Review: Step Response of 1St Order Systems

Review: step response of 1st order systems

Step response in the s—domain c(t) 1 steady state Initial slope = = a (final value) a time constant ; s(s + a) 1.0 0.9 in the time domain 0.8 at 1 − e− u(t); 0.7 63% of final value ¡time constant¢ 0.6 at t = one time constant 0.5 1 τ = ; 0.4 a 0.3 rise time (10%→90%) 0.2 2.2 0.1 Tr = ; a 0 t 1 2 3 4 5 settling time (98%) a a a a a Tr 4 T = . Ts s a

Figure 4.3 Figure by MIT OpenCourseWare.

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Review: poles, zeros, and the forced/natural responses

jω jω system pole – input pole – system zero – natural response forced response derivative & amplification σ σ −5 0 −5 −2 0

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Goals for today

• Second-order systems response – types of 2nd-order systems • overdamped • underdamped • undamped • critically damped – transient behavior of overdamped 2nd-order systems – transient behavior of underdamped 2nd-order systems – DC motor with non-negligible impedance • Next lecture (Friday): – examples of modeling & transient calculations for electro-mechanical 2nd order systems

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 DC motor system with non-negligible inductance

Recall combined equations of motion

LsI(s)+RI(s)+KvΩ(s)=Vs(s) ⇒ JsΩ(s)+bΩ(s)=KmI(s) )

LJ Lb KmKv Km s2 + + J s + b + Ω(s)= V (s) R R R R s ⎧ · µ ¶ µ ¶¸ ⎨⎪ (Js + b) Ω(s)=KmI(s)

⎩⎪ Including the DC motor’s inductance, we ﬁnd

Ω(s) Km 1 = Vs(s) LJ b R bR + KmKv ⎧ s2 + + s + Quadratic polynomial denominator ⎪ J L LJ Second—order system ⎪ µ ¶ µ ¶ ⎪ ⎪ ⎪ b ⎪ s + ⎨ I(s) 1 J = µ ¶ ⎪ Vs(s) R b R bR + K K ⎪ 2 m v ⎪ s + + s + ⎪ J L LJ ⎪ µ ¶ µ ¶ ⎪ ⎩⎪

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Step response of 2nd order system – large R/L

2 L =0.1H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , Overdamped 2 R =6Ω, b =4kg· m · Hz; vs(t)=30u(t)V. response ↔ dissipation > energy storage

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Step response of 2nd order system – large R/L

2 L =0.1H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , Overdamped 2 R =6Ω, b =4kg· m · Hz; vs(t)=30u(t)V. response

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Step response of 2nd order system – large R/L

2 L =0.1H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , Overdamped 2 R =6Ω, b =4kg· m · Hz; vs(t)=30u(t)V. response

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Comparison of 1st order and 2nd order overdamped

1st order 2nd order (L≈0) (L=0.1H)

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Step response of 2nd order system – small R/L

2 L =1.0H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , Underdamped 2 R =6Ω, b =4kg· m · Hz; vs(t)=30u(t)V. response ↔ dissipation < energy storage overshoot

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Comparison of 1st order and 2nd order underdamped NO overshoot overshoot 1st order 2nd order (L≈0) (L=1.0H)

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Overdamped DC motor: derivation of the step response

Using the numerical values L =0.1H, Kv =6V· sec, Km =6N· m/A, J = 2kg · m2, R =6Ω, b =4kg· m2 · Hz we ﬁnd K rad b R bR + K K m =30 ; + =62rad/sec; m v =300(rad/sec)2 . LJ sec · V J L LJ Therefore, the transfer function for the angular velocity is nd Ω(s) 30 A2 —order system is overdamped = 2 . if the transfer function denominator Vs(s) s +62s +300 has two real roots. We ﬁnd that the denominator has two real roots,

Ω(s) 30 s1 = −5.290Hz,s2 = −56.71Hz ⇒ = . Vs(s) (s +5.290)(s +56.71) To compute the step response we substitute the Laplace transform of the voltage source Vs(s)=30/s and carry out the partial fraction expansion: 900 3 3.3 0.3 Ω(s)= = − + ⇒ s(s +5.290)(s +56.71) s s +5.290 s +56.71

5.29t 56.71t ω(t)= 3 − 3.3e− +0.3e− u(t). This is the function whose£ plot we analyzed in slides #5—8.¤

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Overdamped DC motor in the s-domain

jω system poles (2nd order overdamped)

input pole σ −56.71 −5.29 −5 0

1st order system pole

jω

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Undamped DC motor: no dissipation

Consider the opposite extreme where the dissipation due to both the resistor and bearings friction is negligible, i.e. R =0andb = 0. Using the same remaining 2 numerical values L =0.1H, Kv =6V· sec, Km =6N· m/A, J =2kg· m ,we ﬁnd K rad b R bR + K K m =30 ; + =0; m v =180(rad/sec)2 . LJ sec · V J L LJ Therefore, the transfer function for the angular velocity is A2nd—order system is undamped Ω(s) 30 if the transfer function denominator has a = 2 . Vs(s) s +180 conjugate pair of two imaginary roots.

The denominator has a conjugate pair of two imaginary roots,

Ω(s) 30 s1,2 = ±j13.42Hz ⇒ = . Vs(s) (s + j13.42)(s − j13.42) Again, the step response is found by partial fraction expansion: 900 5 5s Ω(s)= = − ⇒ s(s + j13.42)(s − j13.42) s s2 +(13.42)2

ω(t)=[5− 5cos(13.42t)] u(t).

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Undamped DC motor in the s-domain

Natural frequency ωn =13.42rad/sec. Period T =2π/ωn =4.24sec.

system poles (2nd order undamped) jω

j13.42

input pole σ −5 0

1st order −j13.42 system pole

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Underdamped DC motor: small dissipation

Finally, let us return to what we previously labelled as “underdamped” case, 2 i.e. L =1.0H, Kv =6V· sec, Km =6N· m/A, J =2kg· m , R =6Ω, b =4kg· m2 · Hz. The values of L, R are such that the dissipation in the system is negligible compared to the energy storage capacity. We then ﬁnd K rad b R bR + K K m =3 ; + = 8rad/sec; m v =30(rad/sec)2 . LJ sec · V J L LJ Therefore, the transfer function for the angular velocity is

Ω(s) 30 A2nd—order system is underdamped = 2 . Vs(s) s +8s +30 if the transfer function denominator has a conjugate pair of two complex roots. This denominator has a conjugate pair of two complex roots,

Ω(s) 30 s1,2 = −4 ± j3.74 rad/sec ⇒ = . Vs(s) (s +4+j3.74)(s +4− j3.74) We will now develop the partial fraction expansion method for this case, aiming to ﬁnd the step response:

90 90 K K s + K Ω(s)= = =90 1 + 2 3 . s(s +4+j3.74)(s +4− j3.74) s(s2 +8s +30) s s2 +8s +30 µ ¶

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Underdamped DC motor: small dissipation

Using the familiar partial fraction method, we can ﬁnd K1 =1/30, K2 = −1/30, K3 = −8/30, therefore

1 s +8 Ω(s)=3 − . s s2 +8s +30 µ ¶ To ﬁndtheinverseLaplacetransform,werewrite the denominator as a complete square plus a constant, and break down the numerator into the sum of the same factor that appeared in the denominator’s complete square plus another constant: 1 (s +4)+4 Ω(s)=3 − 2 . Ã s (s +4) +14! Ifthecompletesquareinsteadof(s+4)2 were of the form s2, the inverse Laplace transform would have followed easily from Nise Table 2.1:

1 s +4 1 s +4 L− = L− =cos(3.74t)+4sin(3.74t) . s2 +14 2 2 · ¸ "s +(3.74) # To take the extra factor of 4 into account, we must use yet another property of Laplace transforms, which we have not seen until now:

at L e− f(t) = F (s + a). (Nise Table 2.2, #4). £ ¤

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Underdamped DC motor: small dissipation

We apply this “frequency shift” property as follows:

1 s +4 L− 2 =cos(3.74t)+4sin(3.74t) ⇒ "s2 +(3.74) #

1 (s +4)+4 4t ⇒ L− 2 2 =e− [cos (3.74t)+4sin(3.74t)] . "(s +4) +(3.74) # Combining all of the above results, we can ﬁnally compute the step response for the angular velocity of the DC motor as

4t ω(t)= 3 − 3 e− [cos (3.74t)+4sin(3.74t)] u(t). ½ ³ ´¾ With a little bit of trigonometry, which we leave to you to do as exercise, we can rewrite the step response as

4t ω(t)= 3 − 4.39e− cos (3.74t − 0.82) u(t). ½ ¾ So the step response of the 2nd—order underdamped system is characterized by a phase—shifted sinusoid enveloped by an exponential decay.

This step response was analyzed in slides #9—10 of today’s notes.

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 What the real and imaginary parts of the poles do c(t)

Exponential decay generated by real part of complex pole pair

Sinusoidal oscillation generated by imaginary part of complex pole pair

t Damping ratio Figure by MIT OpenCourseWare. 1 Undamped (“natural”) period ζ ≡ . 2π Time constant of exponential decay

Note: the underdamped oscillation frequency is not the same as Figure 4.8 the natural frequency!

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Underdamped DC motor in the s-domain

overshoot system poles (2nd order undamped) jω

−4+j3.74 j3.74

−4 0 σ

input pole −j3.74 −4 − j3.74

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 The general 2nd order system

We can write the transfer function of the general 2nd—order system with unit steady state response as follows:

2 ωn 2 2 , where s +2ζωns + ωn

• ωn is the system’s natural frequency,and • ζ is the system’s damping ratio.

The natural frequency indicates the oscillation frequency of the undamped (“natural”) system, i.e. the system with energy storage elements only and without any dissipative elements. The damping ratio denotes the relative con- tribution to the system dynamics by energy storage elements and dissipative elements. Recall,

1 Undamped (“natural”) period ζ ≡ . 2π Time constant of exponential decay Depending on the damping ratio ζ, the system response is

• undamped if ζ =0; • underdamped if 0 < ζ < 1;

• critically damped if ζ =1; • overdamped if ζ > 1.

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 The general 2nd order system

c (t)

Undamped 2.0 1.8 1.6 1.4 Underdamped 1.2 Critically 1.0 damped 0.8 0.6 Overdamped 0.4 0.2 t 0 0.5 1 1.5 2 2.5 3 3.5 4

Figure by MIT OpenCourseWare.

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 The general 2nd order system

Images removed due to copyright restrictions.

Please see: Fig. 4.11 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.

Nise Figure 4.11

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 The underdamped 2nd order system 2 ωn 2 2 , 0 < ζ < 1 (s +2ζωns + ωn) The step response’s Laplace transform is

1 ωn K1 K2s + K3 × 2 2 = + 2 2 . s s +2ζωns + ωn s s +2ζωns + ωn We ﬁnd 1 1 2ζ K1 = 2 ,K2 = − 2 ,K3 = ωn ωn ωn Substituting and applying the same method of completing squares that we did in the numerical example of the DC motor’s angular velocity response, we can rewrite the laplace transform of the step response as ζ 2 (s + ζωn)+ ωn 1 − ζ 1 1 − ζ2 − . 2 2 p2 s (s + ζωn)p+ ωn 1 − ζ ¡ ¢ Using the frequency shifting property of Laplace transforms we ﬁnally obtain the step response in the time domain as

ζ ζωnt 2 2 1 − e− cos ωn 1 − ζ t + sin ωn 1 − ζ t . " 1 − ζ2 # ³ p ´ ³ p ´ 2.004 Fall ’07 Lecture 07 –p Wednesday, Sept. 19 The underdamped 2nd order system 2 ωn 2 2 , 0 < ζ < 1 (s +2ζωns + ωn) Finally, using some additional trigonometry and the deﬁnitions

2 ζ σd = ζωn, ωd = ωn 1 − ζ , tan φ = 1 − ζ2 p we can rewrite the step response as p 1 σdt 1 − × e− × cos (ωdt − φ) 1 − ζ2 p jω

2 The deﬁnitions above can be re—written X + jωn 1 − ζ = jωd

σ ωn s-plane ζ = d , ωn θ σ ω − ζ ωn = − σd 1 − ζ2 = d , ωn p − jω 1 − ζ2 = −jω ω 1 − ζ2 X n d ⇒ tan θ = d = . σd p ζ Figure 4.10 Figure by MIT OpenCourseWare.

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 The underdamped 2nd order system 2 ωn 2 2 , 0 < ζ < 1 (s +2ζωns + ωn) Finally, using some additional trigonometry and the deﬁnitions

2 ζ σd = ζωn, ωd = ωn 1 − ζ , tan φ = 1 − ζ2 p we can rewrite the step response as p 1 σdt 1 − × e− × cos (ωdt − φ) 1 − ζ2 p c(t)

forced response, Exponential decay generated by real sets steady state part of complex pole pair

Sinusoidal oscillation generated by imaginary part of complex pole pair

Figure 4.10 Figure by MIT OpenCourseWare. t 2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Transients in the underdamped 2nd order system

Peak time π Tp = . 2 ωn 1 − ζ

c(t) Percent overshootp (%OS)

cmax ζπ 1.02cfinal %OS = exp − × 100 c final Ã 1 − ζ2 ! 0.98cfinal −lnp (%OS/100) 0.9cfinal ⇔ ζ = π2 +ln2 (%OS/100) qSettling time (to within ±2% of steady state) 0.1cfinal t 2 Tr Tp Ts ln 0.02 1 − ζ 4 Ts = − ≈ . Figure by MIT OpenCourseWare. ³ ζωpn ´ ζωn (approximation valid for 0 < ζ < 0.9.) Figure 4.14

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19 Transient qualities from pole location in the s-plane

Images removed due to copyright restrictions.

Please see: Fig. 4.19 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.

Recall σ ζ = d , ωn ω 1 − ζ2 = d , ωn p ω 1 − ζ2 ⇒ tan θ = d = . σd ζ Nise Figure 4.19 p

2.004 Fall ’07 Lecture 07 – Wednesday, Sept. 19