The Transfer Function (TF) of a System is the ratio of the output to the input of a system, in the Laplace domain considering its initial conditions and equilibrium point to be zero.
If we have an input function of X(s), and an output function Y(s), we define the transfer function H(s).
H(s) = Y(s)/X(s)
Since we know that in the time domain, generally, we define the input to a system as x(t), and the output of the system as y(t). The relationship between the input and the output is represented as the impulse response, h(t).
We can use the following equation to define the impulse response:
h(t) = y(t) / x(t)
Convolution in Time Domain
If we have the system input and the impulse response of the system, we can calculate the system output using the convolution operation as such as
y(t) = h(t) ∗ x(t).
Time-Invariant System Response
There are major two ways of obtaining a transfer function for the control system. The ways are:
Generally, a function can be represented to its polynomial form. For example,
Now similarly transfer function of a control system can also be represented as
Where K is known as the gain factor of the transfer function.
Now in the above function if s = z1, or s = z2, or s = z3,….s = zn, the value of transfer function becomes zero. These z1, z2, z3,….zn, are roots of the numerator polynomial. As for these roots the numerator polynomial, the transfer function becomes zero, these roots are called zeros of the transfer function.
Now, if s = p1, or s = p2, or s = p3,….s = pm, the value of transfer function becomes infinite. Thus the roots of denominator are called the poles of the function.
Let’s represent the transfer function in its polynomial form:
Now, let us consider s approaches to infinity as the roots are all finite number, they can be ignored compared to the infinite s. Therefore
Hence, when s → ∞ and n > m, the function will have also value of infinity, that means the transfer function has poles at infinite s, and the multiplicity or order of such pole is n – m. Again, when s → ∞ and n < m, the transfer function will have value of zero that means the transfer function has zeros at infinite s, and the multiplicity or order of such zeros is m – n.
Example: Impulse Response
Consider a linear input/output system described by the controlled differential equation
where u is the input and y are the output.
To investigate how a linear system responds to the exponential input u(t) = est we consider the state space system
dx/dt = Ax + Bu
y = Cx + Du
Let the input signal be u(t) = est and assume that s ≠ λi(A), where i = 1,...,n, where λi(A) is the ith eigenvalue of A. then
Since s ≠ λ(A) the integral can be evaluated and we get
& Output y(t) = Cx(t) + Du(t)
= CeAt{x(0) − (sI − A)−1B} + [D + C(sI − A)−1B]est
Note: One term of the output is proportional to the input u(t) = est. This term is called the pure exponential response.
Effects of Poles and Zeros
Note: If the Poles or Zero lie on the imaginary axis then it must be simple (order only 1) then the system is said to be Marginally Stable, If the Order of multiplicity of Poles or Zeros on Imaginary Axis is more than 1 in that case system will become Unstable.
The transfer function of the system, G(s) = I(s)/V(s), the ratio of output to input.
1) Let us explain the concept of poles and zeros of transfer function through an example.
Solution
The zeros of the function are, -1, -2 and the poles of the functions are -3, -4, -5, -2 + 4j, -2 – 4j.
Here n = 2 and m = 5, as n < m and m – n = 3, the function will have 3 zeros at s → ∞. The poles and zeros are plotted in the figure below2) Let us take another example of transfer function of control system
Solution
In the above transfer function, if the value of numerator is zero, then
These are the location of zeros of the function.
Similarly, in the above transfer function, if the value of denominator is zero, then
These are the location of poles of the function.
As the number of zeros should be equal to number of poles, the remaining three zeros are located at s →∞.
3)
Solution
In the above network it is obvious that
Let us assume,Taking the Laplace transform of above equations with considering the initial condition as zero, we get,
Block diagram of Closed Loop SystemE(s) = Actuating or Error Signal
X(s) = Reference Input Signal.
G(s) = Forward Path Transfer Function.
Y(s) = Output Signal.
H(s) = Feedback Transfer Function.
B(s) = Feedback Signal.
So, the transfer function of the closed loop system is Y(s)/X(s).
From the block diagram,
Y(s) = G(s).E(s) .........1
B(s) = H(s).Y(s) .........2
E(s) = X(s) + B(s) .........3a (For positive feedback)
= X(s) - B(s) ..........3b (For negative feedback)
Y(s) = G(s).[X(s)-B(s)]
Y(s) = G(s).X(s) - G(s).B(s) .........4
Put the value of B(s) from eq.2 in eq.4
Y(s) = G(s).X(s) - G(s).H(s).Y(s)
Y(s) + G(s).H(s).Y(s) = G(s).X(s)
Y(s){1 + G(s).H(s)} = G(s).X(s)
Y(s)/X(s) = G(s) / {1 +G(s).H(s)} .........5
For positive feedback system we will use eq.3a and repeat the all the same steps and we will get the transfer function as:
Y(s)/X(s) = G(s) / {1 - G(s).H(s)} .........6
For unity feedback H(s) = 1, the eq.5 & eq.6 will become
Y(s)/X(s) = G(s)/{1 + G(s)} .......for negative unity feedback
Y(s)/X(s) = G(s)/{1 - G(s)} ........for positive unity feedback
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1. What is the Convolution Theorem in the context of transfer functions for linear systems? |
2. What are Poles and Zeros in the concept of transfer functions for linear systems? |
3. How can the transfer function of a closed-loop system be determined in electrical engineering? |
4. How do transfer functions play a crucial role in analyzing and designing linear systems in electrical engineering? |
5. What is the significance of the Convolution Theorem in signal processing and control systems design? |
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