Transfer Functions, Poles and Zeros
MEEN 364 Lecture 13
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
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