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

3

4

-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 [ ].