Tutorial 1 Solution - German University in...
Transcript of Tutorial 1 Solution - German University in...
Department of Communications Engineering
Digital Signal Processing [COMM 602]
Prof. Ahmed El-Mahdy
Tutorial 1 Solution Problem 1:
The arrow (brace) indicates the zeroth sample ( ).
[ ] [ ⏟ ]
a) [ ]
[ ] means shifting [ ] by samples to the left if , or to the right if .
b) [ ⁄ ]
[ ] means scaling the time axis by a factor of so that the signal samples are brought closer together if | | , and spaced further apart if | | (with the insertion of zeros in-between samples). The signal is mirrored relative to the y-axis if .
-2
-1
1
2
3
4
-3 -2 -1 1 2 3 4
x[n]
-2
-1
1
2
3
4
1 2 3 4 5 6
x[n-2]
c) [ ]
Obtaining [ ] from [ ] involves 3 consecutive steps:
1- Shifting the signal by samples to the left if , or to the right if
2- Scaling the time axis by a factor of so that the signal samples are brought closer together if | | , and spaced further apart if | | (with the insertion of zeros in-between samples)
3- Mirroring the signals samples relative to the y-axis if .
-2
-1
1
2
3
4
-5 -4 -3 -2 -1 1 2 3 4 5 6 7
x[n/2]
-2
-1
1
2
3
4
-6 -5 -4 -3 -2 -1
x[n+3]
-2
-1
1
2
-6 -5 -4 -3 -2 -1
x[2n+3]
d) [ ] [ ]
Multiplying a signal [ ] by [ ] zeros all samples except that at .
e) [ ] [ ]
First, [ ] is obtained from [ ] through a left shift by 2 samples then mirroring the signal relative to the y-axis. The two signals [ ] [ ] are then multiplied together on a sample-by-sample basis.
-2
-1
1
2
1 2 3 4 5 6
x[-2n+3]
1
2
-5 -4 -3 -2 -1 1 2 3 4 5
u[2-n]
1
2
3
4
1 2 3 4 5 6
3x[n-2]δ[n-4]
f) [ ]
Through simple substitution:
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
g) [ ]
The even part of [ ] is obtained as
[ ] [ ] [ ]
1
2
3
4
-5 -4 -3 -2 -1 1 2 3 4 5
x[n]u[2-n]
1
2
3
4
-3 -2 -1 1 2 3
x[n2]
-2
-1
1
2
3
4
-4 -3 -2 -1 1 2 3 4
x[n]
h) [ ]
The odd part of [ ] is obtained as
[ ] [ ] [ ]
Note that [ ] [ ] [ ]
-2
-1
1
2
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-4 -3 -2 -1 1 2 3 4
x[-n]
-1
1
2
3
4
-4 -3 -2 -1 1 2 3 4
xe[n]
-1
1
-4 -3 -2 -1 1 2 3 4
xo[n]
Problem 2:
a) [
]
Since the result in its simplest form is a rational number, the signal is periodic with the fundamental period being the denominator of the rational fraction .
b) [ ]
Since the result in its simplest form is not a rational number, the signal is non-periodic.
c)
The signal is periodic with the fundamental period .
d) [
] [
]
This is the summation of two sinusoids with different frequencies.
The first sinusoid is periodic with a period of , while the second one is non-periodic. Therefore, the whole signal is non-periodic.
e) [
] [
]
Through the use of trigonometric identities, the signal can be expressed as the summation of two sinusoids.
[
] [
]
( [
] [
])
( [
] [
])
The first sinusoid has a period of , while the second has a period of . Thus, the whole signal is periodic with the fundamental period being the lowest common multiple of the two periods .
Problem 3:
a) [ ] [ ]
With memory because, for example, [ ] depends on a past input sample [ ].
Causal because the output at any time doesn’t depend on future input samples.
[ ] [ ]
[ ] [ ]
[ ] [ ] [ ]
[ ]
[ ] [ ] [ ]
[ ] [ ]
( [ ] [ ])
[ ] [ ] [ ] [ ]
[ ] [ ] [ ]
Non-linear.
[ ] [ ]
[ ] [( ) ]
[ ] [ ]
[ ] [ ]
[( ) ]
[ ] [ ]
Time invariant.
Stable because a bounded input | [ ]| produces a bounded output | [ ]| .
b) [ ] ( [ ])
Memoryless because the output at any time doesn’t depend on past/future input samples.
Causal because the output at any time doesn’t depend on future input samples.
[ ] ( [ ])
[ ] ( [ ])
[ ] [ ] ( [ ]) ( [ ])
[ ] [ ] [ ]
[ ] ( [ ])
( [ ] [ ])
[ ] [ ] [ ]
Non-linear.
[ ] ( [ ])
[ ] ( [ ])
[ ] [ ]
[ ] ( [ ])
( [ ])
[ ] [ ]
Time invariant.
Stable because a bounded input | [ ]| produces a bounded output | [ ]| .
c) [ ] [ ] ( )
Memoryless because the output at any time doesn’t depend on past/future input samples.
Causal because the output at any time doesn’t depend on future input samples.
[ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
( [ ] [ ]) [ ]
[ ] [ ] [ ]
Linear.
[ ] [ ] [ ]
[ ] [ ] [ ( )]
[ ] [ ]
[ ] [ ] [ ]
[ ] [ ]
[ ] [ ]
Time variant.
Stable because a bounded input | [ ]| produces a bounded output | [ ]| .
d) [ ] | [ ]|
Memoryless because the output at any time doesn’t depend on past/future input samples.
Causal because the output at any time doesn’t depend on future input samples.
[ ] | [ ]|
[ ] | [ ]|
[ ] [ ] | [ ]| | [ ]|
[ ] [ ] [ ]
[ ] | [ ]|
| [ ] [ ]|
[ ] [ ] [ ]
Non-linear.
[ ] | [ ]|
[ ] | [ ]|
[ ] [ ]
[ ] | [ ]|
| [ ]|
[ ] [ ]
Time invariant.
Stable because a bounded input | [ ]| produces a bounded output | [ ]| .
e) [ ] [ ]
With memory because, for example, [ ] depends on a future input sample [ ].
Non-causal because, for example, [ ] depends on a future input sample [ ].
[ ] [ ]
[ ] [ ]
[ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ]
[ ] [ ]
[ ] [ ] [ ]
Linear.
[ ] [ ]
[ ] [ ( ) ] [ ]
[ ] [ ]
[ ] [ ]
[( ) ] [ ]
[ ] [ ]
Time variant.
Stable because a bounded input | [ ]| produces a bounded output | [ ]| .
f) [ ] [ ]
With memory because, for example, [ ] depends on a past input sample [ ].
Non-causal because, for example, [ ] depends on a future input sample [ ].
[ ] [
]
[ ] [
]
[ ] [ ] [
]
[ ]
[ ] [ ] [ ]
[ ] [
]
[
] [ ]
[ ] [ ] [ ]
Linear.
[ ] [
]
[ ] [( )
]
( )
[ ] [ ]
[ ] [
]
[(
) ]
[ ] [ ]
Time variant.
Not stable because a bounded input | [ ]| produces a non-bounded output at [ ].