Mathematics of nyquist plot [autosaved] [autosaved]
-
Upload
asafak-husain -
Category
Education
-
view
96 -
download
2
Transcript of Mathematics of nyquist plot [autosaved] [autosaved]
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