On Phase Margin

On Phase Margin – From the Laplace Transform and Kirchhoff’s Laws By Yong-Nien Rao In this note, I try to link some concepts together to let readers get a global view of stability, especially the phase margin, which is a routine check for feedback amplifier design to circuit design engineers. These concepts are introduced in common electrical engineering degree program and should not be strange to any EE college student. I hope this brief note could give my reader a whole picture quickly without getting strayed by too many details. We focus our discussion in the linear time-invariant (LTI) circuit. And the topics are listed as underlying: Concept 1: Laplace transform Concept 2: Kirchhoff’s Laws, branch equations and passive devices Concept 3: Network function (Transfer function) Concept 4: Engineering Interpretation of Convolution Concept 5: Stability Definition and Basic Criterion – from Partial Fraction Expansion and Laplace transform Concept 6: Nyquist Analysis Concept 7: Phase Margin: From Nyquist Analysis to Bode Plot

© Copyright 2009 by Yong-Nien Rao

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Concept 1 Laplace transform Several techniques used in solving engineering problems are based on the replacement of functions of a real variable (usually time or distance) by certain frequency-dependent representations, or by functions of a complex variable dependent upon frequency. This replacement of function is called “transform”. And Laplace transform is

L{ f ( t )} = F ( s ) ≡

∞

∫ f ( t )e

− st

0

dt

…(eq.1.1)

We could turn the differential equations problems into algebra problems which are easier to manipulate. Some useful properties listed here 1. Uniqueness property: Laplace transform establishes a one-to-one correspondence between the time function

f ( t ) , defined on [0, ∞ ) , and its Laplace transform F ( s ) .

2. Linearity property:

L{ c1 f1 ( t ) + c2 f 2 ( t )} = F1 ( s ) + F2 ( s )

…(eq.1.2)

3. Differential Rule:

 d  L f ( t )  = sF ( s ) − f ( 0 − )  dt 

…(eq.1.3)

4. Integration Rule:

{

}

1 F ( s) s

L ∫ f ( u )du = t

0−

…(eq.1.4)

5. Convolution Rule:

L{ f1 ( t ) ∗ f 2 ( t )} = F1 ( s ) F2 ( s )

…(eq.1.5)

And convolution operator “ ∗ ” is defined as

f1 ( t ) ∗ f 2 ( t ) ≡

∫ f ( t − τ ) f (τ ) dτ t

0

1

2

(Table 1.1) Some useful transform table number

F ( s)

1

1

2

1 s

3

1 s+ a 1 ( s + a) m b ( s + a ) 2 + b2

4 5

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…(eq.1.6)

f ( t ), t ≥ 0 δ ( t ) (impulse function) 1( t ) e − at 1 t m− 1e − at ( m − 1)!

e − at sin ( bt )

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6

s+ a ( s + a ) 2 + b2

7

sk

e − at cos( bt ) d kδ ( t ) dt k

From the uniqueness property, there exists Inverse Laplace Transform which is denoted as

L− 1{ F ( s )} = f ( t )

…(eq.1.7)

(eq.1.5) could be rewritten as

f1 ( t ) ∗ f 2 ( t ) = L− 1 { F1 ( s ) F2 ( s )}

…(eq.1.8)

Another useful theory is the initial value theory

f ( t = 0 + ) = lim sF ( s ) s− > ∞

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…(eq.1.9)

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Concept 2 Kirchhoff’s Laws, branch equations and passive devices

∑ i (t) =

j For all j

And

0

…(eq.2.1)

all the

i with the direction of leaving the node.

j denotes the branch connected to this node.



Vk − j ( t ) = ek ( t ) − e j ( t ) branch voltage. k ,

…(eq.2.2)

e is the node-to-datum voltage; V is the

j denote the two endpoints (nodes) of a branch.

Branch equations are describing the v-i characteristics among the branches. The individual branch equation is defined by the device characteristic which forms the branch. The typical passive devices v-i characteristics is as underlying. For resistor

v j = ij Rj

For capacitor For inductor

ij = Cj

v j = Lj

…(eq.2.3)

dv j dt di j dt

…(eq.2.4) …(eq.2.5)

From these relations, we can write a specified time-domain (differential) equations for a specified circuit N with n nodes and b branches. Solve these equations and derives the at each nodes branch

ei ( t )

i (0 < i ≤ n - 1) , V j ( t ) across each branch j (0 < j ≤ b) and ik ( t ) at each

k (0 < k ≤ b) in this specified circuit N. A method called tableau analysis is a

complete general approach to get the behavior of a circuit. Algorithm: Tableau equations Step 1: Choose a datum node and draw a connected digraph Step 2: Write KCL for n-1 nodes except the datum node in matrix form

Ai(t) = 0

…(eq.2.6)

1

A is the reduced-incidence matrix ,which is (n-1)xb. i(t) is the branch current vector, bx1 Step 3: Write KVL for the b branches in matrix form

1

Leon O. Chua et al, “Linear And Nonlinear Circuits”, p24~p26.

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…(eq.2.7)2 v(t) − A Te(t) = 0 v(t) is the branch voltage vector, bx1. e(t) is the node-to-datum voltage vector, (n-1)x1 Step 4: Write the b branches equations of the circuit

( M0 D +

M 1 ) v(t) + ( N 0 D + N 1 ) i(t) = u s(t)

…(eq.2.8)

M 0 , M 1 , N 0 , N1 are bxb matrices with constant real element, D denotes the operator and

d dt

u s(t) denotes the independent sources.

We could write (eq.2.6), (eq.2.7), (eq.2.8) as (eq.2.9)3

0 A  0   e(t)   − AT   v(t) = 1 0     0 M 0 D + M1 N 0 D + N1   i(t)      T(D) w(t)

 0   0     u s(t)  u(t)

…(eq.2.9)

T(D) is the matrix of circuit system N, w(t) is the circuit variable vector, and u(t) is the independent source vector. <Example> For the linear circuit as underlying, we’d like to derive its tableau equations here. 1

2

L

i3

i6

3

i1

i4

C

i2

4

i5

n1 : n 2

Linear Circuit

Em cos ωt

R

Ideal Transformer

5

6

First we hinge the node 5 and node 6, the digraph of this circuit is as underlying. And we assign the hinged node 5 and 6 as the datum node. 2

3

1

2

1

4 6

5

5

2 3

Digraph of the linear circuit

4

3

6

Leon O. Chua et al, “Linear And Nonlinear Circuits”, p26~p27. Leon O. Chua et al, “Linear And Nonlinear Circuits”, p226~p229; p462~466.

© Copyright 2009 by Yong-Nien Rao

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We could write the tableau equations as underlying:

node 1 KCL:

node 2

i1

node 3

i2

node 4 v1 = v2 = KVL:

v3 = v4 = v5 = v6 =

+ i6 = 0  - i3 = 0   + i4 = 0 - i 4 + i5 = 0  i3

  e3   e1 − e 2  e3 − e 4  e4    e1

Ai (t ) = 0

e2

v (t ) = A T e(t )

Branch equations:

transforme r :

n2 v 1 − n1v 2 n1i1 + n2i2

transforme r : inductor :

v3

capacitor :

= 0

-L C

Resistor :

d i3 dt

d v4 dt

= 0 = 0

- i4

- Ri5 = 0

v5

Ind. Source : This is in the form of

= 0

v6

( M0 D +

= Emcosω t

M 1 ) v(t) + ( N 0 D + N 1 ) i(t) = u s(t)

© Copyright 2009 by Yong-Nien Rao

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Concept 3 Network function (Transfer function) Apply the Laplace transform to (eq.2.9), we get

0 A   e(s)   0  − AT   v(s) = 1 0     0 M 0 s + M 1 N 0 s + N 1   i(s)    T(s) w(s)

 0   0     u s(s)  u(s)

…(eq.3.1) 4

Consider a linear-time invariant circuit with only one independent source transform of

u j (t ) . The Laplace

u j (t ) is u j (s ) .For any circuit variable wk(s) (element of circuit variable vector

w (s ) ) of its zero-state response, we could solve (eq.3.1) with Cramer’s rule5. ) det[ Tk (s)] det[ T(s)]

wk (s) =

…(eq.3.2)

Tk (s) is the matrix formed by replacing the kth column of T(s) by the column vector u(s) We could do the inverse Laplace transform to (eq.3.2) and get the circuit variable wk(t) in The

time domain. Since the

u(s) has only one non-zero element, the det[ Tk (s)] of (eq.3.2) holds for

det[ Tk (s)] = C jk (s)u j ( s )

…(eq.3.3)

C jk (s) is the cofactor6 of ( j , k ) entry of T(s)

We define that the network (transfer) function

H (s) ≡

C jk (s) wk ( s) = u j ( s) det[ T(s)]

In special case, output voltage

H (s) ≡

H (s ) as …(eq.3.4)

u j (t ) could represent an input voltage vi (t ) . And wk(t) could represent an

vo (t ) . We could see the network function in the form

L{ vo (t )} vo ( s ) = L{ vi (t )} vi ( s )

…(eq.3.5)

4

Leon O. Chua et al, “Linear And Nonlinear Circuits”, p595~p596 http://en.wikipedia.org/wiki/Cramer%E2%80%99s_rule 6 http://en.wikipedia.org/wiki/Cofactor_(linear_algebra) 5

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vo ( s ) = H ( s )vi ( s )

© Copyright 2009 by Yong-Nien Rao

…(eq.3.6)

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Concept 4 Engineering Interpretation of Convolution7 If we apply inverse Laplace transform to (eq.3.6), from (eq.1.8) and No. 1 of (Table 1.1)

vo ( t ) = h( t ) ∗ vi ( t ) =

∫ h( t − τ )v (τ ) dτ t

…(eq.4.1)

i

0

vo ( t ) is zero-state response, h( t ) is impulse response and vi ( t ) is input. We’d like to give a time-domain interpretation of (eq.4.1) as underlying. If we define the input pulse component

1 [1( t ) − 1( t − ∆ ) ] …(eq.4.2) ∆ And as ∆ → 0 , p∆ ( t ) → δ ( t ) (called impulse). Thus the vo ( t ) → h( t ) . If we multiply p∆ ( t ) by ak and delay this pulse by τ k , we get the underlying relations. p∆ ( t ) =

vi

v0

ak

akp Δ(t-τk)

akh Δ(t-τk)

Δ

0

For any input

∑ v ( τ ) • ∆ • p (τ k= 0

i

0

t

τ k τk+Δ

vi ( t ) , it could be approximated to

N

vi (τ ) ≅ and

t

τ k τ k+Δ

∆

k

− τ k)

for 0 ≤ τ ≤ t

…(eq.4.3)

t N+1

∆ ≡

…(eq.4.4)

vi

Δ

0

τ1

τk τk+Δ

τN t

τ

From the linearity and time-invariant, we see that

vo ( t ) ≅

N

∑ v (τ ) • ∆ • h ( t − τ ) k= 0

i

k

∆

k

…(eq.4.5)

Let ∆ → 0 , i.e., N → ∞ , (eq.4.5) becomes (eq.4.1)

7

Leon O. Chua et al, “Linear And Nonlinear Circuits”, p620~p623

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Concept 5 Stability Definition and Basic Criterion – from Partial Fraction Expansion and Laplace transform A circuit is stable iff (1) the circuit with bounded response (output) to a bounded input. And (2)

h( t ) approaches zero as t approaches infinity. And we could analysis the transfer function H (s ) to figure out the circuit is stable or not by its impulse response

these criterions and Laplace transform table. (eq.3.5) could be rewritten as m

H(s) =

∑

bi s i

i= 0 n

…(eq.5.1)

∑

ai s

i= 0

i

an = 1, and n ≥ m

First we deal with the case of If

H (s ) has ni poles at − pi , m

H(s) =

∑

i= 0

r

bi s i

∏ (s + p )

…(eq.5.2) ni

i

i= 1

r

Where

∑

i= 1

ni = n

(eq.5.2) could be expanded by the partial fraction expansion891011

H(s) = bn +

Where

cik =

r

ni

cik

∑ ∑ (s + p ) i= 0 k = 1

…(eq.5.3)

k

i

[

1 d ni − k ( s + pi ) ni H ( s ) ni − k ( ni − k )! ds

]

s = − pi

…(eq.5.4)

If we do the inverse Laplace transform to (eq.5.3), from (Table 1.1) we get

h(t) = bnδ ( t ) + If

ni

cik

∑ ∑ ( k − 1)!t

k − 1 − pi t

e

…(eq.5.5)

i= 0 k = 1

pi is a real number, pi must be positive to avoid H (t ) → ∞ when t → ∞ (Hint: expand

the If

r

e − pi t into Taylor series12)

pi is a complex number, from the Euler's formula13,

8

Joseph J. DiStefano, III et al, “Theory and Problems of Feedback and control systems” 2nd ed. P83~p86 http://en.wikipedia.org/wiki/Partial_fraction 10 http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/PartialFraction/PartialFraction.html 9

11

http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/SolutionMethods/LaplaceXform/InvLaplace/InvLaplaceXform.html

12

http://en.wikipedia.org/wiki/Taylor_series 13 http://en.wikipedia.org/wiki/Euler%27s_formula

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e − pi t = e − Re{ pi } t * e − Im{ pi } t = e − Re{ pi } t [ cos(− Im{ pi } t ) + i sin( − Im{ pi } t )] It is obvious that pole

…(eq.5.6)

Re{ pi } must be positive to avoid H (t ) → ∞ when t → ∞ . That is, the

− pi must have negative real part to let the circuit stable.

i sin(− Im{ pi } t ) of (eq.5.6). It will be canceled out by the accompanied complex conjugate of pi ) (Note: Don’t worry about the interpretation the imaginary term

an = 1, and n < m If we let the bounded input vi (t ) = 1(t ) , the output vo (t ) could be determined by (eq.3.6), Second we’d like to deal with the (eq.5.1) for (eq.5.1) and (Table 1.1) as m

1 v o ( s ) = vi ( s ) H ( s ) = s

∑

i= 0 n

∑

i= 0

Because

vo(s) =

m

bi s i ai s

∑

=

i= 0 n

∑

i

i= 0

bi s i …(eq.5.7)

ai s

i+ 1

n < m , the (eq.5.7) could be rewritten as

m− n− 1

∑

j= 0

d js j +

ni

r

d ik

∑ ∑ (s + p ) i= 0 k = 1

…(eq.5.8)

k

i

Apply inverse Laplace transform to (eq.5.8), from (Table 1.1)

vo(t) =

m− n− 1

∑

j= 0

d jδ

j

r

ni

( t ) + ∑ ∑ d ik t k − 1e − p t i = 0 k = 1 ( k − 1)! i

…(eq.5.9)

The 1st term of the right side of (eq.5.9) is not a bounded value. It means the bounded input

1(t ) has unbounded output. This circuit is not stable.

We conclude that the circuit is stable iff (1) n≧m in (eq.5.1) (2) its transfer function

H (s ) has

all its poles in the open Left Half Plane (LHP). In other words, there is no pole in the close right half plan (RHP).

© Copyright 2009 by Yong-Nien Rao

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Concept 6 Nyquist Analysis14 Nyquist analysis is a graphical procedure for determining absolute and relative stability of a circuit transfer function. It is important for two reasons: first, it gives a test for stability and a suggestion to improve it; second, the test relies on easily obtainable experimental data, namely, the sinusoidal steady-state measurements. From concept 5, we already know that the poles must be in LHP if the circuit is stable. But how could we know where the poles are? Nyquist analysis helps us to handle it. Firstly we get to review the concept of mapping. Underlying is the mapping and

H between s

H (s ) . s = σ + jω

Im(s) or jω

Im(H(s)) H Mapping

s0

H(s0) Re(s) or σ

Re(H(s))

s plane

H(s) plane

And we have an important principle “Cauchy's argument principle1516” as

N= Z− P

…(eq.6.1)

N is the net number of clockwise encirclement by curve CA about the origin in the H(s) plane,

Z is the number of zeros encircled by curve C in the s plane, and P is the number of

poles encircled by curve C in the s plane. Underlying is the graphic representation. Im(s)

Im(H(s)) H Mapping CA

C

Re(s)

s plane

Re(H(s))

H(s) plane

From the stability criterion, we know that a circuit is stable iff there is no pole in the close RHP.

14

Joseph J. DiStefano, III et al, “Theory and Problems of Feedback and control systems” 2nd ed. P246~p263 http://en.wikipedia.org/wiki/Argument_principle 16 http://www.pdfcoke.com/doc/14530659/Introduction-of-Cauchy-Argument-Principle-in-Nyquist-Stability-Analysis15

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Page 12 of 16

Thus we could let the curve C enclose the whole RHP (called Nyquist path, sketched as underlying), and check the net number of clockwise encirclement by the mapping curve CA (called Nyquist plot) about the origin in the H(s) plane. That is, we could get N in (eq.6.1). If we know Z in (eq.6.1), we could get P and understand the circuit is stable or not. Im(s)

Nyquist Path

R->infinite Re(s)

One of the most popular application of Nyquist analysis is feedback amplifier design. Let an open loop amplifier circuit with its transfer function as

H (s) =

n( s ) d (s)

…(eq.6.2)

H (s ) has no RHP pole. And the degree of polynomial n(s ) is smaller than the degree of polynomial d (s ) . And this open loop amplifier circuit is stable. That is,

And the unity gain feedback close loop transfer function of this feedback amplifier circuit is

G ( s) = If

H ( s) 1 + H ( s)

…(eq.6.3)

G (s) is stable, G (s) should have no RHP pole. That is, 1 + H ( s) should have no RHP

zero. Following is our analysis for this proposition. Since

H (s ) has no RHP pole, 1 + H ( s) has no RHP pole from (eq.6.2), either. That is

P = 0 in the (eq.6.1). Thus Z = N in this case. What we need to do is only to check the encirclement about the origin of 1 + H ( s ) (in other words, N). If it is not equal to 0, there should be a zero in the RHP for 1 + H ( s ) and G (s ) would be unstable. And please be noted that encirclement about the origin of about (-1,0) of

1 + H ( s) means encirclement

H (s ) .

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This analysis could be represented as underlying Im(s)

Im(H(s)+1)

Im(H(s))

1+H Mapping

Re(s)

Re(H(s)+1)

(0,0)

The most right graph shows that we could check the

=

Re(H(s))

(-1,0)

H (s ) encirclement about (-1,0) to

determine the Z. And we could understand the close loop amplifier is stable or not. We could summarize the Nyquist criterion for a unity gain feedback circuit as “This unity feedback circuit with its open loop transfer function plot of

H (s ) is stable iff the Nyquist

H (s ) does not encircle the (-1,0).”

Another important advantage is that the Nyquist plot could be derived easily from the experiment data. It is because of the underlying characteristics

s = jw 0 < w < ∞ , and apply a sinusoid wave sin ( wt ) to the input of a circuit h( t ) , the sinusoidal steady state output of h( t ) is Asin ( wt + θ ) . Then17 1. For

H ( s ) |s = jw = A and ∠ H ( s ) |s = jw = θ 2. For

s = R∠ ϕ

R → ∞ ,− π

2

≤ϕ ≤π

2 , then

H (s) → 0 (Because of the higher degree of denominator) 3. For s = − jw 0 < w < ∞ then Since the Nyquist plot is symmetric to the real axis18. The

H ( s ) |s = − jw = H ( s ) |s = jw = A and

∠ H ( s ) |s = − jw = − ∠ H ( s ) |s = jw = − θ

We complete a Nyquist plot graphically!

17 18

http://www.pdfcoke.com/doc/16250512/Transfer-Function-vs-Sinusoid-Freqency-Response http://www.pdfcoke.com/doc/16250603/Symmetry-of-Nyquist-Plot

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Concept 7 Phase Margin: From Nyquist Analysis to Bode Plot

H ( jw) decrease as w increases. This is the case of OP amp. As a result of this simple behavior of H ( jw) , it is easy to draw In many applications, the magnitude and the phase of

the conclusions of the Nyquist criterion on the basis of the Bode plot. Basically, Bode plot is another representation of Nyquist plot for s=jω. The Nyquist plot and Bode plot is shown as underlying. And we could find the phase margin, unity-gain frequency and some other popular terms in Bode plot and corresponding Nyquist plot intuitively. Im(H(s))

(-1,0)

ω 180

ω=∞ ω=0

Re(H(s))

ω ug Phase margin

A(jω) ω increases

Nyquist Plot of A (s) (Half) Im(H(s))

20log|A(jω)|

Bode plot of A (jω)

0

ω (log scale)

ωug

Phase(A(jω)) 0˚

ω180

-90˚ -180˚ -270˚

ω (log scale)

Phase margin

We could see that if the phase margin is larger than 0, the Nyquist plot does not encircle the © Copyright 2009 by Yong-Nien Rao

Page 15 of 16

(-1,0) point and the OP amp is stable. The larger the phase margin is, the more unlikely the Nyquist plot encircle the (-1,0) and the more stable this design be. And if the phase margin is less than 0, we could see that the corresponding Nyquist plot encircle the (-1,0) and the OP amp is unstable. Hope you could enjoy the world of margins!

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