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Next we use a list of forward/inverse transforms to replace the terms in the partial fraction expansion.

	f( t)					f( s)
	A					A
						--
						s
	At					A
						----
						s2
	Ae	–α t				A
						-----------
						s + α
	A sin ( ω		t)			Aω
					-----------------
					s2 + ω 2
–ξω	nt		2	ω	n	1 – ξ 2	for( ξ < 1)
e	sin ( ω nt		1 – ξ )	---------------------------------------
				s2 + 2ξω ns + ω n2

etc.

To finish the problem, we simply convert each term of the partial fraction back to the time domain.

vactual =	99.9---------	+	---------------------0.264	–	---------------------1.16
	s		s + 0.113		s + 0.877

vactual = 99.9 + 0.264e–0.113t – 1.16e–0.877t

7.1.5 System Response

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•There are two very common systems assumed - first and second order.

•First order systems are very simple, as is shown below.

	G( s)	A	A first order system, and a typical response to
		= -----------	a stepped input.
		s + B

			A	A second order system, and a typical response to
	G( s)			a stepped input.
			= ---------------------------------------
			s2 + 2ξω ns + ω n2

ω n

ξ

Natural frequency of system, approximate frequency of control system oscillations.

Damping factor of system. If < 1 underdamped, and system will oscillate. If =1 critically damped. If < 1 overdamped, and never any oscillation (more like a first order system). As damping factor approaches 0, the first peak becomes infinite in height.

7.1.6 A Motor Control System Example

• Condsider the example of a DC servo motor controlled by a computer. The purpose of the con-

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troller is to position the motor. The system below shows a reasonable control system arrangement. Some elements such as power supplies and commons for voltages are omitted for clarity.

			2.2K
Computer Running Labview			1K	12Vdc motor
gain K	aquisitioncard		-
		Instruments	+
			LM675
X			op-amp
				shafts are coupled
	1200-PCI data	fromNational
-
+
			+5V	-5V
				5K potentiometer
desired position
voltage Vd

• This system can then be redrawn with a block diagram.

desired
position
voltage	+	gain K	op-amp	motor	shaft
Vd
		-
			potentiometer

• The block diagram can now be filled out with actual values for the components. Do this below.

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Given values:

-desired potentiometer voltage

-gain

For the op-amp:

For the potentiometer:

- assume that the potentiometer has a range of 10 turns

For the motor:

- use the differential equation below

d	ω	K2	ω		K
----		-----		=	-----	Vs
dt		+ JR			JR

- use the speed curve below from rest when 10V is applied

1400 RPM
1s	2s	3s

For the shaft:

- it turns angular velocity into position

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• Convert the block diagram into a transfer function for the entire system.