TIME DOMAIN ANALYSIS

TIME DOMAIN ANALYSISEmail: hasansaeedcontrol@gmail.com

URL: http://shasansaeed.yolasite.com/

1SYED HASAN SAEED

1. AUTOMATIC CONTROL SYSTEM KUO & GOLNARAGHI

2. CONTROL SYSTEM ANAND KUMAR

3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED

SYED HASAN SAEED 2

DEFINITIONS

TIME RESPONSE: The time response of a system is the output (response) which is function of the time, when input (excitation) is applied.

Time response of a control system consists of two parts

1. Transient Response 2. Steady State Response

Mathematically,

Where, = transient response

= steady state response

SYED HASAN SAEED 3

)()()( tctctc sst

)(tcss

TRANSIENT RESPONSE: The transient response is thepart of response which goes to zero as timeincreases. Mathematically

The transient response may be exponential oroscillatory in nature.

STEADY STATE: The steady state response is the part ofthe total response after transient has died.

STEADY STATE ERROR: If the steady state response ofthe output does not match with the input then thesystem has steady state error, denoted by .

SYED HASAN SAEED 4

tcLimit tt

TEST SIGNALS FOR TIME RESPONSE:

For analysis of time response of a control system,

following input signals are used

1. STEP FUNCTION:

Consider an independent voltage source in series witha switch ‘s’. When switch open the voltage atterminal 1-2 is zero.

SYED HASAN SAEED 5

Mathematically,

When the switch is closed at t=0

Combining above two equations

A unit step function is denoted by u(t) and defined as

SYED HASAN SAEED 6

0)( tv 0 t

Ktv )( t0

0)( 0 t

Laplace transform:

£f(t)=

2. RAMP FUNCTION:

Ramp function starts from origin and increases ordecreases linearly with time. Let r(t) be the rampfunction then,

r(t)=0 ; t<0

=Kt ; t>0

SYED HASAN SAEED 7

edtedtetu

ststst 1

LAPLACE TRANSFORM:

£r(t)

For unit ramp K=1

SYED HASAN SAEED 8

KdtKtedtetr stst

3. PARABOLIC FUNCTION:

The value of r(t) is zero for t<0 and is quadratic functionof time for t>0. The parabolic function represents asignal that is one order faster than the ramp function.

The parabolic function is defined as

For unit parabolic function K=1

SYED HASAN SAEED 9

LAPLACE TRANSFORM:

£r(t)

SYED HASAN SAEED 10

Ktdtetr stst

IMPULSE RESPONSE: Consider the following fig.

The first pulse has a width T and height 1/T, area of thepulse will be 1. If we halve the duration and doublethe amplitude we get second pulse. The area underthe second pulse is also unity.

SYED HASAN SAEED 11

We can say that as the duration of the pulseapproaches zero, the amplitude approaches infinitybut area of the pulse is unity.

The pulse for which the duration tends to zero andamplitude tends to infinity is called impulse. Impulsefunction also known as delta function.Mathematically

δ(t)= 0 ; t ≠ 0

=∞ ; t = 0

Thus the impulse function has zero value

everywhere except at t=0, where the amplitude

is infinite.

SYED HASAN SAEED 12

An impulse function is the derivative of a step function

δ(t) = u(t)

£δ(t) = £

SYED HASAN SAEED 13

INPUT r(t) SYMBOL R(S)

UNIT STEP U(t) 1/s

UNIT RAMP r(t) 1/s2

UNIT PARABOLIC - 1/s3

UNIT IMPULSE δ(t) 1

THANK YOU FOR

ATTENTION

SYED HASAN SAEED 14

TIME RESPONSE

FIRST ORDER SYSTEM

Email : hasansaeedcontrol@gmail.com

1SYED HASAN SAEED

1. AUTOMATIC CONTROL SYSTEM KUO &GOLNARAGHI

SYED HASAN SAEED 2

RESPONSE OF FIRST ORDER SYSTEM WITH UNIT STEPINPUT:

For first order system

SYED HASAN SAEED 3

Input is unit step

After partial fraction

Take inverse Laplace

Where ‘T’ is known as ‘time constant’ and defined asthe time required for the signal to attain 63.2% offinal or steady state value.

Time constant indicates how fast the system reachesthe final value.

Smaller the time constant, faster is the systemresponse.

SYED HASAN SAEED 4

632.011)(

When t=T

RESPONSE OF FIRST ORDER SYSTEM WITH UNIT RAMPFUNCTION:

SYED HASAN SAEED 5

Input is unit Ramp

We know that

Take inverse Laplace, we get

The steady state error is equal to ‘T’, where ‘T’ is thetime constant of the system.

For smaller time constant steady state error will besmall and speed of the response will increase.

SYED HASAN SAEED 6

TTeTLimit

TeTttte

tctrte

TeTttc

)()()(

Error signal

Steady state error

RESPONSE OF THE FIRST ORDER SYSTEM WITH UNIT IMPULSE FUNCTION:

Input is unit impulse function R(s)=1

SYED HASAN SAEED 7

Inverse Laplace transform

SYED HASAN SAEED 8

SYED HASAN SAEED 9

ssRFor unit Ramp Input

For Unit Step Input

For Unit Impulse Input

TtTeTttc /)(

Ttetc /1)(

tc /1)(

It is clear that, unit step input is the derivative of unit ramp input and unit impulse input is the derivative of unit step input. This is the property of LTI system.

Compare all three responses:

THANK YOU

SYED HASAN SAEED 10

TIME RESPONSE

SECOND ORDER SYSTEM

1SYED HASAN SAEED

REFERENCE BOOKS:

1. AUTOMATIC CONTROL SYSTEM KUO & GOLNARAGHI

SYED HASAN SAEED 2

SYED HASAN SAEED 3

Block diagram of second order system is shown in fig.

R(s) C(s)_

sRsssR

For unit step input

SYED HASAN SAEED 4

Replace by )1()( 222 nns

Break the equation by partial fraction and put )1( 222 nd

)3()()(

ss dndn

)2()1()(

SYED HASAN SAEED 5

22)( dns Multiply equation (3) by and put

Equation (1) can be written as

SYED HASAN SAEED 6

Laplace Inverse of equation (4)

)5(sin.cos.1)(

tetetc d

21 ndPut

SYED HASAN SAEED 7

sincos.11

cos1)(

)sin(1

SYED HASAN SAEED 8

tan)1(sin1

Put the values of d &

tan)1(sin1

)()()(

tctrte

Error signal for the system

The steady state value of c(t)

tcLimitet

Therefore at steady state there is no error betweeninput and output.

= natural frequency of oscillation or undampednatural frequency.

= damped frequency of oscillation.

= damping factor or actual damping ordamping coefficient.

For equation (A) two poles (for ) are

SYED HASAN SAEED 9

Depending upon the value of , there are four cases

UNDERDAMPED ( ): When the system has two complex conjugate poles.

SYED HASAN SAEED 10

From equation (6):

Time constant is

Response having damped oscillation with overshoot andunder shoot. This response is known as under-dampedresponse.

SYED HASAN SAEED 11

UNDAMPED ( ): when the system has two imaginary poles.

SYED HASAN SAEED 12

From equation (6)

Thus at the system will oscillate.

The damped frequency always less than the undampedfrequency ( ) because of . The response is shown infig.

SYED HASAN SAEED 13

cos1)(

)2/sin(1)(

SYED HASAN SAEED 14

CRITICALLY DAMPED ( ): When the system has two real and equal poles. Location of poles for critically damped is shown in fig.

SYED HASAN SAEED 15

ssssss

Take the inverse Laplace

)8()1(1)(

SYED HASAN SAEED 16

From equation (6) it is clear that is the actual damping. For , actual damping = . This actual damping is known as CRITICAL DAMPING.The ratio of actual damping to the critical damping is known as damping ratio . From equation (8) time constant = . Response is shown in fig.

OVERDAMPED ( ): when the system has two realand distinct poles.

SYED HASAN SAEED 17

Response of the system

From equation (2)

SYED HASAN SAEED 18

)9()1()(

)1( 222 ndPut

)10()(

We get

Equation (10) can be written as

)11())((

After partial fraction of equation (11) we get

SYED HASAN SAEED 19

Put the value of d

)1(112

Inverse Laplace of equation (13)

From equation (14) we get two time constants

SYED HASAN SAEED 20

)14()1(12)1(12

)1( 22

tt nn eetc

SYED HASAN SAEED 21

)15()1(12

From equation (14) it is clear that when is greater thanone there are two exponential terms, first term has timeconstant T1 and second term has a time constant T2 . T1 <T2 . In other words we can say that first exponential termdecaying much faster than the other exponential term.So for time response we neglect it, then

)16()1(

THANK YOU

SYED HASAN SAEED 22

TIME DOMAIN SPECIFICATIONS OF SECOND ORDER SYSTEM

1SYED HASAN SAEED

SYED HASAN SAEED 2

SYED HASAN SAEED 3

Consider a second order system with unit step input andall initial conditions are zero. The response is shown in fig.

1. DELAY TIME (td): The delay time is the time requiredfor the response to reach 50% of the final value infirst time.

2. RISE TIME (tr): It is time required for the response torise from 10% to 90% of its final value for over-damped systems and 0 to 100% for under-dampedsystems.

We know that:

SYED HASAN SAEED 4

Where,

Let response reaches 100% of desired value. Put c(t)=1

SYED HASAN SAEED 5

01sin1

0 tneSince,

)sin())1sin((

0))1sin((

Put n=1

SYED HASAN SAEED 6

3. PEAK TIME (tp): The peak time is the time requiredfor the response to reach the first peak of the timeresponse or first peak overshoot.

For maximum

SYED HASAN SAEED 7

11)(Since

11cos1

Since,

Equation can be written as

Equation (2) becomes

SYED HASAN SAEED 8

1sin11cos

Put cosand

cos1sinsin1cos 22 tt nn

))1cos((

))1sin((

SYED HASAN SAEED 9

))1tan((

The time to various peak

Where n=1,2,3,…….

Peak time to first overshoot, put n=1

First minimum (undershoot) occurs at n=2

4. MAXIMUM OVERSHOOT (MP):

Maximum overshoot occur at peak time, t=tp

in above equation

SYED HASAN SAEED 10

ptPut,

1.1sin

SYED HASAN SAEED 11

)sin(1

sin)sin(

SYED HASAN SAEED 12

100*%21

5. SETTLING TIME (ts):

The settling time is defined as the time required for thetransient response to reach and stay within theprescribed percentage error.

SYED HASAN SAEED 13

SYED HASAN SAEED 14

Time consumed in exponential decay up to 98% of the input. The settling time for a second order system is approximately four times the time constant ‘T’.

6. STEADY STATE ERROR (ess): It is difference betweenactual output and desired output as time ‘t’ tends toinfinity.

)()( tctrLimitet

EXAMPLE 1: The open loop transfer function of a servosystem with unity feedback is given by

Determine the damping ratio, undamped natural frequencyof oscillation. What is the percentage overshoot of theresponse to a unit step input.

SOLUTION: Given that

Characteristic equation

SYED HASAN SAEED 15

)5)(2(

0)()(1 sHsG

SYED HASAN SAEED 16

0)5)(2(

Compare with 02 22 nnss We get

%92.1100*

7826.0

7472.4**2

sec/472.420

22 )7826.0(1

7826.0*

7826.0

sec/472.4

EXAMPLE 2: A feedback system is described by thefollowing transfer function

The damping factor of the system is 0.8. determine theovershoot of the system and value of ‘K’.

SOLUTION: We know that

SYED HASAN SAEED 17

016)164(

16)164(

is the characteristic eqn.

Compare with

SYED HASAN SAEED 18

.sec/4radn

K1644*8.0*2 15.0K

100*100*22 )8.0(1

EXAMPLE 3: The open loop transfer function of a unityfeedback control system is given by

By what factor the amplifier gain ‘K’ should be multiplied sothat the damping ratio is increased from 0.3 to 0.9.

SOLUTION:

SYED HASAN SAEED 19

Characteristic Eq.

Compare the characteristic eq. with

Given that:

SYED HASAN SAEED 20

02 22 nnss

We get

SYED HASAN SAEED 21

Hence, the gain K1 at which 3.0 Should be multiplied

By 1/9 to increase the damping ratio from 0.3 to 0.9

ANALYSIS

1SYED HASAN SAEED

SYED HASAN SAEED 2

STEADY STATE ERROR:

The steady state error is the difference between theinput and output of the system during steady state.

For accuracy steady state error should be minimum.

We know that

The steady state error of the system is obtained by finalvalue theorem

SYED HASAN SAEED 3

sHsGsR

)(.lim)(lim0

sEsteest

SYED HASAN SAEED 4

)(.lim

For unity feedback

Thus, the steady state error depends on the input and open loop transfer function.

STATIC ERROR COEFFICIENTS

STATIC POSITION ERROR CONSTAN Kp: For unit step input R(s)=1/s

SYED HASAN SAEED 5

)()(lim

)()(lim1

KsHsGe

sHsGsse

Where is the Static position error constantpK

Steady state error

STATIC VELOCITY ERROR CONSTANT (Kv):

Steady state error with a unit ramp input is given by

R(s)=1/s2

SYED HASAN SAEED 6

0 sHsGsRsLime

KsHssGe

sHssGssHsGsse

Where )()(lim0

sHssGKs

Static velocity error coefficient

STATIC ACCELERATION ERROR CONSTANT (Ka):

The steady state error of the system with unit parabolic input is given by

where,

SYED HASAN SAEED 7

KsHsGse

sHsGsssHsGsse

)()(lim 2

0sHsGsK

Static acceleration constant.

STEADY STATE ERROR FOR DIFFERENT TYPE OF SYSTEMS

TYPE ZERO SYSTEM WITH UNIT STEP INPUT:

Consider open loop transfer function

SYED HASAN SAEED 8

)1().........1)(1(

.).........1)(1()()( 21

m sTsTs

sTsTKsHsG

KsHsGK

)()(lim

0 Hence , for type zero system the static position error constant Kp is finite.

TYPE ZERO SYSTEM WITH UNIT RAMP INPUT:

TYPE ZERO SYSTEM WITH UNIT PARABOLIC INPUT:

For type ‘zero system’ the steady state error is infinite for ramp and parabolic inputs. Hence, the ramp and parabolic inputs are not acceptable.

SYED HASAN SAEED 9

sTsTKssHssGK

0)....1)(1(

)...1)(1(.lim)()(lim 21

sTsTKssHsGsK

0)....1)(1(

)...1)(1(.lim)()(lim 212

TYPE ‘ONE’ SYSTEM WITH UNIT STEP INPUT:

Put the value of G(s)H(s) from eqn.1

TYPE ‘ONE’ SYSTEM WITH UNIT RAMP INPUT:

SYED HASAN SAEED 10

)()(lim0

sHsGKs

)()(.lim0

sHsGsKs

SYED HASAN SAEED 11

TYPE ‘ONE’ SYSTEM WITH UNIT PARABOLIC INPUT:

Hence, it is clear that for type ‘one’ system step input and ramp inputs are acceptable and parabolic input is not acceptable.

)()(lim 2

0sHsGsK

Similarly we can find for type ‘TWO’ system.

For type two system all three inputs (step, Ramp,Parabolic) are acceptable.

SYED HASAN SAEED 12

INPUT SIGNALS

TYPE ‘0’ SYSTEM

TYPE ‘1’ SYSTEM

TYPE ‘2’ SYSTEM

UNIT STEP INPUT 0 0

UNIT RAMP INPUT 0

UNIT PARABOLIC

DYNAMIC ERROR COEFFICIENT:

For the steady-state error, the static error coefficients gives the limited information.

The error function is given by

For unity feedback system

The eqn.(2) can be expressed in polynomial form (ascending power of ‘s’)

SYED HASAN SAEED 13

)1()()(1

sHsGsR

)2()(1

SYED HASAN SAEED 14

)3(........111

)4().......(1

Take inverse Laplace of eqn.(4), the error is given by

)5(.......)(1

Steady state error is given by

)(lim0

SYED HASAN SAEED 15

.......1

Similarly, for other test signal we can find steady state error.

.......,, 321 KKK are known as “Dynamic error coefficients”

EXAMPLE 1: The open loop transfer function of unity feedback system is given by

Determine the static error coefficients

SOLUTION:

SYED HASAN SAEED 16

)10)(1.01(

avp KKK ,,

0)10)(1.01(

50.lim

)()(.lim

5)10)(1.01(

)()(lim

sHsGsK

EXAMPLE 2: The block diagram of electronic pacemaker isshown in fig. determine the steady state error for unitramp input when K=400. Also, determine the value of Kfor which the steady state error to a unit ramp will be0.02.

Given that: K=400,

SYED HASAN SAEED 17

sR 1)( sH

)20()()(

SYED HASAN SAEED 18

)(.lim

Now, 02.0sse Given

20lim02.0

THANK YOU

SYED HASAN SAEED 19

BASIC CONTROL ACTION AND CONTROLLER CHARACTERISTICS

1SYED HASAN SAEED

SYED HASAN SAEED 2

INTRODUCTION:

The automatic controller determines the value ofcontrolled variable, compare the actual value to thedesired value, determines the deviations andproduces a control signal that will reduce thedeviation to zero or to a smallest possible value.

The method by which the automatic controllerproduces the control signal is called control action.

The control action may operate through eithermechanical, hydraulic, pneumatic or electro-mechanical means.

SYED HASAN SAEED 3

SYED HASAN SAEED 4

ELEMENTS OF INDUSTRIAL AUTOMATIC CONTROLLER:

The controller consists of :

Error Detector

Amplifier

The measuring element, which converts the output variable to another suitable variable such as displacement, pressure or electrical signals, which can be used for comparing the output to the reference input signal.

Deviation is the difference between controlled variable and set point (reference input).

SYED HASAN SAEED 5

CLASSIFICATION OF CONTROLLERS:

Controllers can be classified on the basis of type ofcontrolling action used. They are classified as

i. Two position or ON-OFF controllers

ii. Proportional controllers

iii. Integral controllers

iv. Proportional-plus-integral controllers

v. Proportional-plus-derivative controllers

vi. Proportional-plus-integral-plus-derivativecontrollers

SYED HASAN SAEED 6

Controllers can also be classified according to the power source used for actuating mechanism, such as electrical, electronics, pneumatic and hydraulic controllers.

TWO POSITION CONTROL: This is also known as ON-OFF or bang-bang control.

In this type of control the output of the controller is quickly changed to either a maximum or minimum value depending upon whether the controlled variable (b) is greater or less than the set point.

Let m= output of the controller

M1=Maximum value of controller’s output

SYED HASAN SAEED 7

M2=Minimum value of controller’s output

E= Actuating error signal or deviation

The equations for two-position control will be

m=M1 when e>0

m=M2 when e<0

The minimum value M2 is usually either zero or –M1

SYED HASAN SAEED 8

BLOCK DIAGRAM OF ON-OFF CONTROLLER

Block diagram of two position controller is shown inprevious slide. In such type of controller there is anoverlap as the error increases through zero ordecreases through zero. This overlap creates a spanof error. During this span of error, there is no changein controller output. This span of error is known asdead zone or dead band.

Two position control mode are used in room airconditioners, heaters, liquid level control in largevolume tank.

SYED HASAN SAEED 9

PROPORTIONAL CONTROL ACTION: In this type ofcontrol action there is a continuous linear relationbetween the output of the controller ‘m’ andactuating signal ‘e’. Mathematically

Where, Kp is known as proportional gain orproportional sensitivity.

SYED HASAN SAEED 10

In terms of Laplace Transform

SYED HASAN SAEED 11

INTEGRAL CONTROL ACTION:

In a controller with integral control action, the outputof the controller is changed at a rate which isproportional to the actuating error signal e(t).

Mathematically,

Where, Ki is constant

Equation (1) can also be written as

Where m(0)=control output at t=0

SYED HASAN SAEED 12

)1()()( teKtmdt

)2()0()()( mteKtm i

Laplace Transform of eqn. (1)

The block diagram and step response is shown in fig.

SYED HASAN SAEED 13

sEKssM

The inverse of Ki is called integral time Ti and is defined as time of change of output caused by a unit change of actuating error signal. The step response is shown in fig.

For positive error, the output of the controller is ramp.

For zero error there is no change in the output of the controller.

For negative error the output of the controller is negative ramp.

SYED HASAN SAEED 14

DERIVATIVE CONTROL ACTION:

In a controller with derivative control action the output of the controller depends on the rate of change of actuating error signal e(t). Mathematically,

Where is known as derivative gain constant.

Laplace Transform of eqn. (1)

SYED HASAN SAEED 15

)1()()( tedt

dKtm d

ssEKsM

Eqn. (2) is the transfer function of the controller.

From eqn.(1) it is clear that when the error is zero orconstant, the output of the controller will be zero.Therefore, this type of controller cannot be usedalone.

SYED HASAN SAEED 16

BLOCK DIAGRAM OF DERIVATIVE CONTROLLER

PROPORTIONAL-PLUS-INTEGRAL CONTROL ACTION:

This is the combination of proportional and integralcontrol action. Mathematically,

Laplace Transform of both eqns.

SYED HASAN SAEED 17

)2()(1

)1()()()(

KteKtm

dtteKKteKtm

)()()(

KsEKsM

SYED HASAN SAEED 18

Block diagram shown below

In eqn (3) both and are adjustable. Is calledintegral time. The inverse of integral time is calledreset rate. Reset rate is defined as the number oftimes per minute that the proportional part of theresponse is duplicate.

Consider the fig.(2a), the error varies at

pK iT iT

Fig. (1)

SYED HASAN SAEED 19

Fig. (2a)

Fig. (2b)

Consider the diagram the output of the controllersuddenly changes to due to proportional action,after that controller output changes linearly withrespect to time at rate

For unit step (t1=0), the response shown in fig.(2b).

From equation (2) it is clear that the proportionalsensitivity affects both the proportional andintegral parts of the action.

SYED HASAN SAEED 20

PROPORTIONAL-PLUS- DERIVATIVE CONTROL ACTION:

When a derivative control action is added in series withproportional control action, then this combination isknown as proportional-derivative control action.

Block diagram is shown in fig.

Mathematically,

SYED HASAN SAEED 21

)1()()()( tedt

dTKteKtm ppp

Laplace Transform of equation(1)

This is the transfer function. is known as derivativetime. Derivative time is defined as the time intervalby which the rate action advances the effect of theproportional control action.

PD control action reduces the rise time, fasterresponse, improves the bandwidth and improves thedamping.

SYED HASAN SAEED 22

)2()1()(

)()()(

ssETKsEKsM

Consider the diagram, if the actuating error signal e(t) is aramp function at t=t1. the derivative mode causes a stepmd at t1 and proportional mode causes a rise of mp equalto md at t2. this is for direct action PD control.

SYED HASAN SAEED 23

For unit ramp input, the output of the controller isshown below.

SYED HASAN SAEED 24

PROPORTIONAL-PLUS-INTEGRAL-PLUS-DERIVATIVECONTROL ACTION:

The combination of proportional, integral andderivative control action is called PID control actionand the controller is called three action controller.Block diagram is shown below.

SYED HASAN SAEED 25

SYED HASAN SAEED 26

)()()()(

)1()()(1

ssETKsEsT

KsEKsM

dTKdtte

TKteKtm

Laplace transform

Equation (2) is the transfer function.

THANK YOU

ATTENTION

SYED HASAN SAEED 27

TIME DOMAIN ANALYSIS

Engineering

Lecture5 Time Domain System Analysis

A Practical Time-Domain Fault Current Analysis using … · A Practical Time-Domain Fault Current Analysis using ... The challenge of this paper is to provide a practical time-domain

Time Domain Analysis of Terahertz Photoconductive Antennas

Time domain analysis of vortex- induced vibrations (VIV)

MESB374 Chapter8 System Modeling and Analysis Time domain Analysis Transfer Function Analysis

Sparse Principal Component Analysis in Frequency Domain for Time Seriesjunweil/papers/FourierPCA.pdf · Sparse Principal Component Analysis in Frequency Domain for Time Series Junwei

ii TIME-FREQUENCY DOMAIN ANALYSIS OF ACOUSTIC

Lecture Time Domain Analysis of 2nd Order Systems

S-Domain Analysis - CAE Usershomepages.cae.wisc.edu/~ece902/LectureNotes/Simulation_1up/lec3c.… · S-Domain Analysis. s-Domain Circuit Analysis Time domain (t domain) Complex frequency

Chapter4 Time Domain Analysis of Control System

Time Domain Analysis of Dielectric Relaxation

Time domain earthquake response analysis method for 2-D ...ssu.ac.ir/cms/fileadmin/user_upload/Moavenatha/Mposhtibani/Mdaftar... · Time domain earthquake response analysis method

Time Domain Analysis of 2nd Order System (1)

Time and frequency domain coupled analysis of deepwater

Lecture 16 Time Domain Analysis

Signal Processing Methods of Diagnosing Alzheimer’s ... · Frequency domain analysis, Time domain analysis, Time-Frequency domain analysis. I. I. ... Signal Processing Methods of

TIME DOMAIN ANALYSIS A VERY SUPERFICIAL APPROACHsus.ziti.uni-heidelberg.de/...06_TimeDomain.pptx.pdf · CCS: Time Domain Analysis P. Fischer, ZITI, Uni Heidelberg, Seite 16 t u(t)

Chapter 7. Introduction to time series analysis in the ...Chapter 7. Introduction to time series analysis in the frequency domain Objectives This course emphasizes time domain analysis

6. Frequency-Domain Analysis of Discrete- Time Signals and ...wnt.sjtu.edu.cn/junfengwang/Fundamentals of Signal Processing/6... · Frequency-Domain Analysis of Discrete-Time Signals

Time Domain and IR Analysis of Ridged Horn Antennas