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Introduction to Transfer Function
The Transfer Function
The transfer function is an essential parameter for an LTI system, like the impulse response
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If we define the transfer function for an arbitrary system, the initial
conditions must be zero
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The transfer functions can be defined as the ratio of Laplace transform of output to
the Laplace transform of input when all initial conditions are assumed to be zero
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But if the system is an LDI system, and we are asked to calculate the
impulse response, we can calculate the impulse response of the system and HS
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The transfer function is the Laplace transform of the output to the Laplace transform of the input
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Transfer Function Formula
The transfer function of this system is given as H(s) = 1 / [(s+1) * (s+2)]
Summary
In summary, the transfer function is a crucial parameter for LTI systems
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We will
present more problems on the transfer function in upcoming lectures
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It describes how a system's output responds to changes in its input, and is often used in
control systems and signal processing
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The transfer function can also be represented in the frequency domain
as a ratio of complex numbers, which provides information about the system's frequency response
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The transfer function of a first-order low-pass filter is given by:
H(s) = 1 / (s + a)
where 'a' is the cut-off frequency of the filter
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PID controller: A PID (proportional-integral-derivative) controller is a control system that adjusts the
output of a system based on the error between the desired setpoint and the actual output
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This transfer
function represents the relationship between the error signal and the control output of the PID
controller
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The transfer function of an ideal
voltage amplifier is given by:
H(s) = Vo/Vi = A
where A is the gain of the amplifier
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These are just a few examples of transfer functions
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Example 1: First-Order Low-Pass Filter
Suppose we have a first-order low-pass filter with a cutoff frequency of 10 Hz
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We
can find the output of the filter using the transfer function as follows:
Vi(s) = 2 / (s + 5) (Laplace transform of the input signal) Vo(s) = Vi(s) * H(s) (multiply the transfer function
by the input signal) Vo(s) = 2 / ((s + 5) * (s + 10)) (simplify) Vo(t) = 0
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143 * e^(-10t)
(inverse Laplace transform)
Therefore, the output of the filter is a damped sinusoidal signal with a frequency of 5 Hz, but with some
of the higher frequency components attenuated
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5
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5s
Suppose the desired setpoint is 10 V and the actual output of the system is 8 V
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5s) * 2 C(s) = 4 + 2/s + s (simplify) c(t) = 4 + 2u(t) + e^(-t) (inverse Laplace
transform)
Therefore, the output of the controller is a step function plus an exponentially decaying signal, which is
used to adjust the system output to reach the desired setpoint
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The transfer function of this amplifier is
given by:
H(s) = 5
Suppose the input voltage is a sinusoidal signal with a frequency of 1 kHz and an amplitude of 1 V
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Zeros and Poles of Transfer Functions
Zeros and poles are important concepts in the analysis and design of transfer functions
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Zeros and poles have significant impact on the frequency response and stability of the system
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Here are some important characteristics of zeros and poles in transfer functions:
Zeros: A zero of the transfer function is a value of s that causes the numerator of the transfer function to
be zero
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If the system has a
zero at a certain frequency, it means that the output will be zero at that frequency, regardless of the
input
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Mathematically, a pole is defined as the roots of the denominator polynomial
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In general, poles limit the bandwidth of the system, and can also affect the
stability of the system
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The magnitude and phase of the transfer function change around the zeros and poles
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Similarly, if a system has a pole at a certain frequency, it will have a resonance or peak in
the frequency response at that frequency
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If all the poles of the transfer
function are in the left half of the complex plane, the system is stable
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Poles on the imaginary axis indicate a marginally stable
system
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They can be used to design filters, controllers, and other types of
systems by appropriately placing them in the complex plane
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Solution: To find the zeros and poles, we need to factorize the numerator and denominator of the
transfer function
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The poles are the values of s that make the denominator zero, which are s=-1 and s=2
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Solution: H(s) = (s+1)(s+2)/(s-1)(s-2) The zeros are the values of s that make the numerator zero,
which are s=-1 and s=-2
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Example 3: Find the zeros and poles of the transfer function H(s) = (s^2+4s+4)/(s^2+6s+9)
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The poles are the values of s that make the denominator zero, which is s=-3 (a double pole)
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Solution: H(s) = s/(s+1)(s+3) The zeros are the values of s that make the numerator zero, which is
s=0
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Example 5: Find the zeros and poles of the transfer function H(s) = 1/(s^2+2s+2)
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The poles are the values of s that make the denominator zero, which are s=-1+j and s=-1-j
(complex conjugate poles)
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The knowledge of zeros
and poles is useful in various areas of engineering, including control systems, signal processing, and
circuit analysis
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The closed-loop transfer function is also known as the feedback transfer function, since
it takes into account the feedback path in the system
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The closed-loop transfer function is a key parameter in the
analysis and design of feedback control systems
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Find
the closed-loop transfer function
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The poles of T(s) are the values of s that make the denominator zero, which are s=-1
and s=-2
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Example 2: Consider a unity feedback system with a forward transfer function of G(s) = 10/(s+1)
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Solution: The closed-loop transfer function is given by T(s) = G(s) / (1+G(s)) Substituting G(s) in the above
equation, we get T(s) = 10/(s+11) The zeros of T(s) are the values of s that make the numerator zero,
which is none
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To
find the steady-state error for a unit step input, we use the final value theorem
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Therefore, the steady-state error for a unit step input is given by lim_{s->0} sT(s) = lim_{s->0} 10/(s+11) *
s = 10/11 Thus, the steady-state error for a unit step input is 10/11
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Find
the closed-loop transfer function and the location of the poles
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The poles of T(s) are the values of s that make the denominator zero, which are s=0
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5-sqrt(3)/2j (complex conjugate poles)
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5+sqrt(3)/2j and s=-0
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These are some examples of closed-loop transfer functions, which are important in the analysis and
design of feedback control systems

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