Describing Function Calculations for Some Common Nonlinearities

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1 | Page Detailed Describing Function Calculations for Some Common Nonlinearities (I) First Nonlinearity: Saturation with Dead Band Figure NL1: A Plot of the Saturation-with-Dead-Band Nonlinearity We are to show that the describing function for the Saturation-with-Dead-Band nonlinearity, with the input assumed to be = (1) is given by , = if < (2) , = −− if <<, (3) , = −− if > (4) where and are respectively given by = (5) and = (6)

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Transcript of Describing Function Calculations for Some Common Nonlinearities

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Detailed Describing Function Calculations for Some

Common Nonlinearities

(I) First Nonlinearity: Saturation with Dead Band

Figure NL1: A Plot of the Saturation-with-Dead-Band Nonlinearity

We are to show that the describing function for the Saturation-with-Dead-Band nonlinearity,

with the input assumed to be � = � ��� �� (1) is given by

��, �� = �⟨�� if � < � (2)

��, �� = ��� ��� − � − ��� ��� � ⟨�� if � < � < �, (3)

��, �� = ��� �� − � − ��� ������� ��� � ⟨�� if � > �

(4)

where � and � are respectively given by

� = ����� !"�#

(5)

and

� = ����� !$�#

(6)

����

%

−% −� −� � �

&���

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and the gradient � is given by � = %� − �

(7)

Solution

Situation 1: ' < �

Figure 1: The Sketch of &��� vs �� for the Saturation-with-Deadband Nonlinearity �� < ��

Using Sketches of &���v ���� (the nonlinearity) and ����v ��

It is clear from the sketches of Figure 1 that when ' < �, the input operates entirely in

the dead-band, meaning that the output will always be zero.

For this, there is no need for any calculation of a describing function since it is clear that

it will always be zero.

Therefore,

����

%

−% −� −� � �

��

&���

&���

� = � ()* ��

����

��

If � < � for Saturation with Dead Band, then ��, �� = �⟨��

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Situation 2: � < ' < +

First Step: Drawing Out the Three Sketches

Figure 2: The Sketch of &��� vs �� for the Saturation-with-Deadband Nonlinearity �� < � < +�

Using Sketches of &���v ���� (the nonlinearity) and ����v ��

Second Step: Writing Out The Equations

From the graph of &��� against ,- of Figure 2, it is easy to write out the equations as

&��� = �, � ≤ �� < ��

= ��� ��� �� − ��, �� < �� < ��

= �, �� < �� < �/

= ��� ��� �� + ��, �/ < �� < �1

= �, �1 < �� < �� (8)

where ��, ��, �/, and �1 are the �� values corresponding to the points A, B, C, and D in Figure

2.

����

%

−% −� −� � �

��

&���

� � ��

B is the point � = � 2�� < ,- < 34

A is the point � = � 2� < ,- < ��4

C is the point � = −� 2� < ,- < /�� 4

D is the point � = −� 2/�� < ,- < 234

6 7

8 9

&���

� = � ()* ��

����

��

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Third Step: Checking for Odd and Half-Wave Symmetries

By inspection, the nonlinearity displays odd symmetry and the output displays half-wave

symmetry with respect to time.

Fourth Step: Calculation of :; Using Expression for Half-Wave-Symmetrical Nonlinearities

��, �� = 7�� ⟨�� 9

where

7� = 1� < &�����

� ��� �� ��� ≠ �

10

&��� = �, � ≤ �� < ��

= ���> ��� �� − ��, �� < ,- < ��

11

Therefore,

7� = 1� < &�����

� ��� �� ���

7� = 1� ?@@A< ���

� ��� �� ��� + <���� ��� �� − ����

��� ��� �� ���BCC

D

7� = 1� ?@@A <���

����

���� �� − �� ��� ������BCCD

12

We recall from our elementary trigonometry that

EF� �G = � − � ���� G 13

or ���� G = � − EF� �G�

14

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Therefore, making appropriate substitutions in (12) gives us

7� = 1� <�����

��H� − EF� ���� I − �� ��� ������

7� = 1� <���� [� − EF� ���]��

��− �� ��� ��� ���

7� = ���� H�� − ��� ���� I |��M����M�� − 1��� [−EF� ��] |��M����M��

7� = ���� N�� − ��� � 2��4� − �� + ��� ���� O + 1��� �EF� �� − EF� ��� 7� = ���� H�� − ��� �� − �� + ��� ���� I − 1��� EF� ��

7� = ���� H�� − �� + ��� ���� I − 1��� EF� ��

7� = �� − ������ + ����� ���� − 1��� EF� ��

15

Now, we know that when

� = �, �� = �� 16

Thus, making substitutions into the input sinusoid equation

� = � ��� �� 17

leads to

� = � ��� �� 18

which eventually leads to

�� = ����� !"�#

19

Also, from the well-known elementary identity

����P + EF��P = � 20

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we know that

EF� P = √� − ����P 21

which means that

EF� �� = R� − ������ = S� − �"��� = R������ 22

Substituting for EF� �� in (15) gives

7� = �� − ������ + ����� ���� − 1��� √�� − ���

23

Using (23), the describing function is therefore given by

��, �� = 7�� ⟨��

i.e.

��, �� = �� T�� − ������ + ����� ���� − 1��� √�� − ��� U ⟨��

��, �� = V� − ����� + ���� ���� − 1��� √�� − ���� W ⟨��

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Now, we can re-write 1��� R������� as

1��� √�� − ���� = 1�� . �� . √�� − ��� = 1�� . ��� �� . EF� ��

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and since we know from elementary trigonometry that

��� ��� = � ��� �� . EF� �� 26

then 1��� √�� − ���� = ��� . ���� �� . EF� �� = �� ��� ����

27

Therefore (24) can be re-written as

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��, �� = H� − ����� + ���� ���� − �� ��� ���� I ⟨��

��, �� = H� − ����� − ���� ���� I ⟨��

28

For good measure, we can factor out ��� in (28) to get ��, �� = ��� H�� − �� − ��� ���� I ⟨��

29

��, �� = ��� H�� − �� − ��� ���� I ⟨��

�� = ����� !"�# ; � = %� − �

If � < � < � for Saturation with Dead Band, then

where �� is given by

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Situation 3: ' > + First Step: Drawing Out the Three Sketches

Figure 3: The Sketch of &��� vs �� for the Saturation-with-Deadband Nonlinearity �� > ��

Using Sketches of &���v ���� (the nonlinearity) and ����v ��

Please let me know if any bit of the sketching of this diagram is not clear to you.

Because of lack of space, I have left out the description of the points ,- = Z[, ,- =Z\, ,- = Z] and ,- = Z^ from the x vs. ,- plot. However, it should be easy to deduce

that these points correspond respectively to the points E, F, G, and H on the y vs ,- plot.

����

%

−% −� −� � �

��

&���

� � ��

A is the point � = � 2� < ,- < ��4

B is the point � = � 2� < ,- < ��4

C is the point � = � 2�� < ,- < �4

D is the point � = � 2�� < ,- < �4

E is the point � = −� 2� < ,- < /�� 4

F is the point � = −� 2� < ,- < /�� 4

G is the point � = −� 2/�� < ,- < 2�4

H is the point � = −� 2/�� < ,- < 2�4

6

7

_ `

&��� � = � ()* ��

����

��

7 8 9 a b

�� = �� �corr. to pt. 6)

�� = �� (corr. to pt. B)

�� = �/ (corr. to pt. C) �� = �1 (corr. to pt. D)

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Second Step: Writing Out The Equations

From the graph of &��� against ,- of Figure 3, we can write out the equations as

&��� = �, � ≤ �� < ��

= ��� ��� �� − ��, �� < �� < ��

= %, �� < �� < �/

= ��� ��� �� − ��, �/ < �� < �1

= �, �1 < �� < �c

= ��� ��� �� + ��, �c < �� < �d = %, �d < �� < �e

= ��� ��� �� − ��, �e < �� < �f

= �, �f < �� < �� (30)

where �), ) = �, �, … . f are the �� values corresponding to the points A, B, C, D, E, F, G, and H

respectively in Figure 3.

Third Step: Checking for Odd and Half-Wave Symmetries

By inspection, the nonlinearity displays odd symmetry and the output displays half-wave

symmetry with respect to time.

Fourth Step: Calculation of :; Using Expression for Half-Wave-Symmetrical Nonlinearities Again, we write

��, �� = 7�� ⟨�� 31

where

7� = 1� < &�����

� ��� �� ��� ≠ �

32

&��� = �, � ≤ �� < ��

= ��� ��� �� − ��, �� < �� < ��

= %, �� < �� < ��

33

Therefore,

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7� = 1� < &�����

� ��� �� ���

7� = 1� ?@@A< ���

� ��� �� ��� + < ���� ��� �� − ������

� ��� �� ��� + < %��

�� ��� �� ���BCC

D

7� = 1� ?@@A < ���� ��� �� − ����

��� ��� �� ��� + < %

����

��� �� ���BCCD

34

Let us make �� = � �� = � 35

(34) then becomes

7� = 1� ?@@A<���� ��� �� − ���

� � ��� �� ��� + < %��

� ��� �� ���BCCD

7� = 1� ?@@A<��� ()*� �� − �� ()* ���������

� + < %��

� ��� �� �����BCCD

36

Again, using the relation ���� G = � − EF� �G�

37

in (36) gives us

7� = 1� ?@@A< !�� H� − EF� ���� I − �� ��� ��# ������

� + < %��

� ��� �� �����BCCD

7� = 1� H!��� H�� − ()* ���� I + �� EF� ��# |��M���M� − [% hi( ��] |��M���M�� I

7� = 1� H!��� H� − ()* ��� I + �� EF� �# − !��� H� − ()* ��� I + �� EF� �# − % � hi( �� − hi( �� I

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7� = 1� H!��� �� − �� + ��1 �()* �� − ()* ��� + �� �EF� � − EF� �# + %hi( � I

38

Let us expand (38).

7� = ���� �� − �� + ��� �()* �� − ()* ��� + 1��� �EF� � − EF� �� + 1%� hi( � 39

Pulling out jklm from the RHS of (39) gives

7� = ���� H�� − �� + ��� ���� �()* �� − ()* ��� + 1��� ���� �EF� � − EF� ��+ 1%� ���� hi( � I

7� = ���� H�� − �� + �� �()* �� − ()* ��� + � �� �EF� � − EF� �� + �%�� hi( � I

40

Now, we know that when

� = �, �� = � 41 � = �, �� = � 42

Thus, as we did for the second case earlier,

� = � ��� �� 43

leads to

� = � ��� � 44

and

� = � ��� � 45

So that ��� � = ��

46

and ��� � = ��

47

Again, using the identity

����P + EF��P = � 48

gives

EF� P = √� − ����P 49

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which means that

EF� � = √� − ����� = S� − �"��� = R������ 50

EF� � = R� − ����� = S� − �$��� = R������ 51

Let us take a part of (40) and re-write it in trigonometric form i.e. let n = � �� �EF� � − EF� �� + �%�� hi( �

52

Then, we can write n = � �� EF� � − � �� EF� � + �%�� hi( �

n = � �� EF� � − � �� EF� � + �. %� ! ��# hi( �

53

and since � = %� − �

54

we have (53) becoming n = � �� EF� � − � �� EF� � + �. %� !� − �% # hi( �

n = � �� EF� � − � �� EF� � + �. !� − �� # hi( �

n = � �� EF� � − � �� EF� � + �. !�� − ��# hi( �

n = � �� EF� � − � �� EF� � + �. �� hi( � − �. �� hi( �

55

Replacing �� and

�� with ��� � and ��� � respectively gives n = � ��� � EF� � − � ��� � EF� � + �. ��� � hi( � − �. ��� � hi( � n = �. ��� � hi( � − � ��� � EF� � n = ��� �� − ��� �� 56

Therefore, (40) becomes 7� = ���� H�� − �� + �� �()* �� − ()* ��� + n I

7� = ���� H�� − �� + �� �()* �� − ()* ��� + ��� �� − ��� �� I

7� = ���� H�� − �� + �� �()* �� − ()* ��� − ���� �� − ��� ���I

7� = ���� H�� − �� − �� �()* �� − ()* ���I

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The describing function is therefore given by

��, �� = 7�� ⟨��

i.e. ��, �� = �� V���� H�� − �� − �� �()* �� − ()* ���IW ⟨��

��, �� = ��� H�� − �� − �� �()* �� − ()* ���I ⟨��

57

Let us now put our results together:

��, �� = ��� H�� − �� − �� �()* �� − ()* ���I ⟨��

� = ����� !"�# ; � = ����� !$�# ; � = %� − �

Therefore, if � > � for Saturation with Dead Band, then

where �, � and � are given by

��, �� = ��� H�� − � − ��� ��� I ⟨��

��, �� = ��� H�� − �� − �� �()* �� − ()* ���I ⟨��

� = ����� !"�# ; � = ����� !$�# ; � = %� − �

For a system with a Saturation-with-Dead-Band nonlinearity (of Figure

NL1) excited by a sinusoidal input � = � ��� ��;

(A) If � < �, then ��, �� = �⟨��;

(B) If � < � < �, then

(C) If � > �, then

where �, � and � are given by

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(II) Second Nonlinearity: Variable-Gain Nonlinearity

Figure NL2: A Plot of the Variable-Gain Nonlinearity

We are to show that the describing function for the Variable-Gain nonlinearity is given

by

o�, �p = %⟨�� if � < �

58

o�, �p = q %# −�� o%# − %p 2� + ()* ��� 4s ⟨�� if � > �

59

where � is given by � = ()*�� !��#

60

−� �

&���

���� %

%#

%#

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Solution:

Situation 1: ' < �

First Step: Drawing Out the Three Sketches

Figure 4: The Sketch of &��� vs �� for the Variable-Gain Nonlinearity �� < �� Using Sketches

of &���v ���� (the nonlinearity) and ����v ��

Since the input t = ' sin ,- stays within the first linear domain defined by the gain x = yzt, then the output is strictly a scaled version of the input sinusoid, as shown in the

graph of y against ,-.

��

&���

� � ��

&���

� = � ()* ��

����

−� �

&���

����

%

%#

%#

��

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Second Step: Writing Out The Equations

From the graph of &��� against ,- of Figure 2, it is easy to write out the equations as

&��� = %�, � ≤ �� < �� 61

Which leads on to

&��� = % � ��� �� ; � < �� < ��

62

Third Step: Checking for Odd and Half-Wave Symmetries

Again, by inspection, the nonlinearity displays odd symmetry and the output displays

half-wave symmetry with respect to time.

Fourth Step: Calculation of :; Using Expression for Half-Wave-Symmetrical Nonlinearities

��, �� = 7�� ⟨�� 63

where

7� = 1� < &�����

� ��� �� ��� ≠ �

64

&��� = % � ��� �� ; � < �� < ��

65

Therefore,

7� = 1� < &�����

� ��� �� ���

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7� = 1� ?@@A<[% � ��� ��] ��� ��

��� �����BCC

D 7� = 1� ?@@

A< % � ���� ����

� �����BCCD

7� = 1%�� ?@@A< H� − EF� ���� I

��� �����BCC

D 7� = �%�� ?@@

A<[� − EF� ���]��

� �����BCCD

7� = �%�� H�� − ��� ���� I |��=���=�� 7� = �%�� H�� − ��� �� − � + ��� �� I

7� = �%�� ���� 7� = %� 66 Therefore (63) can be re-written as

��, �� = �� [%�]⟨��

��, �� = %⟨��

67

If � < � for the Variable-Gain Nonlinearity, then ��, �� = %⟨��

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Situation 2: ' > �

First Step: Drawing Out the Three Sketches

Figure 5: The Sketch of &��� vs �� for the Variable-Gain Nonlinearity �� > �� Using Sketches

of &���v ���� (the nonlinearity) and ����v ��

Since the input t = ' sin ,- goes outside the first linear domain defined by the gain x = yzt, and into the region defined by the graphs with slope %#, the output is going to

be a combination of two different sinusoids, scaled by the two gradients yz and %#, as

shown in the graph of y against ,-.

��

&���

� � ��

&���

� = � ()* ��

����

−� �

&���

����

%

%#

%#

��

�� �� �/ �1

��

�� �/ �1

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Second Step: Writing Out The Equations

I’ll take some time to explain how the equations can be derived.

Let us first write out the input-output equations (i.e. y against x).

For convenience, I will redraw the graph of Figure NL2 as Figure 6 below:

Figure 6: A Repeated Plot of the Variable-Gain Nonlinearity

Let us split the graph into 3 domains i.e. the domains t < −�, −� < t < � and t > �

Domain 1: � < −�

We recall that, for a straight line & = >� + h 68

Thus, for the line of gradient yz# in the domain t < −�, we can replace { with yz# in

(68) to get

& = %#� + h� 69

−� �

&���

����

%

%#

%#

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I have used h� here to distinguish it from that of Domain 3 that we will get to shortly.

Our task now is to get the value of h�.

If we look at Figure 6, we will see that at the point of intersection of domains 1 and 2 (i.e.

the point t = −�), the value of y is given by relating the gradient % with x and – � i.e.

% = [� − & |�M��][� − �−��]

& |�M�� = −%� 70

Therefore, when � = −�, then & = −%�.

Replacing & with −%� and � with – � in (69) gives

−%� = %#�−�� + h�

h� = �%# − %�� 71

Thus, for domain 1, the equation of the line segment is

& = %#� + �%# − %�� 72

Domain 2: −� < � < �

The equation here is very straightforward and is given by & = %� 73

Domain 3: � > �

We use the same method used for the calculation of the equation of the line segment of

domain 1 i.e. we have

& = %#� + h� 74

Again, at the point of intersection of domains 2 and 3 (i.e. the point t = �), the value of y

is given by relating the gradient % with x and � i.e.

% = [& |�M� − �][� − �]

& |�M� = %� 75

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Therefore, when � = �, then & = %�.

Replacing & with %� and � with � in (74) gives

%� = %#��� + h�

h� = �% − %#�� 76

Thus, for domain 3, the equation of the line segment is

& = %#� + �% − %#�� 77

Let us put our results together.

Now, back to our sketches of Figure 5, the relevant equations are found to be

&��� = %� = %� ()* ��, � ≤ �� < ��

&��� = %#� + o% − %#p�

= %#� ()* �� + o% − %#p�, �� ≤ �� < �� &��� = %� = %� ()* ��, �� ≤ �� < �/

&��� = %#� + �%# − %�� = %#� ()* �� + o%# − %p�, �/ ≤ �� < �1

&��� = %� = %� ()* ��, �1 ≤ �� < ��

78

& = %#� + �%# − %��

& = %�

& = %#� + �% − %#�� & = %�

(A) If t < −� then

(B) If −� < t < � then

(C) If t > � then

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Third Step: Checking for Odd and Half-Wave Symmetries

Again, by inspection, the nonlinearity displays odd symmetry and the output displays

half-wave symmetry with respect to time.

Fourth Step: Calculation of :; Using Expression for Half-Wave-Symmetrical Nonlinearities

��, �� = 7�� ⟨�� 79

where

7� = 1� < &�����

� ��� �� ��� ≠ �

80

&��� = %� ()* ��, � ≤ �� < ��

&��� = %#� ()* �� + o% − %#p�, �� ≤ �� < �� 81

Therefore,

7� = 1� < &�����

� ��� �� ���

7� = 1� ?@@A< [% � ��� ��] ��� ����

� ����� + <[%#� ()* �� + o% − %#p�] ��� ����

�������BCC

D 7� = 1� ?@@

A< [% � ���� ��]��� ����� + <[%#� ���� �� + o% − %#p� ��� ��]

����

�����BCCD

7� = 1� ?@@A< H% � �� − hi( ����� I��

� �����+ <[%#� �� − hi( ����� + o% − %#p� ��� ��]

����

�����BCCD

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7� = 1� V%�� H �� − ��� ���� I |��M���M��

+ V%#�� !�� − ��� ���� # − o% − %#p� EF� ��W |��M����M�� W 7� = 1� }%�� H �� − ��� ���� − � + ��� ����� I

+ N%#�� ~�� − ��� � 2��4� � − o% − %#p� EF� 2��4− %#�� !�� − ��� ���� # + o% − %#p� EF� ��O�

7� = 1� N%�� H �� − ��� ���� I+ V%#�� 2��4 − %#�� !�� − ��� ���� # + o% − %#p� EF� ��WO

7� = 1� N�� [% − %#] H �� − ��� ���� I + V%#��1 + o% − %#p� EF� ��WO

82

Let us expand (82) 7� = ��� �% − %#� H �� − ��� ���� I + %#� + 1�� o% − %#pEF� ��

83

Let us now “pull” X out of (83) i.e. 7� = � !�� �% − %#� H �� − ��� ���� I + %# + 1��� o% − %#p EF� ��#

84

Let n = 1��� o% − %#p EF� ��

85

We can re-write (85) as n = 1� �� o% − %#p EF� ��

86

and remembering that

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()* �� = ��

87

makes (86) to become n = 1� ��� ��o% − %#p EF� ��

n = �� o% − %#p [���� �� EF� ��] n = �� o% − %#p ��� ���

88

Therefore, (84) becomes 7� = � !�� �% − %#� H �� − ��� ���� I + %# + n#

7� = � !%# + �� �% − %#� H �� − ��� ���� I + n#

7� = � !%# + �� �% − %#� H �� − ��� ���� I + �� o% − %#p ��� ���#

7� = � !%# + �� �% − %#� H �� − ��� ���� + ��� ���I#

7� = � !%# + �� �% − %#� H �� + !� − ��# ��� ���I#

7� = � !%# + �� �% − %#� H �� + �� ��� ���I#

7� = � !%# − �� �%# − %� H �� + �� ��� ���I#

89

Therefore (79) can be re-written as

��, �� = �� H� !%# − �� �%# − %� H �� + �� ��� ���I#I ⟨��

��, �� = !%# − �� �%# − %� H �� + �� ��� ���I# ⟨��

90

Let us put all our results for the Variable-Gain Nonlinearity together. The results are summarized

on the next page:

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N.B.

[1] These solutions are not really as long as you see them. I have only made them this

long for the sake of clarity. You need not be intimidated by the length.

[2] I will try to send the solutions of the others before your mid-semester test. However, I

cannot guarantee that I’ll be able to do so. I have a lot to deal with now. I’ll advise us to

try these things out on our own. However, if you have challenges, please feel free to let

me know.

[3] All the best, as you “stretch your mathematical nerves a bit”! I’m sure you will be fine

in the end.

!%# − �� �%# − %� H �� + �� ��� ���I# ⟨��

� = ����� !"�#

For a system with a Variable-Gain nonlinearity (of Figure NL2)

excited by a sinusoidal input � = � ��� ��;

(A) If � < �, then ��, �� = %⟨��

(B) If � > �, then

where � is given by