Zeros And Poles Of Transfer Functions
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Zeros And Poles Of Transfer Functions

Determine the zeros and poles of the following transfer functions:
(ai) H(s) = 10s/(s2 + 2s + 26)
(ii) Determine the expressions for the network function, the magnitude |H(s)| and phase angle of H(s)
(iii) Illustrate the zero-pole plot
(iv) Determine the ω - |H| distribution
(bi) H(s) = 8((s + 4)/(s2 + 6s + 40))
(ii) Determine the network function, the magnitude |H(s)|and phase angle of H(s)
(iii) Illustrate the zero-pole plot
(iv) Determine the ω - |H| distribution
(c) What is the usefulness of zeros and poles?

The strings: S7P5A51 (change - physical).

The math:
Pj Problem of Interest is of type change (physical - change).


(ai) The zeros (z) of H(s) are the values of s for which the numerator of H(s) = 0.
The poles (p) of H(s) are the values of s for which the denominator of H(s) = 0.
H(s) = 10s/(s2 + 2s + 26)
So, 10s = 0 implies s = 0.
So H(s) has a zero at s = 0. Lets called this zero z1
s2 + 2s + 26 = 0 implies:
s =[- b ± (b2 - 4ac)1/2]/2a. where a = 1, b = 2, c = 26
So, s = -1 ± (4 - 104)1/2/2 = -1 ± j5, where j is the imaginary number √-1
So H(s) has poles at s = -1 + j5 and s = -1 - j5.
Lets called these poles p1 and p2.

(ii) Network function of H(s) = 10(s - z1)/(s - p1)(s - p2)
Since both numerator and denominator are vectors:
Network function of H(s) = 10A/(B)(C)
Where A = s - z1, B = s - p1, C = s - p2

Magnitude, |H(s)| = 10|A|/(|B|)(|C|)

Phase angle, <H(s) = <A - <B - <C

(iii) Zero - Pole Plot:
Zero Pole Plot

(iv) ω - |H| distribution:
Magnitude Plot Of Transfer Function
The coordinates of the magnitude plot is determined for selected ω as follows:
Let ω = 3
So, s = j3
So, |A| = |s - z1| = |j3 - 0| = √9 = 3
|B| = |s - p1| = |j3 - (-1 + j5)| = |(1 - j2)| = √5 = 2.24
|C| = |s - p2| = |j3 - (-1 - j5)| = (1 + j8) = √65 = 8.06
So, |H(s)| = 10|A|/(|B|)(|C|) = 10(3)/((8.06)(2.24)) = 1.66
So for ω = 3, |H| = 1.66.
From the magnitude plot we see that max |H| = 5 and occurs at ω = 5.1 which is roughly the ω associated with the poles. The magnitude of each of the poles is √26 = 5.1.

(bi) Transfer function, H(s) = 8((s + 4)/(s2 + 6s + 40))
Similar steps in (a) give:
zero at s = -4
poles at s = 3 ± j5.5678.

(c) zeros and poles indicate the stability of transfer functions. Manipulation of the positions of zeros and poles can also be used for circuit control.
poles are also called the natural frequency of the circuit and they are equal to the roots of the natural system response.

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