Step Response Analysis. Frequency Response, Relation Between Model Descriptions

Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control

Content. Step Response Analysis 2. Frequency Response 3. Relation between Model Descriptions

Step Response Analysis

Step Response From the last lecture, we know that if the input u(t) is a step, then the output in the Laplace domain is Y (s) = G(s)U(s) = G(s) s It is possible to do an inverse transform of Y (s) to get y(t), but is it possible to claim things about y(t) by only studying Y (s)? We will study how the poles affects the step response. (The zeros will be discussed later). 2

Initial and Final Value Theorem Let F (s) be the Laplace transformation of f (t), i.e., F (s) = L(f (t))(s). Given that the limits below exist, it holds that: Initial value theorem Final value theorem lim t f (t) = lim s + sf (s) lim t + f (t) = lim s sf (s) For a step response we have that: lim y(t) = lim sy (s) = lim sg(s) t + s s s = G() When can we NOT apply the Final value theorem? 3

First Order System Singularity Chart Step Response (K=) Im.5.5 T = T = 2T = 5 y(t).5.5.5.5 Re One pole in s = /T Step response: Y (s) = G(s) s = K s( + st ) G(s) = L K + st 5 5 t y(t) = K ( e t/t ), t 4

First Order System Singularity Chart Step Response (K=) Im.5.5 T = T = 2T = 5 y(t).5.5.5.5 Re 5 5 t Final value: G(s) = K + st lim y(t) = lim sy (s) = lim s K t + s s s( + st ) = K 4

First Order System Singularity Chart Step Response (K=) Im.5.5 T = T = 2T = 5 y(t).5.5.5.5 Re T is called the time-constant: G(s) = K + st 5 5 t y(t ) = K( e T /T ) = K( e ).63K i.e., T is the time it takes for the step response to reach 63% of its final value 4

First Order System Singularity Chart Step Response (K=) Im.5.5 T = T = 2T = 5 y(t).5.5.5.5 Re 5 5 t Derivative at zero: lim ẏ(t) = t G(s) = lim s sy (s) = lim s + K + st s + s 2 K s( + st ) = K T 4

Second Order System With Real Poles Singularity Chart Step Response (K=) Im.5.5 T = T = 2 y(t).5.5.5.5 Re G(s) = K ( + st )( + st 2 ) 5 5 t Poles in s = /T and s = /T 2. Step response: ( ) K Te t/t T 2e t/t 2 T y(t) = T 2 T T 2 K ( e t/t ) t T e t/t T = T 2 = T 5

Second Order System With Real Poles Singularity Chart Step Response (K=) Im.5.5 T = T = 2 y(t).5.5.5.5 Re 5 5 t G(s) = K ( + st )( + st 2 ) Final value: lim = lim sk sy (s) = lim t + s s s( + st )( + st 2 ) = K 5

Second Order System With Real Poles Singularity Chart Step Response (K=) Im.5.5 T = T = 2 y(t).5.5.5.5 Re 5 5 t Derivative at zero: lim ẏ(t) = t G(s) = lim s sy (s) = lim s + K ( + st )( + st 2 ) s + s 2 K s( + st )( + st 2 ) = 5

Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Relative damping ζ, related to the angle ϕ ζ = cos(ϕ) Singularity Chart Im.5 ω ϕ.5 Re 6

Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Inverse transformation for step response yields: y(t) = K = K ( ( ( ) ) ζ 2 e ζωt sin ω ζ2 t + arccos ζ ( ζ 2 e ζωt sin ω ζ2 t + arcsin( ζ )) ) 2, t Exercise: Check of correct starting point of step response. y() = K = K ( ( ( ζ 2 e sin ω ζ 2 + arcsin( ) ) ζ 2 ) ) ζ 2 ζ 2 = 5 5.5.5 Step Response t 6

Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Singularity Chart Step Response (K=) Im ω =.5 ω = ω =.5 y(t).5 Re 5 5 t 6

Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Im Singularity Chart ζ =.3 ζ =.7 ζ =.9 Re y(t).5.5 Step Response (K=) 5 5 t 6

Frequency Response

Sinusoidal Input Given a transfer function G(s), what happens if we let the input be u(t) = sin(ωt)?.5 y(t) u(t).5 5 5 2 t.5.5 5 5 2 t 7

Sinusoidal Input It can be shown that if the input is u(t) = sin(ωt), the output 2 will be where y(t) = A sin(ωt + ϕ) A = G(iω) ϕ = arg G(iω) So if we determine a and ϕ for different frequencies ω, we have a description of the transfer function. 2 after the transient has decayed 8

Bode Plot Idea: Plot G(iω) and arg G(iω) for different frequencies ω. Magnitude (abs) Phase (deg) 2 3 2 2 45 9 35 8 2 2 Frequency (rad/s) 9

Sinusoidal Input-Output: example with frequency sweep (chirp) Resonance frequency of industrial robot IRB2 visible in data.

Bode Plot - Products of Transfer Functions Let G(s) = G (s)g 2 (s)g 3 (s) then log G(iω) = log G (iω) + log G 2 (iω) + log G 3 (iω) arg G(iω) = arg G (iω) + arg G 2 (iω) + arg G 3 (iω) This means that we can construct Bode plots of transfer functions from simple building blocks for which we know the Bode plots.

Bode Plot of G(s) = K If G(s) = K then log G(iω) = log( K ) arg G(iω) = (if K >, else + 8 or 8 deg) 2

Bode Plot of G(s) = K Magnitude (abs) Phase (deg) K = 4 K = K =.5 2 2 45 45 9 35 8 2 2 Frequency (rad/s) 2

Bode Plot of G(s) = s n If G(s) = s n then log G(iω) = n log(ω) arg G(iω) = n π 2 3

Bode Plot of G(s) = s n Magnitude (abs) Phase (deg) n = 2 2 n = n = 2 3 4 2 9 45 45 9 35 8 2 2 Frequency (rad/s) 3

Bode Plot of G(s) = ( + st ) n If then G(s) = ( + st ) n log G(iω) = n log( + ω 2 T 2 ) arg G(iω) = n arg( + iωt ) = n arctan (ωt ) For small ω log G(iω) arg G(iω) For large ω log G(iω) n log(ωt ) arg G(iω) n π 2 4

Bode Plot of G(s) = ( + st ) n Magnitude (abs) Phase (deg) n = 2 n = 2 n = 9 45 45 9 35 8 T Frequency (rad/s) 4

Bode Plot of G(s) = ( + 2ζs/ω + (s/ω ) 2 ) n G(s) = ( + 2ζs/ω + (s/ω ) 2 ) n For small ω log G(iω) arg(iω) For large ω ( ) ω log G(iω) 2n log ω arg G(iω) nπ 5

Bode Plot of G(s) = ( + 2ζs/ω + (s/ω ) 2 ) n Magnitude (abs) Phase (deg) 2 ζ =.2 ζ =. ζ =.5 2 45 9 35 8 Frequency (rad/s) 5

Bode Plot of G(s) = e sl G(s) = e sl Describes a pure time delay with delay L, i.e, y(t) = u(t L) log G(iω) = arg G(iω) = ωl 6

Bode Plot of G(s) = e sl Magnitude (abs) Phase (deg) 2 L =. 2 L = 5 L =.5 4 6 2 Frequency (rad/s) 6

Bode Plot of G(s) = e sl Same delay may appear as different phase lag for different frequencies! Example Delay.52 sec between input and output..5.5 2 2.5 3 3.5 4 4.5 5 5.5 6 (Upper): Period time = 2π 6.28 sec. Delay represents phase lag of.52 36 3 deg 6.28 (Lower): Period time = π 3.4 sec. Delay represents phase lag of.5 36 6 deg. 3.4.5.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6

Bode Plot of G(s) = e sl Check phase in Bode diagram for e.52s for sin(t) ω =. rad/s sin(2t) ω = 2. rad/s >> s=tf( s ) >> G=exp(-.52*s); >> bode(g,.,5) % Bode plot in frequency-range [... 5] rad/s 6

Bode Plot of Composite Transfer Function Example Draw the Bode plot of the transfer function G(s) = (s + 2) s(s + 2) 2 First step, write it as product of sample transfer functions: G(s) = (s + 2) s(s + 2) 2 =.5 s ( +.5s) ( +.5s) 2 Then determine the corner frequencies: w c2 =2 w c =2 (s + 2) {}}{{}}{ G(s) = s(s + 2) 2 =.5 s ( +.5s) ( +.5s) 2 7

Bode Plot of Composite Transfer Function G(s) = (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Magnitude (abs) 2 3 4 2 3 Frequency (rad/s) 8

Bode Plot of Composite Transfer Function G(s) = (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Magnitude (abs) 2 3 2 4 2 3 Frequency (rad/s) 8

Bode Plot of Composite Transfer Function G(s) = 45 (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Phase (deg) 45 9 35 8 2 3 Frequency (rad/s) 9

Nyquist Plot By removing the frequency information, we can plot the transfer function in one plot instead of two..5 Im G(iω) arg G(iω) G(iω).5.5 Re G(iω) 2

Nyquist Plot By removing the frequency information, we can plot the transfer function in one plot instead of two..5 Im G(iω) arg G(iω) G(iω).5.5 Re G(iω) Split the transfer function into real and imaginary part: G(s) = G(iω) = + s + iω = + ω 2 i ω + ω 2 Is this the transfer function in the plot above? 2

From Bode Plot to Nyquist Plot Magnitude (abs) Phase (deg) 2 3 4 2 2 9 8 27 Frequency (rad/s) 2 2 Frequency (rad/s) Im G(iω).5.5.5.5 Re G(iω) 2

Relation between Model Descriptions

Single-capacitive Processes K st +.5.5 Singularity chart.5 Step response.5.5.5 Nyquist plot 2 4 Bode plot.2.2.4.6.5 9 22

Multi-capacitive Processes K (st +)(st 2 +).5.5 Singularity chart.5 Step response.5.5.5 Nyquist plot 2 4 6 8 Bode plot.2.2.4.6.5 2 9 8 23

Integrating Processes s.5.5 Singularity chart 4 2 Step response.5.5.5 Nyquist plot 2 4 Bode plot.5.5.5 9 24

Oscillative Processes Kω 2 s 2 +2ζω s+ω 2, < ζ < Singularity chart.5 Step response.5.5.5 Nyquist plot 2 3 2 2.5 2 2 3 9 Bode plot 8 25

Delay Processes K st + e sl Step response.5 Nyquist plot 2 4 6 8 Bode plot 3 6 2 26

Process with Inverse Responses sa+ (st +)(st 2 +) Singularity chart Step response.5.5.5 Nyquist plot 2 4 6 8 Bode plot.5.5.5 9 8 27 27