Engineering
Applications Electrical with the TI-89
David R. Voltmer
Rose-Hulman Institute of Technology
Terre Haute, Indiana
Mark A. Yoder
Rose-Hulman Institute of Technology
Terre Haute, Indiana
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Contents
Do This First
Chapter 1:
Topic 1:
Topic 2:
Topic 3:
Chapter 2:
Topic 4:
Topic 5:
Topic 6:
Topic 7:
Topic 8:
Chapter 3:
Topic 9:
Topic 10:
Topic 11:
Topic 12:
Chapter 4:
Topic 13:
Topic 14:
Topic 15:
Topic 16:
Topic 17:
Topic 18:
Chapter 5:
Topic 19:
Topic 20:
Topic 21:
Topic 22:
Topic 23:
Topic 24:
Topic 25:
Topic 26:
Chapter 6:
Topic 27:
Topic 28:
Topic 29:
Topic 30:
Topic 31:
1
DC Circuit Analysis
11
Nodal Equations Using solve( )
Nodal Equations with Voltage Sources
Nodal Equations Using simult( )
11
12
14
Transient Circuit Analysis: Symbolic
17
RC First-Order Circuit
Graphing First-Order Solutions
First-Order Circuit with an Initial Condition
First-Order Circuit with a Time Varying Source
RLC Second-Order Circuit
17
18
19
21
23
Transient Circuit Analysis: Numeric
27
RLC Circuit: Direction Field
RLC Circuit: Time Domain
RLC Circuit: Multiple Initial Conditions
RLC Circuit: Adjusting the Circuit Parameters
27
30
30
31
Steady-State Circuit Analysis and Filer Design
33
Phasor Analysis
Graphing Frequency Response
Filter Design Overview
Butterworth Filter
Chebyshev Filter
Logarithmic Frequency Plots
33
35
38
39
42
44
Power Engineering
47
Phasor Algebra
Average Power
Complex Power
Power Factor
Power Factor Correction Using Impedances
Power Factor Correction Using Power Triangle
Y- and -Y Transformations
Unbalanced Three-Phase Systems
47
49
51
53
54
55
56
57
Laplace Analysis: The s-domain
61
RLC Circuit
Critical Damping
Poles and Zeros in the Complex Plane
Frequency Response
3D Poles and Zeros
61
64
65
68
69
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Chapter 7:
The Convolution
Topic 32: Convolution Integral
Topic 33: Piecewise Convolution
Topic 34: Graphing Piecewise Convolution Results
Chapter 8:
Fourier Series
Topic 35: Square Wave: Computing the Coefficients
Topic 36: Square Wave: Constructing the Wave from the Coefficients
Chapter 9:
Topic 37:
Topic 38:
Topic 39:
Topic 40:
Topic 41:
Topic 42:
Topic 43:
Topic 44:
83
85
86
87
88
90
91
92
Gradient
Surface Normal
Divergence
Curl
Laplacian
Line Integrals
Surface Integrals
Separation of Variables
3D Potential Graphs
Relaxation Method
3D Graphs of Tabular Data
Characteristic Impedance
Reflection Coefficient
Phase Shift
Input Impedance/Admittance
VSWR
Impedance Matching
Chapter 13: Antennas
Topic 62: Incremental Dipole
Topic 63: Antenna Patterns
Topic 64: Phased Arrays
Chapter 14: Manipulating Lab Data: The Diode
Topic 65: The Diode Equation
Topic 66: Lab Data
Chapter 15: Financial Calculations
Topic 67:
Topic 68:
Topic 69:
Topic 70:
Simple Interest
Compound Interest
Loans
Annuities
Index
iv
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77
80
83
Chapter 12: Transmission Lines
Topic 56:
Topic 57:
Topic 58:
Topic 59:
Topic 60:
Topic 61:
77
Coordinate Systems and Coordinate Transformations
Vector Components
Angle between Vectors
Parallel and Perpendicular Vectors
Rectangular to Cylindrical Vector Transformation
Cylindrical to Rectangular Vector Transformation
Rectangular to Spherical Vector Transformation
Spherical to Rectangular Vector Transformation
Chapter 11: Electromagnetics
Topic 52:
Topic 53:
Topic 54:
Topic 55:
71
71
75
Vectors
Chapter 10: Vector Calculus
Topic 45:
Topic 46:
Topic 47:
Topic 48:
Topic 49:
Topic 50:
Topic 51:
71
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95
97
98
99
100
101
103
105
105
109
112
115
117
117
121
122
122
123
124
127
127
130
132
135
135
137
141
141
142
143
144
147
Preface
To Students:
This book is written for electrical engineering students. It is a collection of examples that
show how to solve many common electrical engineering problems using the TI-89. It is not a
textbook; if you do not know how to solve the problem, look it up in your textbook first. If you
do know how to solve the problem, this book will show you how to use the TI-89 to get the
answer with more insight and less tedium. We show you how to use the TI-89 in class, in lab,
on homework, and so forth.
Many of you may now use Maple, Mathematica, MATLAB, Mathcad, or other symbolic or
numeric software. You will be pleasantly surprised to find that the TI-89 can solve many of the
same problems as the big boys, but it will boot up in only a second or two, it rarely crashes, it
fits your pocket book (even if you have a small one), and can fit in your pocket (if you have a
big one).
You should find this book easy to use. Although we show how to use many of the features of
the TI-89, we assume you already know your way around it. First read Do This First, then
jump to the section discussing the problem you want to solve.
To Instructors:
Read the To Students section.
When writing this book, we resisted the temptation to show how the TI-89 can be used to solve
problems in ways that differ from standard electrical engineering texts. Although it has the
power and ability to approach many problems in new ways, that was not our focus.
Our focus is to help students learn the basic material better by showing them how to use the
TI-89 to do the tedious things so they dont get lost in the details. Our approach was best
summed up by Gottfried Wilhelm Leibniz when, in the 17th century, he said,
It is unworthy of excellent men to lose hours like slaves in the labor of calculation.
David Voltmer
Mark Yoder
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About the Authors
DAVID VOLTMER (AKA Smilin Dave) loves teaching electrical engineering at Rose-Hulman
Institute of Technology. He claims to be good in the areas of electromagnetics, microwaves,
antennas, communications and design. A few of his many projects designed to assist student
learning include PC-based instruments, SPICE48, and Visual Electromagnetics (VEM). The
writing of this book was accompanied by the sounds of clawhammer banjo music and with
regular training breaks for long-distance cycling.
MARK A. YODER, Associate Professor of Electrical and Computer Engineering, received his
B.S. degree in 1980 and Ph.D. in 1984, both in Electrical Engineering and both at Purdue
University. While there, he did research in speech and in image processing, in addition to
studying the dynamics of a disk passing through a non-viscous medium (that is, playing a lot of
Frisbee).
In 1988 he discovered that teaching was where the fun is, so he headed for Rose-Hulman in
Terre Haute, Indiana. Here he pioneered the use of symbolic algebra systems in electrical
engineering education and helped develop a class on computer vision. He developed an expert
system for diagnosing a fiber-optic communications system for the International Centers for
Communication Technology, and he hopes to work on computer assisted Bible translation. He
has co-authored a book on digital signal processing for sophomores.
Dr. Yoders biography is not complete without mention of his family. His wife Sarah has her
Ph.D. in Electrical Engineering from Purdue. They have nine children aged 15, 15, 12, 10, 8, 6,
4, 2, and -0.5 two boys, six girls, and one on the way.
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Features Used
cSolve( ), , , ,
abs( ), angle( ), Numeric
Solver, when( ), log( ),
DelVar, DrawFunc, DrawInv,
NewProb, a, #,
%, $
Chapter
4
Setup
1, NewFold steady,
.
setMode(Angle,Degree)
Steady-State
Circuit Analysis
And Filter Design
This chapter shows how the TI-89 implements phasors to
perform sinusoidal steady-state analysis. The focus is on
how to enter and display complex numbers. This chapter
also shows a typical steady-state applicationhow to use
the Numeric Solver to find the required order of lowpass
Butterworth and Chebyshev filters in making a standard
handbook filter design.
Topic 13: Phasor Analysis
Given the circuit shown in Figure 1, find v, the voltage across the current source.
Figure 1. A circuit in steady-state
The first step is to convert the actual circuit to its phasor equivalent. The circuit shown in Figure 2
includes these conversions.
Figure 2. The phasor equivalent of the Figure 1 circuit
Only one nodal equation is needed to solve for v
80+
v
v
v
+
+
=0
10 6 + j8 j5
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34
ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89
1.
Clear the TI-89 by pressing 2 2:NewProb .
2.
Be sure the Complex Format mode is set to Rectangular.
Be sure the Graph mode is set to Function.
3.
Enter the equation as shown in screen 1.
8 v e 10 v e c 6 2 ) 8 d v e c
2 ) 5 d 0 n1
4.
(1)
Note: The usual imaginary number j
used in electrical engineering is
entered as i which is 2 ).
Using solve(n1,v) will return false since it is valid for
real solutions only. To get a complex solution, enter
cSolve( ) as shown in screen 2.
(2)
Note: To enter cSolve(, press
A:Complex 1:cSolve( or
cSolve(.
5.
Phasors are expressed as a magnitude at an angle, M.
There are a couple of ways to obtain this form.
The first way is to use the functions abs( ) (top of screen
3) and angle( ) (middle of screen 3). In this example,
is used to get the approximate values for the second
angle( ) command (bottom of screen 3).
This shows that the phasor form of the voltage is
40 -36.87 in Degree mode.
(3)
Note: To see the values on the
right end of a solution line in the
history section, press C to get to the
line and then press B to move to the
right.
Note: Press 3 to switch to
degree mode if it isnt already set. If
in radian mode the angle would have
been given in radians.
The second approach is to put the TI-89 in Polar mode.
Press 3 and select Complex Format 3:Polar (screen 4).
(4)
6.
Using cSolve( ) (and with for a second approximate
solution) gives the same results in the polar mode as
abs( ) and angle( ) in the rectangular mode, as shown in
the bottom two lines of screen 5.
(5)
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CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN
35
Topic 14: Graphing Frequency Response
It is easy to find the voltage across the current source as a function of using symbolic
expressions. Table 1 shows the variations of phasor circuit elements with radian frequency .
Element
Element
Equation
Phasor Result
jt
8cos10 t
v(t)=Re[Ve ]
V=80
10
ZRa = Ra
ZRa = 10
6
ZRb = Rb
ZRb = = 6
80 H
ZL = jL
ZL= j8010-6
5
2 F
ZC =
1
j C
ZC =
1
j 2 10 6
Table 1: Frequency of Dependence Phasor Circuit Elements
The nodal equation then becomes
80o +
v
v
v
+
+
=0
ra rb + zell zc1
Note that ra, rb, and zc1 are used because r1, r2, and zc are TI-89
system variables.
1.
Switch back to Rectangular Complex Format mode and
enter the equations as shown in screen 6.
8 v e ra v e c rb zell d v e zc1 0
n1
as shown in screen 6.
2.
(6)
Note: zell is used to avoid confusing
zl with z1 (z followed by a 1), a
reserved name.
Define the element values from the table (screens 7 and
8). For convenience, w ( j w) is used instead of
( c j w).
(7)
(8)
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36
3.
ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89
With cSolve( ), the solution in screen 9 shows that the
voltage varies with frequency.
(9)
The complete answer is
v=
4000000.00 (iw + 75000.00)
w 125000.00iw 10000000000.00
2
The answer is a bit of a mess. To check it with the
previous solution, enter eqn | w=100000 (screen 10).
Its the same answer as Topic 13, screen 2.
(10)
4.
To view the variation of the voltage magnitude versus
frequency, graph v versus w. Since the original problem
used w=100,000, graph from w=0 to w=200,000. Press
$ and set xmin to 0, xmax to 200000, ymin to 0,
and ymax to 50. Press # and set y1 to graph the
magnitude of v (screen 11).
5.
Press % to see the magnitude graph (screen 12).
6.
This graph takes a long time to complete because the
with substitutions are made over and over again for
each pixel. One way to speed it up is to do the withs
once before graphing and save the result in another
variable name which is then graphed.
(11)
(12)
To do this, press " and enter the expression:
abs( v eqn d eqn2 as shown in screen
13.
(13)
Then press #, deselect y1, and enter y2 (screen 14).
(14)
Note: Deselect equation y1(x) with
.
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CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN
7.
37
Press %. The same result as the previous graph
appears much more quickly (screen 15).
(15)
8.
The phase can be graphed defining the phase angle of
the voltage on the Home screen as eqn3 (screen 16).
(16)
9.
Press #, deselect y2, and enter eqn3 as plot variable
y3 (screen 17).
10. Press $ and set ymin to -90 and ymax to 0 (since
the calculations have been in the degree mode). xmin
and xmax can remain the same.
(17)
11. Press % to see the phase graph as shown in
screen 18.
(18)
12. Usually the magnitude and phase plots are shown
together. This can be done using the split screen mode.
To do this, press 3 B D. Screen 19 presents the
Split Screen options.
(19)
13. Press . Move down to Split 2 App, and select
4:Graph. Finally, set Number of Graphs to 2 (screen 20).
(20)
14. Press to view the split screen plots (screen 21).
The top graph is the phase plot shown before; the
bottom graph contains no data yet.
(21)
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ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89
15. Convention says the magnitude plot should be on top.
To do this, press # and use to select y2 and
deselect y3. Next, press $ and set ymin to 0 and
ymax to 50. Finally, press % to see the magnitude
plot in the upper graph as shown in screen 22.
(22)
16. To set up the phase plot in the lower window, change to
the other half of the screen by pressing 2a and set
up the graph as before. The following operations will
give the phase plot in the bottom window (screen 23).
#, select y3, $, set xmin to 0, xmax to
200000, ymin to -90 and ymax to 0, and finally, %
(23)
17. You can use 2 7:Text to add magnitude and phase
labels to the graphs. To do this, press 2 7:Text and
position the cursor where the text should start. The
characters will appear below and to the right of the
crosshairs. Be careful; once a letter is placed it cant be
erased except by 2 2:Eraser.
(24)
18. To return to a single screen, press 3 and set Split
Screen to 1:FULL.
Topic 15: Filter Design Overview
A class of realizable frequency responses for lowpass filters has the form
H(f) =
2
1
1 + 2 (f)
2
where (f) is a polynomial in f. If
f
(f ) =
f
p
n
the filter is a Butterworth filter. An alternative is to make (f) = Cn(f/fp) where Cn is a Chebyshev
polynomial, the filter is a Chebyshev filter. The next topic deals with a Butterworth filter, the
following topic with a Chebyshev filter.
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CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN
39
The design of lowpass filters requires specification of passband and stopband responses often
given in dB. |H(f)|2 in dB is calculated as
1
| H ( f )|2 = 10 log
dB
1 + 22 (f )
which becomes
1
2
| H ( f )|dB = 10 log
2n
f
2
1+
f
p
for Butterworth and
1
2
| H ( f )|dB = 10 log
1 + 2C2 f
n
fP
for Chebyshev.
Topic 16: Butterworth Filter
The performance specifications of a filter are often given in graphical form as shown in Figure 3.
The design of a Butterworth filter with these performance specifications is described here.
1k
2k
f (Hz)
Passband
-3dB
-40dB
|H(f)|2dB
Stopband
Figure 3. Filter design specifications for a Butterworth filter
Suppose a filter with the maximum passband ripple is -3 dB, and the passband edge is at f p=1kHz is
to be designed. Additionally, the stopband gain is to be no more than -40 dB with a stopband edge
at fs=2kHz.
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40
1.
ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89
From the Home screen, press 1 2 2:NewProb to
reset the TI-89 to a known state. Then enter the
Butterworth equation as shown in screen 25.
10 p log( 1 e c 1 eps Z 2 p c f e fp d Z
c 2n d d d butter
(25)
2.
Press O 9:Numeric Solver exp=butter (screen 26).
(26)
3.
4.
Note that the Numeric Solver listed each of the
variables for values to be entered. Find the value of eps
by entering the data for the passband edge with a -3 dB
response at 1000 Hz. n is unknown but at the passband
edge all values give the same result, so for now enter 1
for n as shown in screen 27.
(27)
Place the cursor on the eps line and press to solve
for eps. After a second or two, the screen shows eps is
about 1 (screen 28).
(28)
5.
(5
Now, find the order of the filter by setting the stopband
edge response (exp) to -40 dB and f to 2000 (screen 29).
(29)
6.
Solve for n by placing the cursor on n and pressing .
After a couple of seconds the solution of n=6.6 is
shown, as in screen 30.
(30)
7.
Since n must be an integer, set n to the next larger
integer value of 7 and solve for exp to find the stopband
gain for this value of n (screen 31).
With a 7th-order Butterworth filter the stopband gain is
- 42 dB, a little better than the minimum needed.
(31)
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CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN
8.
41
Now, plot the Butterworth equation to see the
frequency response. To do this, press # and enter
butter with f replaced by x (screen 32).
(32)
9.
Since the stopband edge of the filter is 2 kHz, plot x
from 0 to 3000. The stopband value is -40 dB, so plot y
from -45 to 0. Enter these values in the Window Editor
( $) as shown in screen 33.
(33)
10. Press 3 to be sure the Graph mode is set to
FUNCTION. Then press % and wait a few seconds
to see screen 34.
(34)
The response in the passband looks very flat, which is
correct for Butterworth, but are the passband and
stopband edges in the right places? These can be
checked graphically by pressing 2 2:DrawFunc to
draw horizontal lines at -3 and -40 dB and
2 3:DrawInv to draw vertical lines at 1000 and
2000 Hz (screen 35).
(35)
11. Press % to plot the results (screen 36).
The curve passes through the -3 dB point at 1000 Hz and
passes below the -40 dB point at 2000 Hz. The filter
meets the required specifications.
(36)
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ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89
Topic 17: Chebyshev Filter
This section shows how to design a Chebyshev lowpass filter with the same specifications as
discussed in Topic 16 and shown in Figure 4.
1k
2k
f (Hz)
Passband
-3dB
-40dB
Stopband
|H(f)|2dB
Figure 4. Filter design specifications for a Chebyshev filter
The Chebyshev equations are
f
f
C n = cos n cos1
f
f
p
p
f
f
C n = cosh n cosh 1
f
f
p
p
f
<1
fp
where n is the order of the polynomial.
f
>1
fp
Therefore HdB is
2
H ( f ) dB
1
= 10 log
f
22
1 + Cn f
p
Follow these steps to enter these three equations.
1.
From the Home screen, clear f, fp, eps, and n using
DelVar.
2.
Note: DelVar can be entered by
pressing 4:DelVar.
Enter Cn for |f/fp| < 1 as shown in screen 37.
2 X n p R f e fp d d cheb1
(37)
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CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN
3.
43
Enter Cn for |f/fp|>1 as shown in screen 38.
cosh( n p cosh-1( f e fp d d
cheb2
(38)
4.
Define cheb to be cheb1 for |f/fp|<1 and cheb2 for |f/fp|>1.
This is done using the when function as shown in
screen 39.
when( abs( f e fp d 2 1 b cheb2
b cheb1 d cheb
(39)
5.
Define hdb to be |H(f)|2dB as shown in screen 40.
10 log( 1 e c 1 eps Z 2 p cheb Z 2 d d
hdb
(40)
6.
To design the filter, use the Numeric Solver. Press
O 9:Numeric Solver exp=hdb (screen 41).
(41)
7.
Proceed as with Butterworth. Find eps for -3dB at the
passband edge by entering the values as shown in
screen 42. The result is the same as with Butterworth.
(42)
8.
Now find n for -40 dB at 2000 Hz. It takes a few seconds
to find the order of the filter (screen 43).
(43)
9.
To meet the design specifications, n must be 5. Enter 5
and calculate the response. See what the gain is
(screen 44).
This filter exceeds the design specifications by more
than 10 dB with a lower order than the Butterworth.
(44)
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44
ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89
10. In the Y= Editor, enter the expressions for the two
filters to compare them (screen 45). Note that n=7 is
added to y1 so that the Butterworth filter uses the order
computed for it instead of using the value of n=5 used
for the Chebyshev filter.
(45)
11. Press % to see the graphs (screen 46).
(See Topic 16 for a review of the instructions for a
graphical comparison of the filters performances.)
As expected, the Chebyshev filter has ripples in the
passband, but it drops more rapidly in the transition
band than the Butterworth.
(46)
12. Compare the filters performances with the lines which
represent the passband and stopband design
specifications (screen 47). Press 2 2:DrawFunc to
draw horizontal lines at -3 and -40 dB. Press 2
3:DrawInv to draw vertical lines at 1000 and 2000 Hz.
Both filters show the required -3 dB response at 1000
Hz; both filters exceed the specifications since they are
below -40 dB at 2000 Hz.
(47)
Topic 18: Logarithmic Frequency Plots
Often frequency responses are plotted on a log frequency scale. Although the TI-89 doesnt directly
support log plots, they are easy to do.
1.
Return to the Y= Editor and alter the with operation
to include the logarithmic relation of frequency with x
as shown in screen 48.
y1: butter f 10 Z x and n 7
y2: hdb f 10 Z x
(48)
The values for x are linearly spaced, but the values of
10x are logarithmically spaced.
2.
Press $ to adjust the range on x. Graph the
functions for f=500 and f=3000. To do this, enter log(500)
for xmin and log(3000) for xmax as shown in screen 49.
(49)
3.
Press % to display screen 50. The logarithmic
plots take on a different appearance than the linear
plots of screen 46.
(50)
1999 TEXAS INSTRUMENTS INCORPORATED
04steady.doc .
Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 44 of 14
CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN
4.
45
The passband and stopband lines are added by using
log() of the f values in DrawInv (screen 51). Press
2 2:DrawFunc for -3 and -40 dB. Press 2
3:DrawInv for log(1000) and log(2000) Hz.
(51)
As expected, the plots in screen 52 show that both
filters meet the design specifications.
(52)
Tips and Generalizations
Topic 13 show how to enter, solve, and display equations with complex numbers. Multiple
equations with multiple complex unknowns can be solved. Also the matrix approach of Topic 3
can be used with complex numbers.
The Numeric Solver works nicely for filter design but can easily solve any equation for an
unknown value.
Once a steady-state response is known, the power dissipated by the various elements can be
found. Chapter 5 explores this topic further.
1999 TEXAS INSTRUMENTS INCORPORATED
04steady.doc .
Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 45 of 14