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Background Review• Elementary functions• Complex numbers• Common test input signals• Differential equations• Laplace transform
– Examples– properties– Inverse transform– Partial fraction expantion
• Matlab
Elementary functions
0
1lnln ln
Exponential Function
, /
( ) , 1
Logrithmic Function ln( )
ln ln ln
ln ln ln
ln ln , ln(1) 0
1
x
x y x y x y x y
x y xy
a
x x x
e
e e e e e e
e e e
x
(xy) x y
x x yy
(x ) a x
e x, e ex
The most beautiful equation
• It contains the 5 most important numbers: 0, 1, i, , e.
• It contains the 3 most important operations: +, *, and exponential.
• It contains equal sign for equations
* 1 0ie
2 2
Even Function: ,
Odd Function: ,
sin sin cos cos
sin cos 1
sin sin cos cos sin
sin sin cos cos sin
sin 2 2sin cos
cos c
f( x) f(x) x
f( x) f(x) x
( x) x, ( x) x
x x
(x y) x y x y
(x y) x y x y
x x x
(x y)
os cos sin sin
cos cos cos sin sin
x y x y
(x y) x y x y
Elementary functions
Elementary functions
x x,
xx,
x
xx,
x
xx
x x) (π
x x) (π
-x)π
()π
( xx
-x)π
()π
( x-x
x)(x
x)(x
xxxxx
sin
1csc
cos
1sec
sin
coscot
cos
sintan
coscos
sinsin2
sin2
sincos
2cos
2cossin
2cos12
1sin
2cos12
1cos
1cos2sin21sincos2cos
2
2
2222
Elementary functions
2sin
2sin2coscos
2cos
2cos2coscos
2cos
2sin2sinsin
22then Substitute
sinsin2
1cossin
coscos2
1coscos
coscos2
1sinsin
vuvuvu
vuvuvu
vuvuvu
vu,y
vuxx-y, y, vxu
y)](xy)(x[yx
y)](xy)(x[yx
y)](xy)(x[yx
Elementary functions
phase, δ magnitudeC
δ)(xBAxBxA
B
Aδ, BAδ, then CCδ, BCA
xBxAxδCxδCδ)(xC
δ)(xBAxBxA
A
Bδ, BACδ, Cδ, BCA
xBxAxδCxδCδ)(xC
xB x A
::
sinsincos
tancossin
sincoscossinsincossin
Also
cossincos
tanthen sincos
sincossinsincoscoscos
sincos gSimplifyin
22
22
22
22
Elementary functions
• F(t)=3sin t +4cos t
• F(t)=Asin(3t-)=Acos sin3t –Asin cos3t
• Acos =3
• Asin =-4
• A2=25, A=5
• tan =-
• F(t)=5sin(3t+)
Complex Numbers
• X2+1=0 x=i where i2=-1
• X2+4=0, then x=2i, or 2j
• If z1=x1+iy1, z2=x2+iy2
• Then z1+ z2= (x1+ x2)+i(y1 + y2)
• z1 z2=(x1+iy1)(x2+iy2)=(x1x2 -y1y2) +i(x1y2
+x2y1)
22
22
211222
22
2121
2222
2211
22
11
2
1
))((
))((
yx
yxyxi
yx
yyxxz
iyxiyx
iyxiyx
iyx
iyx
z
zz
Polar form of Complex Numbers• z=x+iy, let’s put x=rcos, y= rsin• Then z = r(cos+i sin = r cisr• Absolute value (modulus) r2=x2+y2
• Argument= tan-1(y/x)
• Example z=1+i
,...2,1,0,24
arg
2
nnz
z
Euler Formula
• z=x+iy
• ez =ex+iy= ex eiy= ex (cos y+i sin y)
• eix =cos x+i sin x = cis x
• | eix | = sqrt(cos2 x+ sin2 x) = 1
• z=r(cos+i sinr ei
• Find e1+i
• Find e-3i
In Matlab>> z1=1+2*i
z1 = 1.0000 + 2.0000i
>> z2=3+i*5
z2 = 3.0000 + 5.0000i
>> z3=z1+z2
z3 = 4.0000 + 7.0000i
>> z4=z1*z2
z4 = -7.0000 +11.0000i
>> z5=z1/z2
z5 = 0.3824 + 0.0294i
>> r1=abs(z1)
r1 = 2.2361
>> theta1=angle(z1)
theta1 = 1.1071
>> theta1=angle(z1)*180/pi
theta1 = 63.4349
>> real(z1)
ans = 1
>> imag(z1)
ans = 2
Poles and zeros• Pole of G(s) is a value of s near which the
value of G goes to infinity• Zero of G(s) is a value of s near which the
value of G goes to zero.
A r-th order pole p:
( )( ) ; 0,lim r
s p
G s s p R R
A zero z:
( ) 0lims z
G s
Poles and zeros in Matlab>> s=tf(‘s’)
Transfer function: s
>> G=exp(-2*s)/s/(s+1)
Transfer function: 1exp(-2*s) * ----------- s^2 + s
>> pole(G)
ans = 0, -1
>> zero(G)
ans = Empty matrix: 0-by-1
2
2( )
2
seG s
s
Test waveforms
used in control
systems
1st order differential equations• y’ + a y = 0; y(0)=C, and zero input• Solution: y(t) = Ce-at
• y’ + a y = (t); y(0)=0, input = unit impulse• Unit impulse response: h(t) = e-at
• y’ + a y = f(t); y(0)=C, non zero input• Total response: y(t) = zero input response +
zero state response = Ce-at + h(t) * f(t)
• Higher order LODE: use Laplace
Laplace Transform
• Definition and examples
ssF
es
es
es
dteLsF
tutf
FLtf
dttfefLsF
sstst
st
1)(
1111)1()(
1)()(
Example
Lapalce inverse );()(
)()()(
0
00
1
0
Unit Step Function u(t)
Laplace Transform
aseL
ssF
se
s
dtedteeeLsF
etf
at
ts
tststt
t
1)(
3
1)(
3
1
3
1
)()(
)(
Example
0
)3(
0
)3(
0
33
3
1lexponentia )(
1 impulseunit )(
1stepunit
t
astu
seat
___;)tan(sin______sincos 2 ji)b
a(xxbxa
The single most important thing to remember is that whenever there is feedback, one should worry about __________
Name:____________
Laplace Transform
320
00
2
0
22
20
20
000
2122
211
)(
111
11)(
)(
ssstdte
s
dttes
ets
dttetL
se
sdte
s
dtes
tes
tdtetL
ttf
st
ststst
stst
ststst
Laplace Transform
)(1
)'(1
)(1
a
ss'Let
)(1
)(
)()(
zatLet ?)}({about what ),()}({ If
0
'
0
0
a
sF
asF
adzzfe
a
dzzfea
atfL
dtatfeatfL
atfLsFtfL
zs
a
zs
st
Laplace Transform
220
20
2
2
02
2
2
0
2
02
0
000
sin
sin)1(
sin
sin)1
(coscos
cos)1
(sin1
sin)(sin
sin)(
stdte
stdte
s
tdtess
tdtess
tes
tdtes
tdtes
tes
tdtetL
ttf
st
st
st
ststst
ststst
Laplace Transform
)(sin)(cos)sin(cos)(
11
11
)(
cos)( ,sin)(
222222
0
)(
0
)(
0
tiLtLtitLeL
s
i
s
s
s
is
is
is
isis
ise
is
dtedteeeL
ttfttf
ti
tis
tististti
Laplace transform table
Laplace transform theorems
Laplace Transform
220
220
20
2
2
0
)(2
2
2
0
2)(
0
)(2
0
)(
0
)(
0
)(
0
)(cos
)(sin
)(sin)
)(1(
sin)()(
sin)1
(cos)(
cos
cos)1
(sin1
sin)sin(
as
astdte
astdte
astdte
as
tdteasas
tdteasas
teas
tdteas
tdteas
teas
tdteeteL
st
st
st
tas
tastas
tas
tastas
atstat
Laplace Transform
?)52
23(
sin2cos3)1
2
1
3()
1
23(
)3sin3
13(cos3sin
3
13cos
)3)2(
1
3)2(
2()
3)2(
3(
3sin3
1)
3
3
3
1()
3
1()
9
1(
21
22221
221
222
22221
221
221
221
21
ss
sL
Find
ttss
sL
s
sL
ttetete
ss
sL
s
sL
ts
Ls
Ls
L
ttt
Laplace Transform
)0(')0()(
)]0()([)0(')(')0('
)(')()(')(")}("{
)0()()'(
)0()()'(
)()0()()0(
)()()()(')}('{
2
0
00
0
0
00
0
fsfsFs
fssFsfdttfesf
dttfestfedttfetfL
yssYyL
xssXxL
ssFfdttfesf
dttfestfedttfetfL
st
ststst
st
ststst
Laplace Transform
• y”+9y=0, y(0)=0, y’(0)=2
• L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2
• L(y)=Y(s)
• (s2+9)Y(s)=2
• Y(s)=2/ (s2+9)
• y(t)=(2/3) sin 3t
Matlab
F=2/(s^2+9)
F =
2/(s^2+9)
>> f=ilaplace(F)
f =
2/9*9^(1/2)*sin(9^(1/2)*t)
>> simplify(f)
ans =
2/3*sin(3*t)
Laplace Transform
• y”+2y’+5y=0, y(0)=2, y’(0)=-4
• L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2s+4
• L(y’)=sY(s)-y(0)=sY(s)-2
• L(y)=Y(s)
• (s2+2s+5)Y(s)=2s
• Y(s)=2s/ (s2+2s+5)=2(s+1)/[(s+1)2+22]-2/[(s+1)2+22]
• y(t)= e-t(2cos 2t –sin 2t)
Matlab
>> F=2*s/(s^2+2*s+5)
F =
2*s/(s^2+2*s+5)
>> f=ilaplace(F)
f =
2*exp(-t)*cos(2*t)-exp(-t)*sin(2*t)
Laplace transform
• Y”-2 y’-3 y=0, y(0)= 1, y’(0)= 7
• Y”+2 y’-8 y=0, y(0)= 1, y’(0)= 8
• Y”+2 y’-3 y=0, y(0)= 0, y’(0)= 4
• 4Y”+4 y’-3 y=0, y(0)= 8, y’(0)= 0
• Y”+2 y’+ y=0, y(0)= 1, y’(0)= -2
• Y”+4 y=0, y(0)= 1, y’(0)= 1
Y”+2 y’+ y=0, y(0)= 1, y’(0)= -2
>> A=[0 1;-1 -2]; B=[0;1]; C=[1 0]; D=0;>> x0=[1;-2];>> t=sym('t');>> y=C*expm(A*t)*x0 y = exp(-t)-t*exp(-t)
Y”+2 y’+ y=f(t)=u(t), y(0)= 2, y’(0)= 3
Partial Fraction
10
3
3
)2()2(
)3(
1
2sput and 2,-sby multiply
6
1
32)3)(2(
1
0sput and s,by multiply
32)3)(2(
1
6
1)(
y(t)? is What , 6
1)(
222
000
23
23
Bs
sC
s
sAB
ss
s
As
Cs
s
BsA
ss
s
s
C
s
B
s
A
sss
s
sss
ssY
sss
ssY
sss
sss
Partial Fraction
tt
sss
eety
ssssss
ssY
C
s
sA
s
sBC
ss
s
32
23
333
15
2
10
3
6
1)(
3
1
15
2
2
1
10
31
6
1
6
1)(
15
2
)3(
)2(
)3(
)2(
1
3sput and 3,sby multiply
Partial fraction; repeated factor
2)23)(3(
124137)(A
?Abut before, as obtained becan D C, B, ,A123
)(
)(
)(
)23)(3(
124137)(
3
142)1(31
-1(0)y' 1,y(0) ,42'3"
0
2
234
02
12
122
22
234
22
3
ss
t
sss
sssssQ
s
D
s
C
s
B
s
A
s
Aty
sH
sG
ssss
sssssY
ssYsYsYs
etyyy
Partial fraction; repeated factor
3)('A
0sput and s, with ateDifferenti
123)23)(3(
124137
obtained becan A 0,sput and ,sby Multiply
123
)23)(3(
124137)(
01
222
122
234
22
122
22
234
ssQ
s
Ds
s
Cs
s
BssAA
sss
ssss
s
D
s
C
s
B
s
A
s
A
ssss
sssssY
But No FUN
Partial fraction; exercise
22 )5
1 )4
44
1411 )3
9
99 )2
23 )1
2
2
23
3
2
2
ss
s
s
ssss
sss
ss
ss
s
>> [r p k]=residue(n,d)
r =
1
2
p =
1
0
k =
[]
>> d=[1 -1 0]
d =
1 -1 0
>> n=[3 -2]
n =
3 -21/(s-1) + 2/s
Matlab
>> n=[1 9 -9]
n =
1 9 -9
>> d=[1 0 -9 0]
d =
1 0 -9 0
>> [r p k]=residue(n,d)
r =
1.5000
-1.5000
1.0000
p =
3
-3
0
k =
[]
1.5/(s-3)-1.5/(s+3)+1/s
Matlab
>> n=[11 -14]
n =
11 -14
>> d=[1 -1 -4 4]
d =
1 -1 -4 4
>> [r p k]=residue(n,d)
r =
2.0000
-3.0000
1.0000
p =
2.0000
-2.0000
1.0000
k =
[]
2/(s-2)-3/(s+2)+1/(s-1)
Matlab
>> b=[1 2 1]
b =
1 2 1
>> a=[1 0]
a =
1 0
>> [r p k]=residue(a,b)
r =
1
-1
p =
-1
-1
k =
[]
1/(s+1)-1/(s+1)2
Matlab
>> Y=(s^4-7*s^3+13*s^2+4*s-12)/s^2/(s-3)/(s^2-3*s+2) Transfer function:s^4 - 7 s^3 + 13 s^2 + 4 s - 12------------------------------------ s^5 - 6 s^4 + 11 s^3 - 6 s^2>> [n,d]=tfdata(Y,'v')n = 0 1 -7 13 4 -12d = 1 -6 11 -6 0 0>> [r,p,k]=residue(n,d)r = 0.5000 -2.0000 -0.5000 3.0000 2.0000p = 3.0000 2.0000 1.0000 0 0k = [ ]
2
0.5 2 0.5 3 2
3 2 1Y
s s s s s
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