Faculty of Engineering
Credit Hours Engineering Programs
COMM 482: Control Systems
MATLAB Quiz G1
Question 1
a) Using MATLAB find Laplace transform of the following function
𝑓 𝑡 = cos 2𝑡 + 0.2 𝑒 −2.5𝑡
>> syms t s
>> f = cos (2* t) + 0.2 * exp(-2.5 *t)
f=
cos(2*t) + 1/(5*exp((5*t)/2))
>> laplace(f)
ans =
1/(5*(s + 5/2)) + s/(s^2 + 4)
b) Using MATLAB find Inverse Laplace transform of the following function
𝑠 𝑠+3 𝑠+5 𝑠+7
𝐹 𝑠 =
𝑠 𝑠 + 8 𝑠 2 + 10𝑠 + 100
>> syms t s
>> f=(s*(s+3)*(s+5)*(s+7))/s*(s+8)*(s^2+10*s+100);
>> ilaplace(f)
ans =
84000*dirac(t) + 75700*dirac(t, 1) + 26670*dirac(t, 2) + 4883*dirac(t, 3) +
521*dirac(t, 4) + 33*dirac(t, 5) + dirac(t, 6)
Question 2
Consider the system shown in Figure 1. Using MATLAB or SIMULINK plot the unit
step response of this system for K =4, 0.25, 1 on the same graph. For each value of K
estimate the rise time, settling time, maximum overshoot & steady state error.
Comment on your results.
+
𝐾
𝑠+2
1
𝑠
Figure 1
Simulink block diagram:
Scope output:
For k=4
Rising time = 1.2
Settling time= 4.5
Maximum overshoot= 0.15
Steady state error = 0
For k=0.25
Rising time = 17.7
Settling time= 40
Maximum overshoot=0
Steady state error = 0
For k=1
Rising time = 3.8
Settling time= 8
Maximum overshoot=0
Steady state error = 0
Comments:
For K = 4 this is an under damped 2nd order system with to conjugate poles.
For K = 1 this is a critically damped 2nd order system with double poles with the same value.
For K=0.25 this is an over damped 2nd order system with 2 real poles.
The under damped case experiences the least rising time & settling time but with an overshoot,
it is also a damping sinusoidal due to the complex conjugate poles.
The critically damped case experiences settling time & rising time less than that of the over
damped one & both without any overshoots & not sinusoidal because the poles are real.
All second order systems have zero steady state error in response to step input.