# Transfer Functions, Poles and Zeros

**Topics:**Control theory, Feedback, Differential equation

**Pages:**6 (1090 words)

**Published:**October 31, 2011

Parasuram August 22, 2001

HADNOUT E.13 - EXAMPLES ON TRANSFER FUNCTIONS, POLES AND ZEROS Example 1 Determine the transfer function of the mass-spring-damper system. The governing differential equation of a mass-spring-damper system is given by m x + c x + kx = F . Taking the Laplace transforms of the above equation (assuming zero initial conditions), we have ms 2 X ( s ) + csX ( s ) + kX ( s ) = F ( s ), X ( s) 1 ⇒ = . 2 F ( s ) ms + cs + k Equation (1) represents the transfer function of the mass-spring-damper system. Example 2 Consider the system given by the differential equation y + 4 y + 3 y = 2r (t ), where r(t) is the input to the system. Assume zero initial conditions. The Laplace transforms yields, s 2Y ( s ) + 4 sY ( s ) + 3Y ( s ) = 2 R( s ), Y ( s) 2 ⇒ = 2 . R(s) s + 4s + 3 Equation (2) represents the transfer function of the system. Example 3 Find the solution of the differential equation y (t ) + y (t ) = 0, where y (0) = α and y (0) = β . The Laplace transform of the above differential equation gives, s 2Y ( s ) − α s − β + Y ( s ) = 0, ⇒ Y ( s) = .. . .. . .. .

(1)

(2)

α s

s +1

2

+

β . s +1

2

1

MEEN 364 Lecture 13

Parasuram August 22, 2001

After looking up in the transform tables, the two terms in the right side of the above equation, we get y (t ) = α cos(t ) + β sin(t ). Example 4 Consider a RLC circuit. The governing differential equation is given by L di 1 + Ri + ∫ idt = V . dt C (3)

But, i= dq . dt

Therefore, equation (3) reduces to L d 2q dq 1 +R + q = V. 2 dt C dt

The Laplace transforms of the above equation yields Ls 2 Q( s ) + RsQ( s ) + ⇒ Q( s) = V (s) 1 1 Q( s ) = V ( s ), C

. 1 C The above equation represents the transfer function of a RLC circuit. Ls 2 + Rs + Example 5 Determine the poles and zeros of the system whose transfer function is given by G (s) = 2s + 1 . s + 3s + 2 2

The zeros of the system can be obtained by equating the numerator of the transfer function to zero, i.e.,

2

MEEN 364 Lecture 13 2s + 1 = 0, 1 ⇒s=− . 2

Parasuram August 22, 2001

The poles of the system can be obtained by equating the denominator of the transfer function to zero, i.e., s 2 + 3s + 2 = 0, ⇒ ( s + 1)( s + 2) = 0, ⇒ s = −1, s = −2.

Therefore s = -1 and s = -2 are the poles of the system and s = -1/2 is the zero of the system. Example 6 Determine the poles and zeros of the system, whose transfer function is given by H ( s) = 30( s − 6) . s ( s 2 + 4 s + 13)

The zeros of the system are given by 30( s − 6) = 0, ⇒ s = 6. Therefore s = 6 is the zero of the system. The poles of the system are given by s ( s 2 + 4s + 13) = 0, ⇒ s = 0, ⇒ s = 0, ⇒ s = 0, ⇒ s = 0, s 2 + 4s + 13 = 0, − 4 ± 16 − 52 , 2 − 4 + i6 − 4 − i6 s= , s= , 2 2 s = −2 + i3, s = −2 − i3. s=

Therefore the poles of the system are s = 0, s = -2 + i3 and s = -2 – i3.

3

MEEN 364 Lecture 13 Example 7

Parasuram August 22, 2001

Consider the mass-spring-damper system. The governing differential equation of motion for the system is given by m x + c x + kx = F . Let the states of the system be defined as x = x1 , x = x2 . From the above relation, it can be concluded that x1 = x 2 . Substituting the relations given by equation (5) in equation (4), we get m x + c x + kx = F , ⇒ m x 2 + cx2 + kx1 = F , . k c 1 ⇒ x 2 = − x1 − x 2 + F . m m m Representing equations (6) and (7) in matrix format, we have . 0 x1 = k . x 2 − m 1 x 0 c 1 + 1 F. − x2 m m (8) . .. . . . .. .

(4)

(5)

(6)

(7)

If the output of the system is the velocity of the mass, then writing the output relation in matrix format, we get y = x =x 2 , x ⇒ y = [0 1] 1 . x2 (9) .

Equations (8) and (9) represent the state-space representation of the mass-spring-damper system. Obtaining the transfer function from the state-space representation Given the ‘A’, ‘B’, ‘C’ and ‘D’ matrices of the state-space equations, the transfer...

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