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1. Continuous- and discrete- time signals 1. Complex numbers For any complex number, z∈ , we have

• Cartesian/algebraic form ; 1z x jy j= + = − where

{ }{ }

Re

Im

x z

y z

=

=

• Polar form ; 1jz r e jθ= ⋅ = −

{ } { } argr z z z= = The complex conjugate of a complex number: jz x jy r e∗ − θ= − = ⋅ The absolute value 2 2z x y= + 2z z z∗⋅ = Relation between polar and cartesian coordinates

2 2

cossin

arctg 0 0, 02 =

arctg 0, 0 0, 02

arctg 0, 0

x ry r

r x yy x x yxy x y x yxy x yx

= θ⎧⎨ = θ⎩⎧ = +⎪⎪ ⎧ π⎧>⎪ ⎪ = >⎪⎪ ⎪⎪ θ⎨ ⎨⎪ πθ = + π < ≥⎨⎪ ⎪− = <

⎪⎪⎪ ⎩⎪⎪ − π < <⎪⎪ ⎩⎩

Euler’s relation cos sinje jθ = θ+ θ

cos

2

sin2

j j

j j

e e

e ej

θ − θ

θ − θ

⎧ +θ =⎪

⎪⎨

−⎪ θ =⎪⎩

Modulus argument/phase

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Problems P1. Determine the Cartesian coordinates of the following complex numbers, depending on 0x and 0y , where we have: 0 0 0z x jy= +

a) 01 0

jz r e− θ= ⋅ b) 2 0z r=

c) ( )03 0

jz r e θ +π= ⋅

d) ( )04 0

jz r e θ +π−= ⋅

e) ( )05 0

2jz r e θ + π= ⋅ Sketch 1 2 5, ,...,z z z in the complex plane for the following cases of 0r and 0θ

0 0

0 0

2 4

2 2

r

r

π= θ =

π= θ =

Solution a)

( ) ( )[ ]

01 0 0 0 0

0 0 0

0 0

cos sin

cos sin

jz r e r j

r jx jy

− θ= ⋅ = ⋅ −θ + −θ⎡ ⎤⎣ ⎦= ⋅ θ − θ

= −

b) 2 2

2 0 0 0z r x y= = + c)

( ) ( ) ( )[ ]

03 0 0 0 0

0 0 0 0 0

cos sin

cos sin

jz r e r j

r j x jy

θ +π= ⋅ = ⋅ θ + π + θ + π⎡ ⎤⎣ ⎦= ⋅ − θ − θ = − −

d) ( )0

4 0 0 0jz r e x jyθ +π−= ⋅ = − +

e) ( )0

5 0 0 02jz r e x jyθ + π= ⋅ = +

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P2. Consider z a complex variable with its complex conjugate jz x jy r e∗ − θ= − = ⋅ Prove the following relations are true, where z , 1z and 2z are complex numbers.

a) 2z z r∗⋅ = b) 2 jz ez∗

θ=

c) { }2 Rez z z∗+ = d) { }2 Imz z j z∗− =

e) ( )1 2 1 2z z z z∗ ∗ ∗+ = + f) ( )1 2 1 2az z az z∗ ∗ ∗=

g) 1 1

2 2

z zz z

∗ ∗

∗

⎛ ⎞=⎜ ⎟

⎝ ⎠

Solution a) 2j jz z r e r e r∗ θ − θ⋅ = ⋅ ⋅ ⋅ =

b) 2j

jj

z r e ez r e∗

θθ

− θ⋅

= =⋅

c) { }2 2 Rez z x jy x jy x z∗+ = + + − = =

d) { }2 2 Imz z x jy x jy jy j z∗− = + − + = =

e) ( ) ( ) ( )1 2 1 1 2 2 1 2 1 2 1 2z z x jy x jy x x j y y z z∗ ∗ ∗ ∗+ = + + + = + − + = +

f) ( ) ( ) ( )1 2 1 21 21 2 1 2 1 2 1 2 1 2

jj j j jaz z a re r e ar r e a re r e az z∗∗ ∗ ∗θ +θ−θ θ − θ − θ= ⋅ ⋅ = = ⋅ =

g) 1 1

2 2

z zz z

∗ ∗

∗

⎛ ⎞=⎜ ⎟

⎝ ⎠

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Continuous-time signals Continuous-time unit impulse

( )0

0 0, t

t, t

∞ =⎧δ = ⎨ ≠⎩

Continuous-time unit step

( )1 00 0, t

t, t

≥⎧σ = ⎨ <⎩

Ramp signal

( ) ( )0

0 0t , t

r t t t, t

≥⎧= = σ⎨ <⎩

Even signal ( ) ( ) ( )2e

x t x tx t

+ −=

( ) ( )e ex t x t− =

Odd signal ( ) ( ) ( )2o

x t x tx t

− −=

( ) ( )o ox t x t− = −

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Any given signal x(t) has an odd part and an even part ( ) ( ) ( )e ox t x t x t= + With

( ) ( ) ( )2e

x t x tx t

+ −= and ( ) ( ) ( )

2ox t x t

x t− −

=

• Energy of a continuous-time signal

( ) 2W x t dt

∞

−∞

= ∫

If the signals has the support [ ]1 2,t t , the energy is:

( )2

1

2t

tW x t dt= ∫

Power of the continuous-time signal (again support [ ]1 2,t t )

( )2

1

2

2 1 2 1

1 t

t

WP x t dtt t t t

= =− − ∫

For periodic signals the power is computed over one period!!!

( )0

2

0

1TP x t dt

T= ∫

Problems P3. Consider the signal ( )x t represented below.

Sketch the following signals: a) ( )2x t − e) ( )1x t− − i) ( ) ( )x t t⋅ δ b) ( )2x t + f) ( )2x t j) ( ) ( ) ( )1 2x t t t⋅ ⎡σ − − σ − ⎤⎣ ⎦ c) ( )x t− g) ( )/ 2x t d) ( )3x t− + h) ( )2 4x t − Solution. a) ( )2x t − . The “new origin” is 2t = : 2 0t − = . This is a delayed version of the signal

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b) ( )2x t + . The “new origin” is 2t = − : ( )2 2 0x t t+ ⇒ + =

c) ( )x t− . The signal is reversed or “mirrored” on the time axis

d) ( )3x t− + same as c) with shifting on the time axis. The origin is shifted on 3t =

e) ( 1)

( 1)x t

t− −− +

f) ( )2x t . The support is compressed; this is signal compression in time.

new signal original signal2 t t′= ; for 2 1t t′= = ; on the new time axis we have 1/ 2t = and so on.

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g) ( )/ 2x t . The support is dilated; this is signal dilation in time.

new signal original signal/ 2t t′= ; for / 2 1t t′= = ; on the new time axis we have 2t = and so on.

h) ( )2 4x t − . 2 4 0 2t t− = ⇒ = . The origin is now in 2t = (time compression + shifting)

i) ( ) ( )x t t⋅ δ

j) ( ) ( ) ( )1 2x t t t⋅ ⎡σ − − σ − ⎤⎣ ⎦

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Discrete-time signals Discrete-time unit impulse (Dirac impulse)

[ ] 1 00 0, n

n, n

=⎧δ = ⎨ ≠⎩

Discrete-time unit step

[ ] 1, 00, 0

nn

n≥⎧

σ = ⎨ <⎩

Any given signal x[n] has an odd part and an even part [ ] [ ] [ ]e ox n x n x n= + with

[ ] [ ] [ ]2e

x n x nx n

+ −= and [ ] [ ] [ ]

2ox n x n

x n− −

=

Even signal: [ ] [ ]e ex n x n− = ; Odd signal: [ ] [ ]o ox n x n− = −

• Energy of a discrete-time signal

2[ ]n

W x n∞

=−∞

= ∑

If the signals has the support { }1 1 1 2, ,...,N N N+ , the energy is:

[ ]2

1

2N

n NW x n

== ∑

Power of the discrete-time signal (again support { }1 1 1 2, ,...,N N N+ )

[ ]2

1

2

2 1 2 1

11 1

N

n N

WP x nN N N N =

= =− + − +

∑

For periodic signals the power is computed over one period!!!

[ ] 21

n NP x n

N ∈= ∑

Problems P4. Consider the signal x[n] sketched below.

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Sketch the following signals: a) [ ]2x n − d) [ ]x n− g) [ ] [ ]1 3x n n− ⋅δ − b) [ ]1x n + e) [ ]2x n− c) [ ]2x n f) [ ] [ ]1x n n⋅σ − Solution. a) [ ]2x n − . The new origin is 2; for 2 0n n= − =

b) [ ]1x n + . The new origin is 1; for 1 0n n= − + =

c) [ ]2x n . We have

new original2 n n′= . Number "n" integer => n' even values: ... -2; 0; 2; 4...

d) [ ]x n−

e) [ ] ( )2 2x n x n− = ⎡− − ⎤⎣ ⎦ . The signal is reversed and then shifted or vice versa.

f) [ ] [ ]1x n n⋅σ −

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P5–Homework. Draw the odd and even signals of the x[n] signal represented below:

P6–Homework. Consider the discrete-time signal x[n] represented below.

Sketch the following signals: a) [ ]x n− d) [ ] [ ] [ ]( )2 1 2 2x n n n+ σ + −σ −

b) [ ]1x n + e) [ ] [ ] [ ]( )2 2x n n n+ σ + − σ − c) [ ] [ ]4 2x n n⋅ δ −

P7. a) If the signal [ ]x n is odd, prove that: [ ] [ ]lim 0N

Nn n NS x n x n

∞

→∞=−∞ =−

= = =∑ ∑

b) If [ ]1x n is even and [ ]2x n is odd, show that their product [ ] [ ]1 2x n x n⋅ is odd

c) For a finite energy signal [ ]x n , with the even and odd parts [ ]ex n and [ ]ox n , prove that:

[ ] [ ] [ ]2 2 2e o

n n nx n x n x n

∞ ∞ ∞

=−∞ =−∞ =−∞

= +∑ ∑ ∑

d) Verify by direct computation the relation from c) for the sketched signal.