LECTURE-CUM-PRESENTATION

ONMATHEMATICS OF

NYQUIST PLOT

ASAFAK HUSAIN

12115026

E-2 BATCH, EE, 3rd year

CONTENT

• Complex Calculus

• Cauchy’s Theorem

• Principle of Augment

• Nyquist criteria

• Nyquist path

• Example

COMPLEX CALCULUS

Limit :: limit of a function complex ƒ(z) is said to exists at z =𝑧0when

𝑓 𝑧 − 𝑧0 < 𝛿 ∀ 0 < |𝑧 − 𝑧0| < 휀

Continuity lim𝑧→𝑧0

𝑓(𝑧) = 𝑓(𝑧0)

Derivative limℎ→0𝑘→0

𝑓 𝑎+ℎ,𝑏+𝑘 −𝑓(𝑎,𝑏)

ℎ2+𝑘2should exists.

DIFFERENTIAL CALCULUS

• Gradient :

• Divergence

• Curl

𝛻𝑓 =𝜕𝑓

𝜕𝑥 𝑖 +

𝜕𝑓

𝜕𝑦 𝑗 +

𝜕𝑓

𝜕𝑧 𝑘

𝛻. 𝑓 =𝜕𝑓𝑥𝜕𝑥

+𝜕𝑓𝑦

𝜕𝑦+𝜕𝑓𝑧𝜕𝑧

𝛻 × 𝑓 =

𝑖 𝑗 𝑘𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧𝑓𝑥 𝑓𝑦 𝑓𝑧

INTEGRATION

• # Line Integral = 𝑓 𝑧 𝑑𝑧 over a path C

𝑓 𝑧 = 𝑢 𝑥, 𝑦 + 𝑖𝑣 𝑥, 𝑦 here 𝑧 = 𝑥 + 𝑖𝑦 and both 𝑢 𝑎𝑛𝑑 𝑣 are real functions

• Green’s Theorem ∅𝑑𝑥 + 𝜓𝑑𝑦 = (𝜕𝜓

𝜕𝑥−

𝜕∅

𝜕𝑦) 𝑑𝑥𝑑𝑦

C-R EQUATIONS DERIVATION

• Derivative of 𝑓 𝑧 = 𝑢 𝑥, 𝑦 + 𝑖𝑣 𝑥, 𝑦 along real axis 𝛿𝑦 = 0, 𝛿𝑧 = 𝛿𝑥

• 𝑓′ 𝑧 = lim𝛿𝑥→0

𝑢 𝑥+𝛿𝑥,𝑦 −𝑢(𝑥,𝑦)

𝛿𝑥+ 𝑖

𝑣 𝑥+𝛿𝑥,𝑦 −𝑣(𝑥,𝑦)

𝛿𝑥

⇒ 𝑓′ 𝑧 =𝜕𝑢

𝜕𝑥+ 𝑖

𝜕𝑣

𝜕𝑥……………(1)

And along imaginary axis 𝛿𝑥 = 0, 𝛿𝑧 = 𝑖𝛿𝑦

𝑓′ 𝑧 = lim𝛿𝑦→0

𝑢 𝑥, 𝑦 + 𝛿𝑦 − 𝑢(𝑥, 𝑦)

𝑖𝛿𝑦+ 𝑖

𝑣 𝑥, 𝑦 + 𝛿𝑦 − 𝑣(𝑥, 𝑦)

𝑖𝛿𝑦)

⇒ 𝑓′ 𝑧 = −𝑖𝜕𝑢

𝜕𝑦+

𝜕𝑣

𝜕𝑦………….(2)

C-R EQUATIONS

• Since limit should be same from each and every path

so from (1) and (2)

𝜕𝑢

𝜕𝑥=

𝜕𝑣

𝜕𝑦and

𝜕𝑣

𝜕𝑥= −

𝜕𝑢

𝜕𝑦

these are known as Cauchy- Riemann equations.

ANALYTIC FUNCTION AND CAUCHY'S THEOREM

Analytic function

• Single valued

• Unique derivative at all the point of the domain

• 𝐶𝑎𝑢𝑐ℎ𝑦′𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑓(𝑧) 𝑑𝑧 = 0

for analytic function over the entire closed path C.

CAUCHY’S THEOREM

• Let 𝑓 𝑧 = 𝑢 + 𝑖𝑣 𝑓𝑜𝑟 𝑧 = 𝑥 + 𝑖𝑦

then 𝑓 𝑧 𝑑𝑧 = 𝑢 + 𝑖𝑣 𝑑𝑥 + 𝑖𝑑𝑦 = 𝑢𝑑𝑥 − 𝑣𝑑𝑦 + 𝑖 (𝑢𝑑𝑦 + 𝑣𝑑𝑥)

= − 𝜕𝑣

𝜕𝑥+

𝜕𝑢

𝜕𝑦𝑑𝑥𝑑𝑦 + 𝑗

𝜕𝑣

𝜕𝑥−

𝜕𝑢

𝜕𝑦𝑑𝑥𝑑𝑦 (Green’ s thm.)

= 0 (𝐶 − 𝑅 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠)

CAUCHY’S INTEGRAL FORMULA

𝑓 𝑧 𝑑𝑧 = − 𝐶

𝑓 𝑧 𝑑𝑧 + 𝐴𝐵

𝑓 𝑧 𝑑𝑧 + 𝐶0

𝑓 𝑧 𝑑𝑧 + 𝐵𝐴

𝑓 𝑧 𝑑𝑧

⇒ 𝐶 𝑓 𝑧 𝑑𝑧 = 𝐶1𝑓 𝑧 𝑑𝑧 (Cauchy’s theorem)

Similarly, for 𝐶𝑓(𝑧)

𝑧−𝑧0𝑑𝑧 = 𝐶0

𝑓(𝑧)

𝑧−𝑧0𝑑𝑧 , put 𝑧 = 𝑧0 + 𝑟𝑒𝑖𝜃

⟹

𝐶0

𝑓 𝑧0 + 𝑟𝑒𝑖𝜃

𝑟𝑒𝑖𝜃𝑖𝑟𝑒𝑖𝜃𝑑𝜃 = 2𝜋𝑖𝑓(𝑧0)

⇒

𝐶

𝑓 𝑧

𝑧 − 𝑧0𝑑𝑧 =2𝜋𝑖𝑓(𝑧0)

RESIDUE’S THEOREM

• 𝐶 𝑓 𝑧 𝑑𝑧 = 𝐶1𝑓 𝑧 𝑑𝑧 + 𝐶2

𝑓 𝑧 𝑑𝑧 +⋯+ 𝐶𝑛𝑓 𝑧 𝑑𝑧

• 𝐶 𝑓 𝑧 𝑑𝑧 = 2𝜋𝑖[𝑓(𝑧1) + 𝑓 𝑧2 +⋯+ 𝑓(𝑧𝑛)

Here 𝑓 𝑧𝑖 are called Residues of function f(z).

Note: residue are also define as the coefficients of

(𝑧 − 𝑧0)−1 in the expansion of Laurent series

That is 𝑛=−∞∞ 𝑎𝑛(𝑧 − 𝑧0)

𝑛

PRINCIPLE OF ARGUMENT

• Let 𝑓 𝑧 =𝑧−𝑧1

∝1……. 𝑧−𝑧𝑛∝𝑛

𝑧−𝑝1𝛽1…..

𝑧−𝑝𝑚𝛽𝑚

𝐹(𝑧)

• Now 𝑓(𝑧)

𝑓(𝑧)= 𝑖=1

𝑛 𝛼1

𝑧−𝑧𝑖− 𝑖=1

𝑚 𝛽𝑖

(𝑧−𝑝𝑖)+

𝐹 (𝑧)

𝐹(𝑧)

• ⇒ 𝐶 𝑓 𝑧

𝑓 𝑧𝑑𝑧 =2𝜋𝑖𝑍 − 2𝜋𝑖𝑃 + 𝐶

𝐹(𝑧)

𝐹(𝑧)𝑑𝑧 …………..(1)

• 𝐶 𝐹(𝑧)

𝐹(𝑧)𝑑𝑧 = 0 (𝐶𝑎𝑢𝑐ℎ𝑦′𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚)

PRINCIPLE OF ARGUMENT

• Let’s consider 𝐶 𝑓(𝑧)

𝑓(𝑧)𝑑𝑧 = 𝐶

𝑑

𝑑𝑧(log(𝑓(𝑧)))

• = 𝐿𝑜𝑔 𝑓(𝑧) |𝐶 + 𝑖𝑎𝑟𝑔𝑓(𝑧)|𝐶

• = 𝑖 arg 𝑓 𝑧 |𝐶

• Thus we can see, value of integral only depends on the net change in the argument

of f(z) as z traverse the contour.

• If N is number of encirclement about Origin in F(s)-plane then

2π𝑖N = 𝑖 arg 𝑓 𝑧 |𝐶 = 2𝜋𝑖𝑍 − 2𝜋𝑖𝑃

N=Z-P

NYQUIST CRITERIA

• If open loop transfer function of a system is

𝐺 𝑠 𝐻 𝑠 =𝐾 𝑖=1

𝑛 (𝑠+𝑧𝑖)

𝑖=1𝑝

(𝑠+𝑝𝑖)=

𝑁(𝑠)

𝐷(𝑠)

Then close loop transfer function

𝑇. 𝐹. =𝐺(𝑠)

1+𝐺 𝑠 𝐻(𝑠)and let 𝐹 𝑠 = 1 + 𝐺 𝑠 𝐻 𝑠 = 1 +

𝑁(𝑠)

𝐷(𝑠)

We consider right half open loop poles only .

We observes that 𝑜𝑝𝑒𝑛 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠 = 𝑝𝑜𝑙𝑒𝑠 𝑜𝑓 𝐹 𝑠 && 𝑐𝑙𝑜𝑠𝑒 𝑙𝑜𝑜𝑝 𝑝𝑜𝑙𝑒𝑠 = 𝑍𝑒𝑟𝑜𝑠 𝑜𝑓 𝐹(𝑠)

Since here 𝐹(𝑠) is replaced by 1 + 𝐹(𝑠), so in this we will consider encirclement about

− 1 + 𝑗0.

NYQUIST CRITERION

𝑁 = 𝑍 − 𝑃 ⇒ 𝑍 = 𝑁 + 𝑃

Here Z =number of close loop poles S-plane

P=number of open loop poles S-plane

N=number of encirclement about -1+ j0 F(s)-plane

Now close loop system to be stable Z must be zero.

P=0⇒ 𝑍 = 𝑁 ⇒ 𝑁 = 0; 𝑡ℎ𝑒𝑟𝑒 𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒 𝑎𝑛𝑦 𝑒𝑛𝑐𝑖𝑟𝑐𝑙𝑒𝑚𝑒𝑛𝑡𝑠.

𝑃 ≠ 0 ⇒ 𝑁 = −𝑃; 𝑡ℎ𝑒𝑟𝑒 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑝 𝑒𝑛𝑐𝑖𝑟𝑐𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛.

NYQUIST PATH

• Section I 𝐶1: 𝑠 = 𝑗𝜔 ∀ 𝜔 ∈ (0+ , +∞)

• Section II 𝐶2∶𝑠 = −𝑗𝜔 ∀ 𝜔 ∈ (−∞ , 0− )

• Section III 𝐶3: s = R𝑒𝑖𝜃 𝑅 → ∞ ∀ 𝜃 ∈ (−𝜋

2,𝜋

2)

• As Detour (singularities)

𝐶4: s = 휀𝑒𝑖𝜃 휀 → 0 ∀ 𝜃 ∈ (−𝜋

2,𝜋

2)

EXAMPLE

• Section 𝐶3:

• 𝐺 𝑅𝑒𝑗𝜃 𝐻 𝑅𝑒𝑗𝜃 = 0

• At detour

• 𝐺 휀𝑒𝑗𝜃 𝐻 휀𝑒𝑗𝜃 → ∞

• 𝐺 𝑠 𝐻 𝑠 =𝐾(𝜏1𝑠+1)

𝑠2(𝜏2𝑠+1), find close loop

stability of the system.

• (1) Section 𝐶1& 𝐶2

• 𝐺 𝑗𝜔 𝐻 𝑗𝜔 =𝐾

𝜔2

𝜏1𝜔2+1

𝜏2𝜔2+1

𝜑𝐺𝐻 = −𝜋 + tan−1 𝜏1𝜔 − tan−1 𝜏2𝜔

NYQUIST PLOT

𝜏1 = 𝜏2

Here plot passes through (-1+j0) that

indicates that roots lie on imaginary

axis.

𝜏1 < 𝜏2

N=-1

Z=-1, Unstable

𝜏1 > 𝜏2

Real axis is not covered by the

encirclement loop

N=0 so Z=0 Stable

THANKS A LOT !!!!!!!!!!!!!!!

Any queries ??????