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Transfer Function of First Order Control Systems

Explain the significance of first order system, plot output and input with respect to time, find c and d using the plots, relate c and d to a and b, explain the effect of increasing b.

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Added on  2022-10-11

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This article explains the transfer function of first order control systems and its significance in determining the type and order of a control system. It also covers the process of drawing frequency response, Nyquist, Root locus, Nichols and bode plots using the transfer function. The article also discusses the effect of increasing b on the stability of the system.

Transfer Function of First Order Control Systems

Explain the significance of first order system, plot output and input with respect to time, find c and d using the plots, relate c and d to a and b, explain the effect of increasing b.

   Added on 2022-10-11

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Question one
Equation 2 is the transfer function of first order control systems. A transfer function describes the
relationship between the input and the output. In equation 2, X(s) is the input to the system and
Y(s) is the output. A transfer function helps in determining the type and the order of a control
system. Furthermore, frequency response of the control system, Nyquist, Root locus, Nichols and
bode plots can be drawn using the transfer function. The plots provide information that is useful
in study and analysis of control systems, for example information on stability of the system. The
denominator of the transfer function gives the characteristic equation, it is from the characteristic
equation that the roots of the system is derived.
Question two
First order transfer function of the system, T(s) is given by
T(s) = Y ( s)
X (s) = c
ds+1 .................................2
Y (s) = T (s) * X (s), but T(s) = c
ds+1 thus
= c
ds+1 * X (s).................................3
Since the input is the unit step function, X(s) = 1/s. substituting for X (s) in equation 3 we get
Y(s) =
1
s c
(ds +1)
.....................................4
Rearranging equation 4 we get
Y(s) =
1
s c
d(s +1/d )
Y(s) =
1
s c /d
( s+1/d )
......................................5
Decomposing equation 5 using partial fraction gives
Y (s) = c / d
s (s +1/d ) ¿ A
s + B
s+1/d ........................6
A = c /d
s (s +1/d )s=¿ { c /d
( s+1 /d ) } s=0 = c
Transfer Function of First Order Control Systems_1
B = c /d
s (s +1/d )( s+1/d )={ c /d
s }s=-1/d= -c
Y (s) = c
s c
s+1/ d = c { 1
s 1
s+ 1
d
}......................7
Taking inverse Laplace transform of equation 7 we get
y(t) = c – c * exp{-(1/d)t},
y (t) = c{1-e-(1/d)t} ................................8
when t = 0, y (t) = c{1-e-0} = 0
when t = 1d, y(t) = c{1-e-1} = 0.632c
when t = 2d, y (t) = c{1-e-2} = 0.864c
when t = 3d, y (t) = c{1-e-3} = 0.950c
when t = 4d, y (t) = c{1-e-4} = 0.982c
when t = 5d, y (t) = c{1-e-5} = 0.993c
when t = , y (t) = c{1-e-} = c
y (t) is explicitly zero when t is less than 1
Tabulating the information above:
t y(t)
0 0
1d 0.632c
2d 0.864c
3d 0.950c
4d 0.982c
5d 0.993c
c
Table 1
u (t) = 1 when t is equal or greater than 1
u (t) = 0 when t is less than 1
Transfer Function of First Order Control Systems_2

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