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Jack H.Dynamic system modeling and control.2004.pdf

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circuits - 7.16

100KHz. When the op-amp is being used for high frequencies or large gains, the model of the op-amp in Figure 7.19 should be used. This model includes a large resistance between the inverting and non-inverting inputs. The voltage difference drives a dependent voltage source with a large gain. The output resistance will limit the maximum current that the device can produce, normally less than 100mA.

V–

typically,

rn > 1MΩ

A > 10

V +

A( V+

– V-)

ro < 100Ω

Figure 7.19 A non-ideal op-amp model

7.3 IMPEDANCE

Circuit components can be represented in impedance form as shown in Figure 7.20. When represented this way the circuit solutions can focus on impedances, ’Z’, instead of resistances, ’R’. Notice that the primary difference is that the differential operator has been replaced. In this form we can use impedances as if they are resistances.

Device	Time domain		Impedance


Resistor	V( t) = RI( t)		Z = R
Capacitor	V( t)	1	Z =	1
		= ---∫I( t) dt		-------
		C		DC
Inductor	V( t)	d	Z =	LD
		= L----I( t)
		dt

Note: Impedance is like resistance, except that it includes time variant features also.

V = ZI

Figure 7.20 Impedances for electrical components

circuits - 7.17

When representing component values with impedances the circuit solution is done as if all circuit components are resistors. An example of this is shown in Figure 7.21. Notice that the two impedances at the right (resistor and capacitor) are equivalent to two resistors in parallel, and the overall circuit is a voltage divider. The impedances are written beside the circuit elements.

	DL
	t=0sec
50VDC		R	+
+		R	Vo
-	1/DC		Vo
-	1/DC		-
			-

Find the equivalent for the capacitor and resistor in parallel.

50VDC

1 =

------------------------

------------------1

RCD---------------------+ 1

-------------

DC + --

--------

Treat the circuit as a voltage divider,

---------------------

1 + DCR

= 50V

-----------------------------------------= 50V

DL +

D2RLC + DL + R

1 + DCR---------------------

Figure 7.21 A impedance example for a circuit

circuits - 7.18

7.4 EXAMPLE SYSTEMS

The list of instructions below can be useful when approaching a circuits problem. The most important concept to remember is that a minute of thinking about the solution approach will save ten minutes of backtracking and fixing mistakes.

1.Look at the circuit to determine if it is a standard circuit type such as a voltage divider, current divider or an op-amp inverting amplifier. If so, use the standard solution to solve the problem.

2.Otherwise, consider the nodes and loops in the circuit. If the circuit contains fewer loops, select the current loop method. If the circuit contains fewer nodes, select the node voltage method. Before continuing, verify that the select method can be used for the circuit.

3.For the node voltage method define node voltages and current directions. For the current loop method define current loops and indicate voltage rises or drops by adding ’+’ or ’-’ signs.

4.Write the equations for the loops or nodes.

5.Identify the desired value and eliminate unwanted values using algebra techniques.

6.Use numerical values to find a final answer.

Note: The units for various	coefficient		units
Note: The units for various	coefficient		units
electrical quantities are
electrical quantities are			As
listed to the right. They may	C		As
listed to the right. They may	C		-----
be used to check equations			V
be used to check equations			Vs
by doing a unit balance.	L		Vs
by doing a unit balance.	L		-----
			A
	R		V
			V
		--
			A

The circuit in Figure 7.22 could be solved with two loops, or two nodes. An arbitrary decision is made to use the current loop method. The voltages around each loop are summed to provide equations for each loop.

circuits - 7.19

		+
		+
	R2	Vo


		-

Note: when summing voltages in a loop remember to deal with sources that increase the voltage by flipping the sign.

First, sum the voltages around the loops and then eliminate I1. ∑VL1 = – V + R1I1 + L( DI1 – DI2) = 0

( R1 + LD) I1 = V + ( LD) I2

------------------- +

------------------- I

R1 + LD

∑VL2

L( DI2

– DI1) +

--------

+ R2I2 = 0

L( DI

– DI ) +

-------- + R

( DL) I

LD +

--------

+ R

1 +

--------------

-------

CLD2

(1)

(2)

(3)

(4)

Figure 7.22 Example problem

The equations in Figure 7.22 are manipulated further in Figure 7.23 to develop an input-output equation for the second current loop. This current can be used to find the current through the output resistor R2. The output voltage can then be found by multiplying the R2 and I2.

circuits - 7.20

First, sum the voltages around the loops and then eliminate I1.

-------------------

= 1 +

--------------

-------

2 +

+ LD

R1 + LD

CLD

-------------------

1 +

--------------

-------

–

-------------------

R1 + LD

CLD2

R1 + LD

( CLD2 + 1 + CDR2) ( R1

+ LD) – CL2D3

-------------------

---------------------------------------------------------------------------------------------------

R1 + LD

CLD2

---------------------------------------------------------------------------------------------------------------------------------

( R

+ LD) ( ( CLD2 + 1 + CDR ) ( R

+ LD) – CL2D3)

CLD2

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

CL( R

+ R ) D3

+ L( CR2

+ 2CR

+ L) D2 + R

( CR

+ 2L) D + ( R 2)

Convert it to a differential equation.

CL( R1 + R2) I2''' + L( CR21 + 2CR1R2 + L) I2'' + R1( CR1R2 + 2L) I2' + ( R12) I2 = CLV''

Figure 7.23 Example problem (continued)

The equations can also be manipulated into state equations, as shown in Figure 7.24. In this case a dummy variable is required to replace the two first derivatives in the first equation. The dummy variable is used in place of I1, which now becomes an output variable. In the remaining state equations I1 is replaced by q1. In the final matrix form the state equations are in one matrix, and the output variable must be calculated separately.

circuits - 7.21

State equations can also be developed using equations (1) and (3).

(1) becomes R1I1 + LI·1 = V + LI·2

LI·1 – LI·2 = V – R1I1

– I

–

-----I

= I1 – I2

–

-----I

= q1 + I2

–

+ I )

-----( q

+ I

–

–-----

-----

(3) becomes

··

LI2 – LI1

---

+ R2I2

V – R1I1 + ---

+ R2I2

= 0

V – R

( q

+ I )

–R

---

– R

+ q

( –R

) + V

+ ---

(10)

(11)

(12)

(13)

These can be put in matrix form,

–-----

----

1 1

–R

– R

---

Figure 7.24 Example problem (continued)

circuits - 7.22

Evaluate the circuit in Figure 7.22 using the node voltage method.

Figure 7.25 Drill problem: Use the node voltage method

circuits - 7.23

Find the equation relating the output and input voltages,

Vin

Vout

Figure 7.26 Drill problem: Find the state equation

circuits - 7.24

The circuit in Figure 7.27 can be evaluated as a voltage divider when the capacitor is represented as an impedance. In this case the result is a first-order differential equation.

t=0

= 3 cos t

As normal we relate the source voltage to the output voltage. The we find the values for the various terms in the frequency domain.

= V

------------------

where,

= R

ZC =

-------

s Z

+ Z

Next, we may combine the equations, and convert it to a differential equation.

------------------

R +

--------

CRD

---------------------

CRD + 1

Vo( CRD + 1)

= Vs( CRD)

( CR) + Vo

= Vs( CR)

+ Vo -------

= Vs

Figure 7.27 Circuit solution using impedances

The first-order differential equation in Figure 7.27 is continued in Figure 7.28 where the equation is integrated. The solution is left in variable form, except for the supply voltage.

circuits - 7.25

First, write the homogeneous solution using the known relationship.

							t
·		1			yields		–-------
							CR
							CR
Vo	+ Vo	-------	=	0		Vh = C1e
Vo	+ Vo	-------	=	0		Vh = C1e
		CR
Next, the particular solution can be determined, starting with a guess.

·		1		d	( 3 cos t) =


Vo	+ Vo	-------	=	----		–3 sin t
		CR		dt
		CR		dt

A sin t + B cos t

Vp'

A cos t – B sin t

( A cos t – B sin t)

+ ( A sin t + B cos t)

-------

–3 sin t

A + B

-------

–1

A =

-------

– B + A

-------

–3

–1

– B + B

-------

–3

CR CR

B ------------

+ 1

C2R2

3C2R2

–1

–3CR

B =

---------------------

A =

---------------------

-------

---------------------

1 + C2R2

+ B

sin

t + atan

The homogeneous and particular solutions can now be combined. The system will be assumed to be at rest initially.

–-------

Vh + Vp =

C1e

+ A

+ B

sin

t + atan

0 = C1e

+ B

sin

0 + atan

A--

–

+ B

sin

0 + atan

---

Figure 7.28 Circuit solution using impedances (continued)

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