MAK333E Homework -1
Lecturer CRN
: Assoc. : Assoc. Prof. Kenan Refah Kutlu : 13721
Prepared By : Semih Gürel- 030110387 Serdar Değirmenci - 030110191 Murat Can Kilinci- 030110106
Doğan Fethi Kaya - 030130711
1) Determine the poles and zeros of the system and show them on the complex plane. (Hint: You may use roots() and pzmap() command of MATLAB.) There is no zeros of system and the poles are; (from matlab) 0 -5.236067977499790e+00 -7.639320225002103e-01
2) Considering the locations of the poles, comment on the stability of the system in open loop. Does the system oscillate? Although one of the roots are “0” that means the system is a little bit of unstable, the other roots are lie on the negative x axis therefore the system is stable and there is any oscillation.
3) Using the partial fraction expansion method and Laplace transforms, find the response y(t) of the system for a) Unit impulse input, b) Unit step step input For both cases, plot the graph of y(t) in MATLAB. Include your code! (Hint: You may use mathematics software to check your results, but the calculations must be clearly shown.)
= ++ = ++ = + ++
= 0 = 0,25
6=0 = 0,25
4=1 = 1,5
1 0,25 0, 0,25 25 1, 1,55 = 6 4 6 4 1 1 6 = = 0,25 0,25 0.25 0.25 6 4 6 4 =
6 3 3 5 = 6 4 3 5 5 3 5 = 0,25 0,25( − ℎ ℎ√ 5) ( − ℎ ℎ√ 5)) √ 5) √ 5) a) Unit impulse input num = 1 den = 1
6
4
0
r= 0.0427 -0.2927 0.2500 p = -5.2361 -0.7639 0 k= [] G= 1 ----------------s^3 + 6 s^2 + 4 s
Continuous-time Continuous-time transfer function. F= 1 ----------------s^3 + 6 s^2 + 4 s
Continuous-time Continuous-time transfer function. F= 1/(s^3 + 6*s^2 + 4*s) f= 1/4 - (exp(-3*t)*(cosh(5^(1/2)*t) + (3*5^(1/2)*sinh(5^(1/2)*t))/5))/4 (3*5^(1/2)*sinh(5^(1/2)*t))/5))/4 num=[1] den=[1 6 4 0] G=tf(num,den) F=G*1 syms s F=1/(s^3 + 6*s^2 + 4*s) f=ilaplace(F) t=0:0.5:2; f= 1./4 - (exp(-3.*t).*(c (exp(-3.*t).*(cosh(5^(1/2).*t) osh(5^(1/2).*t) + (3.*5^(1/2).*sinh(5^(1/2).*t))./5))./4 plot(t,f)
f(t) 0.2 0.18 0.16 0.14 0.12 t u p t u o
0.1 0.08 0.06 0.04 0.02 0
0
b) Unit step input
0. 5
1 t(s)
1. 5
= ++
num=[1] den=[1 6 4 0 0] G=tf(num,den) syms s Y=1/(s^4 + 6*s^3 + 4*s^2) y=ilaplace(Y) t=0:0.5:4; y= t./4 + (3.*exp(-3.*t).* (3.*exp(-3.*t).*(cosh(5^(1/2).*t (cosh(5^(1/2).*t) ) + (7.*5^(1/2).*sinh(5^(1/2).*t))./1 (7.*5^(1/2).*sinh (5^(1/2).*t))./15))./8 5))./8 - 3/8 plot(t,y)
G= 1 ----------------s^3 + 6 s^2 + 4 s
2
Continuous-time transfer function. Y= 1 ------------------s^4 + 6 s^3 + 4 s^2
Continuous-time transfer function. num = 1 den = 1
6
4
0
0
r= -0.0082 0.3832 -0.3750 0.2500 p = -5.2361 -0.7639 0 0 k= [] Y= 1/(s^4 + 6*s^3 + 4*s^2) y= t/4 + (3*exp(-3*t)*(cosh(5^(1/2)*t) (3*exp(-3*t)*(cosh(5^(1/2)*t) + (7*5^(1/2)*sinh(5^(1/2)*t))/15))/8 (7*5^(1/2)*sinh(5^(1/2)*t))/15))/8 - 3/8
y(t) 0.7
0.6
0.5
t u p t u o
0.4
0.3
0.2
0.1
0
0
4 b.
>> num1=[1]
num1 =
1
>> den1=[1 6 4 0]
den1 =
1
6
4
0
0. 5
1
1. 5
2 t(s)
2. 5
3
3. 5
4
>> sys1=tf(num sys1=tf(num1,den1) 1,den1)
sys1 =
1 ----------------s^3 + 6 s^2 + 4 s
Continuous-timee transfer function. Continuous-tim
>> step(sys1) >> grid
4 a.
>> num1=[1]
num1 =
1
>> den1=[1 6 4 0]
den1 =
1
6
4
0
>> sys1=tf(num sys1=tf(num1,den1) 1,den1)
sys1 =
1 ----------------s^3 + 6 s^2 + 4 s
Continuous-timee transfer function. Continuous-tim
>> impulse(sy impulse(sys1) s1) >> grid
5 a) Impulse Response :
lim = l→ im
→
∞ = l→ im .
1 6 4
num=[1] den=[1 6 4] finalvalue=polyval(num,0)/polyval(den,0) num = 1
den = 1
6
4
finalvalue = 0.2500
b) Step Response :
lim = l→ im
→
∞ = l→ im .
1 6 4
num=[1] den=[1 6 4 0] finalvalue=polyval(num,0)/polyval(den,0)
num =
1
den =
1
6
4
finalvalue =
Inf
0
