ATOMATIC CONTROL LABS

Thus, the asymptotic Bode diagram of real differentiating sys-

tem is composed of two segments of straight lines.
At s ,	it is straight line	with	positive slope
20 dB/dec, and at	s it is horizontal	line	with ordinate of

20lg KDR T decibels. Asymptotic Bode plot of real differentiating circuit LDRa is shown in Fig. 11.

Fig. 11. Bode plots of ideal (DI) and real (DR) differentiating systems

Dependence of output signal phase angle versus frequency is calculated as:

	DR arctan	ImGDR j	.	(66)
		ReGDR j

Graph of DR is shown in Fig. 12.

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Fig. 12. Phase characteristic of the frequency response of ideal (DI) and real (RD) differentiating system in semi-log scale

In	the same Fig. 12 the asymptotic phase angle curve
DRa	is presented. It approximates real phase angle characteris-

tic and is convenient for design of desired controller characteristics. The simple circuit, realizing real differentiating system is

shown in Fig. 13.

Fig. 13. Real differentiating circuit

Differential equation, describing the circuit in Fig. 13 is:

1	idt iR1 u1.	(67)
C
1

System output signal is:

62

u2 iR1.

Substituting (68) to (67) yields:

(68)

1

u

dt u

u ;

C R

du2

u

C R du1

.

(69)

C R

2

2

1

1 1

dt

2

1 1 dt

1

1

DenotingT C1R1 ;

KDR C1R1 ;

y u2

and

u u1

gives the

differential equation (52) of real differentiating system.

Content of the report

1.Elaboration of the mathematical model of the system indicated by teacher.

2.Step responses and asymptotic Bode plots.

3.Actual frequency response magnitude and phase angle characteristics calculated and plotted by Matlab.

Control questions

1.What are the first order systems?

2.Why proportional system is not ideal system?

3.What slope has the magnitude straight of integrating system in Bode plot?

4.What is asymptotic Bode diagram of the first order system? How exactly does it approximate the actual?

5.What systems comprise the real differentiating system?

63

6.How is constructed Bode plot of real differentiating system?

7.How does look like the frequency response magnitude and phase characteristics of ideal differentiating system?

64

THE SECOND ORDER SYSTEMS

Laboratory work No 4

Objectives: get acquainted with the second order system and learn to analyze its characteristics in time domain and frequency domain.

Task of the work:

1.Elaborate mathematical model of the system, indicated by teacher.

2.Construct simulation model of the system.

3.Use MATLAB to plot step response, frequency response and Bode plot.

Theoretical part

In general case the second order system is described by differential equation:

a

d 2 y

a

dy

a

y b

d 2u

b

du

b u ;

(1)

2 dt

2

1 dt

0

2 dt2

1 dt

0

where: u and y are system input and output signals, a0, a1, a2, b0 and b1 are constants. For practical cases it is convenient to assume b1 b0 0 and Eq. 1 becomes:

a2 d 22y a1 dy a0 y b0u . dt dt

Demoting b0/a0 = K, a1/a0 = 2ρ/ω0 rewritten in this way:

(2)

ir a2/a0 = 1/(ω0)2, Eq. 2 is

65

1		d 2 y			2 dy		y Ku.	(3)
2		dt	2			dt
					0
0

Notations, chosen in this way, take the physical meaning: ω0 is natural frequency of the system, ρ is damping ratio and K is gain.

Solution of Eq. (3) is equal of sum of solutions of homogenous equation (if right hand side of Eq. 3 is equal to zero), called free movement of the system and solution of non-homogenous system (called forced movement). General solution of homogenous system depends on roots of characteristic equation. It is obtained replacing

d	in Eq. 3:
dt
	1			2		2	1	0.	(4)
		02				0

Characteristic equation (4) is the same polynomial of transfer function denominator, therefore its roots in automatic control theory are called system poles:

			2						4 2			4
p			0						02			02				2 1.	(5)
1,2								2					0		0
								2
								0
If 2	1 0;							1			and		p R ,		the	general	solution of
													1,2

homogenous equation has the form:

yh C1 exp p1t C2 exp p2t .

(6)

66

Mathematical analysis also considers cases when ρ <- 1, but

then						and p						2 1 0 , therefore the first
0					0	1				0	0

term of the Eq. 6 increases to infinity with rising of t. The automatic control theory discusses just the systems with ρ ≥ 0.

The particular solution depends on the system input u. If

u = 1(0), then solution of non-homogenous system has the form:

ynh A .

(7)

Substituting (7) and its derivatives, whose are equal to zero into Eq. 3 and assuming u 1 0 yields:

ynh A K .

(8)

Summing Eq. (6) and Eq. (8), the general solution of differential equation (3) at unit step input is:

y yh ynh C1 exp p1t C2 exp p2t K.

(9)

Constants C1 and C2 can be calculated from initial conditions, i. e. y = 0 and y‘ = 0 at t = 0, then the set of equations is obtained:

C1 C2 K 0;C1 p1 C2 p2 0.

Solution of Eq. 10 is obtained as:

C				Kp2			;
1			p2		p1
			Kp1
C2		p2		p1		.

(10)

(11)

67

Finally, if δ > 1 and system has zero initial conditions, step response is described as:

y	Kp2 exp p1t			Kp1 exp p2t			K;	(12)
				p		p
	p	p			2
2		1				1

where: p1,2 are calculated from Eq. (5).

While p1,2 < 0, the solution is composed from sum of two terms, decaying by exponential low and gain K. At zero initial condi-

tions there is no any overshoot in this case therefore y has deadbeat shape and with time approaches to gain value K.

The set of second order system responses with ω0 = 5 and K = 2,5 at ρ = [0,5; 0,75; 1; 1,25; 1,5]. Two graphs are presented for ρ = 1,25 and ρ = 1,5. Increase damping ratio increases settling time.The

shape of step response remains the same. If 2 1 0; 0 , then p ≡ p1= p2 and p R . Then the general solution of homogenous system is described as this:

yh C1 exp pt C2 exp pt .

(13)

The particular solution of non-homogenous equation depends only on right hand side of Eq. 3, and is equal to the same expression (8), therefore the general solution of Eq. (3) at δ = 1 is:

y C1 exp pt C2 exp pt K .

(14)

Constants C1 and C2 can be calculated from initial conditions. If the zero initial conditions are assumed, i. e. y = 0 and y‘ = 0 at t = 0, the obtained set has a form:

C K 0;		C K;	(15)
1		1
C1 p C2 0;		C2 Kp.
		68

Finally, if δ = 0, step response at initial zero conditions is expresses as:

y K exp pt 1 pt K.

(16)

There is no overshoot in this case also and step response is deadbeat.

Step response of the second order system at ω0 = 5; K = 2,5 and ρ = 1 is shown in Fig. 1. The settling time is shorter, than for system with ρ > 1.

Fig. 1. Set of second order system step responses at different ρ values

The next case to be considered is matching condition

2 1 0; 1 an p		C and roots are equal:
	1,2
p		j	1 2 .	(17)
1,2	0	0

For this case the general solution of homogenous equation has the form:

69

yh C1 exp 0t sin 0 1 2	t
C2 exp 0t cos 0 1 2 t .	(18)

If the input is unit step, then the particular solution of nonhomogenous system again will have the form (7). While yh → 0 and t → ∞, then expression (8) is valid also and rhe general solution is:

y C1 exp 0t sin 0	1 2	t
C2 exp 0t cos 0		(19)
	1 2 t K.

Constants C1 and C2 are calculated from initial conditions. If the zero initial conditions are assumed , i. e. y = 0 and y‘ = 0 at t = 0, then the set of equations for calculation constants looks like this:

C K 0;						C2 K 0;
2
	Im p1,2 C2	Re p1,2		0;			1 2	C	0.
C1						C
						1 0		2	0
	Solutions of Eq. (18) yields:
			K
	C1				;
			1 2
		K.
	C2

(20)

(21)

Thus, finally, solution of differential equation (3) at ρ < 1 is expresses as:

70