Question

Problem 1 (Problem Solving Workshop 1) For a parallel RL circuit R-10, L 1H Determine 1) 21 3) 4) The transfer function H(s) = (s), the pole-zero map, and the step response. Let L(0) - OA The state and output equations. Let Lt) be the state variable The block diagram of this system. Let (O) = -1 The response (t) due to a step input (t) = (t) A) using a known software. Problem #2 (Problem Solving Workshop 1) For a parallel RLC circuit + RCVCD3L R50L1HC1F 1) 2) 3) 4) The transfer function HS) = 6), the pole zoro map, and the step response. Let (O)=0Aand Ve(O) = OV. The state and output equations. Let (!) and Vel) be the state variables The block diagram of this system. Let (0) - OA, and Vc(O) = OV. The sketch of the response (1) due to a step input (U) - A) using a known software. Problem # 3 (Problem Solving Workshop 1) For the electrical system R = 50, L = 5H, C = .01F Determine 1) The transfer function Hy's) loyvis), the pole-zero map and the step response with L:(0) - OA and Veo) - OV. 2) The state equations. Let Vel) and 1.) be the state variables 3) The output equation in terms of the variable (U). 4) The block diagram of this system. Assume 10) - OA and V0) - Ov 5) The sketch of the response due to a step input (V ) using a known software

Answer 1

Problem # 01

Question 1

Given

R = 12

L = 0.11

The impedance of the inductor in the s domain is

Zz(s) = Ls

Zz(s) = 0.15

The circuit can be drawn in s domain as follows

I(s) 30.1s

So the current flowing through the inductor can be found using current division rule as

1,5) = 1 +0.15

So the transfer function is

1,(s) 19) 1 1+ 0.1s

Another way of finding the transfer function:

The voltage across the inductor is

V/ (t) = L.“ diz(t) dt

The current flowing through the resistor is

Voltage across the resistor ir(t) =

Here the voltage across the resistor is same as the voltage across the inductor. So

(t) ir(t) =

if(t) = ??p 7 dt

Now applying Kirchoff’s Current Law we get

(1)*? + (+)?? = ()

p_ (?)??p 7 +()?? =(:)?

Substituting the values of L and R we get

1P (9):1p του + 0) = 3)!

diz(t) i(t) = iz(t) + 0.14 dt

Taking Laplace Transform on both sides under zero initial conditions, we get

1(s) = 1; (s) + 0.15l, (s)

1(s) = [1 + 0.1s]I, (s)

1,(s) 19) 1 1+ 0.1s

So the transfer function is

1,(s) 19) 1 1+ 0.1s

We got the same answer

To find the poles and zeros

1,(s) 19) 1 1+ 0.1s

The poles are at

1 + 0.1s = 0

0.1

s= -10

There are no zeros. So the pole - zero map will be

Pole-Zero Map Imaginary Axis (seconds) HY -14 -12 -10 Real Axis (seconds)

To find the step response

1,(s) 19) 1 1+ 0.1s

1+0.1 X 1(s)

Step response is the response of the system when a unit step input is applied. So

i(t) = u(t)

Taking Laplace Transform, we get

So

1 1,() = 5(1 + 0.15)

Taking Partial Fraction expansion, we get

1,(s) 1 - (1 +0.15) = A B - +- S'1 + 0.1s

1-- xrot Dir

10--0-- |sro+1)x (T0 + 1) = &

So

1,(9) = 1_ 0.1 2 1 + 0.15

z 4(9) = 1__0.1 s 0.1(10+5)

-

Taking Inverse Laplace Transform, we get

iz(t) = u(t)- e -10€ u(t)

(1)n (201-2 - 1] = (?)??

iz(t) = [1 – е-10] for t o

So the step response of the circuit is

iz(t) = [1 – е-10] for t o

Question 2

To find the state and output equation.

The state is

x= i(t)

The input is

(+)? = 1

We have already shown the current equation in question 1 as

p_ (?)??p 7 +()?? =(:)?

Substituting the state variables and input we get

u = x +-

So

RR i=7*+

Substituting the values of R and L

11 As - 0.17 +o.au

* = -10x + 10

The output is

y = ij(t)

So the state equation is

* = -10x + 10

The output equation is

Question 3

From the equation,

* = -10x + 10

We can write

(x – not = ?

The output is

So the block diagram is

Initial Current + u + Output input Integrator

Question 4

Simulation Using MATLAB

MATLAB Code

clc;
clear all;
close all;

n = [1];
d = [0.1 1];
sys = tf(n, d);

step(sys);
grid

After executing we get

Step Response Amplitude 0.2 0.4 0.6 0.7 0.8 0.5 Time (seconds)