Function Analysis v.2

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CALCULUS
FUNCTION ANALYSIS
Arun Umrao
www.sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING

0.1. FUNCTION 1
0.1 Function
0.1.1 Set
A set is collection of elements, placed between curly brackets. For example, a
set of four numbers is shown below:
A = {5, 7, 10, 14}
The elements of the set may be symbols, names or numbers. This type of set
denomination is called “roster method”.
0.1.2 Ordered Pair
(a1, b1) and (a2, b2) are said to be ordered pairs. In ordered pair, first element
of an ordered pair is belong to set A and second element of the ordered pair is
belong to set B. The set of all ordered pairs whose first entry is in some set A
and whose second entry is in some set B is called the Cartesian product of A
and B. Some time the second element of the ordered pair has a unique relation
with the first element of the ordered pair. For example, in set
X = {(1, 2), (3, 6), (5, 10)}
second element of the ordered pair is twice of the first element of the ordered
pair.
0.1.3 Set Builder Notation
Set-builder notation is used to describe a set that is defined by a logical relation
or a logical formula (i.e. predicate) which is true for an element of the set, and
false otherwise. In other words, set builder notation is used to describe a huge
discrete data in one or two mathematical word lines. Set-builder notation has
three parts: a variable, a colon, and a rule. For example, a set
A = {1, 2, 3, . . .}
is written as
A = {x : x > 1}
If set is an ordered pair and it may be written as set building form, then logical
relation or logical rule may form the function for the given set. Take ordered
paired set
X = {(1, 2), (3, 6), (5, 10)}
Take, first element of ordered pair as x and second element of the ordered pair
as y, then we can write it in set building form as
X = {y : y = 2x, x ∈ {1, 3, 5}}

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Now, y = 2x or y = f(x) = 2x is a logical relation or function for the given set
X.
x y = f(x)
1 2
3 6
5 10
Table 1: Tabulated form of ordered pairs of set X.
In mathematics, ordered pairs are also arranged in tabulated data having
one column for independent variable and second column for dependent variable
and both columns have a logical relation as given in table ??.
Solved Problem 0.1 Write the set X = {(2, 4), (3, 9), (4, 16), (5, 25)} in tabu-
lated form and construct its set building form.
Solution The tabulated data of the given set X is given below:
x y = f(x)
2 4
3 9
4 16
5 25
Table 2: Tabulated form of ordered paired set X =
{(2, 4), (3, 9), (4, 16), (5, 25)}.
The second column value is square of the first column value, hence each
second element of the ordered pairs of set X is square of the first element of the
corresponding ordered pair of the set X. So, its set building form is
X =

y : y = x2
, x ∈ [2, 5]

Solved Problem 0.2 Construct ordered pair set from the tabulated data given
below and represent in set building form. Deduce the logical relation used in
this tabulated data. Find the value of second element of the ordered pair if first
element is 4.5.
Solution

0.1. FUNCTION 3
x y = f(x)
3 9
4 12
5 15
6 18
Table 3: Tabulated data.
The ordered pair set of the tabulated data is given as
X = {(3, 9), (4, 12), (5, 15), (6, 18)}
The second column value is thrice of the first column value, hence each second
element of the ordered pairs of set X is three times of the first element of the
corresponding ordered pair of the set X. So, its set building form is
X = {y : y = 3x, x ∈ [3, 6]}
The logical relation used in this set is y = 3x. Now the value of the second
element of ordered pair for first element x = 4.5 is
y = 3 × 4.5 = 13.5
0.1.4 Discrete Continuous Function
A discrete function accepts only discrete inputs and give result based on the
function logic. The input size of the discrete function is finite. A continuous
function accepts discrete values as well as intermediate values. The input size of
continuous function is infinite. A continuous function is also a discrete function
but a discrete function may or may not be continuous function.
x y = f(x)
3 9
4 12
5 15
6 18
Table 4: Tabulated data.
From the above tabulated data discrete function can be written as
f(x) = 3x; x ∈ [3, 6] and x ∈ I

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It means, the input size of this discrete function is only 4 and input values
would be only integer from 3 to 6. A continuous function accepts the input
values which are in [3, 6], like 3, 3.5, 5 and 5.55 etc. So, a continuous function
can be written as
f(x) = 3x; x ∈ [3, 6] and x ∈ R
This shows that continuous function has infinite input size.
Discrete Values
Take function f(x) of variable x for reference purposes. Take upper and lower
limits as x ∈ [1, 2]. Now, we can find many numbers of function values within
the given limit, by selecting appropriate type of domain of x. For example, if
x ∈ I then there are only two function values at x = 1 and x = 2. If domain of x
is selected as x ∈ R then there are infinite numbers of function values as x may
have infinite numbers of values within given limits as x = 1, x = 1.1, x = 1.11,
x = 1.111 and so on. To avoid the limitless function values, we take limited
numbers of x values either separated in equal distances or unequal distances
within the given limits. The finite function values at these limited x points are
called discrete values. Discrete values are identified by their positions, i.e. first,
second, third, fourth etc, rather that order of variable x and position of discrete
value is identified by the number inside square bracket, [ ]. First discrete value
and function value at point x = 1 both are different.
Solved Problem 0.3 Find the four discrete values of the function f(x) = 2x
for
x ∈ [1, 3] and x ∈ I.
Solution From the given problem, we have only three x values which are
x = 1, x = 2 and x = 3. The function values at these three points are f(1) = 2,
f(2) = 4 and f(3) = 8. If we start counting function values from one then three
discrete function values are f[1] = 2, f[2] = 4 and f[3] = 8.
Discrete Summation
Summation of definite discrete function values is called discrete summation. For
example, consider function values of function
f(x) = 2x
as f(1) = 2, f(2) = 4, f(3) = 6, f(4) = 8, f(5) = 10. Discrete summation of
these function values is given by
X
f(x) = f(1) + f(2) + f(3) + f(4) + f(5) = 30
The symbol
X
is use to symbolised the discrete summation.

0.1. FUNCTION 5
for
x ∈ [1, 3] and x ∈ I and find their sum.
discrete function values are f[1] = 2, f[2] = 4 and f[3] = 8. The sum of these
three (n = 3) discrete values is
X
f[n] = f[1] + f[2] + f[3] = 2 + 4 + 8 = 14
This is desired result.
Cumulative Summation
Cumulative summation of definite discrete function values is defined as sum
of current function value with the total sum of all previous function values.
Cumulative summation is written for all current function values. For example,
consider function values of function
f(x) = 2x
as f(1) = 2, f(2) = 4, f(3) = 6, f(4) = 8, f(5) = 10. Their cumulative
summation (csum) is given by
x 1 2 3 4 5
f(x) 2 4 6 8 10
csum[f(x)] 2 6 12 20 30
for
x ∈ [1, 3] and x ∈ I, and find their cumulative sum.
discrete function values are f[1] = 2, f[2] = 4 and f[3] = 8. The cumulative
sum of these three (n = 3) discrete values is
x 1 2 3
f(x) 2 4 8
csum[f(x)] 2 6 10

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This is cumulative sum of the function values.
Indexed Summation
Indexed summation of definite discrete function values is defined as sum of
function values located at the given indices only. For example, consider function
values of function
f(x) = 2x
as f(1) = 2, f(2) = 4, f(3) = 6, f(4) = 8, f(5) = 10. The indexed form of these
function values is given below:
n 1 2 3 4 5
x 1 2 3 4 5
f(x) 2 4 6 8 10
Table 5: Indexed Discrete Data - n ≥ 1.
In this form, index counting is started from ‘1’. Indexed summation is
represented by symbol
3
X
1
, where lower suffix shows the initial index value
while upper suffix shows the final index value. Indexed summation of the given
data will be
3
X
1
f(x) = f[1] + f[2] + f[3] = 2 + 4 + 6 = 12
Sometime index counting is also started from ‘0’ as shown below:
n 0 1 2 3 4
x 1 2 3 4 5
f(x) 2 4 6 8 10
Table 6: Index Discrete Data - n ≥ 0.
Now indexed summation will be changed as discrete value at index 1 is
different in ?? than the discrete value at index 1 in ??. So
3
X
1
f(x) = f[1] + f[2] + f[3] = 4 + 6 + 8 = 18

0.1. FUNCTION 7
for
x ∈ [1, 3] and x ∈ I, and find
2
X
1
f[x].
discrete function values are f[1] = 2, f[2] = 4 and f[3] = 8. Indexed summation
of the given function is
2
X
1
f[x] = f[1] + f[2] = 2 + 4 = 6
This is indexed summation of the function values.
for
3
X
2
f[x].
3
X
2
f[x] = f[2] + f[3] = 4 + 8 = 12
for
X
[1,3]
f[x].
X
[1,3]
f[x] = f[1] + f[3] = 2 + 8 = 10

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0.1.5 Function
A function f relates with each element of x of a set, say Df , with exactly one
element y of the another set, say Rf . Here Df is called domain of function f
and Rf is called range of function f. Here x is independent variable while y is
called dependent variable. f has single and definite value for every value of x
in its domain.
Solved Problem 0.9 Assume a function f(x) =
√
x − 2, where x ∈ I
Solution Let x ∈ I then for real value of f(x),
√
x − 2 must be positive
and x − 2 ≥ 0. Which gives x ≥ 2. The domain of all real number of x is
x = {2, 3, 4, . . ., ∞}
The range of function is
R =
n
0, 1, 2,
√
3, . . . , ∞
o
respectively. The table is given below.
x 2 3 4 5 . . . ∞
f(x)
√
0
√
1
√
2
√
3 . . . ∞
Table 7: The data table of the function y =
√
x − 2.
This data is plotted in the graph in Polar form as well as Cartesian form as
shown below.
1
2
1 2 3 4 5 6 7 8 9
x
y
Figure 1: Polar (blue) Cartesian (red) plot of the function y =
√
x − 2.
Solved Problem 0.10 Assume a function f(x) =
x
x2 − 4
, where x ∈ I.
Solution Let x ∈ I then for real and finite value of f(x), x2
− 4 must not
be zero. The domain of all real number of x is −∞ to ∞ except x = 2 and
x = −2. At x = ±2 the value of f(x) is tends to ±∞. The range of function is
−∞ to ∞. The table is given below.

0.1. FUNCTION 9
x −∞ . . . -3 -2 -1 0 1 2 3 . . . ∞
f(x) 0 . . . -0.6 −∞ -0.33 0 0.33 ∞ 0.6 . . . 0
The data is plotted as show below.
1
2
3
−1
−2
−3
1 2 3 4 5
−1
−2
−3
−4
−5
x
y
Figure 2: Plot of the function y =
x
x2 − 4
.
Solved Problem 0.11 Assume a function f(x) = x4
− x2
, where x ∈ I.
Solution Let x ∈ I then for real and finite value of f(x), the domain of all
real number of x is −∞ to ∞. The range of function is −0.25 to ∞. The table
is given below.
x −∞ . . . -1 -0.5 0 0.5 1 . . . ∞
f(x) −∞ . . . 0 -0.19 0 0.19 0 . . . ∞
This data is plotted as shown below.
1
1
−1
x
y
Figure 3: Plot of the function y = x4
− x2
.

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Single Variable Function
A single variable function has one independent variable and one dependent
variable. For example,
f(x) = mx2
+ c
is a single variable function of x. Here, m and c are constants.
Solved Problem 0.12 Draw a xy-plane for a function y that depends only on
x and take four auxiliary points for x numerically. Find the function value at
these points and show them in xy-plane. Function is y(x) = x2
− 2.
Solution The given single variable function is
y(x) = x2
− 2
Take four auxiliary points in x-axis, i.e. say horizontal number lines as
x = {−1.75, −0.5, 0.5, 1}
1
−1
1
−1
−2
x
y
b
b b
b
Now function values at these x-points are
y(−1.75) = (−1.75)2
− 2 = 1.0615
y(−0.5) = (−0.5)2
− 2 = −1.75
y(0.5) = (0.5)2
− 2 = −1.75
y(1) = (1)2
− 2 = −1
Now four coordinate points (x, y(x)) are (−1.75, 1.0615), (−0.5, −1.75), (0.5, −1.75)
and (1, −1). These four points are shown in above figure.
Solved Problem 0.13 Draw a tr-plane for a function t that depends only on t
and take four auxiliary points for t numerically. Find the function value at these
points and show them in tr-plane. Function is r(t) = t − 2.
r(t) = t + 0.5

0.1. FUNCTION 11
Take four auxiliary points in t-axis, i.e. say horizontal number lines as
t = {−1, −0.5, 0.5, 1}
1
−1
1
−1
t
r
b
b
b
b
Now function values at these t-points are
r(−1) = (−1) + 0.5 = −0.5
r(−0.5) = (−0.5) + 0.5 = 0
r(0.5) = (0.5) + 0.5 = 1
r(1) = (1) + 0.5 = 1.5
Now four coordinate points (t, r(t)) are (−1, 0.5), (−0.5, 0), (0.5, 1) and (1, 1.5).
These four points are shown in above figure.
Multi Variables Function
A single variable function has more than one independent variables and one
dependent variable. For example,
f(x, y) = mx2
+ cy
is a multi variables function of x and y. Here, m and c are constants.
Solved Problem 0.14 Draw a xy-plane for a function f that depends on x and y
both, and take four auxiliary coordinate points numerically. Find the function
value at these points and show them in xy-plane. Function is f(x, y) = x2
− 2y.
f(x, y) = x2
− 2y
This is two variable function, hence we need pairs of two input values for each
independent variable. Four two values auxiliary ordered pairs in xy-plane are
(−0.5, 0.5), (−0.5, 0), (0, 1) and (1, 1). These four ordered pairs are coordinate
points in xy-plane.

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1
1
−1
x
y
b
[−1.25]
b
[−0.25]
b
[−2]
b
[−1]
bcb
x
y b
−1.25
b
−0.25
b
−2
b
−1
z
Now function values at these points are
f(−0.5, 0.5) = (−0.5)2
− 2 × 0.5 = −1.25
f(−0.5, 0) = (−0.5)2
− 2 × 0 = −0.25
f(0, 1) = (0)2
− 2 × 1 = −2
f(1, 1) = (1)2
− 2 × 1 = −1
Now four function values are obtained which are −1.25, −0.25, −2 and −1.
These four points are shown in above figure. Numerical value inside square
bracket are function values at that (x, y) coordinate point.
Solved Problem 0.15 Draw a ij-plane for a function f that depends on i and j
both, and take four auxiliary coordinate points numerically. Find the function
value at these points and show them in ij-plane. Function is f(i, j) = i2
− 2ij.
Here i and j are real variables, NOT imaginary numbers as used in complex
number system.
f(i, j) = i2
− 2ij
This is two variable function, hence we need pairs of two input values for each
independent variable. Four two values auxiliary ordered pairs in ij-plane are
(−0.5, 0.5), (−0.5, 0), (0, 1) and (1, 1). These four ordered pairs are coordinate
points in ij-plane.
1
1
−1
i
j
b
[0.75]
b
[0.25]
b
[0]
b
[−1]
bcb
x
y
z
b
0.75
b
0.25
b
0
b
−1
Now function values at these points are
f(−0.5, 0.5) = (−0.5)2
− 2 × −0.5 × 0.5 = 0.75

0.1. FUNCTION 13
f(−0.5, 0) = (−0.5)2
− 2 × −0.5 × 0 = 0.25
f(0, 1) = (0)2
− 2 × 0 × 1 = 0
f(1, 1) = (1)2
− 2 × 1 × 1 = −1
Now four function values are obtained which are 0.75, 0.25, 0 and −1. These
four points are shown in above figure. Numerical value inside square bracket
are function values at that (i, j) coordinate point.
0.1.6 f 0, f 0 and f = 0
A two dimensional function, like f(x) = ax2
+ bx + c is plotted in xy-plain,
with variable x in x-axis and f(x) in y-axis. A function f 0 represents that
all values of ax2
+ bx + c for ∀x ∈ R are above the x-axis. For example,
f(x) = x2
+ x + 1 is plotted above the x-axis irrespective of the x values.
1
2
1
−1
−2
x
y
f
(
x
)
=
x
2
+
x
+
1
A function f 0 represents that all values of ax2
+ bx + c for ∀x ∈ R are
below the x-axis. For example, f(x) = −x2
+ x − 1 is plotted above the x-axis
irrespective of the x values.
−1
−2
−3
1
−1
−2
x
y
f
(
x
)
=
−
x
2
+
x
−
1
Case f = 0 is special case. At this condition, there are few values of x for
which f is zero. If other x values are put in the function, function may be
either above or below the x-axis. This condition is used to find the roots of the

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function. Note that, function intersects to x-axis if it has more than one real
roots. If it has only one real root, then function will only touch the x-axis.
1
2
1 2 3
−1
x
y
f
(
x
)
=
x
2
−
3
x
+
2
For example, f(x) = x2
− 3x + 2 is plotted above and below of the x-axis.
At x = 1 and x = 2, function intersects to the x-axis. For 1 x 2, function
f(x) is plotted below the x-axis and ∀x ∈ {(−∞, 1) ∪ (2, −∞)}, function f(x)
is plotted above the x-axis.
0.1.7 Square Root
All of we know about the square root of a number. For example, square root of
the number k2
would be written as
y =
√
k2
Here, k2
is a perfect square root number, say like 4, 9 etc. Solution of above
relation is
y =
√
k2 = ±k
Now which value is exactly the real answer, +k or −k or both ±k? The answer
depends on the problem where we are using k. For example, if k is used only
as positive numbers then
√
k2 would be equal to +k not −k. Similarly, if k is
within −∞ ≤ k ≤ ∞ then its both values (±k) would be the answers. In short,
the result value of k would be based on the domain of function y. Square root
values which are beyond the acceptable range are discarded from results.
Solved Problem 0.16 In the number series 1, 2, 3, . . ., which value of the relation
y =
√
16 would be perfectly fitted?
Solution From the given relation y =
√
16, y = ±4. Now, only +4 is fall
within the given series 1, 2, 3, . . ., hence only +4 value would be fitted in the
given series.
Solved Problem 0.17 In the number series . . ., −1, 0, +1, . . ., which value of
the relation y =
√
4 would be perfectly fitted?

0.1. FUNCTION 15
Solution From the given relation y =
√
4, y = ±2. Now, both −4 and
+4 are fall in the given series . . ., −1, 0, +1, . . ., hence both −4 and +4 values
would be fitted in the given series.
0.1.8 Decreasing/Increasing Function
A function f(x) is said to be decreasing function if its values are decreasing when
we moves from left to right in number line for x values. The function values
may be above the x-axis or below the x-axis. If xi and xi+1 are two consecutive
x values in x ∈ R, then f(xi) f(xi+1). Take example f(x) = x2
− 3x + 2,
which is plotted above and below of the x-axis.
1
2
1 2 3
−1
x
y
f
(
x
)
=
x
2
−
3
x
+
2
Function values are constantly decreasing when we move from −∞ to 1.5. A
function f(x) is said to be increasing function if its values are increasing when
we moves from left to right in number line for x values. The function values may
be above the x-axis or below the x-axis. If xi and xi+1 are two consecutive x
values in x ∈ R, then f(xi) f(xi+1). For the given function f(x) = x2
−3x+2,
function values are constantly increasing when we move from 1.5 to ∞.
0.1.9 Types of Function
There are two types of functions. Algebraic and Transcendental. An
equation, f(x) is said to be algebraic if it is purely a polynomial in x
and is said to be transcendental if f(x) contains trigonometric, loga-
rithmic or exponential function.
Algebraic/Polynomial Function
A function is said to be algebraic or polynomial function, if its all terms have only
exponents and constants, and it is a purely polynomial of independent variable.
For example, f(x) = ax2
+bx+c is an algebraic function. A polynomial function
may have any degree and there may be missing of a particular term. The
linearity, homogeneity of a polynomial function depends on the characteristics

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of polynomial. A polynomial function is given as
f(x) =
n
X
i=0
aixi
If n = 0 then polynomial function is constant function. If n = 1 then polynomial
function is line function. If n = 2 then polynomial function is quadratic function
which is also known as parabolic function and so on.
Trigonometric Function
A function is said to be trigonometric function, if its all terms have only trigono-
metric function and constants. For example, f(x) = sin x+cos x+c is a trigono-
metric function.
Phase in Trigonometric Functions A trigonometric function is in its general
form y1 = a sin θ. Another trigonometric function y2 is said to have a phase of
φ with the function y1 if it is written as
y2 = a sin(θ + φ)
It means that y2 is always lead to y1 by an angle φ to y1. The function value is
instantaneous1
value at any angle θ. The relative peak values of trigonometric
functions, which are in phase, are measured by graph method or by simple
computation method.
Solved Problem 0.18 Plot the trigonometric function y = sin x within 0 ≤ x ≤
π.
Solution To plot the function y = sin x, we shall construct a data table,
in which we shall take radian values of x, converts them into equivalent degrees
and find the sine values. Though the sine values may be find by either using
radian form of angle or degree form of angle yet be careful while you putting
angle values and don’t be confused in radian and degree. Then we will plot the
graph between x and sin x.
xd =
180
π
× xr
The tabulated form of data table is given below:
1
Independent from previous values, memoryless.

0.1. FUNCTION 17
xr xd (◦
) sin(xd)
0.000 0 0.000
0.314 18 0.309
0.628 36 0.588
0.942 54 0.809
1.256 72 0.951
1.570 90 1.000
1.884 108 0.951
2.198 126 0.810
2.512 144 0.589
2.826 162 0.310
3.140 180 0.000
π
3
2π
3 π
x
y
b
b
b
b
b b b
b
b
b
b
sin x
π
3
2π
3 π
x
y
b
b
b
b
b b b
b
b
b
b
sin x
First we will plot the graph in dotted form taking ordered pairs (x, sin x) for
each row, in which x represents to abscissa2
in x-axis and sin x represents to
ordinate3
in y-axis. This plot is shown in first part of above figure. After that
we draw a smooth hand curve as shown in second part of the above figure.
Solved Problem 0.19 Plot the trigonometric function y = cos x within the limits
of cosine argument 0 ≤ x ≤ π.
Solution To plot the function y = cos x, we shall construct a data table,
in which we shall take radian values of x, converts them into equivalent degrees
and find the sine values. Though the sine values may be find by either using
radian form of angle or degree form of angle yet be careful while you putting
angle values and don’t confuse in radian and degree. Then we will plot the
2
abscissa is measurement of a point parallel to the x-axis from origin, i.e. horizontal
distance of point from y-axis.
3
ordinate is measurement of a point parallel to the y-axis from origin, i.e. vertical distance
of point from x-axis.

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graph between x and cos x.
xd =
180
π
× xr
The tabulated form of data table is given below:
xr xd (◦
) cos(xd)
0.000 0 1.000
0.314 18 0.951
0.628 36 0.809
0.942 54 0.588
1.256 72 0.310
1.570 90 0.001
1.884 108 -0.308
2.198 126 -0.587
2.512 144 -0.808
2.826 162 -0.951
3.140 180 -1.000
1
−1
π
3
2π
3 π
x
y
b b
b
b
b
b
b
b
b
b b
cos x
1
−1
π
3
2π
3 π
x
y
b b
b
b
b
b
b
b
b
b
b
cos x
First we will plot the graph in dotted form taking ordered pairs (x, cos x)
for each row, in which x represents to abscissa in x-axis and cos x represents to
ordinate in y-axis. This plot is shown in first part of above figure. After that
we draw a smooth hand curve as shown in second part of the above figure.
Solved Problem 0.20 Plot the trigonometric function y = sin(θ + π/4) within
−π ≤ θ ≤ π.
Solution In the given function y = sin(θ + π/4), π/4 is phase. It means,
when measured angle θ is zero, argument of sine operator is not zero but argu-
ment starts with π/4. Thus the argument of sine operator is shifted by angle
π/4. To plot the function y = sin(θ + π/4), we shall construct a data table, in

0.1. FUNCTION 19
which we shall take radian values of θ within given domain, i.e. −π ≤ θ ≤ π,
and find their corresponding sine values. Note that, here we can’t take degree
values of θ as phase value is given in radian unit. If you want to take degree
values of θ, first convert the phase into degree unit. Remember the sum rule for
different units. Now, we will plot the graph between θ and sin(θ + π/4). The
tabulated form of data table is given below:
θ θ + π/4 y
-3.140 -2.355 -0.708
-2.512 -1.727 -0.988
-1.884 -1.099 -0.891
-1.256 -0.471 -0.454
-0.628 0.157 0.156
0.000 0.785 0.707
0.628 1.413 0.988
1.256 2.041 0.891
1.884 2.669 0.455
2.512 3.297 -0.155
3.140 3.925 -0.706
1
−1
π
3
2π
3 π
−π
3
−2π
3
−π
x
y
b
b
b
b
b
b
b
b
b
b
b
y = sin(θ + π/4)
First we will plot the graph in dotted form taking ordered pairs (θ, y) for
each row, in which θ represents to abscissa in x-axis and y represents to ordinate
in y-axis. After that we draw a smooth hand curve as shown in above figure.
Solved Problem 0.21 Two trigonometric basic sine functions are y1 and y2.
Function y2 leads to y1 by a phase of 120◦
. Plot these two functions in same
graph, within −π ≤ θ ≤ π.
Solution Here, y1 and y2 are basic sine functions, i.e.
y1 = sin θ; y2 = sin θ

20
But, according to the question, y2 leads to y1 by a phase 120◦
. Note that, we are
taking θ in radian, hence the phase should also converted into radian equivalent.
φ =
π
180
× 120 =
2π
3
Now, the trigonometric functions are
y1 = sin θ; y2 = sin(θ + 2π/3)
To plots these functions, we shall construct a data table, in which we shall
take radian values of θ within given domain, i.e. −π ≤ θ ≤ π, and find their
corresponding function values as shown in the below table:
θ y1 y2
-3.140 -0.000 -0.866
-2.512 -0.589 -0.407
-1.884 -0.951 0.208
-1.256 -0.951 0.743
-0.628 -0.588 0.994
0.000 0.000 0.867
0.628 0.588 0.408
1.256 0.951 -0.206
1.884 0.951 -0.742
2.512 0.589 -0.994
3.140 0.000 -0.867
1
−1
π
3
2π
3 π
−π
3
−2π
3
−π
x
y
b
b
b b
b
b
b
b b
b
b
b
b
b
b
b
b
b
b
b
b
b
y1 = sin(θ)
y2 = sin(θ + 2π/3)
First we will plot the graph in dotted form taking ordered pairs (θ, y1) and
(θ, y2) for each row, in which θ represents to abscissa in x-axis and y1 or y2
represents to ordinates in y-axis. After that we draw a smooth hand curve as

0.1. FUNCTION 21
shown in above figure. There may be alternate functions like:
y1 = sin(θ − 2π/3); y2 = sin θ
θ y1 y2
-3.140 0.867 -0.000
-2.512 0.994 -0.589
-1.884 0.742 -0.951
-1.256 0.206 -0.951
-0.628 -0.408 -0.588
0.000 -0.867 0.000
0.628 -0.994 0.588
1.256 -0.743 0.951
1.884 -0.208 0.951
2.512 0.407 0.589
3.140 0.866 0.000
And the corresponding plot is
1
−1
π
3
2π
3 π
−π
3
−2π
3
−π
x
y
b
b
b
b
b
b
b
b
b
b
b
b
b
b b
b
b
b
b b
b
b
y1 = sin(θ − 2π/3) y2 = sin(θ)
This is the desired plot.
Solved Problem 0.22 Two trigonometric sine functions are y1 and y2. Function
y2 leads to y1 by a phase of 60◦
. Plot these two functions in same graph, within
−π ≤ θ ≤ π. Find the function value of y2 when y1 is at θ = 0◦
.
Solution Here, y1 and y2 are basic sine functions, i.e.
y1 = sin θ; y2 = sin θ
But, according to the question, y2 leads to y1 by a phase 60◦
. Note that, we are

22
taking θ in radian, hence the phase should also converted into radian equivalent.
φ =
π
180
× 60 =
π
3
Now, the trigonometric functions are
y1 = sin θ; y2 = sin(θ + π/3)
To plots these functions, we shall construct a data table, in which we shall
take radian values of θ within given domain, i.e. −π ≤ θ ≤ π, and find their
corresponding function values as shown in the below table:
θ y1 y2
-3.140 -0.000 -0.867
-2.512 -0.589 -0.994
-1.884 -0.951 -0.743
-1.256 -0.951 -0.207
-0.628 -0.588 0.407
0.000 0.000 0.866
0.628 0.588 0.995
1.256 0.951 0.744
1.884 0.951 0.209
2.512 0.589 -0.405
3.140 0.000 -0.865
Now plot the ordered pairs (θ, y1) and (θ, y2) of each row as shown in the
below figure.
1
−1
π
3
2π
3 π
−π
3
−2π
3
−π
x
y
b
b
b b
b
b
b
b b
b
b
b
b
b
b
b
b
b
b
b
b
b
y1 = sin(θ)
y2 = sin(θ + π/3)
Now, we have to find the function value of y2 when y1 is at θ = 0◦
, i.e. y2
at θ = 0◦
. The value of y2 = 0.866 as seen in the above plot.

0.1. FUNCTION 23
Logarithmic Function
A logarithmic function contains logarithmic function of any base, like ln, log
or exponential of base ‘e’. For example, f(x) = ln x + ex
+ c is a logarithmic
function.
Integral Function
A function that represents a polynomial equation that has only positive integral
powers to its independent variable, say t or x or θ, is called an integral function
of t or x or θ respectively. For example f(x) = 2 − x + x2
− 4x3
is an integral
function of x.
Rational Fraction
A rational fraction of a variable t (say), is a fraction whose numerator and
denominators are rational functions of the same variable t. For example
f(x) =
t2
− 4
t3 − 2
is a rational fraction. Again, note that, the degree of numerator in a rational
fraction should not be equal to or greater than the degree of its denominator.
Rational Function
A rational function of a variable x, is that algebraic function which has neither
single variable nor fractional powers. For example
f(x) = 2 − x +
x2
− 4
x3 − 2
is a rational function. While
f(x) = 2 − x +
√
x − 4
x3 − 2
is not a rational function. Again, note that, the degree of numerator should not
equal to or greater than the degree of its denominator in its fractional part, i.e.
f(x) = 2 − x +
x4
− 4
x3 − 2
In this case, first we decrease degree of numerator by simplifying it.
f(x) = 2 − x + x +
2x − 4
x3 − 2
= 2 +
2x − 4
x3 − 2
Now this is rational function.

24
Explicit Implicit Function
Explicit functions are those functions, in which dependent variable is directly
related to the independent variable. For example, in y = x2
−4x−3, y is explicit
function of x, as y is directly related to x. Values of y can be found by solving
this relation. In implicit functions, dependent variable is not directly related to
the independent variable. For example, in the group of equations
y = x2
− 4x − 3; y + sin(y) = x
y is implicit function of x. Here, y is directly related to x. Values of y can no
be found by just solving these relations.
0.1.10 Unique Values of Function
A function gives one unique value of one or more different input values or one or
more unique outputs of one input value. It depends on the function. A function
which has root part, may give two or more unique outputs which function free
from root parts gives one unique output for two or more inputs.
Solved Problem 0.23 Find the unique outputs of the function y = x2
− 1 for
x ∈ {2, 3}.
Solution As x ∈ {2, 3}, hence, the unique values of x shall be only 2 and
3. Therefore, unique value of the function y = x2
− 1 are y = 3 at x = 2 and
y = 8 at x = 3 respectively. These unique values of the function are obtained
by just substituting the value of x.
Solved Problem 0.24 Find the unique outputs of the function y =
p
x2 + 2x + 1
for x ∈ {2, 3}.
Solution As x ∈ {2, 3}, hence, the unique values of x shall be only 2 and
3. Therefore, unique value of the function y =
p
x2 + 2x + 1 are y = 3 at
x = 2 and y = 4 at x = 3 respectively. These unique values of the function are
obtained by just substituting the value of x.
0.1.11 Monotone Function
A function which always increases or decreases can be called monotonically
increasing or monotonic decreasing function. If x1 x2 then f(x1) f(x2) is
abide for monotonically increasing function in the domain of x. A monotonic
function is strictly increasing or strictly decreasing function if f(x1) f(x2)
and f(x1) f(x2) respectively for values x1, x2 in domain of x.

0.1. FUNCTION 25
0.1.12 Odd Even Function
Odd Function A function is said to be odd function if it becomes negative
when its independent variable is substituted by negative independent variable.
For example if f(x) is a function of x then it said to be odd function if
f(x) = f(−x) = −f(x) (1)
Odd functions do not plot symmetrical plots about any axis, this is why, it is
also called anti-symmetric function.
Even Function A function is said to be even function if it remains same
when its independent variable is substituted by negative independent variable.
For example if f(x) is a function of x then it said to be even function if
f(x) = f(−x) = f(x) (2)
Even functions are plotted symmetrical about the axes, this is why, it is also
called symmetric function.
Properties of Odd-Even Functions
I. Product of odd and even function is an odd function. For example, if
f(x) and g(x) are odd and even function respectively, then F(x) = f(x) ∗ g(x)
is an odd function.
II. Product of two odd or two even function is an even function. For example,
if f(x) and g(x) are either both odd or both even functions respectively, then
F(x) = f(x) ∗ g(x) is an even function.
III. If f(x) is a function of x then it can be transformed into its odd and
even parts. Function f(x) can be written as its odd and even parts by
f(x)o =
f(x) − f(−x)
2
(3)
and
f(x)e =
f(x) + f(−x)
2
(4)
IV. An algebraic function is sum of its odd and even parts. ie
f(x) = f(x)o + f(x)e (5)
An even part of the pure linear algebraic function has only algebraic terms
of even degree while odd part of the pure linear algebraic function has only
algebraic terms odd degrees. But this case is not true for fractions and trigono-
metric functions.

26
Solved Problem 0.25 Show that y = sin(x) is an odd function.
Solution From the definition of the odd function, independent variable ‘x’
is replaced by ‘-x’. Now
y = sin(−x) = − sin(x)
Or
− sin(x) ⇒ −y
Hence it is an odd function.
x
y
sin x
x
y
− sin x
Solved Problem 0.26 Show that y = cos(x) is an odd function.
y = cos(−x) = cos(x)
Or
cos(x) ⇒ y
Hence it is an odd function.
x
y
cos x
x
y
− cos x
Solved Problem 0.27 Check whether y = x2
− 2x is an odd function.
y = (−x)2
− 2(−x) = x2
+ 2x
Or
x2
+ 2x ; y ; −y

0.1. FUNCTION 27
Hence it is neither odd nor even function.
x
y
y = x2
− 2x
x
y
y = x2
+ 2x
Solved Problem 0.28 Check whether function y = 10 − 6x − x2
is odd or even.
Solution To check the function being odd or even, replace ‘x’ by ‘-x’ in the
given function y = 10 − 6x − x2
.
y = 10 − 6(−x) − (−x)2
= 10 + 6x − x2
Now
10 + 6x − x2
; y ; −y
Hence it is neither odd nor even function.
Solved Problem 0.29 Check whether function y = 9x + x3
is odd or even.
given function y = 9x + x3
.
y = 9(−x) + (−x)3
= −9x − x3
Now
− 9x + x3

⇒ −y
Hence it is odd function.
Solved Problem 0.30 Check whether function y = x4
− 6x2
− 9 is odd or even.
given function y = x4
− 6x2
− 9.
y = (−x)4
− 6(−x)2
− 9 = x4
− 6x2
− 9
Now
x4
− 6x2
− 9 ⇒ y
Hence it is even function.

28
Solved Problem 0.31 Check whether function y = sin(x) cos(x) is odd or even.
given function y = sin(x) cos(x).
y = sin(−x) cos(−x) = − sin(x) cos(x)
Now
− sin(x) cos(x) ⇒ −y
Hence it is odd function.
Solved Problem 0.32 Transform the functions y = x4
− 2x3
+ x + 4 in their odd
or even parts.
Solution To transform the function into odd and even form, relation
f(x)o =
f(x) − f(−x)
2
and
f(x)e =
f(x) + f(−x)
2
are used. Now, odd part of the function is
yo =

x4
− 2x3
+ x + 4

−

(−x)4
− 2(−x)3
+ (−x) + 4

2
It gives
yo =
x4
− 2x3
+ x + 4

− x4
+ 2x3
− x + 4

2
yo =
x4
− 2x3
+ x + 4 − x4
− 2x3
+ x − 4
2
yo = −2x3
+ x
Ans-1.
Again for even part of the function is
ye =

x4
− 2x3
+ x + 4

+

(−x)4
− 2(−x)3
+ (−x) + 4

2
It gives
ye =
x4
− 2x3
+ x + 4

+ x4
+ 2x3
− x + 4

2
ye =
x4
− 2x3
+ x + 4 + x4
+ 2x3
− x + 4
2
ye = x4
+ 4

0.1. FUNCTION 29
− 5x2
− x + 7 in their odd
or even parts.
Solution Odd and even parts of the given function are
yo =
y(x) − y(−x)
2
and
ye =
y(x) + y(−x)
2
Hence odd part of the function is
yo =
(x4
− 5x2
− x + 7) − [(−x)4
− 5(−x)2
− (−x) + 7]
2
yo =
x4
− 5x2
− x + 7 − x4
+ 5x2
− x − 7]
2
yo = −x
Ans-1.
We know that a function can be written as the sum of its odd and even
parts. It means the even part of the function is subtraction of the function by
odd part. Hence
ye = x4
− 5x2
+ 7
− 2x7
+ x4
+ 5 in its odd
and even parts.
Solution Using the direct method, odd part of the function is
yo = −2x7
and even part of the function is
ye = x8
+ x4
+ 5
− 2x7
− x3
+ x4
+ 5 in its
odd and even parts.
yo = −2x7
− x3
ye = x8
+ x4
+ 5

30
Solved Problem 0.36 Express f(x) into its odd and even parts. Function f(x)
is given by
f(x) = x4
− 2x3
+ x2
− 5x + 7
Also show that the function is sum of its odd and even parts.
Solution We know that the odd part of the function can be found as
f(x)o =
f(x) − f(−x)
2
f(x)o =

x4
− 2x3
+ x2
− 5x + 7

−

(−x)4
− 2(−x)3
+ (−x)2
− 5(−x) + 7

2
=

x4
− 2x3
+ x2
− 5x + 7 − x4
− 2x3
− x2
− 5x − 7

2
= −2x3
− 5x
Clearly f(x)o is an odd part of the given function and contains some of the
terms of the given function f(x). Similarly even part of the function can be
found by using relation
f(x)e =
f(x) + f(−x)
2
f(x)e =

x4
− 2x3
+ x2
− 5x + 7

+

(−x)4
− 2(−x)3
+ (−x)2
− 5(−x) + 7

2
=

x4
− 2x3
+ x2
− 5x + 7 + x4
+ 2x3
+ x2
+ 5x + 7

2
= x4
+ x2
+ 7
f(x)e is an even function and contains remaining terms of the given function
f(x). Now the function f(x) can be written as
f(x) = f(x)o + f(x)e
Solved Problem 0.37 Show that f(x) = x3
is a one-to-one function.
Solution If f(x) = x3
is an one to one function then the range of f(x) at
two distinct domain integer values must not be equal. For real number assume
ranges of function for two distinct integers are same. Substituting the two
domain values x1 and x2, where x1 6= x2 and equalling ranges
f(x1) = f(x2)
f(x1) − f(x2) = 0

0.1. FUNCTION 31
Substituting the function values
x3
1 − x3
2 = 0
(x1 − x2)(x2
1 + x2
2 + x1x2) = 0
Here x1 − x2 6= 0 as x1 6= x2. And (x2
1 + x2
2 + x1x2) = 0 gives
x1 =
−x2 ±
p
x2
2 − 4x2
2
2
x1 =
−x2 ±
p
−3x2
2
2
Here x1 is a complex number. Hence ranges of function are same when one
domain value is integer and other is complex number. Hence function f(x) has
no two same ranges f(x1) 6= f(x2) for two domain integer values x1 6= x2. Hence
function is one-to-one.
Solved Problem 0.38 Find the inverse of f(x) = x3
.
Solution Let y = x3
and for x, x = y1/3
which is unique solution. It is a
function of y and can be represented by
g(y) = y1/3
And transforming it for x
g(x) = x1/3
This is the inverse of the function f(x) and written as
f−1
(x) = x1/3
The graph is shown in figure (??).
1
−1
1 2 3
−1
−2
x
y
y = x3
y−1
= x1/3
Figure 4: Inverse of x3

32
Solved Problem 0.39 Separate the function f(x) = x2
− 12x − 4 into its odd
and even parts.
yo = −12x
ye = x2
− 4
Solved Problem 0.40 Separate the function f(x) = x4
− 3x3
− 8 into its odd
and even parts.
yo = −3x3
ye = x4
− 8
Solved Problem 0.41 Separate the function f(x) =
x3
− 1
x + 1
into its odd and even
parts.
Solution To transform the function into odd and even form, relation
f(x)o =
f(x) − f(−x)
2
and
f(x)e =
f(x) + f(−x)
2
are used. Now, odd part of the function is
yo =
h
x3
−1
x+1
i
−
h
(−x)3
−1
(−x)+1
i
2
It gives
yo =
x3
−1
x+1 − −x3
−1
−x+1
2
yo =
x3
+ x
1 − x2
Ans-1.

0.1. FUNCTION 33
Again for even part of the function is
ye =
h
x3
−1
x+1
i
+
h
(−x)3
−1
(−x)+1
i
2
It gives
ye =
x3
−1
x+1 + −x3
−1
−x+1
2
ye =
−x4
− 1
1 − x2
0.1.13 Homogeneous Function
A homogeneous function is function in which value of the function is scaled by
a certain factor then function is itself scaled to that factor. This property of the
function is called homogeneity (scaled everywhere not in portion of function).
A function f(x) is said to be homogeneous if
f(Cx) = Cf(x)
Why is this so? It is because, if we scaled the independent variable of a function
f(x), by a factor C, then whole function must be scaled to the same factor
C not a portion of the function. For example, we have a function f(x) = x.
Now, x is scaled by a factor 0.5 and the function becomes f(0.5x) = 0.5x. The
function graph of f(0.5x) must be scaled in same manner for domain of x.
1
2
3
1 2 3
x
f
f(x)
1
2
3
1 2 3
x
f
f(x)
f(0.5x)
In above figure, it is clear that f(0.5x) is half of the f(x) everywhere. This
is why, f(x) is homogeneous function. In other words, homogeneous function
does not has constant term.
Solved Problem 0.42 Write a homogeneous function which exhibits the proper-
ties of homogeneity.

34
Solution The function shows this property is f(x) = ax, where a is any
constant. Replacing x → cx, the result is
f(cx) = a × cx
Taking c as common in right hand side
f(cx) = c(ax) = cf(x)
Hence the given function is homogeneous.
Solved Problem 0.43 Show that f(x) = ax + b does not show homogeneity.
Solution The given function is f(x) = ax+b. Replacing x → cx, the result
is
f(cx) = a × cx + b
Taking c as common in right hand side
f(cx) = c

ax +
b
c

6= cf(x)
Hence the given function is not homogeneous. To prove this numerically, take
a function f(x) = x + 0.5 and x is scaled by 0.5x. Now the scaled function
is f(0.5x) = 0.5x + 0.5. These two functions are graphed and they are given
below:
1
2
3
1 2 3
−1
x
f
f(x)
1
2
3
1 2 3
−1
x
f
f(x)
f(0.5x)
In above figure, it is clear that f(0.5x) is NOT half of the f(x) everywhere.
This is why, f(x) is NOT homogeneous function. We can prove it by taking
x = −0.5 where f(x) = 0 but f(0.5x) = 0.25 and f(x) 6= 2f(0.5x).
0.1.14 Linear Non Linear Functions
A function is said to be linear, if it satisfy the additive and homogeneity of a
function. If f(x1) and f(x2) are two functions of variables x1 and x2 respectively

0.1. FUNCTION 35
and a and b are scaling constants then those functions which satisfy following
two conditions will be said linear functions.
f(x1 + x2) = f(x1) + f(x2)
and
f(ax1 + bx2) = f(ax1) + f(bx2) = af(x1) + bf(x2)
This condition of additivity and homogeneity is called superposition. If the
condition of superposition is met for a function then function is linear otherwise
it will said non linear.
Solved Problem 0.44 Show that function f(x) = x is a linear function.
Solution The given function is f(x) = x. Let two points are x1 and x2 for
which f(x1) = x1 and f(x2) = x2. Now, from the condition of linearity of the
function
f(x1 + x2) = x1 + x2 = f(x1) + f(x2)
Choose two constants m and n (arbitrary) then
f(mx1 + nx2) = mx1 + nx2 = mf(x1) + nf(x2)
Two conditions for linearity of function are met, hence the function is linear
function.
Solved Problem 0.45 Show that function f(x) = x2
is NOT a linear function.
Solution The given function is f(x) = x2
. Let two points are x1 and x2
for which f(x1) = x2
1 and f(x2) = x2
2. Now, from the condition of linearity of
the function
f(x1 + x2) = (x1 + x2)2
= x2
1 + x2
2 + 2x1x2 6= f(x1) + f(x2)
As first condition is not met, hence given function is NOT a linear function.
0.1.15 Time Variant Time Invariant Functions
Time variant functions are directly dependent to the time. For time variant
functions
∂ft
∂t
= x(t) 6= 0
Time invariant function are not directly dependent to the time. For time in-
variant functions
∂ft
∂t
= 0
For example, let f(t) = x(t) is a function. At t = 2s input value is x(2) at
time t = 2s and corresponding output at time t = 2s is y(2). Hence output is
directly dependent to inputs at an instant of time t. Therefore, this function is
time invariant (time independent or function not varies with time).

36
If this function is modified as f(t) = t x(t). At t = 2s input value is x(2) at
time t = 2s and corresponding output is y(2) at time t = 2s. But the same time,
input is scaled by 2 at t = 2s (function scaling depends on the time instant,
hence function varies with time). Again if we change the time t = 3s then
y(3) = 3 x(3)
Now, input value is scaled by 3 at t = 3s. It means function is scaling to time.
Hence this function is time variant (time dependent or function varies with
time). Simply saying that if variable is a scaling factor then function is said to
be time variant. Similarly, in invariant function, variable is function parameter
(argument of function).
Solved Problem 0.46 Show that whether f(x) = sin(x) is time variant or time
invariant.
Solution The function f(x) = sin(x) depends only on the instant value of
x and x is parameter (argument) of sin function. For example at x = 2, function
value is
f(2) = sin(2)
It means f(2) is function value at x = 2. Here, x is not scaling factor. Hence it
is time invariant function.
0.1.16 Function Delay
Let we have a time function as
f(t) = 2t
Where t ∈ [−5, 5]. Now function values in the given time interval are

0.1. FUNCTION 37
t f(t)
-5 -10
-4 -8
-3 -6
-2 -4
-1 -2
0 0
1 2
2 4
3 6
4 8
5 10
Table 8: Data values for f(t).
Now shift the time by 1 as t → t + 1, the function becomes
f(t + 1) = 2t + 2
Now the tabulated function values for the given time are
t f(t) f(t + 1)
-5 -10 -6
-4 -8 -4
-3 -6 -2
-2 -4 0
-1 -2 2
0 0 4
1 2 6
2 4 8
3 6 10
4 8 12
5 10 -
Table 9: Data values for f(t) and f(t + 1).

38
Comparing tables ?? and ??, we find that same function values of f(t) and
f(t + 1) are obtained at different time events. f(t + 1) is time lagging to the
f(t) for a particular function value (take 0 here). Hence f(t + 1) is delays to
the f(t).
0.1.17 Function Forward
Let we have a time function as
f(t) = 2t
Where t ∈ [−5, 5]. Now function values in the given time interval are
t f(t)
-5 -10
-4 -8
-3 -6
-2 -4
-1 -2
0 0
1 2
2 4
3 6
4 8
5 10
Table 10: Data values for f(t).
Now shift the time by 1 as t → t − 1, the function becomes
f(t − 1) = 2t − 2
Now the tabulated function values for the given time are

0.1. FUNCTION 39
t f(t) f(t + 1)
-5 -10 -
-4 -8 -12
-3 -6 -10
-2 -4 -8
-1 -2 -6
0 0 -4
1 2 -2
2 4 0
3 6 2
4 8 4
5 10 6
Table 11: Data values for f(t) and f(t − 1).
Comparing tables ?? and ??, we find that same function values of f(t) and
f(t − 1) are obtained at different time events. f(t − 1) is time leading to the
f(t) for a particular function value (take 0 here). Hence f(t − 1) is advanced to
the f(t).
0.1.18 Static Dynamic Functions
Output of the static function always depends on the current input of the func-
tion. For example,
f(x) = sin(x)
is a static function as function value at x = 2 output at x = 2.
y(2) = sin(2)
In case of time functions like
y(t) = e−t
also called static function as output at t = 9s is equal to the value of e−t
at
t = 9s. i.e.
y(9) = e−9
A dynamic function depends on the current input value as well as the previous
inputs. For example
y(t) = e−t
+ sin(t − 2)
is a dynamic function. At t = 9s, output is
y(9) = e−9
+ sin(9 − 2)

40
Or
y(9) = e−9
+ sin(7)
If t is time variable then output y depends on the value of e−t
at t = 9s
(current value) and on the value of sine function at t = 7s (previous value).
For dynamic functions, it is require to memorise the previous inputs to find
the current output. This is why, dynamic functions are also called memory
functions. Functions like
y(t) = y(t − 4) + x(t)
is also a dynamic function. If we take t = 6 then
y(6) = y(2) + x(6)
Here, current value of function, y(6), also depends on the previous function
value, i.e. y(2).
0.1.19 Causal Non Causal Functions
A function is said to be causal function (also called as non-anticipatory - means
which can not be predicted) if its output depends upon present and past inputs
and not on future inputs. For example,
y(t) = e−t
+ sin(t − 2)
is causal function. A function is said to be non-causal function (also called as
anticipatory or predictable function) if it depends on the future inputs also. For
example
y′
(t) = e−t
+ sin(t − 2) + sin(t + 2)
is non-causal function as sine value at t + 2 (future time) is future value. Note
that y(t) and y′
(t) both are functions of time and both get same time inputs
without any time shifting. The time argument in right hand sides are computed
time instances of function terms rather than function’s time shifting. Again, if
there were functions as y(t) and y′
(t−1) then there is a time shifting in y′
(t−1).
Because, at t = 4 second, in function y(t) time input for right hand side terms
will be 4 second while in function y′
(t − 1), time input for right hand side terms
will be 3 second.
0.1.20 Stable Non Stable Functions
A function is said to be stable (have definite limit values as bounded limits, i.e.
converging function) if its output is bounded for bounded inputs. If function
output is unbounded then function is unstable (i.e. diverging). For example
f(x) = e−x

0.1. FUNCTION 41
is stable function in [0, ∞) as at any point x ∈ [0, ∞), e−x
is finite. While
f(x) = ex
is unstable function in [0, ∞) as x → ∞, ex
→ ∞.
Solved Problem 0.47 Show that function f(x) = x2
− 4x is unstable function.
Solution A function is said to be unstable function if it has no bounded
limit at x → ∞. So,
lim
x→∞
f(x) = lim
x→∞
x2
− 4x

= ∞
It means this function is unstable function.
Solved Problem 0.48 Show that function f(x) =
x2
− 4
x2
is a stable function.
Solution A function is said to be stable function if it has bounded limit at
x → ∞. So,
lim
x→∞
f(x) = lim
x→∞

x2
− 4
x2

= lim
x→∞

1 −
4
x2

= 1
It means this function is stable function.
0.1.21 Inverse Function
If a function f(x) with domain Df and range Rf is one-to-one, then there is an
inverse of function f(x) with domain Rf and range Df for each values of x in
Df , i.e.
Ψ[f(x)] = x
Generally Ψ is represented by f−1
and it is called inverse of function f(x).
Solved Problem 0.49 Find the inverse of the function f(x) = 3x + 4.
Solution Here f(x) is a function of x. The inverse of the function is
obtained by solving this equation for x. It gives
3x = f(x) − 4
Or
x =
f(x) − 4
3
‘x’ represents the inverse of function and f(x) is the independent variable for
this inverse function ‘x’. Now transforming this relation for easy explanation
and readability
f−1
(X) =
X − 4
3
It is inverse function of the given function f(x).

42
Solved Problem 0.50 Find the inverse of the function f(x) = x2
+ 4.
Solution Here f(x) is a function of x. The inverse of the function is
obtained by solving this equation for x. It gives
x2
= f(x) − 4
Or
x =
p
f(x) − 4
‘x’ represents the inverse of function and f(x) is the independent variable for
this inverse function ‘x’. Now transforming this relation for easy explanation
and readability
f−1
(x) =
√
x − 4
It is inverse function of the given function f(x).
0.1.22 Sampling of Function
Value of a function at a point is called function value. It is also known as
sampled value of the function at that point. For example, function y = sin(x)
has sample value y = sin(90◦
), i.e. ‘1’ at x = 90◦
. Sampling of function is base
of calculus and communication engineering. Mathematically, a sampled value
of a function, f(x), is given by
y[x] = f(x) (6)
Solved Problem 0.51 Find the four sampled values of the function y = x2
within
0 ≤ x ≤ 1.
Solution The four points within 0 ≤ x ≤ 1 are 0.0, 0.3, 0.6 and 1.0. The
function values at these points are
y[0] = y0 = 02
= 0.00
y[0.3] = y0.3 = 0.32
= 0.09
y[0.6] = y0.6 = 0.62
= 0.36
y[1] = y1 = 12
= 1.00
These are four sampled values of the function.
1
x
y
b
b
b
b

0.1. FUNCTION 43
Solved Problem 0.52 Find the three sampled values of the function y = x2
−
3x + 1 within 0 x 4.
Solution Three auxiliary points taken within 0 x 5 are 1, 2 and 3.
The function values at these points are
y[1] = y1 = 12
− 3 × 1 + 1 = −1
y[2] = y2 = 22
− 3 × 2 + 1 = −1
y[3] = y3 = 32
− 3 × 3 + 1 = 1
These are three sampled values of the function.
−1
1 2 3
x
y
b b
b
Solved Problem 0.53 Extract four sampled values of the function y = sin(x)+1.
Solution The five points are 0◦
, 15◦
, 30◦
, 45◦
and 60◦
in degree abscissa.
The function values at these points are
y[0◦
] = y0◦ = sin(0◦
) + 1 = 1.00
y[15◦
] = y15◦ = sin(15◦
) + 1 = 1.26
y[30◦
] = y30◦ = sin(30◦
) + 1 = 1.50
y[45◦
] = y45◦ = sin(45◦
) + 1 = 1.70
y[60◦
] = y60◦ = sin(60◦
) + 1 = 1.87
These are four sampled values of the function.
1
15 30 45 60
x
y
b
b
b
b
b

44
0.1.23 Piece-wise Function
A function which satisfy different equations in different region of independent
variable, is called piece-wise function. See the following function,
f(x) =
(
px + q where a ≤ x ≤ b
rx where b x ≤ c
Above function is a piece-wise function as it uses relation px + q in domain
a ≤ x ≤ b and relation rx in domain b x ≤ c.
Existence of Piece-wise Function A function f(x) has a unique value at any
given point whatever method of solution is applied. For example, f(x) = x2
− 2
has unique value at x = 2, and it is f(2) = 22
−2 = 2. In a piece-wise functions,
if break point (say x = a) is applicable for all pieces of the function, then the
values of pieces of the function at that point shall be same. Consider a piece-wise
function as given below:
f(x) =
(
2x + 2 where 0 ≤ x ≤ 2
3x where 2 ≤ x ≤ 4
Here, x = 2 is applicable for both pieces of the function. So, f(x) = 2x + 2 and
f(x) = 3x shall give same values at x = 2. Here,
f(2) = 2 × 2 + 2 = 6
and
f(2) = 3 × 2 = 6
Hence the function exists. Again, if this function is modified as
f(x) =
(
2x + 2 where 0 ≤ x ≤ 2
3x + 1 where 2 ≤ x ≤ 4
then f(x) = 2x + 2 and f(x) = 3x + 1 do not give same values at x = 2.
f(2) = 2 × 2 + 2 = 6
and
f(2) = 3 × 2 + 1 = 7
Therefore, it is not a function.
Solved Problem 0.54 Show that
f(x) =
(
2x + 2 where 0 ≤ x ≤ 3
3x where 3 ≤ x ≤ 4
is not a function. Give suitable explanation.

0.1. FUNCTION 45
Solution The piecewise relation is defined for 0 ≤ x ≤ 4. The piecewise
relation has two different expressions at the point x = 3. If this relation is perfect
function then it should have same values for its two different expressions. So
y(3) = 2 × 3 + 2 = 8
and
y(3) = 3 × 3 = 9
Here, both values are not same, hence it is not a piecewise function.
0.1.24 Discrete Sum Product of Function Values
Discrete sum of a function is sum of all function values taken at equidistant
points between the given limits. For example take function f(x) = x2
that is
defined in [0, 1]. Let we have 10 discrete points in [0, 1] (i.e [L, U], here L is
lower limit and U is upper limit) where function is evaluated and consequently
we shall get 10 corresponding function values. The equidistant width between
two consecutive x values is
h =
Upper Limit − Lower Limit
Number of Points − 1
And the points are
xi = Lower Limit + i × h
Where, i = 0 to i = 9 (total ten points). These ten points can be symbolised as
x0, x1, . . ., x9. Note that here our count is started from i = 0 not i = 1 so that
x0 must equal to 0 (lower limit) while using relation
xi = Lower Limit + i × h
Now, we can compute h as
h =
1 − 0
9
≈ 0.111
It shall give x0 = 0, x1 = 0.111, x2 = 0.222, . . ., x9 = 0.999 ≈ 1. In natural
count n started from 1 and onward, we have first discrete function value is at
f(x0), second discrete function value is at f(x1), . . . and tenth discrete function
value is at f(x9). Finally, we are ready to construct tabulated data as given
below:

46
n xi f(xi) f[n]
1 0.000 0.000 0.000
2 0.111 0.012 0.012
3 0.222 0.049 0.049
4 0.333 0.111 0.111
5 0.444 0.197 0.197
6 0.555 0.308 0.308
7 0.666 0.444 0.444
8 0.777 0.604 0.604
9 0.888 0.789 0.789
10 1.000 1.000 1.000
Here, n is started from 1 to 10 and you should not confuse between counting
of i and n. Now, we have 10 discrete values.
Discrete Sum of Function Values Sum of these values can be written as
Sn =
10
X
n=1
f[n] (7)
In ‘i’ counting
Si =
9
X
i=0
f[xi] (8)
Both equations (??) and (??) are called discrete sum of function.
Discrete Product of Function Values Product of these values can be written
as
Pn =
10
Y
n=1
f[n] (9)
In ‘i’ counting
Pi =
9
Y
i=0
f[xi] (10)
Both equations (??) and (??) are called discrete product of function.
0.1.25 Probability Density Function
As we knew that, probability of an event and probability of all events is always
positive and lies within [0, 1], irrespective of times the event is repeated. A func-
tion that represents to probabilities of events E is called probability function.

0.1. FUNCTION 47
A function is said to be Probability Density Function, if total area covered by
probability function and horizontal axis is in [0, 1]. Probability Density Function
(PDF) must obey two conditions:
1. Function should be greater than zero, i.e. f(x) 0. It means, if function
is plotted in xy-plane, then function graph should be above x-axis, i.e. in
+y-axis.
1
1 2 3 4
−1
−2
−3
−4
x
y
f(x) = e−x2
2. If the range of continuous random variable is (a, b), then function integral
(say area) for this range should be in [0, 1]. So,
0 ≤
b
Z
a
f(x) dx ≤ 1
1
1 2 3 4
−1
−2
−3
−4
x
y
f(x) = e−x2
Here, integral domain, i.e. (a, b) (lower and upper limits) may be definite
numbers or it may be infinite. This depends on the observation about
function.
For infinite continuous Probability Density Function (pdf) function, its integral
(or area between function and horizontal axis) should be in [0, 1]. A non pdf
function can also converted into pdf function by normalising it. For example,
let integral of a function f(x) in (−∞, ∞) is k. If this function is divided by
k then its integral never be greater than 1 for any limits of integration. So,
function
f(x)
k
becomes pdf function in (−∞, ∞).
Solved Problem 0.55 Show that function f(x) = e−x2
is a probability density
function.
Solution A function is said to be probability density function if f(x) 0
and its integral in whole range, say (−∞, ∞) is within [0, 1]. Now, for first
condition we check whether the function is negative.

48
1
1 2 3
−1
−2
−3
x
y
f(x) = e−x2
As we see that whatever is the value of x, either it is positive or it is negative,
argument of exponential is always negative. From the series expansion, e−x2
=
1/ex2
0. So, first condition is satisfied. Now, for second condition,
∞
Z
−∞
e−x2
dx = 1.77 1
Hence, it is not a probability density function.
Solved Problem 0.56 Show that function f(x) =
1
2
e−x2
is a probability density
function.
Solution A function is said to be probability density function if f(x) 0
and its integral in whole range, say (−∞, ∞) is within [0, 1]. Now, for first
condition we check whether the function is negative.
1
1 2 3
−1
−2
−3
x
y
f(x) =
1
2
e−x2
As we see that whatever is the value of x, either it is positive or it is negative,
argument of exponential is always negative. From the series expansion, 0.5 ×
e−x2
= 0.5/ex2
0. So, first condition is satisfied. Now, for second condition,
∞
Z
−∞
1
2
e−x2
dx = 0.88 1
Hence, it is a probability density function.
Solved Problem 0.57 Convert f(x) = sin(x) into a probability density function
in local domain, i.e. 0 to π.

0.1. FUNCTION 49
Solution Function f(x) = sin(x) in domain [0, π], function has positive
shape. Now, taking integral of the function, we have
I =
π
Z
0
sin(x) dx
1
π
3
2π
3 π
x
f(x)
f(x) = sin(x)
I =
π
Z
0
sin(x) dx
on integral, it gives
I = 2 = k
This is normalisation factor (k) for the given function f(x) to make it probability
density function (pdf) F(x). So, pdf of the function f(x) will be found when it
is divided by factor k.
π
3
2π
3 π
x
F(x)
F (x) =
sin(x)
2
π
3
2π
3 π
x
P(x)
P (x)
Now, the probability density function (pdf) of given function will be
F(x) =
1
2
× sin(x) =
sin(x)
2
Using this pdf function, we can find the probability within an interval of x, say
in limit [π/3, 2π/3] as shown in second part of above figure.
P(x) =
2π/3
Z
π/3
sin(x)
2
dx

50
Solved Problem 0.58 Show that
e− x
λ
λ
is a probability density function (pdf)
function for x 0 and λ 0. Find whether it is also a pdf function or not in
−∞ x ∞.
Solution A function is said to be probability density function (pdf) function
for the given domain in which integral of this function is in [0, 1]. So,
P =
∞
Z
0
e− x
λ
λ
dx
Here x is positive and greater than zero. Limits of integration for x is in 0
x ∞.
x
y
f(x) =
e− x
λ
λ
x → ∞
P =
1
λ
× e− x
λ ×
1
−1/λ

∞
0
= 1
Hence this is pdf function for x 0. For limit of x in −∞ x ∞,
P =
∞
Z
−∞
e− x
λ
λ
dx
P =
1
λ
× e− x
λ ×
1
−1/λ

∞
−∞
≈ ∞
x
y
f(x) =
e− x
λ
λ
x → ∞
x → −∞

0.1. FUNCTION 51
Integral of the given function is not conversing in [0, 1] when limits of inte-
gration for x are −∞ x ∞. Hence for these limits of x, function is not a
pdf function.
0.1.26 Moment Generating Function
From the definition of moment of force, we knew that, if ~
F is force acting on
a body at distance ~
x from axis of rotation, then moment of force is given by
~
F × ~
r. Similarly, if f(x) is function at random variable x in respect of origin
(say x = 0), then its moment is given by
µ =
Z ∞
−∞
(x − 0) f(x) dx
Moment may be of any order. rth
order moment is define as
µr =
Z ∞
−∞
xr
f(x) dx
0.1.27 Unit Type Functions
Unit type functions are those functions whose final outcome is exactly one within
the given domain and zero outside the given domain. Using this property, there
are different unit functions which are given in the subsequent sections.
Unit Sample
Unit sample is used to get the instantaneous value of a function. It is also
known as sampling function. A function that is fast varying with time. To
get the “instantaneous” function value, very short width time so that we just
sample only one function value. This instantaneous time is called unit sample.
It is denoted by δ(n) and represented as
δ(n) =

1 when n = 0
0 Otherwise
(11)
20 40 60
t
y
b
f(a) b
f(b)
a b
In above figure, time width is large, therefore there are two function values
which are sampled. If we reduce width of time, we again get two closure function
values.

52
20 40 60
t
y
b
f(a)
b
f(b)
a b
If b − a → 0 then f(a) and f(b) will be coincide and assume that f(a) =
f(b) = 1 in this case. These conditions are fulfilled by unit sample.
20 40 60
t
y
b
f(a)
b
f(b)
a b
So, at point t = a (actually a is equal to 20 units of times), unit sample is
given by
1
20 40 60
t
δ(t)
b
1
Figure 5:
Time moves continue but we start sample counting at any instant of time (at
noon, at evening at 1030 Hrs etc). When we start sample counting, counting
is said to events of sampling and samples are counted as n = 0 (sample at
event zero), n = 1 (first sample at next event), n = 2 etc. Sample events are
also counted in backward direction as n = −1, n = −2 and so on to find the
sample just before one sampling event, just before two sampling events and
so on respectively. Sample counting used here is similar to the number line
(zero at origin, negative counts in left side and positive counts in right side in
−∞ → +∞). In continuation to the graph of figure (??), if we take n = 0 event
at time t = 20 then unit sample function will be as
1
20 40
t
δ(t)
b
1
1
0 20
n
δ(n)
b
1
Suppose we are taking infinite numbers of events where function samples

0.1. FUNCTION 53
are being taken. For unit sample function, we have only one sample at only
one instant of event. Assume we are taking unit sample at event that occurs at
n = 0 (at origin and for time symmetrical use at time t = 0) then unit sample
function will be plotted as given below:
1
5 10
−5
−10
t
δ(t)
b b b b b b b b b b
b
b b b b b b b b b b
−∞ ∞
1
5 10
−5
−10
n
δ(n)
b b b b b b b b b b
b
b b b b b b b b b b
−∞ ∞
Again if we are taking unit sample at event that occurs at t = 5 (other than
origin) then unit sample function will be plotted as given below:
1
5 10
−5
−10
t
δ(t)
b b b b b b b b b b b b b b b
b
b b b b b b
−∞ ∞
1
0 5
−10
−15
n
δ(n)
b
b b b b b b
−∞ ∞
Using this graph, we can conclude a relation of unit sample in time domain
and event domain (counts). We know that at time t = 5 unit sample is obtained.
Let it is zero counted unit sample. It shows that time is 5 unit ahead to n. To
keep both at origin, root of time is t = 5 ⇒ t − 5 = 0. So, the unit sample
relation between time and count domain is
δ(t − 5) = δ(n); where t ≥ 5, n ≥ 0 (12)
From above discussion, it is find out that there is “only one sample” of unit
sample function in infinite duration or infinite numbers of sampling events.
Unit sample is active for infinitely small (approximately zero) time period
and is unity for that period of time. Hence, unit sample is also known as
function sample switch. For example,
f(x) = x2
− 3
is continuous function in −∞ x ∞. When δ(x) is multiplied with function
f(x) then we have
f(x) × δ(x) = x2
− 3

× δ(x)
δ(x) = 1 at x = 0, so
f(x) × δ(x) = 02
− 3

× 1 = −3; x = 0

54
Note that when values of independent variable are taken from domain of
real numbers then function is said to be continuous type. Similarly, when
values of independent variable are taken from domain of integer numbers
then function is said to be discrete type. Discrete type function has “square
bracket” symbol. Here we have used “parentheses” and discrete values of
independent variable.
Solved Problem 0.59 Find the sum of unit samples of δ(t) function defined in
time domain as δ(t) = 1 at t = 0 for −∞ t 3.
Solution The unit sample function is defined as
δ(0) = 1
1
5
−5
−10
t
δ(t)
b b b b b b b b
b
b b b b b b b b
−∞ ∞
−∞ t 3 1
−5
−10
t
δ(t)
b b b b b b b b
b
b
−∞
2
X
k=−∞
δ(t)
1
−5
t
−4
X
k=−∞
δ(t)
b
We have to find the sum of samples for −∞ t 3. Sample events are
whole integers. Therefore, −∞ t 3 means −∞ t ≤ 2. So,
2
X
k=−∞
δ(t) = δ(−∞) + . . . + δ(−1) + δ(0) + δ(1) + δ(2)
It gives
2
X
k=−∞
δ(t) = 0 + . . . + 0 + 1 + 0 + 0
Or
2
X
k=−∞
δ(t) = 1
This is desired result. This sum is for right side limit of t. So
2
X
k=−∞
δ(t) will be
placed at t = 2 point.

0.1. FUNCTION 55
Solved Problem 0.60 Find the sum of unit samples of δ(t) function defined in
time domain as δ(t) = 1 at t = 0 for (i) −∞ t −3, (ii) −∞ t 0, (iii)
−∞ t 1 and (iv) −∞ t 4.
Solution The unit sample function is defined as
δ(0) = 1
It means unit sample is unity at t = 0 and zero otherwise. The unit sample
graph is given below:
1
5 10
−5
−10
t
δ(t)
b b b b b b b b b b
b
b b b b b b b b b b
−∞ ∞
1. We have to find the sum of samples for −∞ t −3. Sample events
are whole integers. Therefore, −∞ t −3 means −∞ t ≤ −4. So,
−4
X
k=−∞
δ(t) = δ(−∞) + . . . + δ(−5) + δ(−4)
It gives
−4
X
k=−∞
δ(t) = 0 + . . . + 0 + 0
Or
−4
X
k=−∞
δ(t) = 0
1
5
−5
−10
t
δ(t)
b b b b b b b b
b
b b b b b b b b
−∞ ∞
−∞ t ≤ −4 1
−5
−10
t
δ(t)
b b b b b
−∞
−4
X
k=−∞
δ(t)
1
−5
t
−4
X
k=−∞
δ(t)
b
2. We have to find the sum of samples for −∞ t 0. Sample events are
whole integers. Therefore, −∞ t 0 means −∞ t ≤ −1. So,
−1
X
k=−∞
δ(t) = δ(−∞) + . . . + δ(−2) + δ(−1)

56
It gives
−1
X
k=−∞
δ(t) = 0 + . . . + 0 + 0
Or
−1
X
k=−∞
δ(t) = 0
1
5
−5
−10
t
δ(t)
b b b b b b b b
b
b b b b b b b b
−∞ ∞
−∞ t ≤ −1 1
−5
−10
t
δ(t)
b b b b b b b b
−∞
−1
X
k=−∞
δ(t)
1
−5
t
−1
X
k=−∞
δ(t)
b
0
X
k=−∞
δ(t) = δ(−∞) + . . . + δ(−2) + δ(−1) + δ(0)
It gives
0
X
k=−∞
δ(t) = 0 + . . . + 0 + 0 + 1
Or
0
X
k=−∞
δ(t) = 1
1
5
−5
−10
t
δ(t)
b b b b b b b b
b
b b b b b b b b
−∞ ∞
−∞ t ≤ 0 1
−5
−10
t
δ(t)
b b b b b b b b
b
−∞
0
X
k=−∞
δ(t)
1
−5
t
0
X
k=−∞
δ(t)
b
3
X
k=−∞
δ(t) = δ(−∞) + . . . + δ(−2) + δ(−1) + δ(0) + δ(1) + δ(2) + δ(3)

0.1. FUNCTION 57
It gives
3
X
k=−∞
δ(t) = 0 + . . . + 0 + 0 + 1 + 0 + 0 + 0
Or
3
X
k=−∞
δ(t) = 1
1
5
−5
−10
t
δ(t)
b b b b b b b b
b
b b b b b b b b
−∞ ∞
−∞ t ≤ 3 1
−5
−10
t
δ(t)
b b b b b b b b
b
b b b
−∞
3
X
k=−∞
δ(t)
1
−5
t
3
X
k=−∞
δ(t)
b
These are desired results.
Unit Function
A unit function is a function whose value is ‘1’ if condition is met and ‘0’
otherwise. Unit function has function value ‘1’ for ≥ or symbols and ‘0’ for
≤ or symbols. Unit function in forms of unit sample is given by
u(x) =
x
X
k=−∞
δ(k) (13)
In other words, unit function at event x is sum of unit samples occurs between
−∞ to x. For example, assume that unit sample is
1
5 10 15 20
−5
−10
−15
−20
k
δ(k)
b b b b b b b b b b b b b b b b b b b b
b
−∞ ∞
Figure 6:
The unit function u(x) is again a function of x, so value of x shall be in
(−∞, ∞) and x ∈ I for simplicity. Now we can find the unit function for
infinite numbers of x. Take x = −2
u(−2) =
−2
X
k=−∞
δ(k)

58
From figure, ??, we have δ(k) = 1 when k = 0 and δ(k) = 0 otherwise. Now,
solving above notation
u(−2) = δ(−∞) + . . . + δ(−3) + δ(−2) = 0
1
5 10 15 20
−5
−10
−15
−20
x
u(−2)
b
Now, take x = −1
u(−1) =
−1
X
k=−∞
δ(k)
From figure, ??, we have δ(k) = 1 when k = 0 and δ(k) = 0 otherwise. Now,
solving above notation
u(−1) = δ(−∞) + . . . + δ(−2) + δ(−1) = 0
1
5 10 15 20
−5
−10
−15
−20
x
u(−1)
b
Now, take x = 0, we shall get the unit function as
u(0) =
0
X
k=−∞
δ(k)
Or
u(0) = δ(−∞) + . . . + δ(−1) + δ(0) = 1
From figure, ??, we have δ(k) = 1 when k = 0 and δ(k) = 0 otherwise. Thus
the unit function will be like
1
5 10 15 20
−5
−10
−15
−20
x
u(0)
b
−∞

0.1. FUNCTION 59
Similarly, the unit function at x = 1 is given by
u(1) = δ(−∞) + . . . + δ(−1) + δ(0) + δ(1) = 1
The unit function atx = 1 will be as given below:
1
5 10 15 20
−5
−10
−15
−20
x
u(1)
b
For −∞ x ≤ 1, combining all above values of u(x), unit function will be
plotted as
1
5 10 15 20
−5
−10
−15
−20
x
u(x)
b b
−∞
From figure, ??, assuming that δ(k) = 1 when k = 0 and δ(k) = 0 otherwise.
From above graph, we see that unit function is 1 for all values of x ≥ 0. Hence
for all value of x ≥ 0, unit function will be one. For −∞ x ∞, unit function
will be plotted as:
1
5 10 15 20
−5
−10
−15
−20
x
u(x)
b b b b b b b b b b b b b b b b b b b b b
−∞ ∞
Figure 7:
In summation form, unit function that is drawn in figure ?? can be written
as
u(x) =
(
1 if x ≥ 0
0 Otherwise

60
1
5 10 15 20
−5
−10
−15
−20
x
δ(x)
b b b b b b b b b b b b b b b b b b b b b b b
b
a
−∞ ∞
Shifted Unit Function Shifted unit function is equal to 1 other than x = 0. If
independent variable is shifted by a, then unit function would shift from zero
to one at x = a, i.e. root of the variable is (x − a). Now, the corresponding
shifted unit step is given by u(x − a). In other words, if unit function u(x) = 1
where x ≥ a, then root (origin) for the unit function is x = a or x−a = 0. Now,
unit function is
u(x − a) =
(
1 if x ≥ a
0 Otherwise
Substituting argument of unit function equal to zero, we shall get the point
where unit function changes from zero to one. Note that a ≥ 0 so that x − a
would not represent origin (initial position) in negative value. For unit function
starting point is always 0 or more. x − a ≥ 0 should be maintained for the
condition of unit function as
u(x) = 1; for x ≥ 0
Unit function is active for given time period and is unity for that period of
time. Hence, unit sample is also known as function ON switch. For example,
f(x) = x2
− 3
is a continuous function in −∞ x ∞. When u(x) is multiplied with
function f(x) then we have
f(x) × u(x) = x2
− 3

× u(x)
u(x) = 1 when x ≥ 0, so
f(x) × u(x) = x2
− 3

× 1
Or
f(x) × u(x) = x2
− 3; for x ≥ 0
It means f(x) × u(x) is zero for x 0.
If u(x) is taken for the purpose of unit function then x is continuous
independent variable and its value is from real domain as x ≥ 0 and x ∈ R. If

0.1. FUNCTION 61
u(x) is under consideration of unit sample then x is a whole number as x ≥ 0
and x ∈ I. When we use it for rectangular unit function, x is real number and
while we use it for unit sample δ(x) then x and integer.
1
5 10 15 20
−5
−10
−15
−20
x
u(x)
b b b b b b b b b b b b b b b b b b
a
−∞ ∞
Above representation of unit function may also be written as u(x−a). Where,
a that point after which the function value is ‘1’. If a = 1 then above unit
function is written as
u(x − 1) =
(
1 if x ≥ 1
0 Otherwise
This function is graphed as
1
1 2 3
x
y
Rectangular Unit Function Unit function that is closely bounded at lower
and upper ends can also defined as
Rect[u(x)] =
(
1 if a ≤ x ≤ b
0 Otherwise
1
5 10 15 20
−5
−10
−15
−20
x
u(x)
b b b b b b b b b b b
a b
−∞
A lower end (at x = a) and upper end (at x = b) bounded unit function is
called rectangular function and it is written as
Rect[u(x)] = u(x − a) − u(x − b)

62
1
1
x
y
a b
−
1
1
x
y
a b
1
1
x
y
a b
Assume a unit function, which is closely bounded in 0.5 and 1, then it shall
be written as
Rect[u(x)] =
(
1 if 0.5 ≤ x ≤ 1
0 Otherwise
Here, unit function is closely bounded in 0.5 to 1. This function is graphed as
1
1
x
u(x)
Relation Between Unit Sample and Unit Function If δ(x) = 1 at x = 0 only
(i.e. δ(0) = 1 and zero otherwise) then unit sample at x = a can also be defined
in terms of unit function as
δ(a) = u(a) − u(a − 1) (14)
To prove it graphically, two functions, u(a) and u(a − 1) functions are plotted
in the figure ??.
1
5 10 15 20
−5
−10
−15
−20
x
u(x)
b b b b b b b b b b b
a
−∞
Plot of u(a) function and a = 10 (let).
1
5 10 15 20
−5
−10
−15
−20
x
u(x)
b b b b b b b b b b
a − 1
−∞
Plot of u(a − 1) function and a = 9 (let)
Figure 8:

0.1. FUNCTION 63
When graph of u(a) is subtracted by graph of u(a − 1), we get unit sample
δ(a) = 1 at x = a and δ(x) = 0 for other values of x.
1
5 10 15 20
−5
−10
−15
−20
x
δ(x)
b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b
b
a
b b b b b b b b b b
−∞ ∞
The relation
δ(a) = u(a) − u(a − 1)
has unique property of backward difference. For any value a, δ(a) is equal to
the backward difference between two consecutive values of unit function u(a)
and u(a − 1). Thus we can say that
δ(a) = ∆u(a)
If a → 0, then
δ(a) =
d
da
u(a) = u̇(a) (15)
Using the properties of antiderivatives, we can also get new relation as
Z
δ(a) da = u(a) (16)
Solved Problem 0.61 A unit function is represented by u(t − 4). Find where
this unit function is transitioned. Draw the graph of the unit function.
Solution The given unit function is u(t − 4). Its transition occurs at
t − 4 = 0, i.e. t = 4s.
u(t) =
(
1 if t ≥ 4
0 Otherwise
Its graph is given below:
1
1 2 3 4 5
−1
t
u(t)

64
Solved Problem 0.62 A unit function is represented by u(t − 6). Find where
this unit function is transitioned. Draw the graph of the unit function.
Solution The given unit function is u(t − 6). Its transition occurs at
t − 6 = 0, i.e. t = 6s.
u(t) =
(
1 if t ≥ 6
0 Otherwise
Its graph is given below:
1
1 2 3 4 5 6 7 8
−1
t
u(t)
Solved Problem 0.63 Draw the graph of unit function u(t) = u(t−0.5)−u(t−3).
Solution This unit function is subtractive nature, i.e. unit function u(t −
0.5) is subtracted by u(t − 3). To draw a graph, we must find the transition
points of the unit function parts, i.e. u(t − 0.5) and u(t − 3). So,
t − 0.5 = 0 ⇒ t = 0.5; t − 3 = 0 ⇒ t = 3
So, equation of u(t − 0.5) is given by
u(t) =
(
1 if t ≥ 0.5
0 Otherwise
Similarly, equation of u(t − 3) is given by
u(t) =
(
1 if t ≥ 3
0 Otherwise
Now, plot of the given unit function is as shown below:
t
u(t)
0.5
−
t
u(t)
3
t
u(t)
0.5 3

0.1. FUNCTION 65
The equation of unit function u(t − 0.5) − u(t − 3) is
u(t) =
(
1 if 0.5 ≤ t ≤ 3
0 Otherwise
This is answer.
Solved Problem 0.64 Draw the unit function regions of the unit functions u(t−1)
and u(t − 2) and subtract the region of u(t − 1) by u(t − 2).
Solution The transition point of the unit function u(t − 1) is t − 1 = 0 ⇒
t = 1. This unit function is graphed as
u(t) =
(
1 if t ≥ 1
0 Otherwise
The graph of this unit function would be as:
t
u(t)
1
Similarly, transition point of the unit function u(t − 2) is t − 2 = 0 ⇒ t = 2.
This unit function is graphed as
u(t) =
(
1 if t ≥ 2
0 Otherwise
t
u(t)
2
On subtracting the region of u(t − 1) by u(t − 2), we shall get region of
u(t − 1) − u(t − 2) as

66
t
u(t)
1
−
t
u(t)
2
t
u(t)
1 2
This is desired graph of u(t − 1) − u(t − 2).
Solved Problem 0.65 A unit function is represented by u(t − 4) − u(t − 6). Find
where this unit function is transitioned. Draw the graph of the unit function.
Solution The given unit function is u(t−4)−u(t−6). Its transition occurs
at two points where t−4 = 0 and t−6 = 0, i.e. at t = 4s and t = 6s respectively.
Note that second term of unit function in right hand side (i.e. u(t − 6) is in
subtraction from first term of unit function in right hand side (i.e. u(t − 4).
Hence this function has two bounded limits. As t = 6s t = 4s and unit
function is defined as
u(t) =
(
1 if 4 ≤ t ≤ 6
0 Otherwise
t
u(t)
4 6
−
t
u(t)
4 6
t
u(t)
4 6
Solved Problem 0.66 Why unit function is said “Switch ON” function? A unit
function u(t−0.5) is applied to the line function f(t) = t−0.5. Find the output
of f(t).
Solution Unit function is also known as Switch ON function as another
function, f, is accepted as input for that period of time in which u(t) = 1 and
rejected for the period in which u(t) = 0. If f(t) = t − 0.5 is an input function
and unit function u(t − 0.5) is applied to it. The transition point for the unit
function is
t − 0.5 = 0 ⇒ t = 0.5
It means
u(t) =
(
1 if t ≥ 0.5
0 Otherwise

0.1. FUNCTION 67
External function is f(t) = t − 0.5. When unit function is multiplied to it, we
get
O(t) = (t − 0.5)u(t − 0.5)
Graph of this output function is
t
f(t)
×
t
u(t)
0.5
t
O(t)
0.5
t
O(t)
0.5
t
O(t)
0.5
Solved Problem 0.67 A unit function u(t − 0.5) − u(t − 1.5) is applied to the
line function f(t) = t − 0.5. Find the output of f(t).
Solution Here f(t) = t − 0.5 is an input function and unit function u(t −
0.5) − u(t − 1.5) is applied to it. The transition points for the unit function are
t − 0.5 = 0 ⇒ t = 0.5; t − 1.5 = 0 ⇒ t = 1.5
It means
u(t) =
(
1 if 0.5 ≤ t ≤ 1.5
0 Otherwise
External function is f(t) = t − 0.5. When unit function is multiplied to it, we
get
O(t) = (t − 0.5) [u(t − 0.5) − u(t − 1.5)]
Graph of this output function is

68
t
f(t)
×
t
u(t)
0.5 1.5
t
O(t)
0.5 1.5
t
O(t)
0.5 1.5
t
O(t)
0.5 1.5
Unit Function as Electrical Switch
R
V (t)
b
b
b
b
b
b
b
b
S
t
V (t)
1
1 2 3
t
u(t)
u(t) = u(t − 0.5) − u(t − 1.0)
+u(t − 2.0) − u(t − 2.5)
+u(t − 3.0) − u(t − 3.5)
u(t) =







1 if 0.5 ≤ t ≤ 1.0
1 if 2.0 ≤ t ≤ 2.5
1 if 3.0 ≤ t ≤ 3.5
0 Otherwise
t
VR(t)
In above figure, a sinusoidal potential is applied to a resistor R. An
electromagnetic switch is connected in series. It switches between ON and
OFF states by a unit function as
u(t) =













1 if 0.5 ≤ t ≤ 1.0
1 if 2.0 ≤ t ≤ 2.5
1 if 3.0 ≤ t ≤ 3.5
0 Otherwise

0.1. FUNCTION 69
We see that when switch is ON (u(t) = 1) we get an output voltage (VR(t))
across the resistor R. VR(t) = 0 when switch is OFF (i.e. u(t) = 0). It is
equivalent electrical example of unit function as electrical switch.
Solved Problem 0.68 A voltage function of +10V is applied to unit function
u(t−1)−u(t−2) and −10V is applied to unit function u(t−2)−u(t−3). Find
the output function.
Solution Here, unit function acts as ON switch for 1 ≤ t ≤ 3 but input
voltage is different for the ON period 1 ≤ t ≤ 2 and 2 ≤ t ≤ 3. Input voltage
is +10V for 1 ≤ t ≤ 2 and −10V for 2 ≤ t ≤ 3. Using the properties of unit
function, output is
O(t) = 10 × [u(t − 1) − u(t − 2)] − 10 × [u(t − 2) − u(t − 3)]
On solving, it we have
O(t) = 10u(t − 1) − 10u(t − 2) − 10u(t − 2) + 10u(t − 3)
Or
O(t) = 10u(t − 1) − 20u(t − 2) + 10u(t − 3)
This is output function.
Solved Problem 0.69 Plot discrete type unit function u(x) for −∞ x ≤ 2
having unit sample function δ(k) = 1 at k = 0 only.
Solution We have to plot the discrete type unit function in −∞ x ≤ 2.
The given unit sample function is
δ(k) =
(
1 if k = 0
0 Otherwise
The discrete unit function is given by
u(x) =
x
X
k=−∞
δ(k); where − ∞ x ≤ 2
As δ(k) = 1 at k = 0 only, so
X
δ(k) = 1; ∀ k ≥ 0
So, from the relation
u(x) =
x
X
k=−∞
δ(k); where − ∞ x ≤ 2

70
u(x) will be zero for each value of x 0 as k in term
x0
X
k=−∞
δ(k) will always be
negative and consequently δ(k) = 0 for each k 0. Thus unit function is one
for each value of x that is zero or positive as for each these values corresponding
k values are either zero or positive. Hence unit function will be as
u(x) =
(
1 if x ≥ 0
0 Otherwise
1
5 10 15 20
−5
−10
−15
−20
x
u(x)
−∞ ∞
We can solve it mathematically. Unit sample is one at k = 0. So, we will
find unit function values about x = 0. Now, take x = −2, we have
u(−2) =
−2
X
k=−∞
δ(k)
Or
u(−2) = δ(−∞) + . . . + δ(−3) + δ(−2)
It gives u(−2) = 0. Similarly, take x = −1
u(−1) =
−1
X
k=−∞
δ(k)
Or
u(−1) = δ(−∞) + . . . + δ(−2) + δ(−1)
It gives u(−1) = 0. Take x = 0
u(0) =
0
X
k=−∞
δ(k)
Or
u(0) = δ(−∞) + . . . + δ(−1) + δ(0)
It gives u(0) = 1. Note that δ(0) = 1 only. Take x = 1
u(1) =
1
X
k=−∞
δ(k)

0.1. FUNCTION 71
Or
u(1) = δ(−∞) + . . . + δ(−1) + δ(0) + δ(1)
It gives u(1) = 1. Take x = 2
u(2) =
2
X
k=−∞
δ(k)
Or
u(2) = δ(−∞) + . . . + δ(−1) + δ(0) + δ(1) + δ(2)
It gives u(2) = 1. For each value of x ≥ 0, u(x) = 1. So, our unit function will
be like:
1
5 10 15 20
−5
−10
−15
−20
x
u(x)
−∞ ∞
Solved Problem 0.70 Simplify the output unit function given as
O(t) = u(t − 2) + u(t − 3) − 2u(t − 2.5)
and write its unit function and input function.
Solution The given output function is
O(t) = u(t − 2) + u(t − 3) − 2u(t − 2.5)
As we know that transition points of unit function are sequential and increasing
in domain of R. So, we will arrange the given unit function terms in right hand
side of the function f(t) in order to transition points. So,
O(t) = u(t − 2) − 2u(t − 2.5) + u(t − 3)
The transition points for this unit function are t = 2, t = 2.5 and t = 3. Solving
it from left to right u(t − 2.5) term is in subtraction, hence it will form a unit
rectangular function with its previous term, i.e. u(t − 2). Expanding the unit
term u(t − 2.5) so that u(t − 2) and u(t − 2.5) has same coefficient (i.e. function
in product to unit function). Now,
O(t) = u(t − 2) − u(t − 2.5) − u(t − 2.5) + u(t − 3)
Now taking two terms in a group, we have
O(t) = +1 × [u(t − 2) − u(t − 2.5)] − 1 × [u(t − 2.5) − u(t − 3)]

72
1
−1
1 2 3 4 5
t
O(t)
The unit function can be written in piecewise function as
u(t) =
(
1 if 2 ≤ t ≤ 3
0 Otherwise
Interpreting the output function assuming that unit function is acting as ON-
OFF switch, it is found that +1 input is applied for rectangular unit function
u(t−2)−u(t−2.5) while −1 input is applied for unit function u(t−2.5)−u(t−3).
Using this interpretation, we can write input function as
I(t) =
(
+1 if 2.0 ≤ t ≤ 2.5
−1 if 2.5 ≤ t ≤ 3.0
Solved Problem 0.71 Simplify the unit function given as
O(t) = u(t − 2) + 2u(t − 3) − 3u(t − 2.5)
and write it in piecewise function form.
O(t) = u(t − 2) + 2u(t − 3) − 3u(t − 2.5)
O(t) = u(t − 2) − 3u(t − 2.5) + 2u(t − 3)
The transition points for this unit function are t = 2, t = 2.5 and t = 3. Solving
it from left to right u(t − 2.5) term is in subtraction, hence it will form a unit
rectangular function with its previous term, i.e. u(t − 2). Expanding the unit
term u(t − 2.5) so that u(t − 2) and u(t − 2.5) has same coefficient (i.e. function
in product to unit function). Now,
O(t) = u(t − 2) − u(t − 2.5) − 2u(t − 2.5) + 2u(t − 3)

0.1. FUNCTION 73
Now taking two terms in a group, we have
O(t) = +1 × [u(t − 2) − u(t − 2.5)] − 2 × [u(t − 2.5) − u(t − 3)]
2
−2
1 2 3 4 5
t
O(t)
u(t) =
(
1 if 2 ≤ t ≤ 3
0 Otherwise
OFF switch, it is found that +1 input is applied for rectangular unit function
u(t−2)−u(t−2.5) while −2 input is applied for unit function u(t−2.5)−u(t−3).
Using this interpretation, we can write input function as
I(t) =
(
+1 if 2.0 ≤ t ≤ 2.5
−2 if 2.5 ≤ t ≤ 3.0
O(t) = 10 + 2u(t − 3) − 2u(t − 2.5)
and write it in piecewise function form.
O(t) = 10 + 2u(t − 3) − 2u(t − 2.5)
O(t) = 10 − 2u(t − 2.5) + 2u(t − 3)
The transition points for this unit function are t = 2.5 and t = 3. Again
O(t) = 10 − 2 × [u(t − 2.5) − u(t − 3)]

74
2
4
6
8
10
−2 1 2 3 4 5
t
O(t)
Figure 9: In above figure, cyan color is unit function, brown line is output of
unit part, green line is constant term and purple line is O(t).
u(t) =
(
1 if 2.5 ≤ t ≤ 3
0 Otherwise
As O(t) = 10 − 2 × [u(t − 2.5) − u(t − 3)] has constant and unit function terms.
OFF switch, only −2 × [u(t − 2.5) − u(t − 3)] term would go ON-OFF while
term 10 would be constant for time being. It is found that −2 input is applied
for rectangular unit function u(t − 2.5) − u(t − 3) and 0 for rest of time. Using
this interpretation, we can write input function as
I(t) =
(
+10 if t 2.5 and t 3
+8 if 2.5 ≤ t ≤ 3
Or
I(t) =
(
+8 if 2.5 ≤ t ≤ 3
+10 Otherwise
f(t) = 10 − 2u(t − 2.5)
and write it in piecewise function form for t ∈ [−∞, ∞].
O(t) = 10 − 2u(t − 2.5)
in domain of R. The transition points for this unit function are t = 2.5. Again
O(t) = 10 − 2u(t − 2.5)

0.1. FUNCTION 75
2
4
6
8
10
−2 1 2 3 4 5
t
O(t)
Figure 10: In above figure, cyan line is unit function, brown line is output of
unit part, green line is constant term and purple line is O(t).
u(t) =
(
1 if t ≥ 2.5
0 if t 2.5
As O(t) = 10 − 2u(t − 2.5) has constant and unit function terms. Interpreting
the output function assuming that unit function is acting as ON-OFF switch,
only −2u(t − 2.5) term would go ON-OFF while term 10 would be constant
for time being. It is found that −2 input is applied for shifted unit function
u(t−2.5) and 0 for time (t 2.5). Using this interpretation, we can write input
function as
I(t) =
(
+10 if t ≥ 2.5
+8 if t 2.5
Step Function
A step function is a function whose value changes when independent variable
passes require conditions. A step function is a constant piecewise function hav-
ing finite pieces. It looks like a staircase. A step function is defined as
s(x) =

1 if [x] is odd
−1 if [x] is even
(17)
Here symbol [x] represents to the greatest integer value. Greatest integer symbol
returns the greatest integer from a real number. Actually greatest number
symbol returns the integer that its argument has crossed (or at) in numberline
while counting from −∞ to +∞. This is similar to the floor function (⌊x⌋)4
,
i.e. lower bounding integer value, of computer programming languages. The
step function given by above function is graphed as
4
opposite to ceil function (⌈x⌉), i.e. upper bounding integer value

76
1
−1
1 2 3 4 5
x
s(x)
Another step function is
s(x) = [x]; when 0 x 4 (18)
Few examples for floor function are ⌊−2.5⌋ = −3, ⌊2.5⌋ = 2, ⌊0.5⌋ = 0 etc.
s(x) = ⌊x⌋; when 0 x 4 (19)
1
2
3
1 2 3
x
s(x)
1
−1
−2
1
−1
−2
x
s(x)
Similarly, ceiling type step function is
s(x) = ⌈x⌉; when 0 x 4 (20)
Few examples for ceiling function are ⌈−2.5⌉ = −3, ⌈2.5⌉ = 2, ⌈0.5⌉ = 0 etc.
1
2
3
1 2 3
x
s(x)
1
−1
−2
1
−1
−2
x
s(x)

0.1. FUNCTION 77
Auxiliary tabulated data for floor and ceiling functions is given below:
x ⌊x⌋ ⌈x⌉
-2.5 -3 -2
-2 -2 -2
-1.5 -2 -1
-1 -1 -1
-0.5 -1 0
0 0 0
0.5 0 1
1 1 1
1.5 1 2
2 2 2
2.5 2 3
Unit Step Function
Assume a function f(t) that is continuous very where in the domain of t. If
conditionally, if f(t) has two discontinuous function at t = a, then f(t − a) is
pole or origin for this conditional function. Now f(t − a) has two functions.
f(t − a) =

f1(t) if t a
f2(t) if t ≥ a
It means ‘a’ is the point at where conditional function is changed. Again if
f1(t) = 0 and f2(t) = 1 then f(t − a) = u(t − a) is known as unit step function
or Heaviside Function.
u(t − a) =

0 if t a
1 if t ≥ a
1
t
u
a
Illustrated Example Two functions, t3
and (t − 2)2
respectively are sketched
as shown below.

78
1
−1
1
−1
t
y
y = t3 1
−1
1 2 3
t
y
y
=
(
t
−
2
)
3
First figure is sketch of relation y = t3
while second figure is sketch of
y = (t − 2)3
. The second function is shifted by ‘2’ in respect to first function.
If y is a boolean then above sketches becomes
1
t
u
1
t
u
a
First figure represents to u(t) and second figure represents to u(t − a).
Solved Problem 0.74 Sketch the graph of function
u(t − 1) =

0 if t 1
1 if t ≥ 1
Solution This is an unit step function. The function value shall be zero
when ‘t’ is less than ‘1’ and shall be one when ‘t’ is equal or larger than ‘1’. The
sketch of the function is
1
1 2 3
t
u
Solved Problem 0.75 Sketch the graph of function
u(t + 1) =

0 if t −1
1 if t ≥ −1

0.1. FUNCTION 79
Solution This is an unit step function. The function value shall be zero
when ‘t’ is less than ‘−1’ and shall be one when ‘t’ is equal or larger than ‘−1’.
The sketch of the function is
1
1 2
−1
t
u
Ramp Function
A ramp function is a line function with slope m 0 within (a, b). Assume a
piece-wise function
f(x) =



0 if x 0
mx if 0 ≤ x ≤ a
1 if x a
The above piece-wise function is a ramp function for 0 ≤ x ≤ a. This function
is graphed as
x
f(x)
a
b
1
α
The slope of a ramp function may positive or negative. Two ramp functions
with positive and negative slops can form a triangular function (trig) functions.
x
f(x)
b
−a
b
a
b
1
α
β
Square Function
A square function is defined according to the following piecewise function.
f(x) =

1 if 0 ≤ x ≤ 2a
0 Otherwise

80
This function is graphed as
x
f(x)
b
0
b
2a
b
1
The symmetric square function may also be defined as
x
f(x)
b
−a
b
a
b
1
Signum Function
A function f(t) is said to be signum function, if it is defined as
sgn(t) =



+1 when t 0
0 when t = 0
−1 when t 0
1
−1
1 2 3 4
−1
−2
−3
−4
−5
t
sgn(t)
Signum function is a conditional function.
Periodic Function
A periodic function is a function that repeats itself after a fix interval of time.
Periodic functions are defined as f(x) = f(x − a) where a is period of the func-
tion. Simplest example of periodic functions are sin θ and cos θ whose periods
are 2π. A unit periodic function has unity amplitude. Periodic functions may
be symmetric or non symmetric. For example, in the following graph, a periodic
function of period 1 is shown.

0.1. FUNCTION 81
1
1 2 3 4 5
t
y
This is non symmetric periodic function. Active duration of this periodic
function is less than the non-active duration.
1
1 2 3 4 5
t
y
t
′
A trigonometric function may be periodic or not, it depends on its argument.
If argument of the trigonometric operator (like sine, cosine, tangent etc) can be
written as rπ (r ∈ I and r 0) then it is periodic function otherwise it is not
periodic.
Solved Problem 0.76 Find which of these (i) cos(2.1πt), (ii) sin(2.1t) is periodic
and not-periodic? Also find the period.
Solution
1.
Solved Problem 0.77 Find which of these (i) cos(0.1πk), (ii) sin(1.01k) is peri-
odic and not-periodic? Also find the period.
Solution
Impulse Function
When a large force acts for a short time, then the product of the force and time
is called impulse in applied mathematics. It t → 0, impulse tends to infinite.
t
δ δ(t)
The unit impulse function is the limiting function and it has unit area but
zero width.

82
t
δ
-1 1
0.5
t
δ
-0.5 0.5
1
t
δ
−
1
3
1
3
1.5
δ(t − a) =



1
k
if a t a + k
0 Otherwise
Here, limit of k → 0. This function represent the force magnitude 1/k, that is
acted for the time t = a to t = a + k. Here, k is very small. The value of the
function becomes infinity when the width of k approach to zero. In this case
area of the rectangle is unity.
δ(t − a) =
(
∞ for t = a
0 for t 6= a
And Z ∞
0
δ(t − a) · dt = 1
t
δ
1
k
t = a t = a + k
In delta function, shape does not matter, as delta function is very large if
t → 0 and very small away from t → 0.
t
δ
−ε ε
0.5ε
t
δ
−0.5ε 0.5ε
ε
t
δ
−
1
3ε
1
3ε
1.5ε
Delta function (δ) is an even function at it is continuous at t → 0 only. It
means, its left side limit and right side limits are equal. Therefore,
δ(−x) = δ(x) (21)

Function Analysis v.2

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