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SOLCAP10


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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: antoniodagama
Type : Application
Page(s) : 60
Taille Size: 2.82 Mo MB
Mis en ligne Uploaded: 14/08/2019 - 17:55:14
Uploadeur Uploader: antoniodagama (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a2295058

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10-6 Chapter 10: Frequency Response Methods


23. For Type zero: Kp = low frequency gain; For Type 1: Kv = frequency value at the intersection of the

initial slope with the frequency axis; For Type 2: Ka = square root of the frequency value at the intersection

of the initial slope with the frequency axis.
24. No change at all
25. A straight line of negative slope, T, where T is the time delay

26. When the magnitude response is flat and the phase response is flat at 0o.




SOLUTIONS TO PROBLEMS
1.
a.

;


;

b.


;


;

c.


;


;
2.
a.
Solutions to Problems 10-7



b.




c.




3.
a.
90°
120° 60°


150° 30°




180° XX
X 0°
X
X 0 0.5 1 1.5
X
X
X
210° 330°

X
240° 300°
270°
10-8 Chapter 10: Frequency Response Methods



b.
90°
120° 60°



150° 30°




180° X
X X 0°
X
0 0.2 0.4
X X
0.6
X X
XX
X XXX
X
X X X
X
210° 330°


240° 300°
270°
c.
90°
120° 60°



150° 30°




180° X 0°
X0
X 5 10 15 20
X
X
X
X
210° 330°


240°
X 300°
270°
Solutions to Problems 10-9


4.
a.




b.




c.
30
-90
20 -20 dB/dec -100
10 -45 deg/dec +45 deg/dec
-110
Phase
dB




0 -40 dB/dec -120
-20 dB/dec
-10 -130
-40 dB/dec
-20
-140
-20 dB/dec
-30 -150
.1 1 10 100 100
v .1 1 10
v
10-10 Chapter 10: Frequency Response Methods


5.
a. System 1




b. System 2




c. System 3
Solutions to Problems 10-11


d.




6.
0.5


0.4


0.3


0.2


0.1
Imag Axis




0


-0.1


-0.2


-0.3


-0.4


-0.5
-0.1 0 0.1 0.2 0.3 0.4 0.5 0.6

Real Axis
10-12 Chapter 10: Frequency Response Methods


7.
0



-50



Gain dB
-100



-150 -1
10 10 0 101 102
Frequency (rad/sec)


0


-90
Phase deg




-180


-270

10-1 10 0 101 102
Frequency (rad/sec)

8.
Program:
numg=[1 5];
deng=conv([1 6 100],[1 4 25]);
G=tf(numg,deng);
'G(s)'
Gzpk=zpk(G)
nyquist(G)
axis([-3e-3,4e-3,-5e-3,5e-3])
w=0:0.1:100;
[re,im]=nyquist(G,w);
for i=1:1:length(w)
M(i)=abs(re(i)+j*im(i));
A(i)=atan2(im(i),re(i))*(180/pi);
if 180-abs(A(i))<=1;
re(i);
im(i);
K=1/abs(re(i));
fprintf('nw = %g',w(i))
fprintf(', Re = %g',re(i))
fprintf(', Im = %g',im(i))
fprintf(', M = %g',M(i))
fprintf(', Angle = %g',A(i))
fprintf(', K = %g',K)
Gm=20*log10(1/M(i));
fprintf(', Gm = %g',Gm)
break
end
end


Computer response:
ans =

G(s)


Zero/pole/gain:
(s+5)
Solutions to Problems 10-13


----------------------------------
(s^2 + 4s + 25) (s^2 + 6s + 100)


w = 10.1, Re = -0.00213722, Im = 2.07242e-005, M = 0.00213732, Angle =
179.444, K = 467.898, Gm = 53.4026

ans =

G(s)


Zero/pole/gain:
(s+5)
----------------------------------
(s^2 + 4s + 25) (s^2 + 6s + 100)


w = 10.1, Re = -0.00213722, Im = 2.07242e-005, M = 0.00213732, Angle =
179.444, K = 467.898, Gm = 53.4026




9.
a. Since the real-axis crossing is at -0.3086, P = 0, N = 0. Therefore Z = P - N = 0. System is stable.
Derivation of real-axis crossing:
2 2
50 50 9 j 18
G( j ) .
s ( s 3)( s 6) s j 81 4
18 3
10-14 Chapter 10: Frequency Response Methods



Thus, the imaginary part = 0 at 18 . Substituting this frequency into G ( j ) , the real part is
evaluated to be -0.3086.
b. P = 0, N = -2. Therefore Z = P - N = 2. System is unstable.
c. P = 0, N = 0. Therefore Z = P - N = 0. System is stable
d. P = 0, N = -2. Therefore Z = P - N = 2. System is unstable.
10.
System 1: For K = 1,




The Nyquist diagram intersects the real axis at -0.0021. Thus K can be increased to 478.63 before
there are encirclements of -1. There are no poles encircles by the contour. Thus P = 0. Hence, Z = P -
N, Z = 0 + 0 if K <478.63; Z = 0 –(-2) if K > 478.63. Therefore stability if 0 < K < 478.63.
System 2: For K = 1,
Solutions to Problems 10-15


The Nyquist diagram intersects the real axis at -0.720. Thus K can be increased to 1.39 before there
are encirclements of -1. There are no poles encircles by the contour. Thus P = 0. Hence, Z = P - N, Z
= 0 + 0 if K <1.39; Z = 0 – (-2) if K > 1.39. Therefore stability if 0 < K < 1.39.
System 3: For K = 1,




Stable if 0<K<1.
11.
Note: All results for this problem are based upon a non-asymptotic fre...

Archive contentsContenu de l'archive

Action(s) SizeTaille FileFichier
3.06 Ko KB readme.txt
3.65 Ko KB lisezmoi.txt
1.09 Ko KB SOLCAP10.hpprgm
1.61 Mo MB SOLCAP10 01_30.hpappdir.zip
1.21 Mo MB SOLCAP10 31_60.hpappdir.zip
168 octets bytes appslist.txt

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