electrical engineering lecture 1

EE102 Lecture Notes

Flavio Lorenzelli

UCLAElectrical Engineering Department

Winter 2015

Flavio Lorenzelli EE102 Lecture Notes

DefinitionsSignals: Functions that describe evolution of physical quantity intime (e.g., voltage, position, sound, stock market, population,etc.)I Continuous-timeI Discrete-timeI Digital

Systems: Component that establishes relationship between inputs(stimuli, excitations) and outputs (responses)

y(t) = T [x(t)], x(t) 2X

I StaticI Dynamic (ODEs)

Models: Descriptions of systemsI VerbalI GraphicalI Mathematical


Example of Dynamic System

+

x(t)

t0 R i(t)

C

+

y(t)

I y(t) = x(t) Ri(t), i(t) = C ddt

y(t), = 1/RC

Idy

dt+ y = x

I Goal: derive an input-output expression:

y(t) = T [x(t)]


Example Solution

I Method 1: First solve the homogeneous equation, x = 0

dy

dt+ y = 0Zdy

y=

Zdt + K1, (separation of variables)

ln y = t + K1y = Ket , K = eK1


Example Solution (Cont.)

Then vary the constant

y(t) = K (t)et

dy

dt=

dK

dtet Ket| {z }

y(t)

(chain rule)

=dK

dtet y

x =dK

dtet

K (t) K (t0) = Z tt0

ex() d, t > t0

) y(t) =K (t0) +

Z tt0

ex() d

et


Example Solution (Cont.)

I Method 2: Integrating Factor

etdy

dt+ y

= etx

d

dt

ety

= etx (chain rule)

ety(t) et0y(t0) = Z tt0

ex() d

) y(t) =y(t0)e

t0 +

Z tt0

ex() d

et


Properties of Systems

I Linearity (Superposition Principle)

x(t) = k1x1(t) + k2x2(t)

) y(t) = T [x(t)] = k1T [x1(t)] + k2T [x2(t)]8k1, k2, 8x1(t), x2(t) 2X

I Time invariance

z(t) = x(t ) and y(t) = T [x(t)]) T [z(t)] = y(t ), 8t, , 8x(t) 2X

I Causality

y(t) = T [x(t)], y(t0) is a function of x(t) for t t0, 8t0


Proving and Disproving a Property

I In order to prove a property:I Use the definition

I In order to disprove a property:I Use the definitionI Or find a single counterexample


Examples Linearity

I

y(t) =

Z 10

tx() d

Iy(t) = a+ x(t)

Iy(t) = 5 + x2(t)


Examples Time Invariance

I

y(t) =

Z 11

h(t ) x() d

y(t ) =Z 11

h(t ) x() d

T [z(t)] =

Z 11

h(t ) z() d

=

Z 11

h(t ) x( ) d

=

Z 11

h(t u)x(u) du (change of variables)= y(t )


Examples Time Invariance (Cont.)

I

y(t) = t

Z 10

x() d

y(t ) = (t )Z 10

x() d

T [z(t)] = t

Z 10

z() d

= t

Z 10

x( ) d6= y(t )

Iy(t) = t x(t)

Find a counterexample (e.g., x(t) = u(t), = 1)


Examples Causality

I

y(t) =

Z t1

x() d

I

y(t) =

Z 10

e(t)x() d

Iy(t) = x(t) + 4

I

y(t) =

Z t0h(t)x() d, t 0, with x(t) = h(t) = 0, t < 0


Diracs DeltaI Define

pw (t) =1

wrect

tw

,

Z 11

pw (t) dt = 1, 8w

I ConsiderZ 11

f (t)pw (t) dt =1

w

Z w/2w/2

f (t) dt f (0)

I Approximation gets better as w ! 0I Define

d(t) = limw!0 pw (t)

In the sense that Z 11

f (t) d(t) dt = f (0)

if f (t) is a smooth function around t = 0


Properties of the Dirac DeltaI Sifting property Z 1

1f (t)d(t ) dt = f ()

I Having the integral definition in mind, it is common to write

f (t)d(t ) = f ()d(t )I Z 1

1d(t) dt = 1

Id(t) = d(t)

I Z t1

d() d = u(t)


The Dirac Delta as Basis Function

Consider a function x(t) defined in t 2 (a, a)

x(t) n1Xk=n

x(kw)pw (t kw)w !Z aa

x() d(t ) d

for n!1 and w ! 0 while nw = a

This is another proof of the sifting property


The Dirac Delta (Cont.)

x(kT )w pw (t kw)

t

x(t)

nw nwkw0


Unit Step Function

Define

uw (t) =

Z t1

pw () d

u(t) = limw!0 uw (t) =

(0, t < 0

1, t > 0

Note that Z 1a

f (t) dt =

Z 11

f (t) u(t a) dtZ b1

f (t) dt =

Z 11

f (t) u(b t) dt


Properties of the Unit Step Function

Id

dtu(t) = d(t)

I Z t1

u() d = t u(t)

Iu(t) = 1 u(t)

Iu(t a) u(t b) = u(t a)u(b t)

I

ut +

a

2

u

t a

2

=: rect

ta


Examples

Solve

y(t) =

Z 11

d(t )e2 d

I Consider the -axis: the delta sits at = t

I When t < 1 the integrand is zero ) y(t) = 0I When t > 1 use the sifting property:Z 1

1d(t )e2 d = e2t

Z 11

d(t ) d| {z }=1

= e2t

I Therefore: y(t) = e2tu(t + 1)


Examples (Cont.)

Alternatively, consider the use of the unit step function:

y(t) =

Z 11

d(t )e2 d

=

Z 11

d(t )e2u( + 1) d= e2tu(t + 1)

by using the sifting property


Examples (Cont.)

Solve

y(t) =

Z ba

f () d( t) d

=

Z 11

f () d( t) [u( a) u( b)] d= f (t) [u(t a) u(t b)]

Again using the sifting property


Examples (Cont.)

Solve

y(t) =

Z 11

hd(t ) e(t)u(t )

ie2u() dZ 1

1d(t )e2u() d = e2tu(t)Z 1

1e(t)e2u(t )u() d =

Z t0e(t)2 d, t > 0

= etZ t0e d = et

et0

= et1 et = et e2t u(t)

) y(t) =2e2t et u(t)


Examples (Cont.)

Compute

y(t) =

Z 11

u(t ) u( ) d

=

8>:Z t

d, t >

0, t <

= (t ) u(t )


Impulse Response

I Consider a linear system S

I When the input x(t) = d(t ), the output y(t) = h(t; ) iscalled the impulse response Note that there is a dierentresponse for each value of

I If the system is linear and time-invariant (LTI) then a singlefunction is required:

h(t; ) = h(t ; 0) =: h(t )

I If the system is LC, h(t; ) = 0 for t < and if the system isLTIC

h(t) = 0, t < 0


ExamplesI LTIC system:

y(t) =

Z t1

x() d

h(t; ) =

Z t1

d( ) d = u(t )) h(t) = u(t)

I LTVC system:

y(t) =

Z 11

t u(t ) x() d

h(t; ) =

Z 11

t u(t ) d( ) d= t u(t )


Examples (Cont.)

I LTINC system:

y(t) = x(t)Z 1t

2 e(t) x() d

h(t; ) = d(t )Z 1t

2 e(t) d( ) d

= d(t )Z 11

2 e(t) d( ) u( t) d= d(t ) 2 et u((t ))

) h(t) = d(t) 2 et u(t)

Note that the response to x(t) = Ket is y(t) = 0(!)


Examples (Cont.)

I LTIC system (RC circuit):

y(t) =

Z t0

e(t)x() d t0 = 0, x(t) = x(t)u(t)

h(t) =

Z t0

e(t)d() d

= et u(t)


Examples (Cont.)

I LTVC system:

y(t) =

Z t1

( + 1)2 x() d

h(t; ) =

Z t1

( + 1)2 d( ) d

=

Z 11

( + 1)2 u(t ) d( ) d= ( + 1)2 u(t )


Examples (Cont.)

I LTIC system:

d

dty(t) + y(t) =

d

dtx(t), x(t) = x(t) u(t), x(0) = 0, y(0) = 0

d

dt

ety(t)

= et

d

dtx(t)

y(t) =he(t)x()

it0Z t0e(t)x() d

= x(t) etx(0)| {z }=0

Z t0 e(t)x() d

h(t; ) = d(t ) e(t) u(t ) t > 0, > 0


Superposition IntegralLet S be a linear system and remember that

x(t) n1Xk=n

x(kw)pw (t kw)w , a < t < a

Compute the response

y(t) = T [x(t)] n1Xk=n

x(kw)T [pw (t kw)]w

As w ! 0 and n!1 with nw = a

x(t) =

Z aa

x()d(t ) d

and

y(t) =

Z aa

x()h(t; ) d


Superposition Integral (Cont.)

In general, for a linear system

y(t) =

Z 11

x()h(t; ) d

If the system is LTI (convolution integral):

y(t) =

Z 11

x()h(t ) d =: (x h) (t)

If the system is LTIC, h(t) = 0 for t < 0, therefore

y(t) =

Z t1

x()h(t ) d


Examples

Compute the output of an LTIC system with impulse responseh(t) = etu(t) when x(t) = e|t| is its input

y(t) =

Z t1

e(t)e| | d

If t < 0:

y(t) =

Z t1

e(t)e d

= etZ t1

e2 d

= ete2

2

t1

=et

2


Examples (Cont.)

If t > 0:

y(t) = etZ 01

e2 d + etZ t0

d

= ete2

2

01

+ ett

= et1

2+ t

In conclusion

y(t) =1

2etu(t) + et

1

2+ t

u(t)


Examples (Cont.)

Compute the impulse response of the linear system described by thefollowing input-output relationship

y(t) =

Z t4e(t)x() d

where x(t) = x(t)u(t 4). Write the output as

y(t) =

Z 11

he(t)u(t )

ix() d

Thereforeh(t; ) = e(t)u(t )

and the system is LTVC


Graphical Computation of the Convolution Integral

y(t) =

Z 11

h(t )x() d = (h x) (t)

h(t) = u(t) u(t 1)x(t) = t [u(t) u(t 1)] (t 2) [u(t 1) u(t 2)]

0 < t < 1

y(t) =t2

2

x()

2t 1

h(t )

t


Graphical Computation of the Convolution Integral

y(t) =

Z 11

h(t )x() d = (h x) (t)

h(t) = (t 1) [u(t) u(t 1)]x(t) = u(t) u(t 2)

2 < t < 3

y(t) =(t 3)2

2

x()

2t 1

h(t )

t


Properties of the Convolution Operator

I Associative: (f g) h = f (g h)

I Commutative: f g = g f

I Distributive: f (g + h) = f g + f h

I Unity: f d = f


Cascades of LTI Systems

x(t)h1(t)

y(t)h2(t)

z(t)

h1,2(t)

z(t) = (h2 y) (t)= (h2 (h1 x)) (t)= ((h2 h1) x) (t) = (h1,2 x) (t) associativity= ((h1 h2) x) (t) = (h2,1 x) (t) commutativity

Only true for LTI systems


Examples

S1 :y(t) =

Z t0(t )x() d, t > 0

S2 :z(t) =

Z t0y() d, t > 0

h1(t) = t u(t)

h2(t) = u(t)

h1,2(t) =

Z 11

u(t ) u() d =Z t0 d =

t2

2u(t)

h2,1(t) =

Z 11

(t ) u(t ) u() d =Z t0(t ) d = t

2

2u(t)


Examples (Cont.)

S1 :y(t) =

Z t0x() d

S2 :z(t) =

Z t0y() d

h1(t; ) = u(t )h2(t; ) = u(t )

h1,2(t; ) =

Z 11

u(t )u( ) d =Z t d = (t )u(t )

h2,1(t; ) =

Z 11

u(t )u( ) d =Z t d =

t2 22

u(t )


Step Response

g(t) = T [u(t)]

= (h u) (t)=

Z 11

u(t )h() d

=

Z t1

h() d

) ddt

g(t) = h(t)


Example

Lete(t+1)u(t + 1) = T [u(t + 1)]

be the response of an LTI system

Thenetu(t) = T [u(t)] = g(t)

The impulse response of the system is found by computing

h(t) = T [d(t)] =d

dt

etu(t)

= etu(t) + etd(t)= etu(t) + e0|{z}

=1

d(t)

= etu(t) + d(t)


electrical engineering lecture 1

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