Solution of ECE 505 Test 2 F06
1.
A continuous-time sinusoidal signal x a ( t ) = 5 cos ( 2000 t ) is sampled at
fs = 1 / T = 5000 samples/second to form the sinusoidal discrete-time
signal x ( n ) = x a ( nT ) . The signal x ( n ) is applied to a discrete-time
0.7z
0.7
system with transfer function H ( z ) =
. The output
=
z 0.3 1 0.3z 1
signal from the system is also sinusoidal of the form
y ( n ) = A cos ( 0 n + ) . Find the numerical values of A, 0 and .
x ( n ) = 5 cos ( 2000 nT ) = 5 cos ( 2 n / 5 )
(
H e
j 2 /5
)
0.7e j 2 /5
= j 2 /5
= 0.7021 j0.2208 = 0.736 0.3047 or -17.46°
e
0.3
So the response amplitude is 5 0.736 = 3.68 and the response phase shift
relative to the excitation is -0.3047 radians and
y ( n ) = 3.68 cos (( 2 / 5 ) n 0.3047 ) .
In the boxes provided, write the letter of the frequency response
magnitude graph that corresponds to each pole-zero plot.
J
F
I
B
G
E
A
H
D
C
Im(z)
Re(z)
Im(z)
Im(z)
[z]
0
ω
0
2
-2
F
0
2
0
ω
2
0.5
0
0
ω
-2
0
ω
0.5
0
2
-2
2
0
2
-2
0
ω
-2
2
2
1
0.5
0
0
ω
J
1
0.5
0
0
ω
0.5
I
1
|H|
|H|
0.5
E
1
H
1
-2
-2
[z]
Re(z)
D
0.5
G
1
|H|
0
ω
Im(z)
[z]
1
|H|
0.5
Re(z)
Re(z)
C
|H|
|H|
|H|
-2
Im(z)
[z]
1
[z]
Re(z)
Re(z)
B
1
0.5
0
Im(z)
[z]
Im(z)
[z]
Re(z)
Re(z)
A
1
Im(z)
[z]
Re(z)
Re(z)
0
Im(z)
[z]
|H|
[z]
|H|
Im(z)
|H|
2.
-2
0
ω
2
0.5
0
-2
0
ω
2
3.
A signal x a ( t ) has a continuous-time Fourier transform for which
X a ( F ) 0 , 200 < F < 230
X a ( F ) = 0 , otherwise
.
What is the numerical minimum sampling rate for which the signal can be
recovered exactly from the samples?
F 230 =7
kmax = H = B 230 200 Fs =
2FH 460
=
= 65.714
7
kmax
4.
An 8-bit ADC with a full range of -10 V to +10V samples and quantizes a
sinusoidal signal with an amplitude of 7 V. Using the usual assumptions about
the distribution and power spectral density of the quantization noise, find the
numerical signal-to-quantization-noise ratio (SQNR) of the ADC output signal
(defined as the signal power of a signal with no quantization noise divided by the
signal power of the quantization noise) in dB.
72
= 24.5
2
20
2
= 8 = 0.078125 Pq =
= 0.000509
2
12
Ps =
(
)
SQNR dB = 10 log10 Ps / Pq = 46.82 dB
Solution of ECE 505 Test 2 F06
1.
A continuous-time sinusoidal signal x a ( t ) = 5 cos ( 2000 t ) is sampled at
fs = 1 / T = 6000 samples/second to form the sinusoidal discrete-time
signal x ( n ) = x a ( nT ) . The signal x ( n ) is applied to a discrete-time
0.7z
0.7
system with transfer function H ( z ) =
. The output
=
z 0.3 1 0.3z 1
signal from the system is also sinusoidal of the form
y ( n ) = A cos ( 0 n + ) . Find the numerical values of A, 0 and .
x ( n ) = 5 cos ( 2000 nT ) = 5 cos ( n / 3)
(
)
H e j / 3 =
0.7e j / 3
= 0.7532 j0.2302 = 0.7876 0.2966 or -16.996°
e j / 3 0.3
So the response amplitude is 5 0.736 = 3.938 and the response phase shift
relative to the excitation is -0.2966 radians and
y ( n ) = 3.938 cos (( / 3) n 0.2966 ) .
In the boxes provided, write the letter of the frequency response
magnitude graph that corresponds to each pole-zero plot.
E
C
J
I
A
G
B
F
H
D
Im(z)
Re(z)
Im(z)
Im(z)
[z]
0
ω
0
2
-2
F
0
2
0
ω
2
0.5
0
0
ω
-2
0
ω
0.5
0
2
-2
2
0
2
-2
0
ω
-2
2
2
1
0.5
0
0
ω
J
1
0.5
0
0
ω
0.5
I
1
|H|
|H|
0.5
E
1
H
1
-2
-2
[z]
Re(z)
D
0.5
G
1
|H|
0
ω
Im(z)
[z]
1
|H|
0.5
Re(z)
Re(z)
C
|H|
|H|
|H|
-2
Im(z)
[z]
1
[z]
Re(z)
Re(z)
B
1
0.5
0
Im(z)
[z]
Im(z)
[z]
Re(z)
Re(z)
A
1
Im(z)
[z]
Re(z)
Re(z)
0
Im(z)
[z]
|H|
[z]
|H|
Im(z)
|H|
2.
-2
0
ω
2
0.5
0
-2
0
ω
2
3.
A signal x a ( t ) has a continuous-time Fourier transform for which
X a ( F ) 0 , 210 < F < 230
X a ( F ) = 0 , otherwise
.
What is the numerical minimum sampling rate for which the signal can be
recovered exactly from the samples?
F 230 = 11
kmax = H = B 230 210 Fs =
2FH 460
=
= 41.82
11
kmax
4.
An 8-bit ADC with a full range of -10 V to +10V samples and quantizes a
sinusoidal signal with an amplitude of 8 V. Using the usual assumptions about
the distribution and power spectral density of the quantization noise, find the
numerical signal-to-quantization-noise ratio (SQNR) of the ADC output signal
(defined as the signal power of a signal with no quantization noise divided by the
signal power of the quantization noise) in dB.
Ps =
82
= 32
2
20
2
= 8 = 0.078125 Pq =
= 0.000509
2
12
(
)
SQNR dB = 10 log10 Ps / Pq = 47.98 dB
Z
[ n ] 1 , All z
z
1
Z
n u [ n ] =
, z >
z 1 z 1
z
z 1
Z
n u [ n ] =
, z >1
( z 1)2 (1 z 1 )2
z sin ( 0 )
sin ( 0 ) z 1
sin ( 0 n ) u [ n ] 2
=
, z >
z 2 z cos ( 0 ) + 2 1 2 cos ( 0 ) z 1 + 2 z 2
n
Z
cos ( 0 n ) u [ n ] n
Z
z z cos ( 0 ) 1 cos ( 0 ) z 1
=
, z >
z 2 2 z cos ( 0 ) + 2 1 2 cos ( 0 ) z 1 + 2 z 2
z
1
Z
n u [ n 1] =
, z <
z 1 z 1
z
z 1
Z
n n u [ n 1] =
, z <
( z )2 1 z 1 2
(
)