Digital Control System

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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:










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 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

Data Distribution:

















Downloaded from www.jayaram.com.np www.jayaram.com.np x(t)

xs(t)= x(t)×g(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 ∞ g (t ) = c o + ∑ n =1 2c n cos(nω s t ) Where, co = τ/ Ts = τf s Cn = f s τ sinc[nf s τ ] = co sinc[nf sτ] ωs =2 πf s The signal xs(t) can be expressed as xs = x(t)×g(t) ∞ = x(t)×[ c o + ∑n =1 2c n cos(nω s t ) ] = 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-f s)+2c2x(f-2f s)+…………+2cn x(f+nf s)+……. The above series can be graphically represented as:










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. f s ≥ f x In this case the original message signal spectra can be recovered by passing the sampled signal through low pass filter with bandwidth equaling to ± f x Distortion will occur while recovering the message spectrum if. f s ≤ f x The distortion in the above case is caused by the overlapping of side bands and message spectra.

The minimum sampling rate: f s min = 2f s 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 = 1/2f x(min) x(min) Where f x(min) x(min) = maximum frequency in the message spectrum.

Quantizing and Quantization error: x(t)

Xs(f)

2 1 (a) o

Xs(f) aliasign ditortion (fs<2fx)

xs(n) 1 2 3

4

5 6 7

t





















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Assume that the average signal power is x , SQNR for uniform quantization will be, SQNR = average power of signal/ average power of noise. n 2 QSNR = xˆ × 3 × 4 ……………………(iii) Where, 2 xˆ =

x 2 x

2

, 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)] =

∑ x(kT ) z

Again , equation (iii) can be reproduce in dB, as 3 (QSNR) dB = P x ( dB) + 10 log 10 + 10 log 10 ( 4) n

…………….(i)

k = 0

∞

= Normalized signal power.

− k

Or , X(z) = z[x(t)] =

∑ x ( k ) z k = 0

−1

…………….(ii), …………….(ii), for T=1

The z-transform defined by equation (i) and (ii) is one-sided z-












Or, z[1(t)] =

z z − 1

,

Or, z[a ] =

2. Unit Ramp Function: The unit ramp function is defined by ⎧⎪t for t ≥ 0 x(t ) = ⎨ ⎪⎩0 otherwise Or, x(KT) = KT for k = 0,1,2,3…………. Thus, ∞

X(z) = z[x(t)] = z[t] =

1

k

z >1

∑ t . z k = 0

− k

1 − az

−1

4. Exponential function: ⎧⎪e at for t ≥ 0 x (t ) = ⎨ otherwise ⎩⎪0

Where, x(kT) = e We have, -at

X(z) = z[e ] =

∞

∑e k 0

-akT

− akT − k

z

, K = 0,1,2………..












Z[e ] = jwt

Z[e- ] =

k

1

jwt

1 − e jwT z 1 1− e

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

−1

− jwT −1

z

jwt

-jwt

Therefore, z[sinwt] = z[ 1/2j(e -e )] jwt -jwt = 1/2j[z(e )-z(e )] 1 1 ⎡ ⎤ = 1/2j ⎢ − − jwt −1 ⎥ jwt −1 1 − e z ⎦ ⎣1 − e z Z[sinwt] =

z

−1

Proof: k

Z[ a x(k)] =

∞

∑

k

a x (k ) z

− k

∞

= ∑ x(k )(a −1 z ) − k

k = 0

k = 0

-1

= X(a z)

sin wt

1 − 2 z −1 cos wt + z − 2

4. Shifting theorem: If x(t) = 0 for t< 0 and











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Z[x(mT)] = z − n . ∑ x(mT ). z − m m=− n

Since ‘m’ must be zero and non-negative integer -n Therefore, z[x(mT)] = z X(z) For equation (ii)

∞

∑ x(k + n).Tz

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

Z[x(t+nT) = z[ x(KT+nT)] = z[x(K+n)T] Or, z[x(t+nT)] =

aT

= X(e z) aT = X (ze )

− k

6. Final Value Theorem:

































































































































































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)] -aT -1 -bT -1 Y(z)/x(z) = 1/(1-e z ). 1/(1-e z ) 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)] B ⎤ ⎡ A = z⎢ + ⎣ s + a s + b ⎥⎦ −1 ⎡1 ⎤ − b a ⎢ =z + b −a⎥ ⎢ s+a s+b ⎥ ⎣ ⎦ ⎡ 1 ⎡ 1 1 ⎤⎤ =z ⎢ .⎢ − ⎥⎥ ⎣ b − a ⎣ s + a s + b ⎦⎦ ⎡ 1 ⎡ 1 1 ⎤⎤

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) ………..(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:










Fig. (v) From fig (v) , let , -Ts 1-e /s. G p(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 digital PID controller: The PID control action in analog controller is t ⎡ de(t ) ⎤ 1 m(t) = k p ⎢e(t ) + ∫ e(t )dt + T d ⎥ ………….(i) T i 0 dt ⎦ ⎣ where e(t) is the i/p to the controller , m(t) is the o/p of the controller, k is the proportionality gain , T is the integral time and

m (kT) = k p ⎡ e( k − 1) + e(kT ) ⎤ ⎤ 1 ⎡ e(0) + e(T ) e(T ) + e(2T ) + + ...... + ........ ⎢e( KT ) + T T ⎢ ⎥⎥ 2 2 2 ⎣ ⎦⎥ i ⎢ ⎢ ⎥ e(kT ) − e(k − 1)T ⎢+ T d . ⎥ 2 ⎣ ⎦ ⎡ ⎤ T k e((h − 1)T ) + e( hT ) T d =k p ⎢e( KT ) + .∑ + [e(kT ) − e(k − 1)T ]⎥ ….(ii) T i h =1 T i 2 ⎣ ⎦ Let, e((h − 1)T ) + e( hT ) = f (hT ) where, f(o) = 0 k

∑

2 e( h − 1)T + e( hT )

h =1

2

k

=∑ f (hT ) ………(iii) h =1

The z-transform of equation (iii) will be, ⎡ k e((h − 1)T ) + e(hT ) ⎤ z ⎢.∑ ⎥ 2 ⎣ h =1 ⎦ k ⎡ ⎤ = z ⎢.∑ f (hT )⎥ ⎣ h =1 ⎦ -1 = 1/(1-z ). [F(z)-f(0)]

=










⎡ ⎤ T T 1 − z − M(z) = z[m(kT)] = K p ⎢ E ( z ) + . E ( z ) + d (1 − z 1 ) E ( z ) ⎥ −1 T i 2(1 − z ) T ⎣⎢ ⎦⎥ -1 -1 -1 M(z) = k p [1+T/2Ti +(1+z )/(1-z )+Td/T .(1-z )].E(z) ⎡ ⎤ T T T 1 M(z) = k p ⎢1 − + . + d .(1 − z −1 )⎥ E ( z ) −1 T ⎣ 2T i T i 1 − z ⎦ −1

K i ⎡ ⎤ M(z) = ⎢kp'+ + k D .(1 − z −1 )⎥ E ( z ) ……….(vi) −1 1 − z ⎣ ⎦ ⎛ ⎞ T ⎟ = proportional gain Where, kp’ = k p ⎜⎜1 − 2 T i ⎠⎟ ⎝ ⎛ T ⎞ K i = k p ⎜⎜ ⎟⎟ = Integral gain ⎝ T i ⎠ ⎛ T ⎞ K d = k p ⎜ d ⎟ = Derivative gain ⎝ T ⎠ Form equation (vi) we can write that K i ⎡ ⎤ GD(z) = M(z)/E(z) = ⎢kp'+ + k D .(1 − z −1 )⎥ ………(vii) −1 1 − z ⎣ ⎦ 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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