Digital Control Systems Lecture Notes PDF

# University of M’hamed BOUGARA of Boumerdes ## Institute of Electrical and Electronic Engineering ### Lecture EE451/Master 1 ## DIGITAL CONTROL SYSTEMS ### Prof. R. BOUSHAKI ## Chapter 3: Design of Discrete-Time Control Systems ### 3.1 Introduction: We will study different methods in order to find...

# University of M’hamed BOUGARA of Boumerdes ## Institute of Electrical and Electronic Engineering ### Lecture EE451/Master 1 ## DIGITAL CONTROL SYSTEMS ### Prof. R. BOUSHAKI ## Chapter 3: Design of Discrete-Time Control Systems ### 3.1 Introduction: We will study different methods in order to find discrete equivalent of analog controllers or filters. Using the following circuit: A circuit diagram consisting of a resistor, a capacitor, an input signal, and an output signal. The transfer function of this circuit has the form: $ \frac{y(s)}{x(s)} = \frac{a}{s+a} $ $ \Rightarrow ax(s) = sy(s)+ay(s) = ax(t) = \frac{dy(t)}{dt} +ay(t) $ We need to derive difference equation whose solution will approximate the solution of the differential equation, and then, we will get the equivalent discrete time filter (controller). ### 3.2 The backward difference method: Let's have: $ \frac{dy}{dt} =-ay+ ax $ Calculate the integral for t : (k-1)T <t≤kT $ \int_{(k-1)T}^{kT} \frac{dy}{dt} dt = -a \int_{(k-1)T}^{kT} y(t)dt+a \int_{(k-1)T}^{kT} x(t)dt $ $ y(kT)-y((k-1)T)=-a \int_{(k-1)T}^{kT} y(t)dt+a \int_{(k-1)T}^{kT} x(t)dt $ We will approximate the areas by taking: $ \int_{(k-1)T}^{kT} y(t)dt = y(kT).T (Backward: t=kT) $ And $ \int_{(k-1)T}^{kT} x(t)dt = x(kT).T $ A graph showing the backward method of integration. $ -1 So: y(kT) = y((k-1)T)-aT[y(kT)+x(kT)] \Rightarrow y(z)=z¯¹y(z)-aT[y(z)+x(z)] $ $ \frac{Y(z)}{X(z)} = G(z) = \frac{ -1 aT + aT a }{ 1- z¯¹ + aT } = \frac{ a }{ 1-z¯¹ + a } $ $ \frac{Y(z)}{X(z)} = \frac{a}{s+a} with: s = \frac { z -1 }{Tz} $ Mapping from S-plane to Z-plane Remark: Let's get interested into the stability region in the Z-plane using the mapping. A graph comparing the S-plane and Z-plane transformation. $ From : Re(S) <0>Re(\frac { 1 -z^{-1} }{T} )<0 ⇒ Re(\frac { z^{-1} }{T} )>0 From: Re(\frac { z^{-1} }{Tz} )<0 $ $ With: z=σ+jw So: Re(\frac { z^{-1} }{Tz} ) = Re[ \frac { σ +jw−1 }{T(σ+jw)} ]<0 $ $ \Rightarrow Re[( σ+jw−1)( σ-jw)]<0 ⇒σ²-σ+w²£0 $ $ \Rightarrow(σ-1/2)^2 +w²-1/4 <0 $ Circle: The center σ = 1/2; w = 0 ; Radius =1/2 ### 3.3 The forward difference method: $ \int_{(k-1)T}^{kT} y(t)dt = y((k-1)T).T (Forward: t=(k-1)T) $ $ \int_{(k-1)T}^{kT} x(t)dt = x((k-1)T).T $ A graph illustrating the forward method of integration. From: y(kT) = y[(k-1)T]-aT.[y((k-1)T]+x((k-1)T) $ \Rightarrow y(kT)=(1-aT)y[(k-1)T]+aT.x((k-1)T) $ $ \Rightarrow y(z) = (1-aT)z^*y(z)+aT.z¹.x(z) \Rightarrow G₁(z) = \frac {y(z)}{x(z)} = \frac {a}{1-z¯¹ + a } $ $ \frac{Y(z)}{X(z)} = \frac{a}{s+a} with:s = \frac { 1 -z^{-1} }{Tz-1 } = \frac {z-1}{T(z-1)} $ Mapping from S-plane to Z-plane Remark: Stability means Re(s) <0⇒ Re[ Z−1 / T ]<0⇒ Re(z) <1 Conclusion: Backward: t = kT $ S=\frac{z -1}{Tz} $ Forward: t = (k-1)T $ s = \frac {z -1}{T} $ A graph with the unit circle with the stability region shaded. ### 3.5 The matched pole-zero mapping method: We follow the given procedure: 1) The transfer function G(s) should be in a factored form. The poles (or zeros) of G(s) are mapped into the Z-plane using : - For finite poles and zeros: z = e^st - For infinite poles and zeros: z = -1 A graph depicting the mapping from the S-plane to the Z-plane. Remark: Adjust the gain of the discrete time function G₁(z) in order to match the gain of the continuous time G(s). Example: Find the discrete PTF: GD (z)? Adjust the gain K for low pass case. $ G(s) = \frac{a}{s+a} \Rightarrow \begin{cases} infinite zero \rightarrow z = -1 \\ finite pole: s = -a \rightarrow z = e^{sT} \Rightarrow z =e^{-aT} \end{cases} $ Solution: $ G_D(z) = K \frac {z + 1 }{ (z - e^{-aT} )} \Rightarrow \begin{cases} infinite zero: z = -1 \\ finite poles: z = e^{-aT} \end{cases} $ for s = 0 ⇒ z = 1 So G_D(1) = Let's adjust K: from z = e^{st} ; we have: When s → 0 then z→ 1 So G_D(1) = G(0) $ \Rightarrow \begin{cases} G_D(1) = G_D(z)|_{z=1} = \frac{2K}{(1-e^{-aT})}\\ G(0) = \frac{a}{s+a} |_{s=0}=1 \end{cases} \Rightarrow \frac{2K}{1-e^{-aT}} = 1 \Rightarrow K = \frac{1-e^{-aT}}{2} $ $ \Rightarrow G_D(z)=(\frac {1-e^{-aT}}{2})*(\frac {z+1}{z-e^{-aT}}) $ Examples: 1) G(s) = (s + b)/(s + a) 2) G(s) = s/(s + a) $ Find G_D(z)? \begin{cases} finite pole: s = -a \rightarrow z =e^{sT} \Rightarrow z = e^{-aT} \\ infinite zero: s \rightarrow +∞ \rightarrow z = -1 \end{cases} $ 1) G(s) = (s + b)/(s + a) →Gp(z) = K(z-e^{-bT})/(z-e^{-aT}) When s → 0 then z→ 1 $ G_D(1) = G(0) \Rightarrow K(\frac {1-e^{-bT}}{1-e^{-aT}}) = \frac{b}{a} \Rightarrow K = \frac{b}{a}(\frac {1-e^{-aT}}{1-e^{-bT}}) $ 2) G(s) = s/(s + a) → G_D(z) = K(z-1)/(z-e^{-aT}) Remark: For G_D(1) = G(0) we get K=0 (it is not required) When s → -∞ then z→ -1 $ G_D(-1) = G(-∞) \Rightarrow K(\frac {-2}{1-e^{-aT}})=-1 \Rightarrow K = \frac {1+e^{-aT}}{2} $ ### 3.5.1 Step Invariance method: G_D(z) has the same step response as G(s) at the sampling instants. $ G_D(z) = (1-z^{-1})Z^{-1}[\frac {1}{s}L^{-1}(G(s))] = (1-z^{-1})Z^{-1}[\frac {G(s)}{s}] $ ### 3.5.2 Impulse Invariance method: The impulse response of G_D(z) is T times the impulse response of G(s) at t=kT With: g_D(kT) = Z^{-1}[G_D(z)] is the impulse response of G_D(z) And: g(t) = L^{-1}[G(s)] is the impulse response of G(s) $ \Rightarrow g_D(kT)=T.g(t) \Rightarrow g_D(kT) = Tg(t) \Rightarrow G_D(z) = Z[g_D(kT)] = T.Z(g(t)) $ $ G_D(z) = T.G(z) With: G(z) = Z[G(s)] $ ### 3.6 Root Locus in Z-plane: The root locus of an (open-loop) transfer function G(s) is a plot of the locations (locus) of all possible closed-loop poles with proportional gain K. A block diagram of a closed loop system. Closed-loop poles are the solutions to the characteristic equation: 1+ K.GH(z) = 0 $ GH(z) = \frac { (z-z_1)(z - z_2).....(z- z_m) }{ (z - P_1)(z - P_2 ).....(z-P_n) } $ with m open-loop zeros z1, ...,zm and n open-loop poles p1, ...,pn (m≤n) How do the “n” closed-loop poles change as K varies from 0 to ∞? ### 3.6.1 Root Locus Overview - The root locus of a characteristic equation: $ 1+K \frac { (z-z_1)(z - z_2.)....(z- z_m) }{ (z - P_1)(z - P_2 ).....(z-P_n) } = 0 $ consists of n branches - Start from the n open-loop poles p1,..., pn at K=0 - m of them converge to the m open loop zeros z1,...,Zm - The other (n-m) diverge to ∞ along of the center of asymptotes: σ = ( Σ_{i=1}^n P_i - Σ_{i=1}^m z_i)/(n-m) with angles: θ = (2l+1)180°/(n-m) = ±90°,±270°,...with: 1=1,....,n-m Examples: A graph depicting the root locus of a system when n-m=0,1,2,3. n-m=2, the branches start at the poles and go to infinity. ### 3.6.2 Phase Angle Condition An arbitrary point z is on the root locus if and only if: $ [∠(z-z₁)+...∠(z−z_m)]-[∠(z-p₁)+...∠(z-p_n)] = (2l+1)180° $ Example: A graph displaying the angles of the root locus. z is on the root locus if and only if: for some integer 1 $ (φ + φ_2) - (θ_1 + θ_2) = (2l+1)180° $ ### 3.6.3 Magnitude Condition: For a point z that is known to be on the root locus, we can find the corresponding K: $ Form: 1+ K \frac { (z - z₁ ) (z - z_2 ).....(z- z_m ) }{ (z - p₁)(z - p_2 ).....(z-p_n) } = 0 \Rightarrow K = _ \frac { (z - p₁)...(z - p_n) }{ (z - z₁)...(z - z_m) } \Rightarrow K = \frac { z - p_1..z-p_n }{ z - z_1..z - z_m } $ A graph displaying the various angles. Characteristic equation: 1 + K(B(z)/A(z)) = 1 + K( (z -z_1)...(z-z_m) )/( (z-p_1)...(z-p_n) ) = 0 ### 3.6.4 Breakin/Breakaway Points σ_b Breakaway points (points of departure from the real axis) correspond to local maxima of K, whereas break-in points (points of arrival at the real axis) correspond to local minima of K. In order to find σ_b we solve 1+K(B(z)/A(z)) = 0,so K= -A(z)/B(z) Break in/Break away (σ_b) points are the solutions of dK/dz = 0 - The angle of departure from a complex pole p_n is given by: 180° – Σ_{i=1}^{n-1} ∠(p_n - p_i) + Σ_{j=1}^m ∠(p_n - z_j) Example: A closed loop system with a transfer function: G(s) = (1-e^-Ts)/s. 1)Solving the characteristic equation for K: $ 1+ K \frac { 0.368z+0.264 }{z^2 -1.368z+0.368} = 1 + K \frac { 0.368(z+0.7174) }{ (z-1)(z-0.368) } = 0 $ Calculate the: dK/dz = 0 and solve the equation for z to find (Breakin/Breakaway points). The Open-loop zeros and poles: $ K = -\frac {(z−1)(z-0.368)}{0.368(z+0.7174)} \Rightarrow \frac { dK }{dz} = \frac {d}{dz}( \frac {(z−1)(z-0.368)}{0.368(z+0.7174) }) = 0 $ The solutions for z (Breakin/Breakaway points) are: z = 0.65(so then K = 0.196) z = −2.08(sothen K =15) ### 3.6.5 Points, where the root cross the Unit Circle - This is the case when the closed-loop system changes from being stable to being unstable (or vice versa): - Correspond to marginal stability and can be found by - Root locus graphically - Routh-Hurwitz criterion - Jury Test How to Determine K using Routh-Hurwitz Criterion A block diagram of a closed loop system. G(z): is open loop Pulse Transfer (FTF) Characteristic equation: $ 1+G(z)=0⇒1+K \frac { 0.368(z+0.7174) }{ (z-1)(z-0.368) } = 0 $ Using the bilinear transformation: z = (1+(T/2)w)/(1-(T/2)w) and we have T=1s, then: $ ⇒(1-0.0381K)w²+(0.924-0.386K)w+0.924K = 0 $ Routh Hurwitz Table | w^2 | 1 - 0.0381K | 0.924K | |-----|------------------------|---------| | w^1 | 0.924 - 0.386K | | | w^0 | 0.924K | | Condition of stability (no change of sign) so: 1-0.0381K > 0 0.924-0.386K > 0 0.924K > 0 So the range of K for stability is: 0<K <2.39 Example: A block diagram of a closed loop system. We have G_12(z) = K/(1-z¯¹) = K(z-1)/(z-1) Draw the root locus: For t=0.5s $ G_p(z) = Z[Z.O.H \frac {1}{s+1}] = (1-z^{-1})Z[\frac{1}{s(s+1)}] $ $ =(1-z^{-1})Z[\frac {1}{s} - \frac{1}{s+1}] = (1 - z^{-1}) [\frac{z}{z-1} - \frac {z - e^{-T}}{z-1}] = \frac {1-e^{-T}}{z-e^{-T}} $ The forward PTF is (open loop) G_p(z) = (1-e^{-T})/(z-e^{-T}) $ G(z) = G_1(z).G_p(z) \Rightarrow G(z) = K \frac {(z-1)}{z-1} \frac{1-e^{-T}}{z-e^{-T}} = K \frac{1-e^{-T}}{z-e^{-T}} $ The characteristic equation is: 1+G(z) = 0 $ \Rightarrow 1+ \frac{K.z.(1-e^{-T})}{(z-1)(z-e^{-T})} =0 $ $ T = 0.5s \Rightarrow e^{-0.5} = 0.666 So: G(z) = \frac{0.393.K.z}{(z−1)(z−0.666)} $ We have: A graphical representation of the root locus of the system. $ 1+G(z) = 0 \Rightarrow K = \frac { (z−1)(z−0.666) }{0.393z} $ $ \frac {dK}{dz} = 0 \Rightarrow z^2 = 0.606 \Rightarrow σ_1,2 = ±√0.606 = ±0.778 $ For_σ_1 = 0.778⇒ K = 0.081 and For σ_2 = -0.778⇒ K =8.398 ### 3.7 Transit response Recall: $ Using: C(s) = \frac{w_n^2}{s^2 + 2ζw_ns + w_n^2} \times R(s) $ $ With: R(s) = \frac{1}{s}, step input. $ ζ :Damping factor; ω_n: Natural radian frequency; For 0< ζ <1> Under damping response. $ 1+√1-ζ^2 $ The roots are complex: S1,2 =-ζwn±jwn√1-ζ^2 $ c(t)=1-\frac {e^{-ζw_nt}}{√1-ζ^2} sin(w_n√1-ζ^2 t + a) $ with: sin(a) = √1-ζ^2 A graph showing the underdamped response of the system. We have: tp = π / (wn√1-ζ^2) Rise time: tr = 2.5/wn Delay time: td Settling time: ts ≈ 4/ζwn Damped natural frequency: ω_d = ω_n√1-ζ^2 $ We have: sin(a) = √1-ζ^2 So C(t)=1-\frac {e^{-ζw_nt}}{√1-ζ^2} sin(w_n√1-ζ^2 t + a) $ Discrete time case We will use the mapping between the S and Z planes. Example: Po ≤ 20% ; Po =100× (peak value of C(t)−1)/(step size) ; step size= 1 ts ≤20s tr≤8s $ P_0 = e^{-\zeta π/√1-ζ^2} ≤ 0.2 \Rightarrow ζ ≥ 0.5 $ $ t_s ≈ 4/σ ≈ 20 \Rightarrow σ ≈ 4/20 = 0.2 $ $ τ = t_r / w_n ≈ 8 \Rightarrow w_n ≈ 0.3125 $ A block diagram of the system. ### 3.8 Steady state errors: We now examine the effect of sampling on the steady state error. Consider the digital system given below: A block diagram of a sampled-data system with a zero-order hold block. From the figure, we have: E(z) = R(z)-HC(z) E(z) = R(z)-HG(z)E(z) $ \Rightarrow E(z) = \frac{R(z)}{1+GH(z)} $ $ With: GH(z)=(1-z^{-1})Z^{-1}[\frac {G_p(s)H(s)}{s}] $ Using the final value theorem, we find the steady state error: $ e_{ss}^* = e^*(∞) = lim_{t \rightarrow ∞} e^*(t) = lim_{k \rightarrow ∞ } e(kT) = lim_{z \rightarrow 1} (1-z^{-1})E(z) = lim_{z \rightarrow 1} (1-z^{-1}).[ \frac{R(z)}{1+GH(z)}] $ $ e_{ss}^* = lim_{z \rightarrow 1} (1-z^{-1}). [\frac{R(z)}{1+GH(z)}] $ ### 3.8.1 Position error: The Input is Unit Step: R(z) = z/(z - 1) Thus, the steady state error is: $ e_{ss}^* = lim_{z \rightarrow 1} (1-z^{-1}) [ \frac{z/(z-1)}{1+GH(z)} ] = lim_{z \rightarrow 1}[ \frac{1}{1+GH(z)} ] $ $ Defining: K_p = lim_{z \rightarrow 1} GH(z) \Rightarrow e_{ss}^* = \frac {1}{1+K_p} $ Static position error Remark: $ e_{ss}^* \rightarrow 0 \Rightarrow K_p \rightarrow ∞, we should have at least one pole at z=1 for GH(z) $ ### 3.8.2 Velocity error: The Input is Unit Ramp R(z) = Tz/(z-1)^2 ⇒ R(z) = (Tz^{-1})/(1-z^{-1})^2 The steady state error for this case is: e_{ss} = e(∞) $ \Rightarrow e_{ss}^* = e^*(∞) = lim_{z \rightarrow 1} (1-z^{-1}).[\frac{(Tz^{-1})/(1-z^{-1})^2}{1+GH(z)}] $ $ \Rightarrow e_{ss}^* = lim_{z \rightarrow 1} \frac { (1-z^{-1}) }{ (1-z^{-1})^2 } [\frac{Tz^{-1}}{1+GH(z)} ] $ $ Defining: K_v = \frac {1}{T} lim_{z \rightarrow 1}(1-z^{-1})^2GH(z): Velocity error const $ $ \Rightarrow e_{ss}^* = \frac {1}{K_v} $ Remark: $ e_{ss}^* \rightarrow 0 \Rightarrow K_v \rightarrow ∞ \Rightarrow GH(z) Should have at least 2 poles at z=1. $ ### 3.8.3 Acceleration error: The Input is Unit Parabolic : R(z) = T^2 z(z+1)/2 (z-1)^3 If we take : r(t) = t^2/2 u(t) ⇒ R(z)= (T^2 (1+z^{-1})z^{-1})/ (2(1-z^{-1})^3) ) The steady state error for this case is: e_{ss} = e(∞) $ \Rightarrow e_{ss}^* = lim_{z \rightarrow 1} (1-z^{-1}). [ \frac {T^2 (1+z^{-1})z^{-1} / (2(1-z^{-1})^3)}{1+GH(z)} ] $ $ \Rightarrow e_{ss}^* = lim_{z \rightarrow 1} [ \frac{T^2}{(1-z^{-1})^2} \frac{1}{1+GH(z)} ] $ Defining K_a : the acceleration error constant ⇒ K_a = \frac {1} {T^2} lim_{z \rightarrow 1} (1-z^{-1})^2GH(z) $ \Rightarrow e_{ss}^* = \frac {1}{K_a} $ Remark: $ e_{ss}^* \rightarrow 0 \Rightarrow K_a \rightarrow ∞ \Rightarrow GH(z) Should have at least 3 poles at z=1. $ ⇒ System type “n” is the system type: GH(z) = \frac {1}{(z-1)^n} \frac {A(z)}{B(z)} ### 3.9 The dead beat response: In the control theory, the dead-beat control problem consists of finding what input signal must be applied to a system in order to bring the output to the steady state quickly. The system response that reaches the desired value as quickly as possible is called minimum control system. Example: A block diagram showcasing the dead beat response of a system. Take: G_1(s) = 10/( (s+1)(s+2) ;T = 0.1s; $ G_{ZOH}.G_1(z) = (1-z^{-1})Z[\frac{10}{s(s+1)(s+2)}] \Rightarrow G_p(z) = \frac{0.0453(z−1)(z +0.904)}{(z - 0.405)(z – 0.819)} $ Suppose now that: G_2(z) = (z – 0.405)(z -0.819)/(0.0453(z−1)(z +0.904)) $ Using pole zero cancellation: G(z).G_{ZOH}.G_1(z) = 1 $ Let's now: R(z) = 1/(1-z¯¹) = (z-1)/(z-1) =(unitstepinput) V $ C(z) = G(z).G_{ZOH}G_1(z)E(z) = G(z).G_{ZOH}.G_1(z)(R(z) - C(z)) \Rightarrow C(z) = (R(z)-C(z)) $ C(z) = (z-1)^2/(z-1) $ \Rightarrow C(z) = \frac{z^2 - 1}{z - 1} = 1 +(z-1) + (z-1)^2 +.... \Rightarrow C(kT)=1 $ Pole zeros cancellation is ideal to bring the system to steady state: It will be hard to implement practically. Let's use another way. A block diagram showcasing the system. $ G(z)→(Z.O.H)× plant C(z) R(z) = M(z) = G_c(z).G(z) 1+G_c(z).G(z) = \frac{ 1 }{ G(z) } \frac { M(z) }{ 1-M(z) } \Rightarrow G_c(z) = \frac{1}{G(z)} \frac {M(z)}{1-M(z)} $ - Writing: E(z) = R(z) - C(z) = R(z)[1-M(z)] ⇒ E(z) = R(z)[1-M(z)] $ With the input: R(z) = \frac{A(z)}{(1-z^{-1})^N} N: positive integer. $ With A(z) polynomial in z^{-1} with no zeros at z=1 For unit step: R(z) = 1/(1-z^{-1}) ⇒ A(z) = 1; N =1 Unit Ramp (t): R(z) = (T.z^{-1})/(1−z−1)2 ⇒ A(z) = T.z¯¹ ; N = 2 Unit parabolic (t^2): R(z) = (T^2z^{-1}(1+z^{-¹}))/(1-z^{-1})^3 ⇒ A(z) = T^2z^{-1}(1+z^{-¹}); N = 3 ### 3.10 Digital Controller Design Design is the process of introducing suitable digital compensators into the sampled-data control system to achieve desired performance specifications ### 3.10.1 Compensation by Digital Controllers a) Original (uncompensated) system: A block diagram of a sampled-data system. Transfer function: C(z)/R(z) = G(z)/(1+GH(z)) b) System with digital controller (compensator): A block diagram of a closed-loop sampled-data system. Transfer function: C(z)/R(z) = D(z)G(z) / (1+D(z)GH(z)) c) Inner Loop Digital Compensation A block diagram of a closed-loop system with an inner loop digital compensator. Transfer function: $ \frac{C(z)}{R(z)} = \frac{Z^{-1}[\frac{G_1G_2R}{1+G_1G_2H_1}]}{1+D(z)Z^{-1}[\frac{G_1G_2}{1+G_1G_2H_1}]} $ Digital compensator D(z) is employed in the inner loop of the feedback system. Design goals - To decrease the steady-state tracking errors - To improve the system stability - To improve the system transient response and make it robust to parameter variations 1) To Decrease Steady-State Tracking Errors A block diagram of a closed-loop system. $ G(z) = \frac {1-e^{-T} }{z- e^{-T}} $ - Type 0 system, with e_{ss} = 0.5 for unit step input, and e_{ss} = ∞ for unit ramp input With a digital compensator: D(z) = K_1 + K_2 (z^{-1})/(z-1) ⇒ D(z)G(z) = (K_1 + K_2 (z^{-1})/(z-1)) (1- e^{-T})/(z - e^{-T}) - Type 1 system, with gain K_de = K_1, with e_{ss} = 0 for unit step input, and e_{ss} = T/K_1 for unit ramp 2) To Improve System Stability Stability is often characterized by the locations of the closed-loop poles and stability margin( gain margin and phase margin) 3) To Improve System Transient Response - Maximum overshoot ; Settling time; Rise time, peak time. 4) For Better Disturbance Rejection A block diagram of a closed loop system with disturbance input, F(s). Due to disturbance By choosing the compensator D(z)=K with a large K, the effect of disturbance on the system output can be significantly reduced without affecting the tracking performance. ### 3.10.2 Digital Controller Design using Root Locus 1) S-Plane Specifications $ H(s) = \frac{ ω_n^2 }{ s^2 + 2ζω_ns } $ A graph of the root locus of the system. - has two poles: p1,2 =-σ±jω_n With: σ=ζω_n and ω_n = √1-ζ^2 ω_n - Damping ratio: ζ and θ = tan^{-1}(√1-ζ^2/ζ) - Natural frequency: ω_n - Settling time (5%): t_s ≅ 3/(ζω_n) ≅ ζπ/√(1-ζ^2) - Maximum overshoot: M = e^(-ζπ/√(1-ζ^2)) 2) z-Plane Specifications Procedures for Deriving z-Plane Specifications - Given a set of continuous-time specifications (maximum overshoot, rise time, settling time) - Determine the desired (ζ and ω_n - From the z-grids, find the locations of the dominant poles of the corresponding discrete system: $ Z_1,2 = e^{s_1,2T}= e^{(-ζω_n±jω_n√1-ζ^2)T}=e^{-ζω_nT}e^{±joT√1-ζ^2}=r.e^{±jθ} $ Where: r = e^(-ζω_nT) and θ = Tω_n√(1-ζ^2) A graph of the z-grid. Example: G(z) = ( 0.0484(z