# sinusoidal signals - bo sinusoidal signals? † physical reasons: - harmonic oscillators generate...     Post on 24-Mar-2018

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• SINUSOIDAL SIGNALS

• SINUSOIDAL SIGNALS

x(t) = cos(2ft + )

= cos(t + ) (continuous time)

x[n] = cos(2fn + )

= cos(n + ) (discrete time)

f : frequency (s1 (Hz)) : angular frequency (radians/s) : phase (radians)

2

• Sinusoidal signals

0 T/2 T 3T/2 2T

0

A

A

A cos( t)A sin( t)

xc(t) = A cos(2ft) xs(t) = A sin(2ft)= A cos(2ft /2)

- amplitude A -amplitude A

- period T = 1/f -period T = 1/f- phase: 0 -phase: /2

3

• WHY SINUSOIDAL SIGNALS?

Physical reasons:- harmonic oscillators generate sinusoids, e.g., vibrating

structures- waves consist of sinusoidals, e.g., acoustic waves or

electromagnetic waves used in wireless transmission

Psychophysical reason:- speech consists of superposition of sinusoids- human ear detects frequencies- human eye senses light of various frequencies

Mathematical (and physical) reason:- Linear systems, both physical systems and man-made

filters, affect a signal frequency by frequency (hence low-pass, high-pass etc filters)

4

• EXAMPLE:

TRANSMISSION OF A LOW-FREQUENCY SIGNAL USING

- A POSSIBLE (CONVENTIONAL) METHOD:

AMPLITUDE MODULATION (AM)

Example: low-frequency signal

v(t) = 5 + 2 cos(2ft), f = 20 Hz

High-frequency carrier wave

vc(t) = cos(2fct), fc = 200 Hz

Amplitude modulation (AM) of carrier (electromagnetic) wave:

x(t) = v(t) cos(2fct)

5

• 8

6

4

2

0

2

4

6

8

v

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.18

6

4

2

0

2

4

6

8

x

t

Top: v(t) (dashed) and vc(t) = cos(2fct).Bottom: transmitted signal x(t) = v(t) cos(2fct).

6

• Frequency contents of transmitted signal

x(t) = v(t) cos(2fct)

= (5 + 2 cos(2ft)) cos(2fct)

= 5 cos(2fct) + 2 cos(2ft) cos(2fct)

Trigonometric identity:

cos cos =12

cos( ) + 12

cos( + )

2 cos(2ft) cos(2fct) = cos (2(fc f)t)+cos (2(fc + f)t)

7

• x(t) = 5 cos(2fct) + cos (2(fc f)t) + cos (2(fc + f)t)

Spectrumof v(t)

Spectrumof vc(t)

0 20 200

5

2

1

-

Frequency

Spectrum of x(t)

-

Frequency0 200

5

220

1

180

1

8

• We see that a low-frequency signal in frequency range

0 fs fmax (baseband signal)can be transmitted as a signal in the frequency range

fc fmax f fc fmax (RF (radio frequency) signal).

Another analog modulation technique is frequency modulation

(FM)

9

• Representation of sinusoidal signal

A. Amplitude and phase

B. Sine and cosine components

C. Complex exponentials

10

• A. Amplitude and phase

x(t) = A cos(2ft + )

or

x(t) = A cos(t + )

where = 2f is the angular frequency (radians/second)

The phase describes translation in time compared to cos(t):

x(t) = A cos(t + ) = A cos((t + /))

11

• B. Sine and cosine components

x(t) = a cos(t) + b sin(t)

Follows from trigonometric identity:

cos( + ) = cos() cos() sin() sin()

x(t) = A cos(t + )

= A cos() cos(t)A sin() sin(t)

Hence:

a = A cos(), b = A sin()and, conversely,

A =

a2 + b2, = arctan(b/a)12

• EXAMPLE

2 0 2 4 6 8 10 12

1

0

1

x(t)

Signal x(t) = cos(t + )Period: T = 10 (s)Frequency: f = 1/T = 0.1 (s1 =Hz (periods/second))Angular frequency: = 2f = 0.628 (radians/s)

13

• A. Representation using amplitude and phase

Amplitude: A = 1From figure we see that x(t) = cos((t 2)).But x(t) = cos(t + ) = cos((t + /)Phase: = 2 = 0.4 radians or 0.4 180/ = 72

14

• B. Sine and cosine components

x(t) = a cos(t) + b sin(t) wherea = cos() = cos(0.4) = 0.3090b = sin() = sin(0.4) = 0.9511

2 0 2 4 6 8 10 12

1

0

1

x(t)

2 0 2 4 6 8 10 12

1

0

1

Cosine componentSine component

x(t) = 0.3090 cos(t) + 0.9511 sin(t)

15

• C. Representation using complex exponentials

Background. A common operation is to combine sinusoidal

components with different phases:

x(t) = A1 cos(t + 1) + A2 cos(t + 2)

One way: first write

A1 cos(t + 1) = a1 cos(t) + b1 sin(t)

A2 cos(t + 2) = a2 cos(t) + b2 sin(t)

Then

x(t) = (a1 + a2) cos(t) + (b1 + b2) sin(t)

16

• Streamlining the above procedure

We have:

cos(t + ) = cos() cos(t) sin() sin(t)and

sin(t + ) = sin() cos(t) + cos() sin(t)

Instead of working with the two signals cos(t) and sin(t) itis convenient to working with the single complex-valued signal

g(t) = cos(t) + j sin(t)

where j =1 (imaginary unit number)

17

• -

6

Re

Im

*

g(t) = cos(t) + j sin(t)j sin(t)

cos(t)t

1

Then:

g(t + ) = cos(t + ) + j sin(t + )

= cos() cos(t) sin() sin(t)+j

(sin() cos(t) + cos() sin(t)

)

=(cos() + j sin()

)(cos(t) + j sin(t)

)

= g()g(t)

18

• In terms of the complex-valued signal, phase shift with angle

is equivalent to multiplication by the complex number g():

g(t + ) = g()g(t) (A)

cos(t + ) = Re (g()g(t))

sin(t + ) = Im (g()g(t))

We will see that using the multiplication (A) to describe

phase shifts results in significant simplification of formulae

and calculations

19

• Properties of the complex function g() = cos() + j sin():

- g(0) = 1

- g(1 + 2) = g(1)g(2)

- dg()d = sin() + j cos() = jg()The only function which satisfies these relations is

g() = ej

e = 2.71828182845904523536 . . . (Eulers number)

20

• Hence we have:

Eulers relation

ej = cos() + j sin()

e = 2.71828182845904523536 . . . (Eulers number)

As ej = cos() j sin() we have

cos() =12

(ej + ej

)

sin() =12j

(ej ej)

21

• It follows that

x(t) = a cos(t) + b sin(t)

=(

a

2+

b

2j

)ejt +

(a

2 b

2j

)ejt

=12

(a jb) ejt + 12

(a + jb) ejt

or

x(t) = c ejt + cejt

where c = 12 (a jb) and c = 12 (a + jb) (complex conjugate)Thus the cosine and sine components are parameterized as the

real and imaginary components of the complex number c

22

• PREVIOUS EXAMPLE:

The signal

x(t) = cos(t 0.4)= a cos(t) + b sin(t)

= 0.3090 cos(t) + 0.9511 sin(t)

can be represented using complex exponentials as

x(t) = c ejt + c ejt

wherec =

12

(a jb) = 0.1545 j0.4755

c =12

(a + jb) = 0.1545 + j0.4755

Hence:

x(t) = (0.1545 j0.4755) ejt + (0.1545 + j0.4755) ejt

23

• SUMMARY

Let us summarize the results so far. We have the following

alternative representations of a sinusoidal signal:

Amplitude and phase:

x(t) = A cos(t + )

Sine and cosine components:

x(t) = a cos(t) + b sin(t)

where

a = A cos(), b = A sin()or

A =

a2 + b2, = arctan(b/a)24

• Complex exponentials:

x(t) = c ejt + c ejt

where

c =12

(a jb) , c = 12

(a + jb)

or

a = 2Re(c), b = 2Im(c)

25

• EXAMPLE - Signal transmission with reflected component

Transmitted signal: x(t) = cos(t)

Received signal: y(t) = x(t) + rx(t )PROBLEM: Determine amplitude and phase of y(t)

We have

y(t) = cos(t) + r cos((t ))The problem can be solved in a straightforward way by

introducing complex exponential:

x(t) =12

(ejt + ejt

)

26

• Then:

y(t) = cos(t) + r cos((t ))=

12

(ejt + ejt

)+ r

12

(ej(t) + ej(t)

)

=12

(1 + rej

)ejt +

12

(1 + rej

)ejt

Here,

12

(1 + rej

)=

12

(1 + r cos() jr sin()

)

and

12

(1 + rej

)=

12

(1 + r cos() + jr sin()

)

27

• Then, defining

a = 1 + r cos(), b = r sin()

we have (cf. above)

y(t) =12

(a jb) ejt + 12

(a + jb) ejt

= a cos(t) + b sin(t)

= Ay cos(t + y)

where

Ay =

a2 + b2, y = arctan(b/a)

28

• SOME SIMPLIFICATIONS

As

a cos(t) + b sin(t) =12

(a jb) ejt + 12

(a + jb) ejt

it follows that it is sufficient to consider the complex-valued

signal12

(a jb) ejt

Here, the complex number

c =12

(a jb)

determines the signals amplitude and phase.

29

• To see how this works, consider a linear dynamical system with

input x(t) and output y(t)

-- G y(t)x(t)

For the sinusoidal input

x(t) = cos(t)

the output takes the form

y(t) = A cos(t + ) = a cos(t) + b sin(t)

and we want to determine the amplitude A and phase .

30

• As the phase shift introduces a combination of cosines and

sines, the problem can be simplified by embedding the input

signal into a larger class of signals involving both a cosine and

a sine component.

It turns out that a convenient way to do this is to consider the

complex input signal

xc(t) = cos(t) + j sin(t)

= ejt

The output then also has real and imaginary components:

yc(t) = A cos(t + ) + jA sin(t + )

= Aej(t+)

= Aejejt

31

• or

yc(

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