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Frequency ResponseFrequency Response

Lecture #11Chapter 10Chapter 10

BME 310 Biomedical Computing -J.Schesser

271

What Is this Course All About ?What Is this Course All About ?

• To Gain an Appreciation of the• To Gain an Appreciation of the Various Types of Signals and Systems T A l Th V i T f• To Analyze The Various Types of Systems

• To Learn the Skills and Tools needed to Perform These Analyses.

• To Understand How Computers Process Signals and Systems


272

Process Signals and Systems

Frequency Response of LTI SSystems

• Let’s review the Frequency Response forLet s review the Frequency Response for continuous-time systems

• First some definitions:• First, some definitions:– The unit impulse function

Th i S f i– The unit Step function


273

A Special Function – Unit Impulse Function

• The unit impulse function δ(t) also known as

δ(t)

The unit impulse function, δ(t), also known as the Dirac delta function, is defined as:δ(t) = 0 for t ≠ 0; δ(t)

0-100 -50 -25 -1 0 1 25 50 100

δ(t) = 0 for t ≠ 0;= undefined for t = 0

and has the following special property:

∫∞

=− )()()( fdtttf ττδ

∫∞

=∴

∫∞−=

1)(

)()()(

dtt

fdtttf

δ

ττδ


274

∫∞−∴ 1)( dttδ

Uses of Delta FunctionUses of Delta Function

• Modeling of electrical mechanical physicalModeling of electrical, mechanical, physical phenomenon:

point charge– point charge,– impulsive force,

point mass– point mass– point light


275

Unit Impulse Function Continued

• A consequence of the delta function is that itA consequence of the delta function is that it can be approximated by a narrow pulse as the width of the pulse approaches zero while thewidth of the pulse approaches zero while the area under the curve = 1

δ(t)

otherwise. 0;2/2/for /1)( lim0

=<<≈→

εt -εt εδε 1

10

1-1

0.5

.5-.5 -.05 .05


276

Unit Impulse Function Continued

τδ −∞

∫ )()( dtttf

εε

τδ

τδ

−

∞−∫

width and 1height of pulse a with )( eapproximat sLet'

)()(

t

dtttf

ε

εετ

≈

=+

∫1)(

have we, where2

dttf

t-τ

ε

εετ

→−

integral original theapproaches eapproximat the ,0 as integral thisoflimit the take weIf

2

τε

τεε

τε ε

ετ

ετε

→→

→→=→

+

−→∫

t0assince

,)(1)(1)( limlim0

2

20ffdttf


277

τε →→ t,0assince

Another Special Function – Unit Step Function

• The unit step function u(t) is defined as:The unit step function, u(t) is defined as:u(t) = 1 for t ≥ 0;

0 f 01

= 0 for t < 0.and is related to the delta function as follows:

t

∫= ∞−t dtu )()( ττδ∫


278

Integration of the Delta FunctionIntegration of the Delta Function

• δ(t) u(t)δ(t) u(t)• u(t) tu(t) 1st order• tu(t) 2nd order)(2 tut• tu(t) 2nd order

.)(!2 tut

.

.)(ttn• nth order)(! tut

nn


279

Signal Representations using the Unit Step Function

• x(t) = e-σ t cos(ωt)u(t) 1.5x(t) e cos(ωt)u(t)

-0.5

0

0.5

1

-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4

• x(t) = t u(t) – 2 (t-1)u(t-1) + (t-2) u(t-2)

-1.5

-1

x(t) t u(t) 2 (t-1)u(t-1) + (t-2) u(t-2)

2

3

4

-3

-2

-1

0

1

0 0.5 1 1.5 2 2.5 3


280

-4

Convolution of the Unit-Impulse RResponse

• As with Discrete-time system, we find that theAs with Discrete time system, we find that the Unit-Impulse Response of the a Continuous-time system, h(t), is key to determining the output of the system to any input:

∫ −=⊗= τττ dtxhtxthty )()()()()(

• Let’s apply the complex exponential (sinusoidal) signal, x(t)=Ae jφe jωt, for all t

τ

∫∫

∫−− ==

−=⊗=

ωφωττωφ

τ

ττττ

τττ

tjjjtjj eAedehdeAeh

dtxhtxthty

})({)(

)()()()()(

)(


281

∫∫ ==τ

φ

τ

φ ττττ jjjjj eAedehdeAeh })({)( )(

Frequency ResponseFrequency Response

• As in Discrete-time systems, the FrequencyAs in Discrete time systems, the Frequency Response of the LTI system, H(jω), is:

tjjj eAedehty ωφωτ ττ })({)( ∫ −

tj

jjj

dtethjH

eAedehty

ω

τ

φ

ω

ττ

)()(

})({)( =

∫

∫∞

−

tjj

j

eAejHty

dtethjH

ωφω

ω

)()(

)()(

=

= ∫∞−

• Again this is true only for sinusoidal signals!!!!


282

ExamplesExamples• Assume we have a system, H(j3π) = 2 - j2 with input x(t) = 10e j3ωt, then

33 10)22(10)3()( πππ tjtj ejejHty −==

34 )10)(22(

10)22(10)3()(

π

ππ

π

tjjee

ejejHty−

=

• Assume h(t) = 2e-2tu(t), calculate the Frequency Response:

)4

3(220

ππ jtje

−=

2 2

0

(2 ) (2 ) (2 )0

( ) 2 ( ) 2

2 2

t j t t j t

j j j

H j e u t e dt e dtω ωω∞ ∞

− − − −

−∞

= =∫ ∫(2 ) (2 ) (2 )0

0

2

2 2| [ ](2 ) (2 )

2 2[0 1 2[ 1]

j t j j

jj

e e ej j

ee e

ω ω ω

ωω

ω ω− + ∞ − + ∞ − −

− ∞− ∞ − ∞

= = −− + − +

−= − = =


283

[ 1](2 ) (2 ) 2

e ej j jω ω ω− + − + +

Plotting the Frequency ResponsePlotting the Frequency Response• Let’s plot the magnitude

d h f th2)( ω =jHand phase of the

Frequency Response vs. Frequency 2

2)(

2)(

22 ωω

ω

+=

+

jH

jj

q y• What kind of filter is this? )

2(tan}2{0)(

21 ωωω

ω−−=+∠−=∠

+

jjHPhase Magnitude

1

0.5

1

1.5

020 10 0 10 20-1.5

-1

-0.5

0-20 -10 0 10 20


284

-20 -10 0 10 201.5

Frequency Response to a Cosine InputFrequency Response to a Cosine Input

• If the input to an LTI system is p yx(t)=A cos(ωt+ ),

• and if the impulse response is real-valued, then the output will bebe

y(t)=AM cos(ωt + +φ)• where the frequency response is

H(jω) = M e jφ• To show this:

(t) A ( t+ ) ½A{ j jωt + j jωt }x(t) =A cos(ωt+ ) = ½A{e j e jωt + e -j e -jωt }• Using superposition

y(t)= ½A{H(jω) e j e jωt + H(-jω) e -j e -jωt }


285

y( ) { (j ) ( j ) }

Frequency Response to a Cosine IInput

• If the impulse response is real-valued thenIf the impulse response is real-valued theny(t)= ½A{H(jω) e j e jωt + H*(jω) e -j e -jωt }( ) ½A{ j j j t ( j )* j j t }y(t)= ½A{Me jψe j e jωt + (Me jψ)* e -j e -jωt }y(t)= ½A{Me jψe j e jωt + Me -jψe -j e -jωt }y(t)= ½A{Me j(ωt+ +ψ) + Me –j(ωt+ +ψ) }

y(t)=AM cos(ωt + +ψ)y(t) AM cos(ωt + +ψ)


286

Proof of Conjugate SymmetryProof of Conjugate Symmetryωωω tjtj dtethdtethjH

∞+

∞− =

⎞⎜⎜⎛

= ∫∫ **

)()()(* ω

j

dtethdtethjH

∞

∞−∞−

=⎠

⎜⎜⎝

=

∫

∫∫ )()()(

)( ω ωtj jHdteth∞−

−− −== ∫*

)()( )(

ψω jMejHththth =

=

*

)( Ifvalued-real is )( isOnly )()(

ψψωω jj MeMejHjH

j−===− *)()(*)(Then

)(


287

ExamplesExamples• Assume we have a system, given by the following H(jω)

and the input x(t) = 3 cos(40πt – π) is applied. Determine p ( ) ( ) ppthe output signal.

4040)(

ωππω+

=j

jH

40

21

40)40(40)40(

22 ππππ =+

=jH

)40cos()1(3)(

4)1(tan)

4040(tan)40( 11

π

ππππ −=−=−=∠ −−

tt

jH

)4

540cos(1213.2

)4

40cos()2

(3)(

ππ

ππ

−=

−−=

t

tty


288

4

An ExampleAn Example

• An LTI system has an impulse response ofAn LTI system has an impulse response of h(t) = δ (t) - 200πe-200πtu(t)

h f ll i i l i li d• The following signal is applied:x(t)=10+20δ (t - 0.1)+40cos(200π t+0.3π) for all t• The input has 3 parts: a constant, an impulse

and a cosine wave. We will take each part pseparately and use the easiest method to find the solution.


289

An ExampleAn Example

• Let’s first find the frequency response of the systemLet s first find the frequency response of the system from the impulse response:

ωπ tjt∞

∫ 200

πδ

πδω

ωπω

ωπ

dtetuedtet

dtetuetjH

tjttj

tjt

−=

−=

∞−−

∞−

∞−

−−

∫∫

∫

)(200)(

)](200)([)(

200

200

ωπππ ωπωπ e

jdte tjtj

++=−= ∞+−

∞+−

∞−∞−

∫

∫∫

|200

20012001

)()(

0)200(

0

)200(

ωπω

ωππ

jj

j

j

+=

+−=

2002002001

0


290

An ExampleAn Example• Now let’s take the first (constant, ω = 0) part and the third (cosine) part and

evaluate the solution using the frequency response: ihffih

0200

)(

10:input theofpart first The

ωπωω+

=

jj

jjH

)(

)0.3t40cos(200 :input theofpart thirdThe

0100200

010)0(10

ωω

πππ

+

=+

=

jjH

jjjH

1

])200(0.3tcos[200)200(40)0.3t40cos(200200

)(

πππππππ

ωπω

∠

∠+++

+=

jHjHj

jjH

]0 55tcos[20040]250 3tcos[200140)0 3t40cos(200

421

42

21

1200200200)200(

πππππππ

ππππ

ππ

++++

∠=∠

∠=

+=

+=

jj

jjjH


291

]0.55tcos[2002

]25.0.3tcos[2002

40)0.3t40cos(200 πππππππ +=+++

An ExampleAn Example• Now for the second part of the input (the impulse p p ( p

function), we will apply the impulse response:

)1.0(20:inputtheofpart second The −tδ

)1.0(4000)1.0(20)1.0(20)]1.0(200)1.0([20)1.0(20)1.0(20

)1.0(200

)1.0(200

−−−−

−−−=−−−−

−−

tuetttuettht

t

t

π

π

πδδ

πδδ

• The Complete solution by superposition is:( ) 0y t =

200 ( 0.1)

( ) 0 20 ( 0.1) 4000 ( 0.1)

40 cos(200 0 55 )

t

y tt e u t

t

πδ π

π π

− −

=

+ − − −

+ +


292

cos(200 0.55 )2

tπ π+ +

Frequency Response to a Sum of C i I tCosine Inputs

• We can extend this to the case when x(t)=Σ Akcos(ωkt+ k)

• By superposition ( ) AM ( + + )yk(t)=AMk cos(ωkt + k +φk)

• where the frequency response isHk(jωk) = Mk e jφkk(jωk) k e

• Then:y(t) =Σ AMk cos(ωkt + k +φk)

• This model can also be used when analyzing periodic input signals since a Fourier Series can be generated which has the similar form: x(t)=Σ Akcos(kωot+ k) where ωo is the


293

fundamental frequency.

HomeworkHomework• Exercises:

– 10.1-10.3• Problems:

– 10.1, 10.2, – 10.4 Use Matlab to plot the frequency response and submit

your codeyour code– Using unit step functions, construct a single pulse of

magnitude 10 starting at t=5 and ending at t=10.i h 2 l h h d i f i d– Repeat with 2 pulses where the second is of magnitude 5

starting at t=15 and ending at t=25.


294

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