(a) One dimensional, multichannel, discrete time, and digital.
(b) Multi dimensional, single channel, continuous-time, analog.
(c) One dimensional,
... [Show More] single channel, continuous-time, analog.
(d) One dimensional, single channel, continuous-time, analog.
(e) One dimensional, multichannel, discrete-time, digital.
1.2
(a) f = 0.01
2 = 1
200 ⇒ periodic with Np = 200.
(b) f = 30
105 ( 1
2 ) = 1
7 ⇒ periodic with Np = 7.
(c) f = 3
2 = 3
2 ⇒ periodic with Np = 2.
(d) f = 3
2 ⇒ non-periodic.
(e) f = 62
10 ( 1
2 ) = 31
10 ⇒ periodic with Np = 10.
1.3
(a) Periodic with period Tp = 2
5 .
(b) f = 5
2 ⇒ non-periodic.
(c) f = 1
12 ⇒ non-periodic.
(d) cos(n
8 ) is non-periodic; cos( n
8 ) is periodic; Their product is non-periodic.
(e) cos( n
2 ) is periodic with period Np=4
sin( n
8 ) is periodic with period Np=16
cos( n
4 +
3 ) is periodic with period Np=8
Therefore, x(n) is periodic with period Np=16. (16 is the least common multiple of 4,8,16).
1.4
(a) w = 2k
N implies that f = k
N . Let
α = GCD of (k,N), i.e.,
k = k′α,N = N′α.
Then,
f =
k′
N′
, which implies that
N′ =
N
α
.
3
(b)
N = 7
k = 0 1 2 3 4 5 6 7
GCD(k,N) = 7 1 1 1 1 1 1 7
Np = 1 7 7 7 7 7 7 1
(c)
N = 16
k = 0 1 2 3 4 5 6 7 8 9 10 11 12 . . . 16
GCD(k,N) = 16 1 2 1 4 1 2 1 8 1 2 1 4 . . . 16
Np = 1 6 8 16 4 16 8 16 2 16 8 16 4 . . . 1
1.5
(a) Refer to fig 1.5-1
(b)
0 5 10 15 20 25 30
−3
−2
−1
0
1
2
3
−−−> t (ms)
−−−> xa(t)
Figure 1.5-1:
x(n) = xa(nT)
= xa(n/Fs)
= 3sin(πn/3) ⇒
f =
1
2π
(
π
3
)
=
1
6
,Np = 6
4
0 10 20
t (ms)
3
-3
Figure 1.5-2:
(c)Refer to fig 1.5-2
x(n) =
n
0, 3 √2
, 3 √2
, 0,− 3 √2
,− 3 √2
o
,Np = 6.
(d) Yes.
x(1) = 3 = 3sin(
100π
Fs
) ⇒ Fs = 200 samples/sec.
1.6
(a)
x(n) = Acos(2πF0n/Fs + θ)
= Acos(2π(T/Tp)n + θ)
But T/Tp = f ⇒ x(n) is periodic if f is rational.
(b) If x(n) is periodic, then f=k/N where N is the period. Then,
Td = (
k
f
T) = k(
Tp
T
)T = kTp.
Thus, it takes k periods (kTp) of the analog signal to make 1 period (Td) of the discrete signal.
(c) Td = kTp ⇒ NT = kTp ⇒ f = k/N = T/Tp ⇒ f is rational ⇒ x(n) is periodic.
1.7
(a) Fmax = 10kHz ⇒ Fs ≥ 2Fmax = 20kHz.
(b) For Fs = 8kHz, Ffold = Fs/2 = 4kHz ⇒ 5kHz will alias to 3kHz.
(c) F=9kHz will alias to 1kHz.
1.8
(a) Fmax = 100kHz, Fs ≥ 2Fmax = 200Hz.
(b) Ffold = Fs
2 = 125Hz.
5
1.9
(a) Fmax = 360Hz, FN = 2Fmax = 720Hz.
(b) Ffold = Fs
2 = 300Hz.
(c)
x(n) = xa(nT)
= xa(n/Fs)
= sin(480πn/600) + 3sin(720πn/600)
x(n) = sin(4πn/5) − 3sin(4πn/5)
= −2sin(4πn/5).
Therefore, w = 4π/5.
(d) ya(t) = x(Fst) = −2sin(480πt).
1.10
(a)
Number of bits/sample = log21024 = 10.
Fs =
[10, 000 bits/sec]
[10 bits/sample]
= 1000 samples/sec.
Ffold = 500Hz.
(b)
Fmax =
1800π
2π
= 900Hz
FN = 2Fmax = 1800Hz.
(c)
f1 =
600π
2π
(
1
Fs
)
= 0.3;
f2 =
1800π
2π
(
1
Fs
)
= 0.9;
But f2 = 0.9 > 0.5 ⇒ f2 = 0.1.
Hence, x(n) = 3cos[(2π)(0.3)n] + 2cos[(2π)(0.1)n]
(d) △ =
xmax−xmin
m−1 = 5−(−5)
1023 = 10
1023 .
1.11
x(n) = xa(nT)
= 3cos
100πn
200
+ 2sin
250πn
200
6
= 3cos
πn
2
− 2sin
3πn
4
T′ =
1
1000 ⇒ ya(t) = x(t/T′)
= 3cos
π1000t
2
− 2sin
3π1000t
4
ya(t) = 3cos(500πt) − 2sin(750πt)
1.12
(a) For Fs = 300Hz,
x(n) = 3cos
πn
6
+ 10sin(πn) − cos
πn
3
= 3cos
πn
6
− 3cos
πn
3
(b) xr(t) = 3cos(10000πt/6) − cos(10000πt/3)
1.13
(a)
Range = xmax − xmin = 12.7.
m = 1 +
range
△
= 127 + 1 = 128 ⇒ log2(128)
= 7 bits.
(b) m = 1 + 127
0.02 = 636 ⇒ log2(636) ⇒ 10 bit A/D.
1.14
R = (20
samples
sec
) × (8
bits
sample
)
= [Show Less]