n-1]
= 2y [n]+4 y [n-1]
=2( x [n-2]+
x [n-3] )+4(x [n-3]+
x [n-4])
=2 x [n-2]+5x [n-3]+ 2 x [n-4]
The input-output relationship for S is once again
1.16 (a)The system is not memory less because y[n] depends on past values of x[n].
(b)The output of the system will be y[n]=
=0
y[n]=2x[n-2]+ 5x [n-3] + 2x [n-4]
(c)From the result of part (b), we may conclude that the system output is always zero for inputs of the
form
, k ґ. Therefore , the system is not invertible .
1.17 (a) The system is not causal because the output y(t) at some time may depend on future values of x(t). For
instance , y(-
)=x(0).
(b) Consider two arbitrary inputs x (t)and x (t).
x (t)
y (t)= x (sin(t))
x (t)
y (t)= x (sin(t))
Let x (t) be a linear combination of x (t) and x (t).That is , x (t)=a x (t)+b x (t)
Where a and b are arbitrary scalars .If x (t) is the input to the given system ,then the corresponding output
y (t) is y (t)= x ( sin(t))
=a x (sin(t))+ x (sin(t))
=a y (t)+ by (t)
Therefore , the system is linear.
1.18.(a) Consider two arbitrary inputs x [n]and x [n].
x [n]
y [n] =
x [n ]
y [n] =
Let x [n] be a linear combination of x [n] and x [n]. That is :
x [n]= ax [n]+b x [n]
where a and b are arbitrary scalars. If x [n] is the input to the given system, then the corresponding output
y [n] is y [n]=
=
=a
+b
Therefore the system is linear.
(b) Consider an arbitrary input x [n].Let
= ay [n]+b y [n]
5
12121121112222122212222]2[][nn][kn12111222312312333312121211][001kxnnnnk22][002kxnnnnk312312333][003kxnnnnk])[][(2100kbxkaxnnnnk][001kxnnnnk][002kxnnnnk121
y [n] =
be the corresponding output .Consider a second input x [n] obtained by shifting x [n] in time:
The output corresponding to this input is
x [n]= x [n-n ]
y [n]=
=
=
Also note that y [n- n ]=
.
Therefore , y [n]= y [n- n ]
This implies that the system is time-invariant.
(c) If
Also note that y [n- n ]= x
[n-2- n ]
Therefore , y [n]= y [n- n ]
This implies that the system is time-invariant.
(c) (i) Consider two arbitrary inputs x [n]and x [n].
x [n]
y [n] = x [n+1]- x [n-1]
x [n ]
y [n] = x [n+1 ]- x [n -1]
Let x [n] be a linear combination of x [n] and x [n]. That is :
x [n]= ax [n]+b x [n]
where a and b are arbitrary scalars. If x [n] is the input to the given system, then the
corresponding output y [n] is y [n]= x [n+1]- x [n-1]
=a(x [n+1]- x [n-1])+b(x [n +1]- x [n -1])
=a x [n+1]+b x [n +1]-a x [n-1]-b x [n -1]
Therefore the system is linear.
(ii) Consider an arbitrary input x [n].Let y [n]= x [n+1]- x [n-1]
= ay [n]+b y [n]
be the corresponding output .Consider a second input x [n] obtained by shifting x [n] in time: x [n]=
x [n-n ]
The output corresponding to this input is
Also note that y [n-n ]= x [n+1- n ]- x [n-1- n ]
y [n]= x [n +1]- x [n -1]= x [n+1- n ]- x [n-1- n ]
Therefore , y [n]= y [n-n ]
This implies that the system is time-invariant.
(d) (i) Consider two arbitrary inputs x (t) and x (t).
x (t)
y (t)=
x (t)
y (t)=
Let x (t) be a linear combination of x (t) and x (t).That is x (t)=a x (t)+b x (t)
where a and b are arbitrary scalars. If x (t) is the input to the given system, then the corresponding output
y (t) is y
(t)=
=
=a
+b
= ay (t)+b y (t)
Therefore the system is linear.
(ii) Consider an arbitrary inputs x (t).Let
y (t)=
=
be the corresponding output .Consider a second input x (t) obtained by shifting x (t) in time:
The output corresponding to this input is
y (t)=
=
x (t)= x (t-t )
Also note that y (t-t )=
Therefore the system is not time-invariant.
=
y (t)
7
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