Control Systems Test 2

Concept:

A transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

Given the differential equation is,

\(\ddot y\left( t \right) - 3\dot y\left( t \right) - 4y\left( t \right) = \ddot u\left( t \right) + 2\dot u\left( t \right) + 2u\left( t \right),\)

By applying the Laplace transform,

\(\Rightarrow \frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{{{s^2} - 2s + 2}}{{{s^2} - 3s - 4}}\)

\(= \frac{{\left( {s - \left( { - 1 + j} \right)} \right)\left( {s - \left( { - 1 - j} \right)} \right)}}{{\left( {s + 1} \right)\left( {s - 4} \right)}}\)

Poles of the system = -1, 4

Zeros of the system = -1 + j, -1 - j

The system is unstable as one pole lies on the right half of the s-plane.

Type of the system indicates the number of poles at origin.

Order of the system indicates the total number of poles.

1) \(T/F = \frac{{2\left( {s + 2} \right)}}{{s\left( {s + 5} \right)}}\)

Type = 1, order = 2

2) \(T/F = \frac{{\left( {s + 2} \right)}}{{\left( {s + 3} \right)\left( {s + 5} \right)}}\)

Type = 0, order = 2

3) \(T/F = \frac{{2\left( {s + 5} \right)}}{{{s^2}\left( {s + 2} \right)}}\)

Type = 2, order = 3

4) \(T/F = \frac{{5\left( {s + 2} \right)}}{{\left( {s + 1} \right)\left( {s + 3} \right)\left( {s + 5} \right)}}\)

Type = 0, order = 3

\(G\left( s \right) = \frac{{K\left( {s + 1.1} \right)\left( {s + 2.2} \right)}}{{s\left( {s + 3.3} \right)\left( {s + 4.4} \right)}}\)

Closed loop poles are the roots of the characteristic equation.

1 + G(s) H(s) = 0

\(\Rightarrow 1 + \frac{{K\left( {s + 1.1} \right)\left( {s + 2.2} \right)}}{{s\left( {s + 3.3} \right)\left( {s + 4.4} \right)}} = 0\)

For K = 0,

\(s\left( {s + 3.3} \right)\left( {s + 4.4} \right) = 0\)

Closed loop poles = 0, -3.3 and -4.4

y(t) = 1 – e^-3t (sin 4t + cos 4t)

By applying Laplace transform,

\(Y\left( s \right) = \frac{1}{s} - \frac{4}{{{{\left( {s + 3} \right)}^2} + {4^2}}} - \frac{{\left( {s + 3} \right)}}{{{{\left( {s + 3} \right)}^2} + {4^2}}}\)

\(= \frac{1}{s} - \frac{4}{{{{\left( {s + 3} \right)}^2} + 16}} - \frac{{\left( {s + 3} \right)}}{{{{\left( {s + 3} \right)}^2} + 16}}\)

\(Y\left( s \right) = \frac{{{{\left( {s + 3} \right)}^2} + 16 - \left( {4 + s + 3} \right)s}}{{s\left( {{{\left( {s + 3} \right)}^2} + 16} \right)}}\)

DC gain of open loop system

\(= \begin{array}{*{20}{c}} {lt}\\ {s \to 0} \end{array}sY\left( s \right)\)

\(= \begin{array}{*{20}{c}} {lt}\\ {s \to 0} \end{array}s \cdot \frac{{{{\left( {s + 3} \right)}^2} + 16 - s\left( {s + 7} \right)}}{{s\left( {{{\left( {s + 3} \right)}^2} + 16} \right)}} = 1\)

System 1:

c₁(t) = 3(1 – e^-2t), r₁(t) = r(t)

By applying Laplace transform,

\({C_1}\left( s \right) = 3\left( {\frac{1}{s} - \frac{1}{{s + 2}}} \right) = \frac{6}{{s\left( {s + 2} \right)}}\)

\(\frac{{{C_1}\left( s \right)}}{{{R_1}\left( s \right)}} = \frac{{\frac{6}{{s\left( {s + 2} \right)}}}}{{\frac{1}{{{s^2}}}}} = \frac{{6s}}{{\left( {s + 2} \right)}}\)

System 2:

c₂(t) = e^-4t, r₂(t) = u(t)

By applying Laplace transform,

\({C_2}\left( s \right) = \frac{1}{{\left( {s + 4} \right)}},\;{R_2}\left( s \right) = \frac{1}{s}\)

\(\frac{{{C_2}\left( s \right)}}{{{R_2}\left( s \right)}} = \frac{{\frac{1}{{\left( {s + 4} \right)}}}}{{\frac{1}{s}}} = \frac{s}{{s + 4}}\)

As the two systems are connected in cascade connection, the transfer functions will get multiplied.

The overall transfer function is,

\(= \frac{{6s}}{{\left( {s + 2} \right)}} \cdot \frac{s}{{\left( {s + 4} \right)}}\)

Concept:

A transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

ÿ(t) + 6ẏ(t) + 5y(t) = u(t)

By applying the Laplace transform,

s²y(s) + 6sy(s) + 5y(s) = U(s)

\( \Rightarrow \frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{1}{{{s^2} + 6s + 5}}\)

\( \Rightarrow \frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{1}{{\left( {s + 1} \right)\left( {s + 5} \right)}}\)

For unit step response, \(U\left( s \right) = \frac{1}{s}\) 

\( \Rightarrow Y\left( s \right) = \frac{1}{{s\left( {s + 1} \right)\left( {s + 5} \right)}}\)

\(= \frac{1}{{5s}} - \frac{1}{{4\left( {s + 1} \right)}} + \frac{1}{{20\left( {s + 5} \right)}}\)

By applying inverse Laplace transform,

y(t) = 0.2 – 0.25 e^-t + 0.05 e^-5t

The steady state value of output is,

x(t) = e^-3t u(t)

\(\Rightarrow X\left( s \right) = \frac{1}{{s + 3}}\) 

y(t) = (2t + 1) e^-2t u(t)

= (2t e^-2t + e^-2t) u(t)

= 2t e^-2t u(t) + e^-2t u(t)

\( \Rightarrow y\left( s \right) = \frac{2}{{{{\left( {s + 2} \right)}^2}}} + \frac{1}{{s + 2}} = \frac{{s + 4}}{{{{\left( {s + 2} \right)}^2}}}\) 

Transfer function,

\(T\left( s \right) = \frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{{\left( {s + 4} \right)\left( {s + 3} \right)}}{{{{\left( {s + 2} \right)}^2}}}\)

Zeros of T(s) = -4 and -3