digital control system

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Date: 2065/4/20 Digital control system: The rapid increase in the use of digital controller in the controlled system is due to its achievement in the optimum performance. Digital control system provides optimal performance in the form of maximum productivity, maximum profit, minimum cost or minimum energy use etc. The application of computer control has made possible the intelligent motion in industrial robots, the optimization of the fuel economy in automobiles and refinement in the operations of house hold appliances and machines such as microwave ovens, washing machine, Air-conditioning. Decision making capability and flexibility in the control programs are major advantages of digital control system. The current trend towards rather then analog control system is mainly due do the availability of low cost digital computers and the advantages found in working with digital signals rather then continuous time signals. Basic Blocks of Digital control system:

S/HandADC

DigitalControlSystem

DAC HoldCircuit Actuator

Plant orProcess

Clock

Transducer

+-

DigitalControl

Figure 1 shows the basic block diagram or principle of DCS . The controller operation is performed or controlled by the clock. In such a DCS points of the systems pass signals of varying amplitude either in continuous time or discrete time or in numerical code.

1. Sample and Hold ( S/H) : It is the circuit that receives an analog input signal and holds this signals at a constant value for a specified period of time. Usually the signal is electrical but it may be optical or mechanical.

2. ADC: ADC also called an encoder is a device that converts an analog signal into a digital signal, usually a numerically coded signal in binary form. Such a converter is need as an interface between an analog component and the digital component. Basically ADC involves sampling , quantizing and encoding.

3. Digital Computer: The digital computer processes the sequences of numbers by mean of an algorithm an produces an new sequences of numbers.

4. DAC: DAC also called an decoder is a device that converts a digital signal ( Numerically coded data) into an analog signal. It acts as an the interfacing device between the digital component and an analog component. The real time clock in the computer synchronizes the events. The output of the hold circuit which is continuous time signal is fed the plant either directly or through the actuator which controls the dynamics of the system ( i.e it smoothens the slope of the signal)

5. Plant or process: A plant is a physical object to be controlled. The examples are a furnace, chemical reactors and a set of machine parts functioning together to perform a particular operations such as servo system etc.

6. Transducer/sensor: A transducer is a device that converts an input signal into an output signal of a another form such as device that converts a temperature into a voltage output ( thermistor or thermocouple ), an optical signal into voltage ( phototransistor )

Discrete times control system: Discrete time control system is control system in which one or more variable can change only at discrete instants of time. These instants which are denoted by KT



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or tk ( k =0, 1, 2, ………), specify the times at which some physical measurements is performed. The time interval between two discrete instants is taken to be sufficiently short that the data for the time between them can be approximated by simple interpolation.

Date: 2065/4/21 Data Acquisition, Conversion and Distribution: Data Acquisition system:



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Fig.1 shows the diagram of data acquisition system. The basic parameters are explained below: 1. Physical variable: The input to the system is a physical

variable such as position, velocity, acceleration, temperature, pressure etc.

2. Transducer amplifier and low pass filter: The physical variables (which are generally in non-electrical form) is first converted into an electrical signal (a voltage or a current signal) by a suitable transducer. Amplifier then amplifies the voltage output of the transducer (i.e the signal have rises to the necessary level). The LPF follows the amplifier which attenuates the high frequency signal components such as noise signals which are random in nature. The o/p of LPF is an analog signal. The signal is then fed to an analog multiplexer.

3. Analog Multiplexer: It is a device that performs the function of time sharing and ADC among many analog channels. It is a multiple switch (usually an electronic switch) that switches sequentially among many analog input channels in some prescribed fashion. The no of channels may be 4,8,16.

4. Sample and hold circuit: A simpler in a digital system converts an analog signal into an train of amplitude modulated pulses. The hold circuit holds the value of the sampled pulse signal over a specified period of time. It is necessary in the AD converter to produce a number that accurately represents the i/p signal at the sampling instant.

Assignment: 1. Draw a circuit of sample and hold and explain its basic operation. 5. ADC: The output of sample and hold is then fed to the AD

converter. The o/p of the converter is the signal in digital form which is fed to the digital controller.

In this way data acquisition system is held.

Data Distribution:



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1. Register: The o/p of digital controller is then stored for a certain period of time in a memory device called register.

2. Multiplexer: The demultiplexer , which is synchronized with the i/p sampling signal, separates the composite o/p signal which is in the from of digital data from the digital controller into the original channels. Each channel is connected to DAC to produce the o/p analog signal for that channel.

3. DAC: At the o/p of the digital controller, the digital must be converted to an analog signal by the process called D/A conversion. For the full range of digital i/p, there are 2n different analog values , including zero.

4. Hold: The sampling operation produces an amplitude modulated pulse signal. The function of hold operation is to reconstruct the analog signal that has been transmitted as a train of pulse samples. The purpose of hold operation is to fill the spaces between the sampling periods and thus roughly reconstruct the original analog input signal which is then fed to the actuator which smoothens the slope of signal.

5. Plant or process: A plant is a physical object to be controlled. The examples are a furnace, chemical reactors and a set of machine parts functioning together to perform a particular operations such as servo system etc.

Data conversion Process: SIGNAL SAMPLING ,QUANTIZING AND ENCODING Signal sampling is the first step of transmission of analog signal over digital signal. 1. Sampling: Sampling is the process of conversion of

continuous time analog signal into discrete time analog signal.

The discrete signal obtained after sampling is called sampled signal.

Sampling Theorem: It states “Analog signal can be reproduce from an appropriate set of its samples taken at some fixed intervals of time.” This theorem has made possible to transmit only samples of analog signal by changing or encoding this samples into block of code words suitable for digital control systems.

If fs = sampling frequency fx = maximum frequency component of the i/p signal, then the distortion less recovery of the signal fs ≥ 2fx

If the signal x(t) to be sampled is band limited , then the sampled signal can be represented as: xs(t) = x(t) × g(t) Where, g(t) is the sampling function (rectangular pulse train) which be represented as shown below.

g(s)

Fig.1 Rectangular pulse train Where , Ts = Sapling period. τ = duration of sampling pulse= pulse width Sampler can be implemented as:

Ts

τ/2 - τ/2 τ



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x(t)

g(t) Fig.2 Implementation of sampler Proof of sampling theorem: The gate function g(t) can be expressed interms of fourier series as ∑∞

=+=

1)cos(2)(

n sno tncctg ω Where, co = τ/ Ts = τfs Cn = fs τ sinc[nfs τ ] = co sinc[nfsτ] ωs =2 πfs The signal xs(t) can be expressed as xs = x(t)×g(t) = x(t)×[ ∑∞

=+

1)cos(2

n sno tncc ω ]

= cox(t)+2c1x(t)cos ωnt+2c2x(t)cos2 ωst+……….+2cnx(t)cosn ωst+………… The fourier transform of above series as xs(f) =co(f)+2c1x(f-fs)+2c2x(f-2fs)+…………+2cn x(f+nfs)+……. The above series can be graphically represented as:

fx-fx

x(f)

Fig.3. Message spectrum

It is clear from fig.4 that the spectrum of the sampled signal contains the spectrum of the sampled signal contains the spectrum of the original message signal.

xs(t)= x(t)×g(t)



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Date:2065/4/26 It is evident that for distortion less recovery of original message signal, from the spectrum of the sampled signal, the following condition should be met. fs ≥ fx In this case the original message signal spectra can be recovered by passing the sampled signal through low pass filter with bandwidth equaling to ± fx Distortion will occur while recovering the message spectrum if. fs ≤ fx The distortion in the above case is caused by the overlapping of side bands and message spectra.

o

Xs(f)

aliasign ditort ion(fs<2fx)

o

Xs(f)

The minimum sampling rate: fs min = 2fs is called Nyquists’s sampling rate for distortion less recovery of one message spectrum. The minimum interval of the sampling for a real signal is Ts min = 1/2fx(min) Where fx(min) = maximum frequency in the message spectrum. Quantizing and Quantization error:

t41 2 3 5 6 7

x(t)

12

n41 2 3 5 6 7

xs(n)

0.90.9

1.8

0.5

1.81.7 1.7

n41 2 3 5 6 7

xq(n)

1

2

(a)

(b)

(c)



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Fig. 1(a) continuous - continuous value signal 1(b) Discrete time

- continuous value signal 1(c) Discrete time - discrete value signal. Quantization is the process of representing the analog sample values by a finite set of levels. The sampling process coverts a continuous time signal to a discrete time signal with amplitude and that can take any values from zero to maximum level and the quantization process converts continuous amplitude samples to a finite set of discrete amplitude values. The output sates of each quantized sample is then described by a numerical code. The process of representing a sampled value by a numerical code is called encoding . Thus encoding is a process of assigning a digital word to each discrete states. In uniform quantization it is assumed that the range of input sample is –xmax to xmax and the number of quantization level, known as Q-level, N=2n. Where, n = is the number of bits per source sample , then the step size △ on length of the Q level is assumed to be, △ = 2xmax / N = 2xmax/2n = 2xmax/ 2n-1 △ = xmax/2n-1 The step size ‘△’ is also called ‘quantum’. Quantization error: Since the number of bits in the digital word is finite , ie the digital output can assume only a finite number of levels and therefore

analog number must be round off to the nearest digital level. Hence any ADC involves quantization error. It is evident that the maximum Q error could be only △/2. In uniform quantization the steps size △ is constant for the entire dynamic range of the input discrete signal level. Q error depends on the fineness of the quantization level and can be made as small as desire by making the quantization level smaller or by increasing the number of bits ‘n’. In practice there is a maximum for n and so there is always some error due to quantization. The uncertainty present in the quantization process results quantization noise. Signal to quantization noise ratio (SQNR): It is evident that the Q-error ( i.e qe) lies between – △/2 to △/2 in random manner , the average power of Q-noise is therefore given by,

∫∆

∆−

∆=

2

2

21eeq dqqP

= 12

2∆=qP …………………..(i)

It is seen form equation (1) that Q-noise is dependent on step-size ‘△’ only. Reducing the step-size or increasing the no of representation level, we can reduce pq and hence Q-error. From equ (1) Pq = △2/12 But, △= xmax /2n-1 Pq = (xmax/2n-1)/12 = xmax

2/3×4n ……….(ii)



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Assume that the average signal power is x2 , SQNR for uniform quantization will be, SQNR = average power of signal/ average power of noise.

nxQSNR 43ˆ 2 ××= ……………………(iii) Where,

2

22ˆ

xxx = = Normalized signal power.

Again , equation (iii) can be reproduce in dB, as n

xdB dBPQSNR )4(log10log10)()( 103

10 ++= 10log108.4)( ndBPx ++= Where, dBxdBPx )ˆ((log10)( 2

10= Thus, (QSNR)dB = Px+4.8+6n………………….(iv) Also, SQNR(dB) = Px +4.8+20 log10

N …………….(v ) [since, N= 2n] And for N>> 1 SQNR = 20 log10N …………….(vi) Which is approximated SQNR for uniform quantization interms of level of quantization “N”. Assignmet:1

1. Simple and hold ckt. Analog multiplexer and Demultiplexer.

Date:2065/5/1 Review of Z-transform: mathematical tool commonly used for the analysis and synthesis of discrete time control system is the z-transform. In considering the z-transform of a time function x(t)

, we consider only the sampled values of x(t) , i.e x(0), x(T) , x(2T),………, where ‘T’ is sampling period. The z-transform of a time function x(t) , where ‘t’ is non –negative or of a sequence of values x(kT) , where k = 0,1,2,3 …

X(z) = z[x(t)] = ∑∞

=

−

0)(

k

kzkTx …………….(i)

Or , X(z) = z[x(t)] = ∑∞

=

−

0

1)(k

zkx …………….(ii), for T=1

The z-transform defined by equation (i) and (ii) is one-sided z-transform . The both sided ( or double sided) z-transform is defined by:

X(z) = z[x(t)] = z[x(k)] = ∑∞

−∞=

−

kzkTx 1)(

z-transform of Elementary function: (i) Step unit function: The unit step function defined by

⎪⎩

⎪⎨⎧ ≥

=otherwise

tfortx

0

01)(

Thus , X(z) = z[x(t)] = z[1(t)]=∑∞

=

−

0.1

k

kz

= 1 + z-1 +z-2+z-3 +………. = 11

1−+ z

Therefore, z[1(t)] = 111

−+ z



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Or, z[1(t)] = 1,1

>−

zz

z

2. Unit Ramp Function: The unit ramp function is defined by

⎪⎩

⎪⎨⎧ ≥

=otherwise

tforttx

0

0)(

Or, x(KT) = KT for k = 0,1,2,3…………. Thus,

X(z) = z[x(t)] = z[t] = ∑∞

=

−

0.

k

kzt

Or, X(z) = ∑∞

=

−

0)(

k

kzKT

= k(z-1+2z-2+3z-3 ……..)

= T. 21

1

)1( −

−

− zz

X(z) = 21

1

)1(.

−

−

− zzT

3. Polynomial Function: The polynomial function is defined by

⎪⎩

⎪⎨⎧

<

==

00

.......3,2,1,0)(

K

katx

k

Where a is constant.

Then, X(z) = z[x(k)] = z[aK] = ∑∞

=

−

0)(

k

kzkx

X(k) = ∑∞

=

−

0k

kk za

= 1 + az-1 +a 2z-2+ a3z-3+…………… Z[ak] = 11

1−− az

Or, z[ak] = 111

−− az

4. Exponential function:

⎪⎩

⎪⎨⎧ ≥

=otherwise

tforetx

at

0

0)(

Where, x(kT) = e-akT, K = 0,1,2……….. We have,

X(z) = z[e-at] = ∑∞

=

−−

0k

kakT ze

Or, X(z) = 1 + e-aTz-1+e-2aTz-2+e-3aTz-3……… X(z) = aTaT ez

zze −−− −

=− 11

1

5. Sinusoidal Function:

⎪⎩

⎪⎨⎧

<

≥=

00

0sin)(

t

tforttx

ω

Noting , ejwt = coswt +jsinwt e-jwt = coswt-jsinwt we have , Sinwt = 1/2j[ejwt-e-jwt] & coswt = 1/2[ejwt+e-jwt] We know that, Z[e-at] = 11

1−−− ze at



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Z[ejwt] = 111

−− ze jwT

Z[e-jwt] = 111

−−− ze jwT

Therefore, z[sinwt] = z[ 1/2j(ejwt-e-jwt)] = 1/2j[z(ejwt)-z(e-jwt)]

= 1/2j ⎥⎦⎤

⎢⎣⎡

−−

− −−− 11 11

11

zeze jwtjwt

Z[sinwt] = 21

1

cos21sin

−−

−

+− zwtzwtz

Similarly,

Z[coswt] = 21

1

cos21cos

−−

−

+− zwtzwtz

Important properties of z-transform: 1. Multiplication by a constant:

If x(t) X(z), then, Z[ax(t)] = az[x(t)] = ax(z) Where ‘a’ is a constant.

2. Linearity of Z-transform: If z[f(t)] = z[f(kT)] = z[f(k)] = F(z) Z[g(t)] = z[g(kT)] = z[g(k)] = G(z) And a and b be scalers then then x(k) formed by liner combination, x(k) = a f(k) + b g(k) Has the z-transfom X(z) = a F(z) +G(z)

3. Multiplication by ak : If z[x(k)] = X(z) , then Z[akx(k)] = X(a-1z) Proof:

Z[ akx(k)] = ∑∑∞

=

−−∞

=

− =0

1

0

))(()(k

k

k

kk zakxzkxa

= X(a-1z) 4. Shifting theorem: If x(t) = 0 for t< 0 and x(t) X(z) Then, z[x(t-nT)] = z-n X(z) ……………(i)

And , z[x(t+nT)] = zn ⎥⎦

⎤⎢⎣

⎡− ∑

−

=

−1

0)()(

n

k

kzKTxzX …………..(ii)

Where, n is zero or +ve integer. Proof: For equation (i) We know that , z[x(t-nT)] = z[x(KT-nT)] = z[x(k-n)T]

Or, z[x(t-nT)] = ∑∞

=

−−0

).(k

kTznkx

= ∑∞

=

−−−−0

)( ..).(k

nnk zzTnkx

= ∑∞

=

−−− −0

)(.).(.k

nkn zTnkxz

Let us define k-n = m , therefore,

Z

Z



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Z[x(mT)] = ∑∞

−=

−−

nm

mn zmTxz ).(.

Since ‘m’ must be zero and non-negative integer Therefore, z[x(mT)] = z-nX(z) For equation (ii) Z[x(t+nT) = z[ x(KT+nT)] = z[x(K+n)T]

Or, z[x(t+nT)] = ∑∞

=

−+0

).(k

kTznkx

= ∑∞

=

+−+0

)( ..).(k

nnk zzTnkx

= ∑∞

=

+−+0

)(.).(.k

nkn zTnkxz

Let us define, K+n = m

Z[x(mT)] = ∑∞

=

−−

nm

mn zmTxz ).(.

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

0

1

0⎥⎦

⎤⎢⎣

⎡+− ∑∑∑

∞

=

−−

=

−−

=

−

nm

mn

m

mn

m

mn zmTxzmTxzmTxz

= .).().(1

00⎥⎦

⎤⎢⎣

⎡− ∑∑

−

=

−∞

=

−n

m

m

m

mn zmTxzmTxz

4. Complex translation Theorem:

If x(t) X(z), then, e-atx(t) X(zeaT)

Proof:

Z[e-atx(t)] = ∑∞

=

−−

0).(

k

kakT zkTxe

= ∑∞

=

−

0)).((

k

kaT zekTx

= X(eaTz) = X (zeaT) 5. Initial value Theorem:

If x(t) has the z-transform X(z) and if )(lim zXz ∞⎯→⎯ exist then, the initial value x(0) of x(t) is given by , x(0) = )(lim zXz ∞⎯→⎯

6. Final Value Theorem:

[ ])()1(1)( 1limlim zXzzkxk −−⎯→⎯=∞⎯→⎯ Assignment # 02: 1. Prove all the important properties of z-transform. Example: 01: Obtain the z-transform of x(s) =

)1(1+ss

Solution: Given, x(s) =

)1(1+ss

= 1+

+s

BsA

Where, A = 0

.)1(

1=+ s

sss

A = 1

And, B = 1

)1.()1(

1−=

++ s

sss

B = -1 Therefore, x(s) =

111+

−ss

Z Z



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Taking inverse Laplace Transform, we have, x(t) = 1 – e-t Again taking z-transform on both sides, Z[x(t)] = z [ 1 – e-t] = z[1] – z[e-t] = 11 1

11

1−−− −

−− zez T

= )1)(1(

1111

11

−−−

−−−

−−+/−−/

zezzze

T

T

= )1)(1(

)1(11

1

−−−

−−

−−−

zezze

T

T

Example: 02 Obtain the z-transform of

⎪⎩

⎪⎨⎧

≤

==

−

00

.......3,2,1,0)(

1

K

kaaf

k

Z[f(a)] = ∑∞

=

−

0

).(k

kzaf

Here, z[x(k)] = z[ak] = )(

11

1 kXaz

=− −

Then, z[x(k-1)] = z[ak-1] = z-1.X(z)

Therefore, z[ak-1] = z-1 ⎥⎦⎤

⎢⎣⎡

− −111az

Z[ak-1] = 1

1

1 −

−

− azz

Example:03 obtained the z-trasform [i] e-at sinωt [ii] e-at cosωt Date:2065/5/3 [i] e-atsinωt We know that

Z[sinwt] = )(cos21sin

11

1

zXzwTz

TZ=

+− −−

− ω

We know that, Z[e-atx(kT)] = X[zeakT]

Therefore , z[e-atsinwT] = 221

1

cos21sin

−−−−

−−

+− zewtzewtze

aTaT

aT

Example 04: Consider the function y(k) , which is sum of functions x(h) where, h = 0, 1, 2 …….k , Such that

y(k) = .....2,1,0,)(0

=∑=

khxk

h

Where y(k) = 0 for k < 0 . Obtained the z-transform of y(k). Solution: Given,

y(k) = ∑=

k

h

hx0

)(

or, y(k) = x(0) + x(1) + x(2) + x(3)+………….+ x(k)---------(i) similarly, y(k-1) = x(0)+ x(2)+ x(3)+ ……….+x(k-1) -------------(ii) Now from (i) and (ii) gives. y(k) – y(k-1) = x(k) Now taking z-transform on both sides, Z[y(k) – y(k-1)] = z[x(k)] Y(z) – z-1Y(z) = X(z) Y(z) [1-z-1] = X(z) Therefore, Y(z) = X(z)/ ( 1-z-1) Ans. Example 05: Obtained the z-transform of t.e-at. Given,



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x(t) = t e-at x(kT) = kT e-at We know z –transform of t is given by, z[t] = Tz-1/(1-z-1)2 Using complex translational theorem, Z[e-at.t] = Te-at.z-1/(1-e-atz-1)2 Example 06: Determine the initial value x(0) if, X(z) = (1-e-T)/(1-z-1)(1-e-Tz-1) Solution: Given, X(z) = (1-e-T)/(1-z-1)(1-e-Tz-1) We know initial value theorem is given by , X(0) = )(lim zXz ∞⎯→⎯

= )1)(1(

)1(11

1lim

−−−

−−

−−−

∞⎯→⎯zez

zez T

T

= 0 Ans. Example 07: Determine the final value x(∞) of X(z) =

11 11

11

−−− −−

− zez aT a > 0

Solution: From final value theorem, x(∞) = )]()1[(1 1lim zXzz −−⎯→⎯

= ( ) ⎥⎦

⎤⎢⎣

⎡⎥⎦⎤

⎢⎣⎡

−−

−−⎯→⎯

−−−

111lim

11

1111

zezzz aT

= ⎥⎦

⎤⎢⎣

⎡−

−−⎯→⎯

−−

−

1

1lim

1111

zezz aT

= ⎥⎦

⎤⎢⎣

⎡−

−⎯→⎯−

−

aTezzz

1lim 11

= 1-0 = 1 Ans.

The inverse z-transform: (1) Direct division method: Example 01: Find x(k) for k = 0, 1,3,4. When X(z) is given by , X(z) = (10z+5)/(z-1)(z-0.2) Solution: X(z) = (10z+5)/(z-1)(z-0.2) = (10z+5)/(z2

- 1.2z+0.2) = (10z-1+ 5z-2)/(1-1.2z-1+0.2z-2)

65

654

54

543

43

432

32

321

21

7362.3736.18

7362.3416.2268.18

68.368.18

68.308.224.18

4.34.18

4.34.2017217

21210510

−−

−−−

−−

−−−

−−

−−−

−−

−−−

−−

−

+−

−

+−

++

+−

−

+−

+

zz

zzz

zz

zzz

zz

zzzzz

zzzzz

Quotient of the x(z) gives the inverse z-transform is given by X(0) = 0 X(1) = 10 X(2) = 17 X(3) = 18. 4 X(4) = 18.68 Example 02: Find x(k) when X(z) is given by X(z) = 1/(z+1)



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Example 03: Obtained x(k) for X(z) = z-1/ (1- az-1) X(z) = z-1/(1-az-1)

= z-1⎥⎦⎤

⎢⎣⎡

− −111az

= z-1[X(z)] Where, Y(z) = 1/(1-az-1) The inverse z-transform of Y(z) is z-1[Y(z)] = ak So, the inverse z-transform of (z) = z-1[Y(z)] z-1[X(z)] = ak-1 = y(k-1) Therefore, x(k) = y(k-1) = ak-1 2. partial fraction method: We use the following formula for

the function written in the form

n

n

pza

pza

pza

ZzX

−+

−=

−= ............)(

2

2

1

1

(i) ai = (z-pi) ipzZ

zX=

)(

Again if,

)()(

)(

2

2

1

1

pzC

pzC

ZzX

−+

−= Then,

C1 = (z-p1)2 1

)(pzZ

zX=

And , C2 =

1

)()( 21

PZzzXpz

dzd

=⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ −

In general , if there are n- multiplication roots then,

Cn = )!1(

1−n

nPZ

nn

n

zzXpz

dzd

=−

−

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ −

)()( 11

1

Example 01: Given X(z) = ))(1(

.1aT

aT

ezzze−

−

−−− Where ‘a’ is a

constant and T is the sampling period. Determine the inverse z-transform x(KT) by the use of partial fraction expansion method. Solution:

X(z) = ))(1(

)1(aT

aT

ezzZe−

−

−−−

))(1(

1)(aT

aT

ezze

ZzX

−

−

−−−

=

aTez

Bz

AZ

zX−−

+−

=1

)(

A = (z-1) 1))(1(

1)1(1

)(=−−

−−=

= −

−

zezzez

zZzX

aT

aT

= 111

=−−

−

−

aT

aT

ee

Threfore, A = 1.

Similarly B = aTaT

aTaT

aT ezezzeez

ezZzX

−−

−−

− =−−−

−== ))(1(

1)()(

= )1(

1−

−−

−

aT

aT

ee

B = -1

aTezzZzX

−−−

−=

11

1)(

aTezz

zzzX

−−−

−=

1)(

11 11

11)(

−−− −−

−=

zezzX aT

Therefore, z-1 [X(z)] = 1-e-aT= 1k-e-akT



15

Therefore, x(t) = 1k – e-akT ans.

Example 02: Obtained the z-transform of )1)(1(

2)( 2

2

+−−++

=zzz

zzzX

by partial fractional expression method. Solution:

1)1()1)(1(

2)( 22

2

+−+

+−

=+−−

++=

zzcBz

zA

zzzzzzX

)1)(1()1)(()1(

)1)(1(2

2

2

2

2

+−−−+++−

=+−−

++zzz

ZCBzzzAzzz

zz

Z2 + z +2 = (A+B)z2- (A+B-C)z +(A-C) Comparing the coefficient of z2 , z and z0 we get, B+A = 1 -(A+B-C) = 1 A-C = 2 Solving the above equation we get, A = 4 , B = -3, C = 2 Thus ,

1)1()( 2 +−

++

−=

zzcBz

zAzX

1

13)1(

4)( 2 +−+−

+−

=zz

zz

zX

21

21

1

1

123

)1(4)(

−−

−−

−

−

+−+−

+−

=zzzz

zzzX

⎥⎦

⎤⎢⎣

⎡+−

−−⎥⎦

⎤⎢⎣⎡

−=

−−

−−

−−

21

11

11

15.013

114)(

zzzz

zzzX

⎥⎦

⎤⎢⎣

⎡+−

+⎥⎦

⎤⎢⎣

⎡+−

−−⎥⎦

⎤⎢⎣⎡

−=

−−

−−

−−

−−

−−

21

11

21

11

11

1.6/13

15.013

114)(

zzzz

zzzz

zzzX

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+−+⎥

⎦

⎤⎢⎣

⎡+−

−−⎥⎦

⎤⎢⎣⎡

−=

−−

−

−−−

−−

−−

21

1

121

11

11

123

31

15.013

114)(

zz

zz

zzzz

zzzX

We know that

z-1 k

z1

11

1 =⎥⎦⎤

⎢⎣⎡

− −

z-1

3cos

15.01

21

1 πkzz

z=⎥

⎦

⎤⎢⎣

⎡+−

−−−

−

z-1

3sin

123.

21

1

πkzz

z=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+− −−

−

Therefore, z-1 = [X(z)] = x(k) = 4.1k-1 – 3 cos(k-1) π/3 + 1/√3 . sin(k-1) π/3 Date: 2065/5/10 3. Inverse Integral method: The inversion integral for the z-transform X(z) is given by z-1[x(z)] = x[kT] = x(k) = ∫ − dzzzx

jk 1)(

21π

………(i)

Where ‘c’ is a circle with its centre at the origin of the z-plane such that all poles of x(z)z-1 are inside it. The equation for giving the inverse z-transform in terms of residues can be derived by using the theory of complex variables. It can be obtained as follows. X(kT) = x1 + x2 + ……….+xn denotes the residues of x(z)zk-1 at poles z1, z2 ……..zm respectively.



16

In evaluating residues, if the denominator of x(z)zk-1 contains a simple pole at z = zi , then the corresponding residue x is given by, ( )[ ]1

1lim

0 )( −−⎯→⎯= kii zzXzzzzk …………….(ii)

If X(z) zk-1 contains a multiple pole zi or order q, then the residue k is given by ,

( )[ ]11

1lim )(2.

)!1(1 −

−

−

−⎯→⎯−

= kqii zzXzz

dzqdqjz

qk ………..(iii)

Example: 01 Obtained x(kT) by using the inversion integral

method when, ))(1(

)1()( aT

aT

ezzezzX

−

−

−−−

=

Solution:

))(1(

)1()( aT

aT

ezzezzX

−

−

−−−

=

))(1(

)1()( 1aT

aTkTk

ezzezzzzX

−

−−

−−−

=

))(1(

)1()( 1aT

aTkk

ezzezzzX

−

−−

−−−

=

For k = 0,1,2……….x(z)zk-1 has simple poles at z1 = 1and z2= e-aT

Hence, x(kT) = x(k) = ∑=

2

1iresidue of x(z)zk-1 at pole z = zi

= k1 + k2 Where k1 = ( )[ ]1

1lim )( −−⎯→⎯ k

i zzXzzzz

= ( ) ⎥⎦

⎤⎢⎣

⎡−−

−−⎯→⎯

−

−

))(1()1(11

limaT

aTk

ezzezzzz

= )()1(1

aT

aTk

eze−

−

−−

K1 = 1 For K2 , K2 = ( )[ ]1

22lim )( −−⎯→⎯ kzzXzzzz

= ⎥⎦

⎤⎢⎣

⎡−−

−−⎯→⎯

−

−−−

))(1()1()(lim

aT

aTkaTaT

ezzezezez

= ⎥⎦

⎤⎢⎣

⎡−

−−

−−

1)1(

aT

aTakT

eee

Therefore , k2 = - e-akT X(kT) = x(k) = x1 + x2 = 1-e-akT which is required inverse z-trasnform. Example 02: Obtained the inverse z-transform of

x(z) = )()1( 2

2

aTezzz

−−− by using the inverse integral method.

Solution:

X(z) = )()1( 2

2

aTezzz

−−−

X(z)zk-1 = )()1( 2

1

aT

k

ezzz

−

+

−−

X(z) zk-1 has simple pole at z = z1 = e-aT and double pole at z = z2

=1 ( i.e q = 2) The inverse z-transform is therefore given by, x(kT) = k1 +k2 Where, k1 = ( )[ ]1lim )( −−− −⎯→⎯ kaTaT zzXezez

= ⎥⎦

⎤⎢⎣

⎡−−

−⎯→⎯

−

+−−

)()1()(

2

1lim

aT

kaTaT

ezzzezez

= ⎥⎦

⎤⎢⎣

⎡−−

+−

2

)1(

)1( aT

Tka

ee

And for K2 , k2 = ⎥⎦

⎤⎢⎣

⎡−−

−⎯→⎯− −

+

)()1()1(.1.

)!12(1

2

12lim

aT

k

ezzzz

dzdz

= ⎥⎦

⎤⎢⎣

⎡−

⎯→⎯−

+

)(.1

1lim

aT

k

ezz

dzdz



17

= ⎥⎦

⎤⎢⎣

⎡−

−+−⎯→⎯

−

+−

2

1lim

)()1)((1 aT

kkaT

ezzzkezz

= 22

1

)1(1)1)(1(

)(1)1()1(

aT

aT

aT

kaTk

eek

ezekk

−

−

−

+−

−−−+

=−

−−+

= 2)1((

aT

aTaT

eekek

−

−−

−−−

Therefore, x(kT) k1 + k2 = 22

)1(

)1()1( aT

aTaT

aT

Tka

eekek

ee

−

−−

−

+−

−−−

+−

X(kT) = 2

)1(

)1( aT

aTaTTka

eekeke

−

−−+−

−−−+

Example 03: Obtain the inverse z-transform of x(z) = 31

2

)1( −

−

− zz

Where k1 = ( ) ⎥⎦

⎤⎢⎣

⎡−

−⎯→⎯−

3

13

2

2lim

)1(.1.1.

!21

zzzz

dzdz

k

= [ ]1

.!2

12

2

=zz

dzd k

= [ ]1

.21 1

=−

zkz

dzd k

= [ ]1

)1(.21 2

=− −

zzkk k

= 2

)1( −kk

Therefore, z-1(x(z)) = x(kT) = k(k-1)/2 Solution of difference equation by z-transform method: Example 01: Solve the following difference equation by use of z-transform method of x(K+2)+ 3x(k+1)+2x(k) = 0 Taking z-transform on both the sides , we have, Z[x(k+2)]+3z[x(k+1)]+3z[x(k+1)] +2z[x(k)] = 0

Z2x(z)-z2x(0)-zx(1) + 3[zx(z)-zx(0)] +2x(z) = 0 X(z) [z2+3z+2] – (z2 +2)x(0) – zx(1) – zx(1) = 0 But, x(0) = 0 and x(1) = 1 Therefore, x(z) [z2+3z+2] – z = 0 X(z) =

21)2)(1()23( 2 ++

+=

++=

++ zB

zA

zzz

zzz

21)2)(1(

1)(+

++

=++

=z

Bz

Azzz

zx

1

)()1(−=

+=zz

zxzA

A = 1. Similarly, B = -1,

21

11)(

+−

+=

zzzzx

21)(

+−

+=

zz

zzzx

11 211

11)(

−− +−

+=

zzzx

11 )2(11

)1(11)(

−− −−−

−−=

zzzx

The inverse z-transform of x(z) is X(z) = -(1)k – (-2)k, k = 0,1,2……….. Example 02: Obtained the solution of the following difference equation in terms of x(0) and x(1). X(K+2) + (a+b)x(k+1) + ab x(k) = 0 Z2x(z) – z2x(0) – zx(1) + (a+b) [z x(z) – z x(0)] + ab x(z) = 0 X(z) [z2 + (a+b)z+ab] = [z2 + (a+b)z] x(0) + z x(1)

X(z) = abzbaz

zxxzbaz+++

+++)(

)1()0(])([2

2

X(z) = bz

Baz

Abzaz

xxbaz+

++

=++

+++))((

)1()0(][



18

Where, A = )(

)1()0(abxbx

−+ and B =

)()1()0(

baxax

−+

)()(

)1()0(

)()(

)1()0()(

bzbaxax

azabxbx

zzx

+−+

++−+

=

11 )(1)(

)1()0(

)(1)(

)1()0(

)(−− −−

−+

+−−−+

=zb

baxax

zaabxbx

zx

kk bbaxaxa

abxbxzX )(

)()1()0()(

)()1()0()( −⎥

⎦

⎤⎢⎣

⎡−+

+−⎥⎦

⎤⎢⎣

⎡−+

= [ for the case a ≠ b]

Now for a = b,

X(z) = 22

2

2)1()0()2(

aazzzxxazz

++++

2)()1()0()2()(

azxxaz

zzx

+++

=

2)()()(

azB

azA

zzx

++

+=

We get , x(z) = 211 )1()1()0(

)1()0()(

−− ++

++

=az

xaxaz

zxzx

Therefore, x(k) = x(0)(-a)k + [ ax(0)+x(1)][(-a)k-1] for case a = b Assignment: 1. Obtained the z-transform of (a) k2 (b) k ak-1

2. Show that )(

1)(

)(1

1)(

1

11

0

10

zxz

zhxz

zxz

hxz

k

n

k

n

−

−−

=

−=

−=⎥

⎦

⎤⎢⎣

⎡

−=⎥

⎦

⎤⎢⎣

⎡

∑

∑

And , ∑∞

=

⎯→⎯=0

lim )(1)(k

zXzkx

Also show that,

⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡ ∑∑−

=

−−

=

1

01 )()()1(

1)(i

h

hk

ihzhxzX

zhx

Where 1 ≤ i ≤ (k-1) 3.Obtained the z-transform of Figure: 4. Obtained the z-transform of

(a) )1()2(

22)( 2

3

−−+

=zz

zzx (b) 2).2()2()(zz

zzx−

+=

(c ) )1)(2(

10)(−−

=zz

zx (d) 2)1()2()(

−+

=zzzzx

(e) )2.01)(1(

61)( 11

32

−−

−−

−−++

=zz

zzzx

(f) 11

1)(21

−−++

=−−

zzzzx

5. Solve the following difference equations: a. 2x(k) – 2x(k-1) +x(k-2) = u(k) Where x(k) = 0 for k< 0

And u(k) = ⎪⎩

⎪⎨⎧

<

=

0,0.....2,1,0,1

kk

b. x(k+2) –x(k+1) + 0.25x(k) = u(k+2) where, x(0) = 1, x(1) = 2 u(k) = 1 for k ≥ 0. c. x(k-2) – x(k-1) + 0.25 x(k) = u(k-2) x(0) =1 x(k) = 0, for k<0. u(k) = 1 for k ≥ 0 (6) Consider the difference equation . X(k+2) = x(k+1) +x(k) Where x(0) = 0 and x(1) = 1. Note that x(2) = 1 , x(3) = 2, x(4) = 3.



19

The series 0,1,2,3,5,8,13,………is known as Fibonacci series. Obtained in general solution x(k) in closed form. Show that the liming value of x(k+1)/ x(k) as k ∞⎯→⎯ is (1+√5)/2 or approx 1.6180. Date: 2065/5/15 Example 05: (c ) x(k-2) –x(k-1) + 0.25x(k) = 4 (k-2) X(0 ) = 1 X(k) = 0 for, K < 0 and u(k) = 1 for k ≥ 0 Solution: X(k-2) - x(l-1) + 0.25x(k) = u(k-2) Z[x(k-2)] – z[x(k-1)] + 0.25 z [ x(k0] = z [ u(k-1)] Or, z-2 x(z) – z-1 x(z) + 0.25 x(z) = z-2 u(z) Or, x(z) [z-2 –z-1 + 0.25] = z-2 . 1/(1-z-1) X(z) = z-2/(1-z-1)(z-2-z-1 +0.25) Or x(z) = z.z2 /z2 ( z-1)(1-z+0.25) = 4z/(z-1)(z-2)2 X(z)/z = 4/(z-1) (z-2)2 = A/(z-1) + B/(z-2) + C/(z-2)2 By partial fraction method we get, A = 4, B = -4, C =4. Therefore, X(z)/z = 4 [ 1/(z-1) – 1/(z-2) + 1 (z-2)2 ] X(z) = 4 [1/(1-z-1) – 1/1-2z-1 + z-1/(1-2z-1)2 Or, X(z) = 4 [ z/z-1 – z/(z-2) + z/(z-2)2 ] Taking inverse z transform on both sides x(k) = 4[1k – 2k + 2k-1.k] , k = 0, 1, 2……. = 4 – 2k+2 + k.2k+1 Example: 06 Given, x(k+2) = x(k+1) + x(k) , x(0) = 0 and x(1) = 1 To get x(k) And to prove,

Lim k ∞ x(k+1) /x(k) = (1+ √5)/2 = 1.6180 We have , X(k+2) = x(k+1) + x(k) Z[x(k+2)= z[z(k+1) +x(k) ] Z2x(z) – z2 x(0) – 2x(1) = z x(z) – 2x(0) + x(z) Or, z2x(z) – 2x(z) – x(z) = z2x(0) + 2x(1) – 2x(0) Or, x(z) =

12 −− zzz

1)(

2 −−=

zzz

zzx

Now, the roots of z2-z -1 are

=a

acbb2

42 −±−

=1.2

)1(1.411 −−±−

=2

51±

=2

51+

=2

51−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

=−−

=

251

251

11

)(2

zzzz

zzzx

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

=

251

251 z

B

z

A

We get, A = 1/√5 , B = -1/√5



20

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

=

251

1

251

15

1)(

zzzzx

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

=−− 11

2511

1

2511

15

1)(

zz

zx

Taking inverse z-transform we get,

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

kk

kx2

512

515

1)( …………….(i)

Similarly we can write,

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=+

++ 11

251

251

51)1(

kk

kx ……….(ii)

From (i) and (ii)

kk

kk

kxkx

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

=+

++

251

251

251

251

)()1(

11

Here,

251− < 1 when k ∞

Lim k ∞ 02

51→⎟⎟

⎠

⎞⎜⎜⎝

⎛ −k

The terms k

⎟⎟⎠

⎞⎜⎜⎝

⎛ −2

51 and 1

251

+

⎟⎟⎠

⎞⎜⎜⎝

⎛ −k

can be neglected when

k ∞ , Thus the equation (iii) become,

k

k

kxkxk

⎟⎟⎠

⎞⎜⎜⎝

⎛ +

⎟⎟⎠

⎞⎜⎜⎝

⎛ +

=+

∞→

+

251

251

)()1(lim

1

= 2

51+ = 1.6180 proved

Impulse Sampling: Let us consider an ideal sampler commonly called an impulse sampler. We assume x(t ) = 0 for t< 0. The sampler o/p is equal to the product of the continuous time i/p x(t) and train of unit impulse δT(t) which means the sampler may be considered as a modular with x(t) as the modulating signal and δT(t) as the carrier as shown in fig1 below.

Modulatorx(t) x*(t)=x(t)×δΤ(t)

δΤ(t)

Figure:1 Let us consider x*(t) to represent the impulse sampled o/p. The sampled signal x*(t), a train of impulses, can thus be represented

by the infinite summation, i.e x*(t) = ∑∞

=

−0

)()(k

kTtkTx δ ……..(i)

x*(t) = ........)2()2()()()()0( +−+−+ TtTxTtTxtx δδδ ------- (ii) Here, we are defining ,

∑∞

=

−=0

)()(k

T kTtt δδ ………….(iii)



21

Now taking the laplace transform of equation (iii) we get ℒ[x*(t)] = x*(s) = x(0) ℒ[δ(t)]+x(T)ℒ[δ(t-T)]+ x(2T)ℒ[δ(t-2T)]+….. X*(s) = x(0).1+x(T).e-Ts + x(2T).e-2Ts +…….

X*(s) = ∑∞

=

−

0)(

k

kTSekTx …………….(iv)

We define esT = z ……….(v) Then equation (iv ) can be rewritten as

X*(s) = ∑∞

=

−

0)(

k

kzkTx = X(z)

i.e )(ln.1)(* zX

zT

ssX =

= ……………(vi)

Data hold:

zero order hold (ZOH)x(t) x(kT) h1(t)

Fig 2(a): Sampler and a zero order hold.

Gho(s)x(t) h2(t)

H2(s)δΤX*(s)

X*(s)

Fig 2(b): Mathematical Model that consist of a sampler and a Transfer function Gho(s) for ZOH.

Data hold is a process of generating a continuous time signal h(t) from a discrete time sequence x(k). A hold ckt covert the sampled signal into a continuous time signal. A hold ckt hold the amplitude of the sampled from one sampling instant to the next. Such a data hold is called zero order hold or clamper or stair case generator. The o/p of the zero order hold is a staircase function. Assuming x(t) = 0 for t<0, The ZOH circuit smoothens the sampled signal to produce the signal h(t) , which is constant for the last sampled value until the next sample is available, i.e , h1(t+kT) = x(kT) for 0 ≤ t ≤ T …………..(i) The o/p h1(t) may be assumed as h1(t) = x(0) [ 1(t) -1(t-1)]+x(T) [ 1(t-1)-1(t-2T)] + x(2t)[1(T-2t) -1(t-3T)]+ ………….

h1(t) = [ ]))1((1)(1)(0

TktkTtkTxk

+−−−∑∞

=

…………..(ii)

We know that ℒ[1(t-kT)] = e-kTs/s Therefore , the laplcae transform of equation (ii) be written as:

ℒ[h1(t)] = H1(s) = ∑∞

=0)(

kkTx ℒ [1(t-kT) -1(t-(k+1)T)]

= ∑∞

=0)(

kkTx

Se

se TSkkTS )1( +−−

−

= ∑∞

=0)(

kkTx

see TSkkTS )1( +−− −

H1(s) = se TS−−1 KTS

kekTx −

∞

=∑ .)(

0………..(iii)

But from fig (ii) b,



22

ℒ[h2(t)] = H2(s) = H1(s) Thus,

H2(s) = se TS−−1 KTS

kekTx −

∞

=∑ .)(

0……….(iv)

From fig (ii)b, we can write , H2(s) = Gho(s) × X*(s) ………(v) But we know

X*(s) = KTS

kekTx −

∞

=∑ .)(

0

Thus from equation (iv) become,

H2(s) = se TS−−1 X*(s)

Gho(s) = se TS−−1 ………..(vi)

Date: 2065/5/17 z-transform by convolution integral method: The convolution integral is defined by

dpe

pXj

sX psT

jc

jc)(1

1)(21)(*

−−

∞+

∞− −= ∫π

……….(i)

Where, the integration is along the line from c-j∞ to c+j∞ and this line is parallel to imaginary axis in the p-plane and separate the poles of X(P) from those of

]1[1

)( psTe −−−

Equation (i) also be written as

∫ −−−= dp

epX

jsX psT )(1

)(21)(*π

The above integral is equal to the sum of residues of X(p) in the closed container i.e

∑ ⎥⎦⎤

⎢⎣⎡

−=

−−)(

1)()(* )( pXofpolesat

epXofresiduesX psT

But we know that eTs = z X*(s) = X(z)

∑ ⎥⎦⎤

⎢⎣⎡

−= )()()( pXofpolesat

ezzpXofresiduezX Tp

By changing the complex variable rotation from p to s obtained

∑ ⎥⎦⎤

⎢⎣⎡

−= )()()( sXofpolesat

eszzsXofresiduezX

Assume that X(s) has poles S1 , s0 ……sm. If a pole at s = sj is a simple pole , then residue.

⎥⎦⎤

⎢⎣⎡

−−⎯→⎯= TSjjj ez

zsXssssk )().(lim

If a pole at s = si is a multiple pole of order ni , then the residue ki

= ⎥⎦⎤

⎢⎣⎡

−−⎯→⎯

− −

−Ts

nj

ni

nii

i ezzsXss

dsd

ssn

)(.)()!1(

1

1

1lim

Example:01: Obtained the z-transform of X(s) =

)1(1

2 +ss by the

use of convolution integral method. Solution: X(s) =

)1(1

2 +ss

))(1()(

2 TSTS ezssz

ezzsX

−+=

−

Thus by convolution integral method,

∑ ⎥⎦

⎤⎢⎣

⎡−+

= )())(1(

)( 2 sXofpolesatezss

zofresiduezX TS



23

= k1 + k2

Where, k1 = ⎥⎦

⎤⎢⎣

⎡−+

+−⎯→⎯))(1(

).1.(1 2lim

TSezsszss

= )(1 Tez

z−−−

K1 = )( Tez

z−−

K2 = ⎥⎦

⎤⎢⎣

⎡−+

⎯→⎯− − ))(1(

.0)!12(

12

2limTsezss

zsdsd

s

= ⎥⎦

⎤⎢⎣

⎡−+

⎯→⎯))(1(

.0 2lim

Tsezssz

dsd

s

= 2

2

)1( −++−

zzTzz

Thus X(z) = k1 +k2

= )( Tez

z−−

+ 2

2

)1( −++−

zzTzz

= Tez −−− 111 + 21

11

)1(1

−

−−

−++−

zTzz

Example:02: Given )(.1)( sGsesX

TS−−= . Prove that

⎥⎦⎤

⎢⎣⎡−= −

ssGzzzX )()1()( 1 OR consider the zero order Hold circuit

succeeded by the plant with transfer function G(s). Explain how would you determine the z-transform of such combined system. Soluotion: Suppose the transfer function G(s) follows the zoH. Then, the product of the transfer function of ZOH and G(s) becomes:

)(.1)( sGsesX

TS−−=

X(s) = ⎥⎦⎤

⎢⎣⎡− −

ssGe TS )()1(

X(s) = )().1( 1 sGeTS− X(s) = )(.)( 11 sGesG TS−− X(s) = )()( 11 sXsG − Where X1(s) = e-TS.G1(s) ℒ -1[X1(s)] = ℒ -1[e-TSG1(s)] = ℒ -1[e-TS]*ℒ -1[G1(s)] ℒ -1[X1(s)] = g0(t) * g1(t)

x1(t) = ∫ −t

dgtg0

10 )(*)( τττ

But, ℒ -1[e-TS] = δ (t-T)

Thus , x1(t) = ∫ −−t

dgTt0

1 )().( τττδ

= g1(t-T) Hence by writing z[g1(t)] = G1(z) Thus, Z[x1(t) ] = z[g1(t-T)] X1(z) = z-T.G1(z) Thus , X(s) = G1 (s) –X1(s) Or, X(z) = z[g1(t)]-z[X1(t)] = G1(z) –z—T.G1(z) = G1(z)[1-z-T] = [1-z-T].z[G(s)/s] [G1(s) = z[(G(s))/s] ∴ X(z) = (1-z-T) z.(G(s)/s) For T = 1, X(z) = (1-z-1).z [G(s)/s] proved Criteria For stability in Z-domain:



24

Example: 01: consider the closed loop control system shown in fig (i) below. Determine the stability of system when k = 1. Fig.

G(s) = k. )1(

1.1+

− −

ssse s

G(s) = .)1(

12 +− −

sse s

[since k = 1]

G(z) = Z[G(s)]

= z ⎥⎦

⎤⎢⎣

⎡+

− −

)1(12 ss

e s

= )1)(3679.0(

2642.03679.0−−

+zz

z

Now the closed loop transfer function of G(z) is C(z)/R(z) = G(z)/[1+G(z)] Therefore , the characteristics equation is given by , 1+G(z) = 0 1+

)1)(3679.0(264.03679.0

−−+

zzzz

Or (z-0.3679)(z-1) + 0.3679z+0.2642 Z2-z+0.6321 = 0 From which we get, Z2 –z+0.6321 = 0 Form which we get, Z1 = 0.5+j0.6181 Z2 = 0.5 – j0.6181 ∴ |z| = |z1| = |z2| = √ (0.52 + (0.6181)2) = 0.7950 Since the |z| < 1 , the system is stable.

The jury stability test: Assume, P(z) = anzn+ an-1zn-1+ an-2zn-2 + ……. a1z+a0 ……..(i) Where, An > 0 , then for stable system the following condition should be specificed. 1. |a0|> an 2. P(z) |z =1 > 0 3. p(z) |z =-1 > 0 if n = even < 0 if n = odd 4. |bn-1| >|bo|

|cn-2| > |co| . . |q2| > |qo|

Where,

bk = ⎥⎦

⎤⎢⎣

⎡

+

−−

10

1

k

knn

aaaa k = 0,1,2…….n-1

ck = ⎥⎦

⎤⎢⎣

⎡

+

−−−

10

21

k

knn

bbab k = 0,1,2…….n-2

.

.

qk = ⎥⎦

⎤⎢⎣

⎡

+

−

10

23

k

k

PPPp k = 0,1,2

Date: 2065/5/22 The jury stability Test: P(z) = a0zn+ a1zn-1+ a2zn-2 + ……. An-1z+an ……..(i) Where,



25

a0 > 0 then, for a stable system the following conditions should be fulfilled.

1. |an| > a0 2. p|(z)|z =1 > 0 3. P(z)|z= -1 > 0 n = even < 0 n = odd 4. |bn-1| > |b0|

|Cn-2| > |c0| |q2| > |qo|

Example:01: construct the Jury stability table for the following equation. P(z) = a0z4+ a1z3+ a2z3 +a3z+a4 ,Where a0 > 0. Write the stability conditions since n = 4 the stability conditions are:

(i) |a4| > a0 (ii) p|(z)|z =1 = a0+ a1+ a2 +a3+a4 (iii) P(z)|z= -1 = a0- a1+ a2 -a3+a4 > 0 [since n = 4 = even ] (iv) |b3| > |b0|

|C3| > |c0| The Jury stability table can be constructed for n = 4 as follows: Row z0 Z1 Z2 Z3 Z4 a4 a0 = b3 a3 a4 a4 a1 = b2 a0 a3 a4 a2 = b1 a0 a2 a4 a3 = b0 a0 a1 b3 b0 = c2 b0 b1 b3 b1 = c1 b0 b2 b3 b2 = c0 b0 b1

Example:02: Examine the stability of following characteristics equation. P(z) = z4-1.2z3+ 0.07z3 +0.3z - 0.08 = 0 Use Jury stability test. Solution: P(z) = z4- 1.2z3+ 0.07z3 +0.3z - 0.08 = 0 Here, a0 = 1 , a1 = -1.2 , a2 = 0.07 a3 = 0.3 , a4 = -0.08 The conditions for stability in Jury tests are as follows.

1. |a4| > a0 |-0.08| < 1 (which is true)

2. p|(z)|z =1 = 14- 1.2×13+ 0.07×12+0.3×1- 0.08 = 0.09 > 0 (which is true) 3. Since n = 4 = even , p|(z)|z =-1 = 1-+ 1.2+ 0.07- 0.3- 0.08 = 1.89> 0 (true) 4. |b3| > |b0| |c3| > |c0|

Row z0 Z1 Z2 Z3 Z4 -0.08 1 = -0.993 1 -0.08 -0.08 -1.2 = 1.176 1 0.3 -0.08 0.07 = -0.0756 1 0.07 -0.08 0.3 = -0.204 1 -1.2 -0.9936 -0.204 = 0.9456 -0.204 -0.0756 -0.9936 -0.0756 = -1.1838 -0.204 -1.176 -0.9936 1.176 = 0.315



26

-0.204 -0.0756 |b3| > |b0| i.e |-0.9936| > |-0.0204| 0.09936 > 0.204 (true) Again |c2| > |c0| |0.9456| > |0.3150| (ture) Example: 03: A control system has the following characteristics equation P(z) = z3 – 1.3z2 – 0.08z + 0.24 = 0 Determine the stability of the system by jury stability test. Example:04: Examine the stability of the characteristics equation given by : P(z) = z3 – 1.1z2

– 0.1z + 0.2 = 0 . Use jury method. Example: 05: Consider the discrete time unity feed back control system whose open loop transfer function is given G(z) =

)1)(3679.0()2642.03679.0(

−−+

zzzK Determine the range of k for

stability by the use of Jury method. Solution: G(z) =

)1)(3679.0()2642.03679.0(

−−+

zzzK

The close loop transfer function is T(z) = G(z)/ 1+G(z) of whose characteristics equation is given by 1+G(z) = 0 Or, 1+

)1)(3679.0()2642.03679.0(

−−+

zzzK = 0

P(z) = z2 + ( 0.3679k – 1.3679)z + 0.3679+0.2642k = 0 Here , a0 = 1

a1 = (0.3679k – 1.379) a2 = 0.3679 + 0.2642k For the system to be stable the conditions are :

(i) |a2| < a0 i.e | 0.3679+0.2642K|< 1 0.2642K < 0.632 K < 2.3925

(ii) p|(z)|z =1 = 1+ 0.3679k- 1.3679 + 0.3679+ 0.262K K > 0 (iii) p|(z)|z =-1 = (1- 0.3679k+ 1.3679 + 0.3679+ 0.262K ) > 0 = 2.7358 – 0.1037k > 0 2.7358 > 0.1037k K < (2.7358)/(0.1037) K < 26.3818 Combining (i) (ii) and (iii) conditions K < 2.3925 K > 0 0 < K < 2.3925 Stability Analysis by use of the Bilinear transformation and Routh stability: The bilinear Transformation is defined by:

11

−+

=wwz

11

−+

=zzw

We choose ‘w’ so that W = σ +jω Example:01: Consider the following characteristics equations P(z) = z3 – 1.3z2 – 0.08z +0.24 = 0 Determine the stability of the system using Bilinear transformation and Routh stability criterion



27

Solution: Here, P(z) = z3 – 1.3z2 – 0.08z +0.24 = 0 Using Bilinear Transformation , i.e

11

−+

=wwz , we get P(z) as,

P(z)|z=(w+1)/(w-1) = P(w) = [(w+1)/(w-1)]3 – 1.3 [(w+1)/(w-1)]2 – 0.08[(w+1)/(w-1)] + 0.24 = 0 Or w3 – 7.571w2 – 36.43w – 14.14 = 0 The Routh stability criteria is checked by routh arrow as follows: W3 1 -36.43 W2 -7.75 -14.14 W1 -38.30 0 W0 -14.14 The above equation and array show that there is a sign change in the following system is not stable. It indicates that there is a pole in the right half of w plane which in turn indicates that there is a pole outside the unit circle in z-plane. Example: 02: Consider y(k) = 0.6y(k-1) – 0.81y(k-2)+0.6y(k-3) – 0.12y(k-4) = x(k) where, x(k) is the input and y(k) is the output of the system. Check the stability of the system using Jury stability method. Solution: Y(z) -0.6z-1y(z) – 0.81z-2 y(z) + 0.6z-3y(z)- 0.12z-4 y(z) = X(z)

4321 12.06.081.06.011

)()(

−−−− −+−−=

zzzzzXzY

12.06.081.06.0 234

4

−+−− zzzzz

For which the characteristics equation is Z4 – 0.6z3 – 0.81z2 + 0.6z – 0.12 = 0

Date: 2065/6/5 S-plane to z-plane mapping: Mapping of the left half of s-plane into z-plane: We know z and s are related by the equation, Z = eTS ………….(i) This means that a pole in s-plane can be located in the z-plane through transformation Z = eTS , Also we know that , S = σ +jω ………….(ii) Thus equation (i) become , Z = eT(σ +jω) = eσ T. ejωT Or in general, Z = eTσ . e(Tw + 2 π k) ……….(iii) We see that the poles and zeroes in s-plane , where frequencies differ in integral multiple of the sampling frequency 2 π /T are mapped into the some locations in the z-plane . This means that there are infinite may values of s for each value of z. Since σ is –ve in the left half of the s-plane , the left arm of the s-plane corresponds to |z| = eT(‐σ ) = e-T σ < 1 The jw- axis in the s-plane corresponding to |z| = 1 (∴ σ =0) i.e the imaginary axis in the s-plane (the line σ =0) corresponding to



28

the unit circle in the z-plane and interior of the unit circle corresponds to the left half of the s-plane.

jw

Complementarystrip

Primarystrip

Complementarystrip

s-plane

Re3

im

z-plane

Re

(i) Mapping of constant –attenuation loci: A constant attenuation line ( a line plotted as σ = constant) is the splane maps into a circle of radius z = eT σ centered at the origin in the z-plane as shown below.

jw

s-plane

0 σ2-σ1σ

im

Re1

eTσ2

e-Tσ1

z-plane

-σ1

jw

s-plane

im

Re1

z-plane

(ii) Mapping of constant frequency Loci: A constant frequency loci w = w1 , in the s-plane is mapped into a radial line of constant angle Tw1 ( in radian) in the z-plane as shown in the fig. below.

σ

jw

w1

w2

-w1

s -plane

w1T

w1Tw2T

im

Re

z -plane (iii) Mapping of constant damping ratio. (ξ ) A constant damping ratio line (a radial line) in the s-plane is mapped into a spiral in the z-plane as shown in figure below.



29

ξ = ξ1

s -plane

Re

z -plane In s-plane a constant damping ratio is related to it by the equation . s = - ξ wn + jwn √ (1+ξ2) = - ξ wn + jwd Where, Wn = natural frequency (rad/sec) Wd = damped frequency (rad/sec) = wn√ (1-ξ2) Thus , Z = eTS = eT(-ξwn+jwd) = e(-ξ Twn+jTwd)

Z = exp. ⎥⎦

⎤⎢⎣

⎡+

− s

d

s

d

ww

jww

πξπ 2

12

2

Hence ,

|z| = exp ⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−

s

d

ww

21

2

ξ

πξ

∠ z = s

d

ww

π2

Example:01: For a given region in s-plane draw the corresponding region in z – plane.

σ=−σ2 σ=−σ1

jw1

-jw2

jw

e−σ1Τ

e−σ2Τ

1

w1T(2π −w2T)

2065/6/5

Reconstruction of original signal from samples: Instantaneous sampling (i.e when the duration of the sampling pulse τs –0 or delta function )is referred to as ideal sampling . In this case the sampling function is the train of impulses, i.e Sδ (t) = ∑ ∞

k= - ∞ δ (t- kTs ) Therefore , the sampled signal for ideal sampling can be written as Xδ(t) = x(t) x sδ (t) = x(t) x ∑ ∞

k=- ∞ δ (t-kTs) The fourier transform of the above expression is the convolution Of the fourier transformation of x(t) and sδ (t) , i.e xδ’(f) = x(f) *sδ (f) where, Xδ’(f) = f[sδ (t)] = fs ∑ ∞

n=- ∞ δ (f-nfs) ∴ xδ(f) = x(f) *fs ∑ ∞

n=- ∞ δ (f-nfs) ∴ xδ(f) = fs ∑ ∞

n=- ∞ δ (f-nfs) Xδ (f) = fsX(f) assuming the condition fs ≥ 2fx is satisfied , let us



30

Pass the sampled signal through ideal LPF with the following parameters HLPF (f) = K for fx ≤ f ≤ fs –fx = 0 , otherwise At the output of LPF , we ge , Xδ(f) HLPF (f) = [ fs ∑ ∞

n= - ∞ x(f- nfs)] HLPF(f) Xδ (f) HLPF(f) = X(f) …………….(i) As HLPF(f) = 0 for the frequency components higher then fx , the spectrum at the output of LPF for K = 1/fs will be x(f) , the spectrum of the message signal . Which is shown in equation (i) In time domain X(t) = f-1[x(f)] = f-1[xδ(f) HLPF(f)] ∴ x(t) = xδ (t) * hLPF(t) Where hLPF(t) is the impulse response of ideal LPF which is expressed as. hLPF (t) = 2BTs sinc [ 2Bt] Where, Ts = 1/fs B = fx , the higher frequency component. Thus signal at the output of low pass filter for

Xδ(t) = x(kTs) ∑∞

−∞=n x(kTs) δ(t- kTs) at t = kTs

Will be,

x(t) = x(kTs) ∑∞

−∞=n x(kTs) δ(t- kTs)*2BTs sinc [2Bt]

x(t) = 2BTs ∑∞

−∞=n x(kTs) sinc [2B(t-kTs)] ……….(ii)

The above equation (ii) shows that the original message signal

Can be reconstructed from its sampled values x(kTs) if the sampling is done at Ts ≤ 1/2fx Chapter: 3 Analysis of control system: The pulse Transfer Function:

G(z)X(z) Y(z)

The transfer function for the continuous time system relates the Laplace Transform of the continuous time output to that of continuous time input while the pulse transfer function relates the z-transform of the output at the sampling instants to that of the sampled input. The pulse transfer function is defined by G(z) = Y(z)/X(z) …………..(i) Of the discrete time system As seen from the above equation we write, Y(z) = G(z) x(z) ……..(ii) Equation (ii) may also be written as, Y*(s) = G*(s) x*(s) ……………..(iii) General procedure for obtaining Pulse transfer function:

x(t)

y(t)G(s)δΤ

δΤ

x*(t)

X*(z)y*(t)Y(z)

Figure (ii) (a)

x(t)X(s)

y(t)

Y(s)G(s)

Figure (ii) (b)



31

The pulse transfer function for the figure (ii) (a) will ve , G(z) = Y(z)/X(z) Also , the transfer function for the system shown in fig (ii) (b) will be , G(s) = Y(s)/X(s) The important fact to be noted is that the pulse transfer function for the system in fig (ii) (b) is not z-transfer of z[G(s)], because of the absence of the instant sampler. The presence or absence of the input sampler is crucial in determining the pulse transfer function of the system. This difference is explained below. Consider the case as shown in fig (ii) (a) ℒ [y(t)] = Y(s) = G(s) X*(s) Taking starred Laplace transform on both sides, we get, Y*(s) = G*(s) X*(s) Or, Y(z) = G(z) X(z) ……………(i) Now let us consider the case for fig (ii) (b) , ℒ [y(t)] = Y(s) = G(s) X(s) Taking Starred LT on both sides, Y*(s) = [G(s)X(s)]* Y*(s) = [GX(s)]* Where, GX(s) = G(s)X(s) Or Y(z) = z[GX(s)] …………………(ii) Thus equation (i) and (ii) clearly describe the presence and absence of instant sampler.

Date:2065/6/12 Pulse Transfer function of cascaded elements:

G(s) H(s)

x(t)x*(t) y(t) y*(t)u(t) u*(t)

δΤ δΤ δΤ Fig. 3(a)

δΤ

y(t) y*(t)G(s) H(s)x*(t)x(t)

Fig. 3(b) Consider the system shown in fig (iii) (a) we can write that G(s) = G(s).X*(s) ……………..(i) Y(s) = H(s).U(*(s) ……….(ii) Now taking starred laplace transform of equation (i) and (ii) U*(s) = G*(s).X*(s) …………(iii) Y*(s) = H*(s) U*(s) …………….(iv) From equation (ii) Y*(s) = H*(s)U*(s) = H*(s).G*(s).X*(s) Or, Y*(s)/X*(s) = G*(s). H*(s) Taking z-transform of the above equation. Y*(z)/X*(z) = G(z).H(z) …………….(v) Which is the pulse transfer function between i/p x*(t) and o/p y*(t) Now consider the cascaded system shown in fig (iii) (b) Y(s) = G(s)H(s).X*(s) Or Y(s) = GH(s).X*(s) Now taking the starred laplace transform of the above equation , we get , Y*(s) = [GH(s).X*(s)]* Y*(s) = GH*(s).X*(s) Or Y*(s)/X*(s) = GH*(s) Now taking z-transform on both sides we get, Y(z)/X(z) = GH(z) …………………….(vi) Which is the required PTF for fig (iii) (b) From equation (v) and (vi) , it is to be noted that, G(z)H(z) ≠ GH(z)



32

Example:01: Consider the system shown in figs (iii) a & b obtained the pulse transfer function (PTF) Y(z)/X(z) for each of these tow systems. G(s) = 1/(s+a) H(s) = 1/(s+b) Y(z)/X(z) = G(z)H(z) = z[G(s)] z[H(s) ] = Z[1/(s+a)] z[1/(s+b)] Y(z)/x(z) = 1/(1-e-aTz-1). 1/(1-e-bTz-1) for fig (iii) (a) For fig (iii) (b) Y(z)/X(z) = GH(z) = z[GH(s)] =z[G(s)H(s)] =z[1/(s+a).1/(s+b)]

= z ⎥⎦⎤

⎢⎣⎡

++

+ bsB

asA

=z⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

+−

−+

+−

bsab

asab

11

= z ⎥⎦

⎤⎢⎣

⎡⎥⎦⎤

⎢⎣⎡

+−

+− bsasab11.1

Y(z)/X(z) = ⎥⎦

⎤⎢⎣

⎡⎥⎦⎤

⎢⎣⎡

−−

−− −−−− 11 11

11.1

zezeab BTAT for fig (iii) (b)

Thus we conclude that G(z).H(z) is not equal to function GH(z). PTF of a closed loop system:

+ -G(s)

H(s)

R(s) C(s)E(s) E*(s)

δΤ

Fig(iv) closed loop system.

From fig. (iv) E(s) =R(s) – C(s)H(s) ………….(i) Where, C(s) = G(s) E*(s) ……….(ii) Thus equation (i) become E(s) = R(s) –G(s)H(s)E*(s) …………….(iii) Taking starred Laplace transform E*(s) = R*(s) – [ G(s)H(s) E*(s)]* E*(s) = R*(s) – GH*(s)E*(s) ……………..(iv) Or, E*(s)[1+GH*(s)] = R*(s) ………..(v) E*(s) = R*(s)/(1+GH*(s)) ………….(vi) Now taking Starred laplace transform of equation (ii) , C*(s) = G*(s) E*(s) = G*(s).R*(s)/(1+GH*(s)) C*(s)/R*(s) = G*(s)/1+GH*(s) …………..(vii) Now taking z-transform of equation of (vii) we get, C(z)/R(z) = G(z) / (1+GH(z)) …………..(viii) which is the required PTF for closed loop system. Assignment # 3:

(1) Prove the given PTF for the closed loop system shown in Table 3.2 (page 112) form k. Ogata book .

Closed Loop PTF of a digital control system:

+ -G*(s)R(s) E(s) E*(s)

δΤD G (s)p

C(s) -TS1-es

U(s)m*(s)



33

Fig. (v) From fig (v) , let , 1-e-Ts/s. Gp(s) = G(s) ……..(i) Also, C(s) = G(s) GD*(s) E*(s) Or C(s) = G*(s) GD*(s) E*(s) ………(ii) In term of z-transform, equation (iii) can be written as: C(z) = R(z) – C(z) ………(iv) Thus , equation (iii) becomes C(z) = G(z) GD(z)[R(z)-C(z)] C(z) = [ 1+G(z)GD(s)] = G(z) GD(z) R(z) Or C(z)/R(z) = G(z)GD(z) / 1+G(Z)GD(z) C(z)/R(z) = GD(z)G(z)/1+GD(z)G(z) ………(v) Which the required closed loop PTF of a Digital control system. PTF of a digital PID controller: The PID control action in analog controller is

m(t) = kp ⎥⎦

⎤⎢⎣

⎡++ ∫

t

di dt

tdeTdtteT

te0

)()(1)( ………….(i)

where e(t) is the i/p to the controller , m(t) is the o/p of the controller, k is the proportionality gain , Ti is the integral time and Td is the derivative time. To obtain the PTF for the digital PID controller , we first discritise equation (i) . Then , by approximating the integral term by the Trapezoidal summation rule and the derivative term by two-point difference formula. Thus equation (i) becomes at t = kT ,

m (kT) = kp

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−+

⎥⎦⎤

⎢⎣⎡ +−

+++

++

+

2)1()(.

2)()1(..............

2)2()(

2)()0(1)(

TkekTeT

kTekeTeTeTeeTT

KTe

d

i

=kp [ ]⎥⎦

⎤⎢⎣

⎡−−+

+−+ ∑

=

k

h i

d

i

TkekTeTThTeThe

TTKTe

1)1()(

2)())1((.)( ….(ii)

Let, )(

2)())1(( hTfhTeThe

=+− where, f(o) = 0

∑∑==

=+− k

h

k

hhTfhTeThe

11)(

2)()1( ………(iii)

The z-transform of equation (iii) will be,

= z ⎥⎦

⎤⎢⎣

⎡ +−∑=

k

h

hTeThe1 2

)())1((.

= z ⎥⎦

⎤⎢⎣

⎡ ∑=

k

hhTf

1)(.

= 1/(1-z-1). [F(z)-f(0)] = 1/(1-z-1). F(z) …….(v) Also ,

F(z) =z[f(hT)]= z ⎥⎦⎤

⎢⎣⎡ +−

2)())1(( hTeThe

F(z) = (1-z-1)/2.E(z) Thus from equation (iv) becomes,

z )()1(2

12

)())1((. 1

1

1zE

zzhTeThek

h−

−

= −+

=⎥⎦

⎤⎢⎣

⎡ +−∑ ……….(V)

Hence equation (ii) becomes by taking z-transform on both sides,



34

M(z) = z[m(kT)] = Kp⎥⎥⎦

⎤

⎢⎢⎣

⎡−+

−−

+ −−

−

)()1()()1(2

1.)( 11

1

zEzTT

zEz

zTTzE d

i

M(z) = kp [1+T/2Ti +(1+z-1)/(1-z-1)+Td/T .(1-z-1)].E(z)

M(z) = kp )()1.(1

1.2

1 11 zEz

TT

zTT

TT d

ii⎥⎦

⎤⎢⎣

⎡−+

−+− −

−

M(z) = )()1.(1

' 11 zEzk

zK

kp Di

⎥⎦

⎤⎢⎣

⎡ −+−

+ −−

……….(vi)

Where, kp’ = ⎟⎟⎠

⎞⎜⎜⎝

⎛−

ip T

Tk2

1 = proportional gain

Ki = ⎟⎟⎠

⎞⎜⎜⎝

⎛

ip T

Tk = Integral gain

Kd = ⎟⎠

⎞⎜⎝

⎛TT

k dp = Derivative gain

Form equation (vi) we can write that

GD(z) = M(z)/E(z) = ⎥⎦

⎤⎢⎣

⎡ −+−

+ −−

)1.(1

' 11 zk

zK

kp Di ………(vii)

Where, GD(z) is the closed loop PIF for digital PID controller equation (vii) is also know as the proportional form of PID control scheme.

digital control system

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