Embed Size (px)
Citation preview
Continuous Time Signals
UNIT – I Signal :- A Signal is defined as a function of one or more variables, that conveys
information about the state or behaviour of a physical system. When the function depends
on a single variable, the signal is said to be one dimensional signal.
Ex:- Speech signal (amplitude varies with time)
When the function depends on two or more variables the signal is said to be
multidimensional signal.
Ex :- IMAGE, Two dimensional signal (Horizontal & Vertical coordinates)
At any instant of time the function is having unique value. The value may be a
real number, in which case we have a real valued signal or it may be a complex number,
in which case we have a complex valued signal. In any case the independent variable
(namely time) is real valued.
Classification of signals :- The signals are classified based on their nature and characteristics. These are
a) Continuous – time signals
b) Discrete – time signals.
a) Continuous – time signal :- ( )x t ↑ t A signal ( )x t is said to be continuous – time signal if it is defined for all time –t. Ex :- Light wave is converted into an Electrical signal by using Transducers.
b) Discrete – time signal :-
( ) ( ) |t nTx n x t == nα α− < <
= ( )x nT if T=1 = ( )x n n=0, ±1, ±2, ±3……. T = Sampling period (Time period between two successive samples) A signal ( )x n is said to be discrete – time signal if it is defined only at discrete instants of time. Analog signal :- ( )x t A signal is said to be analog signal if both amplitude and time are continuous.
An analog signal can be converted into discrete –time signal by sampling. The Discrete – time signal is converted into Digital signal by quantization & coding. Both continuous – time and discrete – time signals are further classified as
1) Even and odd signals 2) Periodic and aperiodic signals 3) Deterministic and stochastic ( random) signals 4) Energy and power signals.
1) Even (symmetric) and odd (anti symmetric) signals.
A continuous time signal x(t) is said to be an even signal if it satisfies the condition
( ) ( )x t x t= − for all t. A Continuous – time signal x(t) is said to be odd signal if it satisfies the condition
( ) ( )x t x t= − − for all t Ex :- ( ) sinx t tω= ,odd signal ( ) cosx t tω= ,even signal Even signal ( ) ( )x n x n= − for all n Odd signal ( ) ( )x n x n= − − for all n x(t) x(n) t n Even signal x(t) x(n) t n Odd signal
A signal can be expressed as a sum of two components, namely even component of the signal and the odd component of the signal. Both can be obtained from signal itself. ( ) ( ) ( )e ox t x t x t= + → (1.1) ( )ex t = even ( )ox t = odd ( )ex t = ( )ex t− & ( )ox t = ( )ox t− − (or) ( )ox t− = ( )ox t− put t t= − in equation (1.1) ( ) ( ) ( )e ox t x t x t− = − + − ( ) ( ) ( )e ox t x t x t− = − → (1.2)
adding equation (1.1) & (1.2) ( )1( ) ( ) ( )2ex t x t x t= + −
subtracting equation (1.1) - (1.2) ( )1( ) ( ) ( )2ox t x t x t= − −
similarly ( ) ( ) ( )e ox n x n x n= + → (1.3) ( )ex n = ( )ex n− , ( )ox n = ( )ox n− − put n=-n in eqn (3) ( ) ( ) ( )e ox n x n x n− = − + − ( ) ( ) ( )e ox n x n x n− = − → (1.4)
(1.3) + (1.4) ( )1( ) ( ) ( )2ex n x n x n= + − e = even
(1.3) – (1.4) ( )1( ) ( ) ( )2ox n x n x n= − − o = odd
This is in the case of real valued signal.
A complex valued signal ( )x t is conjugate anti symmetric signal of its real part is odd and its imaginary part is even. In the case of complex valued signal, we may speak of conjugate symmetry. ∴A complex valued signal x(t) is said to be conjugate symmetry signal, If it satisfies the condition *( ) ( )x t x t= − (or) *( ) ( )x t x t= − * = conjugate. Let ( ) ( ) ( )x t a t j b t= + a(t) = Real part of x(t) *( ) ( ) ( )x t a t j b t= − b(t) = Imaginary part of x(t) 2 1j = − , 1j = − ( ) ( ) ( )x t a t j b t− = − + −
∴A complex valued signal ( )x t is conjugate symmetric signal, if its real part is even and its imaginary part is odd.
A complex valued signal ( )x t can also be expressed as sum of conjugate symmetric and conjugate anti symmetric components. Let ( )x t is complex valued and satisfying complex conjugate symmetry. i.e. *( ) ( )x t x t= − (or) *( ) ( )x t x t= − ( ) ( ) ( )cs casx t x t x t= + → (1.5) ( )csx t is conjugate symmetric signal if it satisfies the condition *( ) ( )cs csx t x t= − ( )casx t is conjugate – anti symmetric signal if it satisfies the condition *( ) ( )cas casx t x t= − − put t t= − in eqn (1.5) and take conjugate on both sides * * *( ) ( ) ( )cs casx t x t x t− = − + − *( ) ( ) ( )cs casx t x t x t− = − → (1.6)
adding equation (1.5) & (1.6) *1( ) [ ( ) ( )]2csx t x t x t= + −
subtracting equation (1.5) - (1.6) *1( ) [ ( ) ( )]2casx t x t x t= − −
Similarly in the case of discrete time *1( ) [ ( ) ( )]2csx n x n x n= + −
*1( ) [ ( ) ( )]2casx n x n x n= − −
P.1# Which of these two signals is even and which one is odd. 1( )x t 2 ( )x t
A A
-A Fig 1.a Fig 1.b 1( )x t is even 2 ( )x t is odd Express 1( )x t & 2 ( )x t in terms of even & odd components.
1 1 11( ) [ ( ) ( )] ( )2ex t x t x t x t= + − = 1 1( ) ( ) 0x t x t∴ = +
1 11( ) [ ( ) ( )] 02ox t x t x t= − − =
2 21( ) [ ( ) ( )] 02ex t x t x t= + − =
2 2 21( ) [ ( ) ( )] ( )2ox t x t x t x t= − − = 2 2( ) 0 ( )x t x t∴ = +
P.2 # if 1 2( ) ( ) ( )x t x t jx t= + find ( )x t is a conjugate symmetric signal or not. *( ) ( )x t x t= − (or) ( ) *( )x t x t= −
*1 2( ) ( ) ( )x t x t jx t= −
1 2( ) ( ) ( )x t x t jx t− = − + −
From fig 1.a 1̀ 1( ) ( )x t x t= − From fig 1.b 2 2( ) ( )x t x t= − − ∴ ( )x t is conjugate symmetric signal. (2) Periodic and aperiodic (Non periodic) signals : A signal is said to be periodic if it repeats after a fixed time period (or) Mathematically a continuous – time signal ( )x t is said to be periodic if it satisfies the following condition . ( ) ( )x t x t T= + for all t (2.1) T is positive constant: ie T > 0 The smallest value of T that satisfies eqn (2.1) is called the fundamental period of x(t).
Fundamental frequency = f = 1T
Hz (or) cycles per second.
ω = angular frequency: 2Tπ radians/second.
Any signal ( )x t for which there is no value of T to satisfy the eqn (2.1) is called an aperiodic (or) non periodic signal. x(t) x(t)
Exponential signal t
t Periodic signal Non periodic signal
Periodic signal ( ) sin(2 ) sinx t A ft A tπ ω= = sin(2 2 ) sin( 2 )A ft A tπ π ω π= + = + The fundamental period of ( )x t is T = 2π
( )x t 1 0 -1
0 .1 .2 .3 .4 .5 t (seconds) P.3 # Find the periodicity of the triangular signal shown above.
Fundamental period is T = 0.2 seconds
f = 10.2
= 5 Hz
(or) ω = 2π f = 10π rad/second. A discrete time signal ( )x n is said to be periodic if it satisfies the condition
( ) ( )x n x n N= + for all n ( n =integer) ( 2.2) N is positive integer
Fundamental period of ( )x n N= Freq 1FN
=
Fundamental angular frequency 2 radiansNπ
Ω =
Any signal ( )x n for which there is no value of N to satisfy the eqn (2.2) is called
aperiodic or non periodic sequence. 1( )x n 2 ( )x n
-3 -2 -1 0 1 2 3 4 5 6 7 8 n -2 -1 0 1 2 3 n Periodic Aperiodic
Fundamental period of 1( ) 3x n = , 13
F =
23
radiansπΩ =
P.4 # Find the fundamental period 1( ) sin15x t tπ= & 2( ) sin 2x t tωπ=
2 2 1 0.13315 7.5
T π πω π
= = = = & 2 1 0.120 10
T ππ
= = =
(3) Deterministic signals & Random signals (Non Deterministic signals) A Deterministic signal is a signal about which there is no uncertainty w.r.t its value at any time. Ex:- A completely specified function time, continuous-time signal or discrete –time signal or sine wave, cosine wave etc. A Random signal is a signal about which there is uncertainty with respect to its value at any time. Ex:- Noise signal. (4) Energy and power signals :- In Electrical systems a signal may represent a voltage or current. Consider a voltage ( )v t developed across a Resistor R , producing a current ( )i t . The instantaneous power developed across resistor R is defined as
2( )V t
PR
= or → (4.1)
2( ) .P i t R= → (4. 2) In both case the instantaneous power P is proportional to the squared amplitude of the signal. If R = 1 Ω The above eqn (4.1) & (4.2) takes same mathematical form so that regardless of whether a given signal ( )x t represents a voltage or current, the instantaneous power
associated with the signal is 2( )P x t=
The total energy dissipated by the signal ( )x t is the integral of the instantaneous
power during interval 2 2T Tto− is defined as
/ 2
2
/ 2
lim ( )T
TT
E x t dt→∞
−
= ∫
2( )E x t dt∞
−∞
= ∫ . Joules (non periodic signals)
The average power dissipated by the signal ( )x t during interval 2 2T Tto− is defined as
/ 2
2
/ 2
1lim ( )T
TT
P x t dtT→∞
−
= ∫
The average power of a periodic signal ( )x t of fundamental period T is
/ 2
2
/ 2
1 ( )T
T
P x t dtT −
= ∫ watts.
Root mean square value of ( )x t P= P = average power The signal ( )x t is an energy signal if and only if the total energy of the signal satisfies the condition 0 E< < ∞ Ex : Non periodic signals. The signal ( )x t is a power signal if and only if the average power of the signal satisfies the condition 0 P< < ∞ Ex: Periodic signals.
&E P are always real & non negative quantities 2( )x t instead of 2 ( )x t to allow for the possibility of complex signal models.
In the case of discrete –time signal, the total energy of ( )x n is defined as
2 ( )n
E x n∞
=−∞
= ∑
21lim ( )2
N
N n NP x n
N→∞=−
= ∑
The average power in a periodic signal ( )x n with fundamental period N is given by
1
2
0
1 ( )N
nP x n
N
−
=
= ∑
Note : The Energy signal has zero average power, whereas power signal has infinite energy. P.5 # What is the total Energy of Rectangular pulse and average power of square wave shown in figure. ( )x t ( )x t 1 A 0 -1 t 0.2 0.4 0.6 t
T1 Amplitude A=1
1/ 2
1/ 2
2 2 2 2 21 11( ) *
2
T
T
T TE x t dt A dt A t A A T∞
−∞ −
+= = = = =∫ ∫
0.2 0.1 0.2
2
0 0 0.1
1 1 1( ) 1 10.2 0.2
P x t dt dt dtT
= = +∫ ∫ ∫
[ ]0.2
0
10.2
t=
= [ ]1 0.20.1 0 0.2 0.1 10.2 0.2
− + − = =
P.6 # What is the average power of the Triangular wave shown in fig. 1 0 -1 0 .1 .2 .3 .4 .5 .6 .7 t
1 2 1 1 2 1( )( ) ( )( )x x y y y y x x− − = − −
20 1y t= − from 0 to 0.1 20 3y t= − + from 0.1 to 0.2
Average power = 2
/ 2 0.1 0.22 2
/ 2 0 0.1
1 1 1( ) (400 1 40 ) (400 9 120 )0.2 0.2
T
T
x t dt t t dt t t dtT −
= + − + + −∫ ∫ ∫
[ ]1 10.133 0.1 0.2 0.933 0.9 1.8 0.330.2 3
watts= + − + + − = =
P.7 # What is the total energy of the discrete –time signal shown in fig. ( )x n 1 -1 0 1 n
12 2
1( ) ( )
nE x n x n
∞
=−∞ −
= =∑ ∑
= 3 joules
P.8 # Prove that the signal 14( ) ( 1)x t t u t
−= − is neither an energy nor a power signal.
1
1
2( )E x t dt∞
−∞
= ∫
1
1
21/ 4 1/ 2
1
( 1)t u t dt t dt∞ ∞
− −
−∞
= − =∫ ∫
1/ 2
12 2t
∞= = ∞ − = ∞
E = ∞ It is not an energy signal.
/ 2
2
/ 2
1lim ( )T
TT
P x t dtT→∞
−
= ∫
/ 2
21/ 2 1/ 2
11
1 1lim lim 2T
T
T Tt dt t
T T−
→∞ →∞
⎡ ⎤= ⎢ ⎥⎣ ⎦∫
1/ 2
1/ 21 2 2lim 2( / 2) 2 lim2T T
TTT T T→∞ →∞
⎡ ⎤⎧ ⎫⎡ ⎤− = −⎢ ⎥⎨ ⎬⎣ ⎦ ⎩ ⎭⎢ ⎥⎣ ⎦
= 0 Since P = 0, it is not a power signal P.9 # Find the energy and power of the signal 10( ) ( )tx t e u t−=
210 ( ) 20
0 0
tu t tE e dt e dt∞ ∞
− −= =∫ ∫
=20 20 0
0
1 1
20 20 20
te e e e∞− − ∞ ∞ −−= =
− − −
= 120
joules
/ 2
210
/ 2
1lim ( )T
t
TT
P e u t dtT
−
→∞−
= ∫
2/ 2 20
20
0 1
1 1lim lim20
TT tt
T T
eP e dtT T
−−
→∞ →∞
⎡ ⎤= = ⎢ ⎥
−⎢ ⎥⎣ ⎦∫
20. 2021lim 0
20
T
T
e eT
− −
→∞
⎡ ⎤−⎢ ⎥ =⎢ ⎥−
⎢ ⎥⎣ ⎦
P.10 # Determine the energy of the signal (2 4)( ) j tx t e +=
22 (2 4)( ) j tE x t dt e dt
∞ ∞+
−∞ −∞
= =∫ ∫
(2 4) (2 4)j t j te e dt∞
+ −
−∞
= ∫
4
4 04 4 4
tt e e ee dt
∞∞ ∞ −∞
−∞ −∞
− ∞ −= = = = = ∞∫
P.11 # Find the average power of a periodic signal ( ) cos(5 )x t tπ=
When T = fundamental period = 2 2 2 0.45 5
π πω π
= = =
152
2
12 5
1 5 (1 cos10 )cos (5 )2 2
T
T
tt dt dtT
πρ π− −
+= =∫ ∫
=1/ 5
1/ 5
5 sin104 10
tt ππ
−
⎡ ⎤+⎢ ⎥⎣ ⎦
= 5 2 sin 2 sin 2 14 5 10 2
wattsπ ππ−⎡ ⎤+ =⎢ ⎥⎣ ⎦
.
P.12 # Find the power of the signal ? ( )x t
A 2π 4π t ( ) sinx t A t=
2
2 2 2
0 0
1 1( ) sin2
T
P x t dt A tdtT
π
π= =∫ ∫
2 22 2
0 0
1 1 1(1 cos 2 ) cos 22 2 2 2 2A AP t dt t dt
π π
π π⎛ ⎞= − = −⎜ ⎟⎝ ⎠∫ ∫
[ ]22 2
2
00
sin 2 2 sin 22 2 4 8A t t AP t t
ππ
π π⎡ ⎤= − = −⎢ ⎥⎣ ⎦
[ ]2 2 2
4 sin 4 ) (0 0) 48 8 2A A AXπ π ππ π
= − − − = =
Basic Operations on signals :
1) Amplitude Scaling : Let x(t) denote a continuous-time signal. The signal y(t) resulting from amplitude scaling applied to x(t) is defined by y(t) = C x(t),
C= scaling factor. 2) Addition : Let x1(t) and x2(t) denote a pair of continuous-time signal. The signal
y(t) obtained by the addition of x1(t) and x2(t) is defined by y(t) = x1(t)+x2(t) . 3) Multiplication : Let x1(t) and x2(t) denote a pair of continuous-time signal. The
signal y(t) resulting from the multiplication of x1(t) by x2(t) is defined by y(t) = x1(t) x2(t) .
4) Differentiation : Let x(t) denote a continuous-time signal. The derivation of x(t)
w.r.t time is defined by ( ) ( )dy t x tdt
= .
Ex : Inductor performs Differentiation. The voltage developed across inductor
L is ( ) ( )dV t L i tdt
=
5) Integration : Let x(t) denote a continuous-time signal. The integral of x(t) w.r.t
time is defined by ( ) ( )t
y t x dτ τ−∞
= ∫ τ = integration variable.
Ex : Capacitor performs integration. i(t)
C The voltage v(t) devoped across capacitor is defined by
1( ) ( ) .t
v t i dC
τ τ−∞
= ∫
6) Time Scaling : Let x(t) denote a continuous-time signal. The signal y(t) is obtained by scaling the independent variable time t by a factor a is defined by
y(t) = x(at). If a > 1 the signal y(t) is a compressed version of x(t). (or) 0 < a < 1 the signal y(t) is an expanded version of x(t).
x(t) y(t) = x(2t) y(t) =x(1/2 t) 1 1 1 -1 0 1 t -½ 0 ½ t -2 0 2
7) Time Shifting Let y(t) denote a continuous-time signal that is derived from another continuous-time signal x(t) through Time Shifting . ( ) ( 3)y t x t= +
x(t) ( ) ( 3)y t x t= + 1 1 -1 0 1 t -4 -3 -2 -1 0 t (a) (b)
8) Precedence rule for time SHIFTING & time SCALING. Let y(t) denote a continuous-time signal that is derived from another continuous-time signal x(t) through a combination of time SHIFTING & time SCALING, ( ) ( )y t x at b= −
(Verification (0) ( )y x b= − , (0)by xa
⎛ ⎞ =⎜ ⎟⎝ ⎠
)
First perform time shifting operation on x(t), [Let an intermediate signal be v(t)] ie ( ) ( )v t x t b= − Next, the time scaling operation is performed on v(t), ( ) ( )y t v at= ( ) ( )y t x at b= −
x(t) v(t)=x(t+3) y(t)=v(2t)
1 1 1 -1 0 1 t -4 -3 -2 -1 0 t -2 -1 0 t (a) (b) (c)
Elementary Signals : (1) Real Exponential signal :- A Real Exponential signal in its general form is
( ) atx t Be= . where B & a are real parameters. Decaying exponential for which a < 0 Growing exponential a > 0 x(t) x(t) atBe− atBe t t A complex exponential signal in its general form ( ) atx t Be= where B & a assumes complex values. Ex : complex exponential signals j te ω (C.T.S), j ne Ω (D.T.S) D.T.S ( ) nx n Br= r eα= =B neα (2) Sinusoidal signal : Its general form ( ) cos( )x t A tω φ= + A = amplitude, ω = frequency in radians per second, φ = Phase angle in radians. A sinusoidal signal is an example of a periodic signal. The period of which is
ω = 2π f 1fT
= 2T πω
∴ = .
Complex exponential signal : General form is x(t) = est s is a complex variable s jσ= + Ω ( )x t = ( )st j t t j te e e eσ σ+ Ω Ω= = cos sinj te t j tΩ = Ω + Ω ( ) (cos sin )tx t e t j tσ= Ω + Ω Depending the values of &σ Ω we get different signals
Relation between sinusoidal and complex exponential signals : Consider the complex exponential signal je φ cos sinje jθ θ θ= + (or) cos sinj te t j tω ω ω= + From the above equation we can express the continuous time sinusoidal signal i.e. A cos ( ( )tω φ+ as the real part of the complex exponential signal j tBe ω . When B is a complex quantity Let B= A je φ ∴A cos ( )tω φ+ = Real part of (
j t
Beω
) = Re{ }j tBe ω
( ) )j t j j t j tBe Ae e Aeω φ ω ω φ+= = = A cos ( ) sin( )t j tω φ ω φ+ + + A sin { }( ) j t
mt I Be ωω φ+ = (3) Exponentially Damped sinusoidal signal : The multiplication of a sinusoidal signal by a real valued decaying exponential signal results in a new signal referred to as an exponentially Damped sinusoidal signal. ( ) sin( )atx t Ae tω φ−= + a > 0. ( )x t 0φ = a > 0. (4) UNIT STEP – FUNCTION : The discrete time version of the unit-step function commonly denoted by u(n), is defined by ( ) 1u n = n ≥ 0 ( ) 0u n = n < 0. The continuous-time version of the step function, commonly denoted by u(t) is defined by
u(t) = 1 t > 0
u(t) = 12
t = 0
u(t) = 0 t < 0 u(t) 1.0 0 t P.13. Express ( )x t in terms of unit step signal.
x (t) = A 0 ≤ 2Tt <
= 0 2Tt >
Let T = 1 sec A 0.5t < is t < 0.5 0.5t >
t > -0.5 t > 0.5 t < -0.5 -1 -0.5 0 0.5 1 t
x (t) = A u 1 12 2
t Au t⎛ ⎞ ⎛ ⎞+ − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
x 1(t) x 2(t) Rectangular pulse x (t) is represented as the difference between two time shifted step functions. x2(t) x1(t) A A -0.5 0.5 Unit Impulse function : The discrete-time version of the impulse- commonly denoted by ( )nδ is defined by ( )nδ = 1 n = 0 ( )nδ =1 = 0 n ≠ 0 0
The continuous-time version of the unit impulse, commonly denoted by ( )tδ is defined by the following pair of relations. ( ) 0tδ = t≠ 0
( ) 1t dtδ∞
−∞
=∫
The impulse ( )tδ is zero every where except at the origin. At the origin the total area under the unit impulse is unity. The impulse ( )tδ is also referred to as the Direct delta function. x (t) x (t) x (t) Area=1 Area=1 Area=1 x (t) ^ 0 t ( ) ( )x t tδ∴ =
( )tδ∴ is a limiting form of a rectangular pulse of unit area. The step function u(t) is the integral of the impulse ( )tδ with respect to time t.
( ) ( )t
u t dδ λ λ−∞
= ∫ t ≥ 0
=0 t < 0 The impulse ( )tδ is the derivative of the step function u(t) with respect to time t.
( ) ( )t dt u tδ∞
−∞
=∫
( ) ( ) 0dt u tdt
δ = = t > 0 (not defined for t = 0 in terms of u(t))
t < 0
Unit Ramp Function : The integral of the step function u(t) is a ramp function of unit slope. r(t) = t t ≥ 0 = 0 t < 0 Which is defined as r(t)
Slope =1 t
5. FOURIER SERIES : A motivation for using the Fourier Series or the Fourier Transform is to obtain the spectrum of a given signal, from the spectrum we can estimate the frequency contents of the signal. The Fourier Series is used to resolve the periodic signal in to an infinite sum of sine and cosine terms. Periodic signal Ex: x (t) A T t -A x (t) x (t) A A T 2T 3T t T 2T t Periodic signals with period = T Let gp(t) denote a periodic signal with period T0. By applying Fourier series
{ }0 0 0
1( ) 2 cos( ) sin(p n n
ng t a a n t b ntω ω
∞
=
= + +∑ → (5.1)
0 02 fω π= , 00
1fT
= ∴Fundamental frequency of gp(t)
an & bn are the coefficients of the cosine and sine terms. The terms cos 0( )n tω & sin 0( )n tω are called BASIS Functions. n = 1,2,3,………… ∞ a0 is the average value of gp(t).
0
nT
is nth harmonic of the fundamental frequency, 00
1fT
=
To determine the average value a0 of gp(t) Integrate equation (5.1) on both sides over the
interval 0
2T− to 0
2T
{ }0 0 0
0 0 0
2 2 2
0 0 01
2 2 2
( ) 2 cos( ) sin( )
T T T
p n nnT T T
g t dt a dt a n t b n t dtω ω∞
=− − −
= + +∑∫ ∫ ∫
=
0 0
0 0
2 2
1 0 2 0
2 2
cos cos 2 ........
T T
T T
a tdt a tdtω ω− −
+ +∫ ∫
= 0 01 2
0 0
2 4sin * sin * .........2 4T Ta t a t
T Tπ π
π π+ +
= { } { }0 01 2sin sin sin 2 sin 2 ..........
2 4T Ta aπ π π ππ π
+ + + +
= 0 01 2(0) (0) .......... 0
2 4T Ta aπ π
+ + =
=
0 0
0 0
2 2
1 0 2 0
2 2
sin sin ....
T T
T T
b tdt b tdtω ω− −
+ +∫ ∫
= 0 01 2
0 0
2 4cos * cos * ......2 4T Tb t b t
T Tπ π
π π⎧ ⎫ ⎧ ⎫
− − +⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭
= { } { }0 01 21 1 1 1 ...... 0
2 4T Tb bπ π
− − + − − + + =
0
0
20 0
0 0 0 0
2
( ) *2 2
T
pT
T Tg t dt a t a a T−
⎛ ⎞∴ = = + =⎜ ⎟⎝ ⎠∫
0
0
2
00
2
1 ( )
T
pT
a g t dtT −
∴ = ∫
To determine an multiply both sides of equation (5.1) by cos(n 0tω ) and integrate over
the interval 0
2T− to 0
2T
0
0
2
0 02
1 2( )cos
T
n pT
ta g t dtT T
πη
−
⎛ ⎞= ⎜ ⎟
⎝ ⎠∫ n=1,2,…….. ∞
To determine bn multiply both sides of equation (5.1) by sin(n 0tω ) and integrate over the
interval 0
2T− to 0
2T
( )0
0
2
00
2
1 ( )sin
T
n pT
b g t n t dtT
ω−
= ∫
0
0
2
00
2
1 ( )
T
pT
a g t dtT −
∴ = ∫
0
0
2
00
2
1 ( )cos( )
T
n pT
a g t n t dtT
ω−
= ∫
Integration of cosine or sine function for a complete period is zero.
0
0
2
00
2
1 ( )sin( )
T
n pT
b g t n t dtT
ω−
= ∫
To apply the Fourier Series the function gp(t) must satisfy the following conditions.
1) The function gp(t) should be single valued with in the interval T0. 2) The function gp(t) may have a finite number of discontinuities in the interval T0. 3) The function gp(t) may have a finite number of maxima & minima in the interval
T0. 4) The function gp(t) should be absolutely integrable.
0
0
2
2
( )
T
pT
g t dt−
∴ < ∞∫
These conditions are known as DIRICHLET’s Conditons.
The above functions from an orthogonal set over the interval T0 in that they satisfy the following set of relations. These relations are used to evaluate an & bn
0
0
20
0 0
2
cos( ) ( )2
T
T
Tm t cos n t dtω ω−
=∫ m = n
0
0
2
0 0
2
cos( ) ( ) 0
T
T
m t cos n t dtω ω−
=∫ m ≠ n
0
0
2
0 0
2
cos( )sin( ) 0
T
T
m t n t dtω ω−
=∫ for all m and n.
0
0
20
0 0
2
sin( )sin( )2
T
T
Tm t n t dtω ω−
=∫ m = n
0
0
2
0 0
2
sin( )sin( ) 0
T
T
m t n t dtω ω−
=∫ m ≠ n
Symmetry Conditions : i) If the function f(t) is even then f(t) = f(-t) Ex :- cos(t), t2, t sin t (t= odd, sin( )t = odd)
Cost = 2 4 6
1 ........2 4 6t t t
− + + +
0
( ) 2 ( )f t dt f t dt∞ ∞
−∞
=∫ ∫ if f(t) is even.
Only cosine terms present (d.c term optional). The sum (or) product of two or more even functions is an even function. ii) f(t) is odd f(t) = -f(-t). Only sine terms are present (d.c term optional) Ex : sin t, t3, t cos t .
f(t) is odd ( ) 0f t dt∞
−∞
=∫
The sum of two or more odd functions is an odd function and the product of two odd functions is even function. P.14 # Obtain the Fourier components of the periodic signal shown below. f(t) -T/2 -T/4 0 T/4 T/2 t
Periodicity = 2 2T Tto T−
= .
f(t) = f(-t) ∴ odd component is zero. a0 = 0, bn=0
2
2
1 ( ) cos
T
nT
a f t n tdtT
ω−
= ∫
( )f t A= − 2 4T Tt− −< <
( )f t A= 4 4T Tt− −< <
( )f t A= − 4 2T Tt< <
4 4 2
2 4 4
cos cos cos
T T T
nT T T
Aa n tdt n tdt n tdtT
ω ω ω
−
− −
⎡ ⎤⎢ ⎥= − + + −⎢ ⎥⎢ ⎥⎣ ⎦∫ ∫ ∫
sin sin sin* * *A n t n t n t
T n n nω ω ω
ω ω ω− −⎡ ⎤= + +⎢ ⎥⎣ ⎦
2 2 2 2 2sin . sin . sin sin . sin sin
2 2 4 4 2 4A T T T T T Tn n n n n n
Tn T T T T T Tπ π π π π πω
ω⎡ ⎤= − − + − − +⎢ ⎥⎣ ⎦
(sin nπ =0)
. sin sin. .2 2AT n n
T nππ
π⎡ ⎤= − +⎢ ⎥⎣ ⎦
2. sin
2. . 2nA nan
ππ
=
a0=0, bn=0
0 0 01
( ) 2 (cos sin )n nn
f t a a n t b tω ω∞
=
= + +∑
01
2 2 cos sin2 2n
A nn tn
πωπ
∞
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
∑
0
1
cos4 sin2n
n tA nnω π
π
∞
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
∑
P.15 # Obtain the Fourier components of the periodic rectangular wave form shown in figure. f(t) A -T/2 -T/4 0 T/4 T/2 t
( ) 0f t = 2 4T Tt− −< <
( )f t A= 4 4T Tt−< <
( ) 0f t = 4 2T Tt< <
f (t) = f (-t) 0nb∴ =
2 4
0
2 4
1 1( )
T T
T T
a f t dt AdtT T− −
= =∫ ∫
1 1* .2 2T AA A
T T= = =
4
4
1 sincos *
T
nT
A nwta A n tdtT T n
ωω−
= =∫
2 2sin sin. 2 4 4nA T T Ta n n
T n T Tπ π
π⎡ ⎤= +⎢ ⎥⎣ ⎦
sin2
A nn
ππ
⎛ ⎞= ⎜ ⎟⎝ ⎠
01
( ) 2 sin cos2 2n
A A nf t n tn
π ωπ
∞
=
⎛ ⎞∴ = + ⎜ ⎟⎝ ⎠
∑
P.16 # Obtain the trigonometric Fourier Series for the Half Wave Rectified sine wave shown in figure. f (t) A 0 T/2 T 3T/2 t no symmetry ∴series may contain both sine & cosine terms. f (t) = A sin ω t
2
00
1 cossin *
T
A ta A tdtT T
ωωω
−⎡ ⎤= = ⎢ ⎥⎣ ⎦∫
2 2cos 1
2 2 2A T A A
Tπ
π π π⎡ ⎤= − − = =⎢ ⎥⎣ ⎦
2 (cos 1)2 (1 )n
Aa nn
ππ
= +−
for n=1 the above equation is infinite and hence we have to integrate separately to evaluate a1. ∴a1 = 0.
bn = 0, b1 = 4A
0 0 02 2( ) 1 sin cos 2 cos 4 .....
2 3 15Af t t t tπ ω ω ωπ⎡ ⎤= + − −⎢ ⎥⎣ ⎦
Note : if ( )2Tf t f t⎛ ⎞+ =⎜ ⎟
⎝ ⎠ only even harmonics are present .
( )2Tf t f t⎛ ⎞+ = −⎜ ⎟
⎝ ⎠ only odd harmonics are present and hence the waveform has
Half Wave Symmetry.
P.17 # Obtain the trigonometric Fourier Series of the triangular wave shown in figure. A t -T -T/2 T/2 T 3T/2 -A The waveform has equal Positive & Negative area in one cycle ∴a0 = 0 1 2 1 1 2 1( )( ) ( )( )y y x x x x y y− − = − − an = 0 odd function
2 2
0 00
2
1 2( )sin ( )sin
T T
nT
b f t n tdt f t n tdtT T
ω ω−
= =∫ ∫
For the region 04Tt< < ( ) 4 Af t t
T=
4 2T Tt< < ( ) 4 2Af t t A
T= − +
0 0 02 2 2
8 1 1( ) sin sin 3 sin 5 ....3 5
Af t t t tω ω ωπ
⎡ ⎤= − + +⎢ ⎥⎣ ⎦.
6. Complex Exponential form of Fourier Series : The trigonometric form of the Fourier series is
( )0 0 01
( ) 2 cos sinp n nn
g t a a n t b n tω ω∞
=
= + +∑ → (6. 1)
0 0
0cos2
jn t jn te en tω ω
ω−+
=
0 0
0sin2
jn t jn te en tj
ω ω
ω−−
=
∴substituting in equation (6.1)
{ }0 0
0 00
1 1( ) 2
jn t jn tjn t jn t
p n nn n
e eg t a a e e bj
ω ωω ω
−∞ ∞−
= =
⎧ ⎫−= + + + ⎨ ⎬
⎩ ⎭∑ ∑
{ }0 00
1( ) ( )jn t jn t
n n n nn
a a jb e a jb eω ω∞
−
=
= + − + +∑
Let n n nc a jb= − c-n is the complex conjugate of cn
n n nc a jb− = +
0
0
2
00
2
1 ( ) cos
T
n pT
a g t n tdtT
ω−
= ∫
0 0c a=
0
0
2
00
2
1 ( ) cos
T
n pT
b g t n tdtT
ω−
= ∫
{ }0
0
2
0 00
2
1 ( ) cos sin
T
n pT
c g t n t j n t dtT
ω ω−
∴ = −∫
0
0
0
2
02
1 ( )
T
jn tp
T
g t e dtT
ω−
−
= ∫
Similarly
0
0
0
2
02
1 ( )
T
jn tn p
T
c g t e dtT
ω−
−
= ∫ [ n nc c− = for negative values of n]
0 0
01 1
( ) jn t jn tp n n
n ng t c c e c eω ω
∞ +∞−
−= =+
∴ = + +∑ ∑
0 00
1 1
jn t jn tn n
n nc c e c eω ω
∞ −∞
= =−
= + +∑ ∑
0( ) jn t
p nn
g t c e ω∞
=−∞
= ∑
Cn is complex Fourier coefficient.
0
0
0
2
02
1 ( )
T
jn tn p
T
c g t e dtT
ω−
−
= ∫ for n=0, ±1, ±2, ±3,……..∞
The complex exponential Fourier Series expansion of gp(t) is
0( ) jn tp n
ng t c e ω
∞
=−∞
= ∑ → (6.2)
In equation (6.1) only positive frequencies occur. In equation (6.2) both positive & negative frequencies occur. Advantage of equation (6.2) is over equation (6.1) is compact mathematical description of the periodic signal. Properties of Fourier Series : (1) If gp(t) is real then *
n nc c−= nc− = complex conjugate of Cn
or *n nc c−=
0
0
0
2
02
1 ( )
T
jn tn p
T
c g t e dtT
ω−
−
= ∫ .
0
0
0
2*
02
1 ( )
T
jn tn p n
T
c g t e dt cT
ω−
−
= =∫
(2) If gp(t) is real & even ie gp(t) = gp(-t) Imaginary part of nc =0
∴ nc is purely real & even symmetry (3) If gp(t) is odd gp(t) = -gp(-t) Real part of nc =0
∴ nc is purely Imaginary & odd symmetry Proof of (2) gp(t) = gp(-t) nc is purely real .
0
0
0
2
02
1 ( )
T
jn tn p
T
c g t e dtT
ω−
−
= ∫
0
0 0
0
0 2
0 02
1 ( ) ( )
T
jn t jn tp p
T
g t e dt g t e dtT
ω ω− −
−
⎡ ⎤⎢ ⎥= +⎢ ⎥⎢ ⎥⎣ ⎦∫ ∫
Put t = -t is the first integral term
0 0
0 0
2 2
0 0 0
1 ( ) ( )
T T
jn t jn tp pg t e dt g t e dt
Tω ω−
⎡ ⎤⎢ ⎥= − +⎢ ⎥⎢ ⎥⎣ ⎦∫ ∫ gp(t) = gp(-t)
0
2
00 0
2 ( ) cos
T
n pc g t n tdtT
ω⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦∫ ∴ nc is purely real .
Proof for (3) gp(t) = -gp(-t) nc is purely Imaginary .
0
0 0
0
0 2
0 02
1 ( ) ( )
T
jn t jn tp p
T
g t e dt g t e dtT
ω ω− −
−
⎡ ⎤⎢ ⎥= +⎢ ⎥⎢ ⎥⎣ ⎦∫ ∫
Put t = -t is the first integral term
0 0
0 0
2 2
0 00 0
1 1( ) ( )
T T
jn t jn tp pg t e dt g t e dt
T Tω ω−= − +∫ ∫ -gp(t) = gp(-t)
( )0
0 0
2
0 0
1 ( )
T
jn t jn tpg t e e dt
Tω ω−
⎡ ⎤⎢ ⎥= − −⎢ ⎥⎢ ⎥⎣ ⎦∫
02
00 0
2 ( )sin( )
T
n pjc g t n t dt
Tω= − ∫ ∴ nc is purely Imaginary .
Spectrum of Signals : Each harmonic in Fourier Series representation has someone amplitude. ∴A spectrum is drawn between amplitude and harmonic frequencies We know that *
n nc c−= ∴Their magnitudes are equal n nc c−= The frequency spectrum of real periodic signals is symmetric. Phase is arg( ) arg( )n nc c−= − Phase spectrum is anti symmetric. P.18# Obtain the complex Fourier Series representation of the train of rectangular pulses of duration T and period T0 as shown in figure. gp(t)
-T0 -T/2 0 T/2 T0 t
( )pg t A= 2 2T Tt− −≤ ≤
( ) 0pg t = for remaining period over one period
0( ) jn t
p nn
g t c e ω∞
=−∞
= ∑
0
0 0
0
2 2
0 022
1 1( )
T T
jn t jn tn p
T T
c g t e dt Ae dtT T
ω ω− −
− −
= =∫ ∫
0
0 0
0
0 2 2
0 0
*2
T Tjn t jn jn
jn
TA e A e eT jT n
ω ω ω
ω π
− − +
−
⎡ ⎤⎡ ⎤= = − −⎢ ⎥⎢ ⎥
⎣ ⎦ ⎣ ⎦
=0 02 2
0sin2 2
T Tjn jnn TA e e A
n j n
ω ωω
π π
−⎡ ⎤− ⎛ ⎞⎢ ⎥ = ⎜ ⎟⎢ ⎥ ⎝ ⎠
⎢ ⎥⎣ ⎦
for n=0, ±1, ±2, ±3,……..
00( ) sin
2jn t
pn
n TAg t en
ωωπ
∞
=−∞
⎛ ⎞∴ = ⎜ ⎟⎝ ⎠
∑ complex Fourier Series
Sinc Function :
sinc sin( )( ) πλλπλ
=
λ = Independent variable It has maximum value at λ =0 and approaches to 0 as λ approaches to infinity. It goes through zero at λ = ±1, ±2, ±3,…….. Sinc (λ )
λ
0sin2n
n TAcn
ωπ
⎛ ⎞= ⎜ ⎟⎝ ⎠
0
0
2sin sin2
n f T nTT
π π=
0 0 0
0 0 0
0
sin. sin
nTT T AT nTA cT T T Tn
T
π
π
⎛ ⎞⎜ ⎟ ⎛ ⎞⎝ ⎠= = ⎜ ⎟
⎝ ⎠
0
nT
= nth harmonic of the fundamental frequency f0.
0
0 0
sinnAT nTc cT T
⎛ ⎞= ⎜ ⎟
⎝ ⎠ Let
0
0.2TT
= (Duty cycle) 0 5TT=
ATp/T0
-10/T0 -5/T0 5/T0 10/T0 n/T0
π
n/T0
-π Phase spectrum.
0
sinnA n Tc
n Tπ
π⎛ ⎞
= ⎜ ⎟⎝ ⎠
1 1. sin 0arg( ). cos 0
mn
I pc tan tanR P
− −⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
=0 nc > 0 0 0
1 5toT T
0
0
sin 0, 5Tn TT Tπ⎛ ⎞
> =⎜ ⎟⎝ ⎠
1 sinarg( )cosnc tan π π
π− ⎡ ⎤= = ±⎢ ⎥⎣ ⎦
nc < 0 0 0
5 10toT T
0 5 : sin 05
nT T π⎛ ⎞= >⎜ ⎟⎝ ⎠
true for 0 0
1 5toT T
0 0
1 5mT T
< <
sin 05
nπ⎛ ⎞ <⎜ ⎟⎝ ⎠
true for 0 0
5 10toT T
0 0
5 10nT T
< <
The above spectrum is Discrete spectrum. The spacing between two successive samples
is 0
1T
.
The amplitude spectrum has zero values at frequencies that are integer multiplies of
1T
. 1 2 3, , ....T T T⎡ ⎤⎢ ⎥⎣ ⎦
0 1 50.2
TT= =
0
1 5T T
∴ =
P.19 # Find the exponential Fourier Series & plot the amplitude & phase spectrums for the periodic waveform shown in figure. x(t) e-t
-1.5 -1 -0.5 0 0.5 1 1.5 t T0=0.5
( ) tx t e−=
0
0
0
2
02
1 ( )
T
jn tn
T
c x t e dtT
ω−
−
= ∫
0 00
2 2 202 40.5 5
fTπ πω π π π= = = = =
0.5 0.54 (1 4 )
0 0
2 2t jn t j n tnc e e dt e dtπ π− − − += =∫ ∫
= (1 4 )2 *(1 4 )
j n tej n
π
π− +
− +
{ }(0.5 2 )2 1(1 4 )
j nej n
π
π− += −
− +
{ }0.5 22 1(1 4 )
j ne ej n
π
π− −−
= −+
2 1j ne π− =
e-0.5=0.606
( 2)( 0.393) 0.791 4 1 4nc
j n j nπ π− −
= =+ +
2 2 20
1 10.79 0.791 16 1 ( )
ncn nπ ω
= =+ +
1arg( ) tan (4 )nc nπ−= −
P.20 # Find the exponential Fourier Series and find an & bn of the Trigonometric series and compare the results. f(t) A -T -T/2 0 T/2 T t -A
( )f t A= − 02T t−< < Half wave symmetry
( )f t A= 02Tt< <
2
00
2 ( )sin
T
nb f t n tdtT
ω∴ = ∫
0
2
2
1 ( )
T
jn tn
T
c f t e dtT
ω−
−
= ∫
0 0
0 2
02
1 ( )
T
jn t jn t
T
A e dt Ae dtT
ω ω− −
−
⎡ ⎤⎢ ⎥= = − +⎢ ⎥⎢ ⎥⎣ ⎦∫ ∫
0
0
0 0( ) ( )1( 1) * *
jn tjn t
jn jnA e eT
ωω
ω ω
−−
− −
⎡ ⎤= − +⎢ ⎥
⎣ ⎦
{ } [ ]0 0 2 22 2
jn jnA Ae e e e jnj n j n
π π ω ππ π
−= − − + + = −− −
( )1jnn
Ac j en
π
π= − 1jne π = if ‘n’ is even
1jne π+ = − if ‘n’ is odd 2
nAc j
nπ⎛ ⎞∴ = − ⎜ ⎟⎝ ⎠
for ‘n’ is odd
0( ) jn t
nn
Af t c en
ω
π
∞
=−∞
= ∑
n n nc a jb= − [ ]n e na R c∴ =
[ ]n m nb I c= − f(t) is Ant symmetrical about the cosine. f(t) is ODD a0 = an =0
2
0
2
1 ( )sin
T
nT
b f t n tdtT
ω−
= ∫
0 2
0 00
2
1 ( )sin sin
T
T
A n tdt A n tdtT
ω ω−
= − +∫ ∫
0 0
0 0
cos cos* *n t n tA AT n T n
ω ωω ω
⎡ ⎤ ⎡ ⎤− −= − +⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
0 0
0 0 0
cos cos1 2 2* 1
T Tn nA AT n n T n
ω ω
ω ω ω
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
= − − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
0 00
1 cos 1 cos2 2
A T Tn nnT
ω ωω
⎡ ⎤= − + −⎢ ⎥⎣ ⎦
[ ]2 21 cos . 1 cos2 2A T An n
T nnTT
π ππ π⎡ ⎤= − = −⎢ ⎥⎣ ⎦
cos nπ = 0 n = even cos nπ = -1 n = odd
[ ] 21 1nA Ab
n nπ π= + = for n = odd.
|Cn| 0 4π 8π 12π 0ω arg Cn 0 4π 8π 12π 0ω P.21 # Determine the complex exponential Fourier Series of the signal shown below. x(t) 5 0 2π 4π 6π 0tω T0=Periodicity= 2π 1 2 1 1 2 1( )( ) ( )( )y y x x x x y y− − = − − 1 1( )x y 2 2( )x y ( 0)(2 0) ( 0)(5 0)y xπ− − = − − (0 0) (2π ,5) 2π y=5x x= 0tω y=x(t)
2π x(t) = 5 0tω 05( )
2x t tω
π∴ =
0
2
0 00
1 52 2
jn tnc te d t
πωω ω
π π−⎛ ⎞= ⎜ ⎟
⎝ ⎠∫ .
=( )
( )0
02 2
5 51 *( ) 22
jn te jn t jjn n
ω
ωππ
−⎡ ⎤− − =⎢ ⎥−⎣ ⎦
7. FOURIER TRANSFORM
A periodic signal may be expressed as a sum of spectral components by using Fourier series. These components have finite amplitudes and represented by finite
frequency intervals 00
1fT
=
To represent a signal g(t) i.e. non periodic in terms of exponential signals. First construct a periodic function ( )pg t of period T0 in such a way that g(t) defines one cycle of the periodic function
0
( ) lim ( )pTg t g t
→∞=
g(t) 0 t gp(t) -T0 0 T0 t The periodic function ( )pg t in terms of the complex exponential form of the Fourier series
0( ) jn tp n
ng t c e ω
∞
=−∞
= ∑ → (7.1)
0
0
0
2
00
2
1 ( )
T
jn tn p
T
c g t e d tdtT
ω ω−
−
= ∫ → (7.2)
Define 0
1 ,fT
Δ = 0
nnfT
= 0( )n nG f c T=
2( ) ( ) .nj f tp n
ng t G f e fπ
∞
=−∞
= Δ∑ → (7.3)
0
0
22
2
( ) ( ) n
T
j f tn p
T
G f g t e dtπ−
−
= ∫ → (7.4)
Let the period T0 approaches infinity, the discrete frequency fn approaches the continuous frequency variable f, and the discrete sum in equation (7.3) becomes an integral defining the area under a continuous function of frequency f, the function ( )pg t approaches g(t).
2( ) ( ) j ftg t G f e dfπ∞
−∞
= ∫ → (7.5)
2( ) ( ) j ftG f g t e dtπ∞
−
−∞
= ∫ → (7.6)
The function G(f) is a Transformed version of g(t) and is referred to as the Fourier transform of g(t). The time function g(t) is similarly referred to as the Inverse Fourier Transform of G(f). The functions g(t)and G(f) are said to constitute a Fourier Transform pair. ( ) ( )g t G f⇔ A signal g(t) to be Fourier Transformable for that g(t) must satisfy the following conditions. 1) The function g(t) should be a single valued, with a finite number of maxima & minima and a finite number of discontinuities in any finite time interval. 2) The function g(t) is absolutely integrable.
( )g t dt∞
−∞
< ∞∫
[ ]( ) ( )F g t G f= [ ]( ) ( )F G f g t= . g(t) is real G(f) = G*(-f).
2( ) ( ) j ftG f g t e dtπ∞
−
−∞
= ∫
* 2( ) ( ) j ftG f g t e dtπ∞
−∞
= ∫
Replace f = -f * 2( ) ( ) j ftG f g t e dtπ∞
−
−∞
− = ∫
( ) ( )G f G f= −
{ } { }( ) ( )Arg G f Arg G f= − −
Amplitude spectrum of real valued signal g(t) is even function of frequency & phase spectrum of real valued signal g(t) is even function of frequency.
2( ) ( ) j ftG f g t e dtπ∞
−
−∞
= ∫ .
Or ( ) ( ) j tG g t e dtωω∞
−
−∞
= ∫ .
2( ) ( ) j ftg t G f e dfπ∞
−∞
= ∫ .
Or 1( ) ( )2
j tg t G e dωω ωπ
∞
−∞
= ∫
Let g(t) is a Real Valued signal then its Fourier Transform satisfies Complex-Conjugate symmetry is *( ) ( )G f G f= − *( ) ( )G f G f= − .
2( ) ( ) j ftG f g t e dtπ∞
−
−∞
= ∫ → (7.7)
( ) ( ) ( )cos 2 ( )sin 2R IG f jG f g t ftdt j g t ftdtπ π∞ ∞
−∞ −∞
+ = −∫ ∫ → (7.8)
2( ) ( ) j ftG f g t e dtπ∞
−∞
− = ∫
( ) ( ) ( ) cos 2 ( )sin 2R IG f jG f g t ftdt j g t ftdtπ π∞ ∞
−∞ −∞
− + − = +∫ ∫ → (7.9)
* 2( ) ( ) j ftG f g t e dtπ∞
−
−∞
− = ∫ → (7.10)
∴7.7 = 7.10 Comparing (7.8) & (7.9) ( ) ( )R RG f G f= − Real component is even function ( ) ( )I IG f G f= − − Imaginary component is ODD function. Magnitude spectrum is even function :
( ) ( )G f G f= − * *( ) ( ) ( ) ( ) ( ) ( )G f G f G f G f G f G f= = − − = − .
Phase spectrum is ODD function: 1 1 1( ) ( ) ( )( ) tan tan tan
( ) ( ) ( )I I I
R R R
G f G f G fArgG fG f G f G f
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤− − −= = = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
P.22 # Find the Fourier Transform of Rectangular pulse shown in figure and draw the Amplitude spectrum. g(t) A -T/2 0 T/2 t (a) Consider a rectangular pulse of unit Amplitude and unit duration centered at t = 0. Its mathematical form is g(t) 1 -1/2 0 1/2 t
rect (t) = 1 1 12 2
t− < <
rect (t) = 0 12
t >
∴g(t) = rect (t)
For fig (a) g(t) = A rect tT⎛ ⎞⎜ ⎟⎝ ⎠
22 2
2
22
( )2
T Tj ft
j ft
TT
eG f Ae dt Aj f
ππ
π
−−
−−
= =−∫
2 2
2 2
2 2
T T j fT j fTj f j fA A e ee ej f f j
π ππ π
π π
−− +⎡ ⎤ ⎡ ⎤−= − =⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎣ ⎦
= sin( )sin( ) sin ( ).
A fTfT AT AT c fTf f T
πππ π
= =
∴A rect tT⎛ ⎞⎜ ⎟⎝ ⎠
⇔ sin ( )AT c fT .
( ) sin ( )G f AT c fT= ( )G f AT
5T− 4
T− 3
T− 2
T− 1
T− 0 1
T 2
T 3
T 4
T 5
T f
The above spectrum is Continuous spectrum. The above example shows that the relation ship between the time domain & frequency domain description of a signal. A pulse narrow in time has a significant frequency description over a wide range of frequencies and vice versa. ( )G f π -3/T -2/T -1/T 0 1/T 2/T 3/T f -π Phase Response ( )G f = 0 ( ) 0G f >
( )G f π= ± ( ) 0G f <
( ) sin( ) sin ( )AG f fT AT c fTf
ππ
= =
If G(f) is positive then its phase is zero, if G(f) is negative then its phase is 1800 or -1800. Choose phase as +π for positive frequencies -π for negative frequencies or vice versa.
Properties of the Fourier Transforms : (1) Linearity : 1 1( ) ( )g t G f⇔ , 2 2( ) ( )g t G f⇔ 1 2 1 2( ) ( ) ( ) ( )ag t bg t aG f bG f+ ⇔ +
[ ] { } 21 2 1 2( ) ( ) ( ) ( ) j ftF ag t bg t ag t bg t e dtπ
∞−
−∞
+ = +∫
= 1 2( ) ( )aG f bG f+ P.23# Find the Fourier Transform of the decaying exponential as shown in figure. g(t)
( ) ( )atg t e u t−=
2( ) ( )at j ftG f e u t e dtπ∞
− −
−∞
= ∫ .
( 2 )
0
a j f te dtπ∞
− += ∫ =( 2 )
0
1( 2 ) 2
a j f tea j f a j f
π
π π
∞− +
=− + +
1( )2
ate u ta j fπ
− ⇔+
2 2
1( )(2 )
G fa fπ
=+
1 12rg[ ( )] fA G f tan tana aπ ω− −⎡ ⎤ ⎡ ⎤= − =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
|G(f)| ( )G f 2
π
-ω 2a ω -ω -2a 0 2a ω 2
π−
Amplitude spectrum Phase spectrum Note : a > 0 then only g(t) is Fourier Transformable.
P.24# Find the Fourier Transform of the Raising exponential pulse shown in figure. g(t) -t -1/a 0
( ) ( )atg t e u t= − a > 0 if a < 0 ? then only g(t) is Fourier Transformable
[ ] 2( ) ( ) ( )at j ftF g t G f e u t e dtπ∞
−
−∞
= = −∫ .
0 02 ( 2 )at j ft t a j fe e dt e dtπ π− −
−∞ −∞
= =∫ ∫
=0
( 2 )1 1 112 2 2
a j fe ea j f a j f a j f
π
π π π− −∞
−∞
⎡ ⎤= − =⎣ ⎦− − −
2 2
1( )(2 )
G fa fπ
=+
phase = 1 2tan faπ−
1/a 2
π Phase
- 2
π
Amplitude spectrum Phase spectrum
P.25# With the help of Linearity property obtain the Fourier Transform of double exponential pulse as shown in fig.
g(t) ( ) atg t e−= t > 0 1 ( ) 1g t = t = 0 ( ) atg t e= t < 0
-t -1/a 1/a t
[ ] 2( ) ( ) ( ) j ftF g t G f g t e dtπ∞
−
−∞
= = ∫
0 0
2 2 2
0 0
1.at j ft j ft at j fte e dt e dt e e dtπ π π+ ∞
− − − −
−∞ − +
= + +∫ ∫ ∫
02
0 0
0
02
j fte e ej f
π
π
−−+ −
+
⎡ ⎤= − =⎣ ⎦−.
2 2
1 1 2( )2 2 (2 )
aG fa j f a j f a fπ π π
∴ = + =− + +
.
Note : ( ) a tg t e−= ,t t= t > 0 ,t t= − t < 0 ( ) atg t e= t < 0 |G(f)| ( ) atg t e−= t ≥ 0
2 2
2(2 )
a t aea fπ
−∴ ⇔+
-f 0 f P.25# Obtain the Fourier Transform of the anti symmetric exponential pulse as shown in figure. g(t) 1 -1/a -t 1/a t
( ) atg t e−= t > 0 ( ) 1g t = t = 0 -1
( ) atg t e= − t < 0 0 0
2 2 2
0 0
( ) 1at j ft j ft at j ftG f e e dt e dt e e dtπ π π− + ∞
− − −
−∞ − +
= − + +∫ ∫ ∫
2 2
1 1 402 2 (2 )
j fa j f a j f a f
ππ π π
−= − + + =
− + +.
(2) Time Scaling Property : ( ) ( )g t G f⇔
Then 1( ) fg at Ga a
⎛ ⎞⇔ ⎜ ⎟⎝ ⎠
a=constant (a>0)
Case (i) a > 0
[ ] 2( ) ( ) j ftF g at g at e dtπ∞
−
−∞
= ∫
Let atλ = taλ
= d adtλ =
21 ( )
fjag e d
aπ λ
λ λ∞
−
−∞
= ∫
1 fGa a
⎡ ⎤= ⎢ ⎥⎣ ⎦
Case (ii) a < 0 [ ] 2( ) ( ) j ftF g at g at e dtπ∞
−
−∞
− = −∫
atλ = −
d adtλ = − taλ−
= Limits + ∞ to - ∞
[ ] ( )21( )
fj taF g at g e d
a
πλ λ
−⎛ ⎞−∞ − ⎜ ⎟⎝ ⎠
+∞
−⎛ ⎞− = ⎜ ⎟⎝ ⎠ ∫
21 ( )
fj tag e d
a
πλ λ
−⎛ ⎞∞ − ⎜ ⎟⎝ ⎠
−∞
= ∫
[ ] 1( ) fF g at Ga a
−⎡ ⎤− = ⎢ ⎥⎣ ⎦
[ ] 1( ) fF g at Ga a
⎡ ⎤= ⎢ ⎥⎣ ⎦
1( ) fg at Ga a
⎡ ⎤∴ ⇔ ⎢ ⎥⎣ ⎦
This property states that the compression of a function g(t) in the time domain is equivalent to its Fourier Transform G(f) in the frequency domain by the same factor or vice versa. P.26 #Find the Fourier Transform of 0.5 ( )te u t− using scaling property.
Let ( ) ( )tg t e u t−= 1( )1 2
g tj fπ
⇔+
1( ) fg at Ga a
⎡ ⎤= ⎢ ⎥⎣ ⎦ 1 1( ) .
1 2g at f aj
aπ
⇔+
[ ] 0.5 1( ) ( )0.5 2
tF g at F e u tj fπ
−⎡ ⎤∴ = =⎣ ⎦ +
(3) Duality property : if [ ]( ) ( )F g t G f= then [ ]( ) ( )F G t g f= −
2( ) ( ) j ftg t G f e dfπ∞
−∞
= ∫
replace t = -t
2( ) ( ) j ftg t G f e dfπ∞
−
−∞
− = ∫
Interchange t & f
[ ]2( ) ( ) ( )j ftg f G t e dt F G tπ∞
−
−∞
− = =∫
( ) ( )G t g f∴ ⇔ − P.26 # Consider a signal g(t) in the form of a sine function is g(t) = A sinc(2ω t) find the Fourier Transform using duality & Time scaling properties.
We know that sin ( )tA rect AT c fTT⎛ ⎞ ⇔⎜ ⎟⎝ ⎠
.
A AT ⇔
2T− 0
2T t
-1/T 0 1/T f g(t) = A sinc(2ω t) g(t) G(f) A A/2ω ⇔ -ω 0 ω f -1/ω 1/ω t
( ) sin ( )rect t c f⇔ (rect (f) = rect (-f)) even function ( ) sin ( )Arect t A c f⇔
sin ( ) ( )c t rect f⇔
sin ( )tArect AT c fTT⎛ ⎞ ⇔⎜ ⎟⎝ ⎠
sin ( ) ( )A c t Arect f⇔
Scaling
sin ( ) fAT c tT ArectT
⎛ ⎞⇔ ⎜ ⎟⎝ ⎠
sin (2 )2 2A fA c t rectωω ω
⎛ ⎞∴ ⇔ ⎜ ⎟⎝ ⎠
Duality.
(4) Time Shifting property : [ ]( ) ( )F g t G f=
[ ] 020( ) ( ) j ftF g t t G f e π−− =
2[ ( )] ( ) j ftF g t g t e dtπ∞
−
−∞
= ∫
20 0[ ( )] ( ) j ftF g t t g t t e dtπ
∞−
−∞
− = −∫
Let 0t tλ = − 0t tλ= + d dtλ =
= 0 02 ( ) 22( ) ( )j f t j ftj fg e d g e e dπ λ ππ λλ λ λ λ∞ ∞
− + −−
−∞ −∞
=∫ ∫
02( ) j ftG f e π−= This property states that a time shift to has no change on the amplitude spectrum but then is a phase shift of -2π ft0. [ ] 02
0( ) ( ) j ftF g t t G f e π+ = P.27 #Find the Fourier Transform of Rectangular pulse (a) and then applying Time shifting property find the Fourier transform of (b) & (c) g(t) A
2T− 0
2T t G(f) = AT sinc(fT)
(a) g(t) A
2
2( ) sin ( )Tj f
TG f AT c f eπ−
= 0 T t sin ( ) j fT
TAT c f e π−= (b) g(t) A
2
2( ) sin ( )Tj f
TG f AT c f eπ
= -T 0 t = sin ( ) j fT
TAT c f e π . (c)
(5) Frequency Shifting : ( ) ( )g t G f⇔ 2 ( ) ( )cj f t
cF e g t G f fπ⎡ ⎤ = −⎣ ⎦ cf is a real constant.
2 2 ( )( ) ( )c cj f t j t f fF e g t g t e dtπ π∞
− −
−∞
⎡ ⎤ =⎣ ⎦ ∫ .
( )cG f f= − Multiplication of a function g(t) by the factor 2 cj f te π is equivalent to shifting its Fourier Transform G(f) in the positive direction by amount fc. This property is called modulation theorem or frequency translation theorem. P.28 #Obtain the Fourier Transform and amplitude spectrum of the RF pulse shown in fig.
( ) cos(2 )ctg t A f t rectT
π ⎛ ⎞= ⎜ ⎟⎝ ⎠
g(t)
A 2 2
cos 22
c cj f t j f t
ce ef t
π π
π−+
= t
( )2 2( )2
c cj f t j f tA tg t rect e eT
π π−⎛ ⎞= +⎜ ⎟⎝ ⎠
1/fc
sin ( )2 2 TA t ATF rect c f
T⎡ ⎤⎛ ⎞ =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
[ ] [ ]( ) sin {( ) } sin {( ) }2 c c
ATF g t c f f T c f f T= − + +
AT/2 -fc fc (a)
π -fc fc -π (b) (a) Amplitude modulation (b) Phase modulation P.29 #Find the Fourier transform of exponentially Damped sinusoidal signal. 1 e-t
( ) sin(2 ) ( )tcg t e f t u tπ−=
F.T 1( )1 2
te u tj fπ
−⎡ ⎤ =⎣ ⎦ + 0 t
( )2 21sin(2 )2
c cj f t j f tcf t e e
jπ ππ −= − -1
[ ] 1 1 1. . ( )2 1 2 ( ) 1 2 ( )c c
F T g tj j f f j f fπ π⎡ ⎤
= −⎢ ⎥+ − + +⎣ ⎦
( ) ( )2 2
1 2 2 1 2 212 1 2 2
c c
c
j f j f j f j fj j f f
π π π ππ π
⎡ ⎤+ + − − += ⎢ ⎥
+ +⎢ ⎥⎣ ⎦
( ) ( )2 2
21 2 2
c
c
fj f f
ππ π
=+ +
.
(6) Area under g(t) : if [ ]( ) ( )F g t G f=
then ( ) (0)g t dt G∞
−∞
=∫
The Area under a function g(t) is equal the value of its Fourier Transform G(f) at f = 0.
2( ) ( ) j ftG f g t e dtπ∞
−
−∞
= ∫
at f = 0 (0) ( )G g t dt∞
−∞
= ∫
P.30 # Find area of ( ) sin ( )g t c t=
A rect sin ( )Tt AT c fT⎛ ⎞ ⇔⎜ ⎟⎝ ⎠
rect ( ) sin ( )t c f⇔ rect (f)
Duality
sin ( ) ( ).c t rect f⇔ 12− 1
2 f
( ) ( )G f rect f= (0) 1G =
P.31 # if ( ) sin (2 )2Ag t c tωω
= find (0)G
sin (2 )2 2A fA c t rectωω ω
⎛ ⎞⇔ ⎜ ⎟⎝ ⎠
A/2ω
( )2 2A fG f rectω ω
⎛ ⎞= ⎜ ⎟⎝ ⎠
-ω ω f
(0)2AGω
=
(7) Area under G(f): if [ ( )] ( )F g t G f=
(0) ( )g G f df∞
−∞
= ∫ .
The area under a function G(f) is equal to the value of its inverse F.T g(t) at t = 0
2( ) ( ) j ftg t G f e dfπ∞
−∞
= ∫
at t = 0 (0) ( )g G f df∞
−∞
= ∫ .
P.32 # Find Area under sinc(f) ( ) sin ( )G f c f= sin ( ) ( )c f rect t⇔ . ( ) ( )g f rect t⇔ (0) 1g∴ = (8) Differentiation in the time domain ( ) ( )g t G f⇔ Assume that the first derivative of g(t) is Fourier transformable then
( ) 2 ( )dF g t j fG fdt
π⎡ ⎤ =⎢ ⎥⎣ ⎦
2( ) ( ) j ftd dg t G f e dfdt dt
π∞
−∞
= ∫ .
= 2( ) j ftdG f df edt
π∞
−∞∫
2( ) .( 2 )j ftG f e df j fπ π∞
−∞
= ∫
[ ]1( ) 2 ( )d g t F j fG fdt
π−=
( ) 2 ( )dF g t j fG fdt
π⎡ ⎤ =⎢ ⎥⎣ ⎦
∴Differentiation of a time function g(t) has the effect of multiplying its Fourier transform G(f) by the factor 2j fπ . Similarly, the result can be extended to the thn derivative
( ) ( 2 ) ( )n
nn
d g t j f G fdt
π⇔ .
(9) Differentiation in the frequency domain: ( ) ( )g t G f⇔
2 ( ) ( )dj tg t G fdf
π− ⇔
2( ) ( ) j ftd dG f g t e dtdf df
π∞
−
−∞
= ∫
= 2( ) ( 2 )j ftg t e dt j tπ π∞
−
−∞
−∫
[ ]( ) 2 ( )d G f F j tg tdf
π= −
For higher derivative ( 2 ) ( ) ( )n
nn
dj t g t G fdf
π− ⇔ .
∴The differentiation on the frequency domain is equal to the multiplication of g(t) by ( 2 )j tπ− in the time domain. (10) Integration in the time domain :
( ) ( )g t G f⇔ then provided G(0) = 0 = ( ) 0 (0)g t dt G∞
−∞
= =∫
We have 1( ) ( )2
t
g d G fj f
λ λπ−∞
⇔∫ .
Integration of a time function g(t) has the effect of dividing its F.T. G(f) by the factor 2j fπ assuming G(0) is zero.
Let ( ) ( )t
g dλ λ φ λ−∞
=∫
Taking differentiation on both sides
( ) ( ) ( )td dg g d
d dλ φ λ λ λ
λ λ −∞
⎡ ⎤= = ⎢ ⎥
⎣ ⎦∫
[ ( )] ( ) 2 ( )t tdF g F g d j fF g d
dλ λ λ π λ λ
λ −∞ −∞
⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦∫ ∫
( ) 2 ( )t
G f j fF g dπ λ λ−∞
⎡ ⎤= ⎢ ⎥
⎣ ⎦∫
1( ) ( ). .
2
t
g d G fj f
λ λπ−∞
⇔∫
If G(0) is non zero then the definite integral of g(t) has a Fourier Transform that includes a dirac delta function or impulse ( )fδ at the origin.
1 (0)( ) ( ) ( )2 2
t Gg d G f fj f
λ λ δπ−∞
⇔ +∫ .
(11) Conjugate factors : If ( ) ( )g t G f⇔ then for a complex valued time function g(t) we have
* *( ) ( )g t G f⇔ − . * denotes the complex conjugate operation.
2( ) ( ) j ftg t G f e dfπ∞
−∞
= ∫ .
Taking complex conjugate on both sides
* * 2( ) ( ) j ftg t G f e dfπ∞
−
−∞
= ∫ .
Replace f = -f
* * 2 * 2( ) ( ) ( )j ft j ftg t G f e df G f e dfπ π−∞ ∞
∞ −∞
= − − = −∫ ∫ .
= 1 *( )F G f− ⎡ ⎤−⎣ ⎦ * *( ) ( )g t G f∴ ⇔ − . (12) Multiplication in the time domain Let 1 1( ) ( )g t G f⇔
2 2( ) ( )g t G f⇔
1 2 1 2( ) ( ) ( ) ( )g t g t G G f dλ λ λ∞
−∞
⇔ −∫
1 2 12( ) ( ) ( )g t g t G f⇔ or 1 2( )* ( )G f G f
212 1 2( ) ( ) ( ) j ftG f g t g t e dtπ
∞−
−∞
= ∫
Substitute the inverse F.T of g2(t) in the above equation
11 2 12 2( ) ( ) j f tg t G f e dfπ
∞
−∞
= ∫
11 2 2 1
12 1 2( ) ( ) ( ) j ft j f tG f g t G f e e dtdfπ π∞ ∞
−
−∞ −∞
= ∫ ∫
11 2 ( ) 1
1 2( ) ( ) j t f fg t G f e df dtπ∞ ∞
− −
−∞ −∞
= ∫ ∫ .
or inner integration in w.r.to df1
1f fλ = − 1d dfλ = − f appears as constant Limits + ∞ to - ∞ and –ve sign = - ∞ to + ∞ & +ve sign
21 2( ) ( ) j tg t G f e d dtπ λλ λ
∞ ∞−
−∞ −∞
= −∫ ∫
= 22 1( ) ( ) j tG f d g t e dtπ λλ λ
∞ ∞−
−∞ −∞
−∫ ∫
12 2 1( ) ( ) ( ) .G f G f G dλ λ λ∞
−∞
= −∫
The above integral is known as the Convolution integral expressed in the frequency domain, and the function 12 ( )G f is referred to as the convolution of 1 2( ) & ( )G f G f . ∴The multiplication of two signals in the time domain is transformed into the Convolution of their individual Fourier Transforms in the frequency domain.
12 1 2( ) ( )* ( )G f G f G f= Convolution is a Commutative 12 21( ) ( )G f G f= . (13) Convolution in the time domain : Let 1 1( ) ( )g t G f⇔ and 2 2( ) ( )g t G f⇔
1 2 1 2( ) ( ) ( ) ( )g g t d G f G fλ λ λ∞
−∞
− ⇔∫ .
[ ]1 21 2 1 2( ) ( ) ( ) ( ) j ftF G f G f G f G f e dfπ
∞−
−∞
= ∫
11 2 1
2 2( ) ( ) j ftG f g t e dtπ∞
−
−∞
= ∫ . or let t1 = τ
=11 2 2 1
1 2( ) ( ) j ft j ftG t g t e e dt dfπ π∞ ∞
−
−∞ −∞∫ ∫ .
11 2 ( ) 1
2 1( ) ( )j f t tg t e G f dt dfπ∞ ∞
−
−∞ −∞
= ∫ ∫
Let 1t tλ = − 1d dtλ = − +∞ to - ∞
22 1( ) ( )j fg t e G f d dfπ λλ λ
∞ ∞
−∞ −∞
= −∫ ∫
= 22 1( ) ( ) .j fg t d G f e dfπ λλ λ
∞ ∞
−∞ −∞
−∫ ∫
= 1 2( ) ( )g g t dλ λ λ∞
−∞
−∫
[ ]1 2 1 2( )* ( ) ( ). ( )F g t g t G f G f∴ = This property states that the Convolution of two signals in the time domain is transformed into the multiplication of their individual Fourier Transforms in the frequency domain. This property is known as the Convolution theorem. Convolution is commutative operation.
1 2 1 2( )* ( ) ( ) ( )g t g t g g t dλ λ λ∞
−∞
= −∫
Let t xλ− = t xλ = − d dxλ = − +∞ to - ∞
1 2( ) ( )g t x g x dx∞
−∞
= −∫
2 1( )* ( ).g t g t=
1 2 1 2( ) ( ) ( ) ( ) .g g t d g t g dλ λ λ λ λ λ∞ ∞
−∞ −∞
= − = −∫ ∫
Dirac Delta function (or) Unit impulse : ( ) 0tδ = t ≠ 0
( ) 1t dtδ∞
−∞
=∫
Properties of ( )tδ : (1) The product of ( )tδ and any time function g(t) that is continuous at t = 0
( ) ( ) (0)t g t dt gδ∞
−∞
=∫
(2) 0 0( ) ( ) ( )t t g t dt g tδ∞
−∞
− =∫
(3) The Convolution of any function with the delta function leaves that function unchanged. ( )* ( ) ( )t g t g tδ =
1 2( ) ( ) ( ). ( )t g d G f G fδ λ λ λ∞
−∞
− =∫
[ ] 21( ) . . ( ) ( ) 1j ftG f F T t t e dtπδ δ
∞−
−∞
= = =∫
[ ]2 2( ) . . ( ) ( )G f F T g t G f= = 1 2 2 2( ) ( ) 1. ( ) ( )G f G f G f G f= = [ ]1
2 ( ) ( )F G f g t−∴ = ( ) 1tδ ⇔
( ) ( ) ( )t g d g tδ λ λ λ∞
−∞
∴ − =∫
g(t) G(f) ^ 0 t -f 0 f The spectrum of the delta function ( )tδ extends uniformly over the entire frequency interval from -∞ to ∞ .
(4) [ ] 1( ) fF at Ga a
δ ⎛ ⎞= ⎜ ⎟⎝ ⎠
[ ]( ) . ( )G f F T fδ=
=1
1fGa
⎛ ⎞ =⎜ ⎟⎝ ⎠
1a
∴=
P.33 # Find the Fourier Transform of the d.c. signal.
g(t) ( ) ( )g t G f⇔ 1.0 ( ) ( )G t g f⇔ − ( ) 1tδ ⇔ 0 t
1 ( )fδ⇔ − or 1 ( )fδ⇔ G(f) ^
0 f ∴A dc signal is transformed in the frequency domain into a delta function
( )fδ occurring at zero frequency.
P.34 # Fourier Transform of Complex exponential signal : 2 cj f te π 1 ( )fδ⇔ 21. ( )cj f t
ce f fπ δ⇔ − fc f Similarly [ ] 02
0. ( ) 1. j ftF T t t e πδ −− =
P.35 # Find the Fourier Transform of cos 2 cf tπ
2 21cos 22
c cj f t j f tcf t e eπ ππ −⎡ ⎤= +⎣ ⎦
2. ( )cj f tcF T e f fπ δ⎡ ⎤ = −⎣ ⎦ G(f)
[ ]1cos 2 ( ) ( )2c c cf t f f f fπ δ δ∴ ⇔ − + + 1/2
-fc 0 fc f Spectrum
G(f) P.36 # ( ) sin 2 cg t f tπ=
1/2
2 212
c cj f t j f te ej
π π−⎡ ⎤= −⎣ ⎦ -fc fc f
[ ]1sin 2 ( ) ( )2c c cf t f f f f
jπ δ δ= − − + -1/2
[ ]1( ) ( ) ( )2 c cG f f f f fδ δ= − − +
Unit Step Function : u(t) = 1 t > 0 g(t) u(t) = 0 t < 0 1
( ) ( )t
d u tδ λ λ−∞
=∫ 0 t
We know that 1 (0)( ) ( ) ( )2 2
t Gg d G f fj f
λ λ δπ−∞
⇔ +∫ .
Where G(f) : F.T[ ]( )g λ
1 ( ). ( )2 2
t fF T dj f
δδ λ λπ−∞
⎡ ⎤∴ = +⎢ ⎥
⎣ ⎦∫ G(f) = F.T[ ( )tδ
=1 G(0)=1
|G(f)| ^ 0 f Amplitude Spectrum. Signum function: The signum function denoted by Sgn(t) is defined as follows. g(t)=Sgn(t) Sgn(t) = 1 t > 0 1.0 Sgn(t) = 0 t = 0 0 t Sgn(t) = -1 t < 0 -1.0 The signum function in terms of unit step function Sgn (t) = 2 u(t)-1 u(t)=1 t > 0 = ½ t = 0 = 0 t < 0
[ ( )] . [2 ( ) 1]F Sgn t F T u t= − |G(f)|
2 ( ) 1.2 ( )2 2
f fj f j f
δ δπ π
= + − =
-f f
1( )Sgn tj fπ
⇔
Duality : 1 ( )Sgn fj tπ
⇔ − .
( ) ( )Sgn t Sgn t= − − Multiplying with ‘j ‘ on both sides
1 ( )jSgn ftπ
∴ ⇔ − Sgn(f) = 1 f > 0
Sgn(f) = 0 f = 0 Sgn(f) = -1 f < 0.
( )* ( ) ( )g t t g tδ =
0 0( )* ( ) ( )g t t t g t tδ − = − [ ]1 2 0( ) [ ( )] : ( ) ( )G f F g t G f F t tδ= = − 0t tλ = −
= 20( ) j ftt t e dtπδ
∞−
−∞
−∫ . 0t tλ= +
= 022( ) j ftj fe d e ππ λδ λ λ∞
−−
−∞∫
= 02j fte π− 02
1 2 1( ) ( ) ( ) j ftG f G f G f e π−= 021
1 0[ ( ) ] ( )j ftF G f e g t tπ−−= = −
P.37 # Evaluate the following expressions with 2( ) ( 3)g t t= −
(a) ( ) ( 4) ( 4)g t t dt gδ∞
−∞
+ = −∫
2( 4 3) 49∴= − − = (b) ( )* ( 4) ( 4)g t t g tδ + = + 2 2( 4 3) ( 1)t t= + − = +
(c) ( )* ?4tg t δ −⎛ ⎞ =⎜ ⎟
⎝ ⎠
[ ] 1 1. ( ) .1F T ata a
δ = =
1 4144
tF δ⎡ − ⎤⎛ ⎞∴ = =⎜ ⎟⎢ ⎥ ⎛ ⎞⎝ ⎠⎣ ⎦ ⎜ ⎟⎝ ⎠
[ ( )] ( )F g t G f= 1[ ( ).4] 4. ( )F G f g t− =
24( 3)t= − .
P.38 # A function of time g(t) has the following Fourier Transform
2
22
12
1( )1
G eω
ωωω
−+=
+
Using the properties of the F.T. Write the Fourier Transforms of (a) g(2t).
Using the Scaling property 1( )g at Ga a
ω⎛ ⎞⇔ ⎜ ⎟⎝ ⎠
1(2 )2 2
g t G ω⎛ ⎞⇔ ⎜ ⎟⎝ ⎠
2
2
2 *44
42
42 4
G e
ω
ωωω
−
+⎛ ⎞ =⎜ ⎟ +⎝ ⎠
( )
2
224
2
2(2 )4
g t eω
ω
ω
−+⇔
+
(b) ( 2) jtg t e− Using the time shifting property 0
0( ) ( ) j tg t t G e ωω −− ⇔ 2( 2) ( ) jg t G e ωω −− ⇔
( )
2
22
212
1( 2)1
jg t e eω
ωω
ω
−−+∴ − ⇔
+
Using the frequency shifting property 0
0( ) ( )j tg t e Gω ω ω⇔ − ( ) ( 1)jtg t e G ω∴ ⇔ −
2( 1)( 2) ( 1)jt jg t e G e ωω − −− ⇔ −
( )
2
22( 1)
2( 1)( 1) 12
11 1
je eω
ωω
ω
− −− −− +⇔
− +
2 ( ) ( )cj f t
ce g t G f fπ ⇔ − 0
0( ) ( )j te g t Gω ω ω⇔ −
(c) 4 ( )d g tdt
Using the differentiation property ( ) ( )d g t j Gdt
ω ω⇔
2
22
12
14 ( ) 41
d g t j edt
ωωω
ω
−+⇔
+
(d) ( )t
g dτ τ−∞∫ .
Using the integration property
( ) (0)( ) ( )2
t G Gg djωτ τ δ ωω−∞
⇔ +∫
2
22
12
1( )1
G eω
ωωω
−+=
+ G(0)=1
2
22
12
1 1 ( )( 1) 2
ej
ωω δ ω
ω ω
−+⇔ +
+
P.39 # Find the F.T of Gate function. g(t)
1
2T− 0
2T t
g(t) =1 2 2T Tt−< <
= 0 otherwise
( ) sin ( )G f T c fT=
P.40 # Find the F.T of the Triangular pulse shown in figure. g(t) A
2T− 0
2T t
2T− to 0 1 1 2 2( , )( , )x y x y
-T ,0 (0, )2
A⎛ ⎞⎜ ⎟⎝ ⎠
1 2 1 1 2 1 1 2 1( )( ) ( )( ) ( )( )x x y y y y x x x x y y− − = − − = − − T(y-0) ( )2 2
Tx A⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
T y2 2
TAx A= + x = t y=g(t)
AT 2g(t)=At+ . 22
At AT T
∴ = +
A=A+2. .T
t for 2T− to 0
Similarly 0 to 2T 1 1 2 2( , )( , )x y x y
T (0, ) ,02
A ⎛ ⎞⎜ ⎟⎝ ⎠
( ) ( )2Ty A x A⎛ ⎞− = −⎜ ⎟⎝ ⎠
2( )y A AxT
− = − 2g(t)=-At +AT
∴
A g(t)=A-2 tT
for 0 to 2T
0 22 2
02
( ) 2 2
T
j ft j ft
T
A AG f A t e dt A t e dtT T
π π− −
−
⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫ .
2 2 2 2
0 0 0 0
2T T T T
j t j t j t j tAA e dt A e dt te dt te dtT
ω ω ω ω− −
⎡ ⎤⎢ ⎥+ − +⎢ ⎥⎢ ⎥⎣ ⎦
∫ ∫ ∫ ∫
2 2
0 0
22cos 2 cos .
T T
AA tdt t tdtT
ω ω= −∫ ∫
2
0
sinsin sin22 * 42
TTt A T tA dt
T
ωω ωω ω ω
⎡ ⎤⎢ ⎥
= − −⎢ ⎥⎢ ⎥⎣ ⎦
∫
2
2 2
4 41 cos 2sin2 4
A T A TT T
ω ωω ω
⎡ ⎤ ⎡ ⎤= − =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
2
8 2 2sin sin4 4
A fT fTT
π πω
=
2
sin sin8 2 2
2 2
fT fTA
fT fTT
π π
ω π π=
2fTπ .
2fTπ
8.2 .2
AT f fπ π
sin sin2 2fT fTc c⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2fTπ .
2fTπ
2sin .
2 2AT fTc ⎛ ⎞= ⎜ ⎟
⎝ ⎠
P.41 # Find the Fourier transform of Gaussian pulse f(t) This is defined as
2
( ) tf t e π−=
2 2
. t t j tF T e e e dtπ π ω∞
− − −
−∞
⎡ ⎤ =⎣ ⎦ ∫ 0 t
Gaussian pulse 2( )( ) t j tF e dtπ ωω
∞− +
−∞
= ∫ .
Substituting 2
t j tπ ω+ = 2 2
.42
jt ω ωπππ
⎧ ⎫+ +⎨ ⎬⎩ ⎭
=2 2
2 4jt
e e dtω ωππ π
⎛ ⎞∞ −− +⎜ ⎟⎝ ⎠
−∞∫
22
24
jte e dt
ωω πππ
⎛ ⎞∞− − +⎜ ⎟⎝ ⎠
−∞
= ∫
Putting 2ju t ωππ
= + du dtπ= dudtπ
=
2
22 24
4
0
u udu ee e e du
ωω ππ
π π
−∞ ∞−
− −
−∞
= = =∫ ∫
We know that 2
0 2ue du π∞
− =∫
2 2 2 22
44 4 42( ) : ( )
2
ffF f G f e e e e
ω ω πππ π ππ
π
− − −−= = = =
The area under the Gaussian pulse if G(0) at f = 0 =
2
1fe π−⎡ ⎤ =⎣ ⎦
2 2t fe eπ π− −⇔
P.42 # Find the F.T of the Gaussian pulse 2 2
( ) a tg t e−=
2 2( )a t j te dtω∞
− +
−∞
= ∫
2 2
2 222 4
ja t j t ata aω ωω ⎛ ⎞+ = + +⎜ ⎟
⎝ ⎠
22
22 4
jata ae e dtω ω⎛ ⎞ −∞ − +⎜ ⎟
⎝ ⎠
−∞
= ∫
2 2
2 24
jataae e dtωω ⎛ ⎞− ∞ − +⎜ ⎟
⎝ ⎠
−∞
= ∫ .
2ju ataω
= + du adt=
22 2
2 2 24 422
aa ae e ea a a
ωω ωπ π π− − ⎛ ⎞− +⎜ ⎟⎝ ⎠= = =
2
2 2f
a t ae ea
ππ ⎛ ⎞−⎜ ⎟− ⎝ ⎠⇔ .
P.43 # Find the Fourier Transform of the signal f(t) shown below.
g(t) 0 to T A
( ) Ag t tT
= 0 < t < T
A= T < t < 2T 0 T 2T t 2
0
( )T T
j t j t
T
AG f te dt Ae dtT
ω ω− −= +∫ ∫
22 sin2 2
Tj j tA Te c ej f
ω ωωπ
− −⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
P. 44 # Obtain the Fourier Transform of the trapezoidal pulse shown in figure. g(t) A
( ) p
p a p a
AtAtg tt t t t
= +− −
p at t t− < < −
-tp- ta 0 ta tp t = A a at t t− < <
p
p a p a
At Att t t t
= −− −
for a pt t t< < .
( )2
2( ) cos cosa pp a
AG f t tt t
ω ωω
⎡ ⎤= −⎣ ⎦−
P. 45 # Find the Fourier Transform of the signal shown in figure g(t) 2 1 -2 -1 0 1 2 t g(t) = 2 -1 < t < 1 = 1 -2 < t < -1 & = 1 1 < t < 2
( ) 2sin (2 ) 4sin (4 )G f c f c fπ π= +
P. 46# Fourier Transform of ( ) cosatg t e bt−=
0
1( )2
at jbt jbt j tG f e e e e dtω∞
− − −⎡ ⎤= +⎣ ⎦∫
( )2 2
a ja j b
ωω+
=+ +
P. 47 # Fourier Transform of ( ) cos .g t t at=
( )
2 2
22 2( ) aG f
a
ω
ω
+= −
−
P. 48 # Fourier Transform of [ ]( ) ( )2j dtx t f
dfπ=
We know that 2 ( ) ( )dj tg t G fdf
π− ⇔
= 2( ) ( )2 2
j ftj d j df x t e dtdf df
π
π π
∞−
−∞
= ∫
( ) 2( ) 22
j ftj x t j t e dtπππ
∞−
−∞
= −∫
2( 2 ) ( )2
j ftj j tx t e dtπππ
∞−
−∞
= − ∫
P. 49 # Fourier Transform of [ ]. (2 )t x t
1. [ ( )] fF T g at Ga a
⎛ ⎞⇔ ⎜ ⎟⎝ ⎠
Similarly, 1(2 )2 2
fx t G ⎛ ⎞⇔ ⎜ ⎟⎝ ⎠
We know that ( ) ( )2j dtx t X f
dfπ⇔
(2 )4 2j d ftx t X
dfπ⎛ ⎞∴ ⇔ ⎜ ⎟⎝ ⎠
2
2( )4
fj tj dx t e dtdf
π
π
∞−
−∞
= ∫
2
2( ) 2.24 2
fj tj tx t j e dtπ
ππ
∞−
−∞
= −∫
21 2 . (2 ) .24
j ftt x t e dtπ∞
−
−∞
= ∫
. [ . (2 )]4 2j d fX F T t x t
dfπ⎛ ⎞ =⎜ ⎟⎝ ⎠
P. 50 # Fourier Transform of [ ] [ ]( 2) ( ) . . ( ) 2 ( )t x t F T t x t x t− = −
( ) 2 ( )2j d x f x f
dfπ= −
P. 51 # Fourier Transform of [ ] [ ]( 2) ( 2 ) . . ( 2 ) 2 ( 2 )t x t F T t x t x t− − = − − −
( )4 2 2j d f fx x
dfπ−⎛ ⎞= − − ⎜ ⎟
⎝ ⎠
P. 52 # Fourier Transform of . ( )dt x tdt
⎡ ⎤⎢ ⎥⎣ ⎦
( )2j d dF x t
df dtπ⎛ ⎞⎧ ⎫= ⎨ ⎬⎜ ⎟⎩ ⎭⎝ ⎠
( )2 ( )2j d j fx f
dfπ
π=
{ }( )d fx fdf
= −
P. 53 # Fourier Transform of [ ] 2(1 ) (1 ) .j ftx t x t e dtπ∞
−
−∞
− = −∫
1 t m− = 1 m t− = dt dm= −
2 (1 )( ) j f mx m e dmπ∞
− −
−∞
= ∫
2 2( ) .j fm j fx m e dm eπ π∞
+ −
−∞
= ∫
2 ( )j fe x fπ−= − .
Since 2( ) ( ) j ftG f x t e dtπ∞
−
−∞
= ∫ .
Replace f = -f
2( ) ( ) j ftG f x t e dtπ∞
−∞
− = ∫ .
P. 54 # Find Fourier Transform of (1 ) (1 )tx t x t− − + − .
{ }2 2( ) ( ).2
j f j fj d e x f e x fdf
π π
π− −−
= − + −
P. 55 # Find Hilbert Transform of 1t
1 1ˆ( )g tt tπ
= −
1. ( )F Tof jSgn ftπ= −
F.T of 1t⇔ -jπ ( )Sgn f
( )[ ( )]jSgnf j Sgn fπ π= − − = −
Inverse Fourier transform of [ ] ( )tπ πδ− = − .
P. 56 # 2 ( ) ( )dj tg t G fdf
π− ⇔
. ( ) ( )2j dt g t G f
dfπ⇔
. ( ) ( )dt g t j Gd
ωω
⇔ put t = 2t
. (2 )2 2j dt g t G
dω
ω⎡ ⎤⎛ ⎞∴ ⇔ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
2( )2
j tj d g t e dtd
ω
ω
∞−
−∞
= ∫
(2 ) .22
j tj d g t e dtd
ω
ω
∞−
−∞
= ∫
(2 )( ) j tj g t jt e dtω∞
−
−∞
= −∫ .
. (2 ) j tt g t e dtω∞
−
−∞
= ∫ : F.T [t g(2t)]
8. Fourier Transform of periodic signals gp(t)
-T0 0 T0 t Let gp(t) a periodic signal of period T0. g(t)
t
0
2T
− 0 0
2T
In terms of Complex Exponential Fourier series
02
( )nj tT
p nn
g t C eπ∞
=−∞
= ∑ → (8.1)
nC is complex Fourier coefficient.
0
0
0
2 2
02
1 ( )
Tnj tT
n pT
C g t e dtT
π−
−
= ∫ → (8.2)
Note : ( )pg t dt∞
−∞
= ∞∫ for a periodic function.
as 0nC → as 0T →∞ ∴equation (8.1) is Fourier Transformable.
The F.T of equation (8.1) is 02
.nj tT nF T e f
πδ
ω
⎡ ⎤ ⎛ ⎞= −⎢ ⎥ ⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
Q
0
. ( )p nn
nF T g t C fT
δ∞
=−∞
⎛ ⎞⎡ ⎤ = −⎜ ⎟⎣ ⎦
⎝ ⎠∑ → (8.3)
Let g(t) be a pulse like function, which equals to gp(t) over one period and is zero else where.
( ) ( )pg t g t∴ = 0 0
2 2T Tt− ≤ ≤
= 0 otherwise The periodic signal ( )pg t in terms of ( )g t
0( ) ( )pm
g t g t mT∞
=−∞
= −∑
Based on above representation we can say g(t) is a generating function which generates the periodic signal ( )pg t . The function ( )g t is a Fourier Transformable.
0
0
0
2 2
02
1 ( )
Tnj tT
n pT
C g t e dtT
π−
−
∴ = ∫ .
0 0
0 0 0
0 0
2 22 2 2
0 0 02 2
1 1 1( ) ( ) ( )
T Tn n nj t j t j tT T T
n pT T
C g t e dt g t e dt g t e dtT T T
π π π
−∞− − −
−−∞
= + +∫ ∫ ∫
02
0
1 ( )nj tTg t e dt
T
π∞ −
−∞
= ∫
0 0
1n
nC GT T
⎛ ⎞= ⎜ ⎟
⎝ ⎠ (or) 0
2
0 0
1( )nj tT
pn
ng t G eT T
π∞
=−∞
⎛ ⎞= ⎜ ⎟
⎝ ⎠∑
∴ equation (8.3) can be written as
00 0 0
1( )m n
n ng t mT G fT T T
δ∞ ∞
=−∞ =−∞
⎛ ⎞ ⎛ ⎞− ⇔ −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∑ ∑
The above relation states that the Fourier Transform of a periodic signal consists of Delta
function occurring at integer multiplies of the fundamental frequency 0
1T
including the
origin, and each Delta function weighted by a factor equal to 0
1T 0
nGT⎛ ⎞⎜ ⎟⎝ ⎠
.
The above relation is alternative way to display the frequency content of a periodic signal ( )pg t . ( )g t has Continuous spectrum of G(f) ( )pg t has Discrete spectrum of G(f) P. 57 # Find the Fourier Transform of the periodic pulse train shown in figure.
0 ( )T tδ
-3T0 -2T0 -T0 T0 2T0 3T0
0 ( )( )p T tg t δ=
( ) ( )g t tδ= [ ] [ ]0
. ( ) . ( ) 1.nG F T g t F T tT
δ⎛ ⎞
∴ = = =⎜ ⎟⎝ ⎠
0 ( ) 0( )T tm
t mTδ δ∞
=−∞
= −∑
0 ( )T tF δ⎡ ⎤⎣ ⎦
00 0
1( )m n
nt mT fT T
δ δ∞ ∞
=−∞ =−∞
⎛ ⎞− ⇔ −⎜ ⎟
⎝ ⎠∑ ∑
0
3T−
0
2T−
0
1T− 0
0
1T
0
2T
0
3T
f
Spectrum
9. RAYLEIGH’S ENERGY THEOREM: The signals for which the Energy E is finite are known as Energy signals. The Energy of the signal ( )g t in time domain is
2( )E g t dt∞
−∞
= ∫ → (9.1)
By using Rayleigh’s Energy theorem, it is possible to define (find) total energy of the signal in the frequency domain
2( )E G f df∞
−∞
= ∫
From Equation 2 *( ) ( ) ( )g t g t g t=
* *( ) ( )g t G f⇔ −
* * 2( ) ( ) j ftg t G f e dfπ∞
−∞
= −∫
* * 2( ) ( ) ( ) ( ) .j ftE g t g t dt g t G f e dfdtπ
∞ ∞ ∞
−∞ −∞ −∞
= = −∫ ∫ ∫
* 2( ) ( ) j ftG f g t e dtdfπ∞ ∞
−∞ −∞
= −∫ ∫
*( ) ( )G f G f df∞
−∞
= − −∫ *( ) ( )G f G f= −
*( ) ( )G f G f= −
*( ) ( )G f G f df∞
−∞
= ∫
2( )G f df∞
−∞
= ∫ .
The above equation satisfies that the energy of a signal is equal to area under the 2( )G f curve.
P. 58 # find the total energy of the sine function given below. ( ) sin (2 ).g t A c tω=
2( )E g t dt∞
−∞
= ∫
2 2sin (2 ) .A c t dtω∞
−∞
= ∫
sin ( ) ( )A c t Arect f⇔
sin (2 )2 2A fA c t rectωω ω
⎛ ⎞⇔ ⎜ ⎟⎝ ⎠
2
2 2( )2 2A fE G f df rect dfω ω
∞ ∞
−∞ −∞
⎛ ⎞ ⎛ ⎞= = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫
[ ]2 2
2 2A Adf
ω
ω
ω ωω ω−
⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫
= 2 2
2 24 2A Aωω ω
= Joules .
Similarly, The average power in the frequency domain
22
2
1lim ( ) .
T
TT
P g t dtT→∞
−
= ∫
The average power of a periodic signal ( )pg t of fundamental period T0 is
0
0
2 2
02
1 ( )
T
pT
P g t dtT −
= ∫
0
0
2*
02
1 ( ) ( ) .
T
p pT
g t g t dtT −
= ∫
02
0 0
1( )nj tT
pn
ng t G eT T
π∞
=−∞
⎛ ⎞= ⎜ ⎟
⎝ ⎠∑
Where 0
nGT⎛ ⎞⎜ ⎟⎝ ⎠
is Fourier Transform of the generating function ( )g t
i.e ( ) ( )pg t g t= 0 0
2 2T Tto−
= 0 otherwise.
02
*2
0 0
1 ( )nj tT
n
nP g t G e dtT T
π∞ ∞
=−∞−∞
⎛ ⎞= ⎜ ⎟
⎝ ⎠∑∫ .
Interchanging the order of summation and integration we get
02
*2
0 0
1 ( )nj tT
n
nP G g t e dtT T
π∞∞
=−∞ −∞
⎛ ⎞= ⎜ ⎟
⎝ ⎠∑ ∫ .
02
0
( )nj tTnG g t e dt
T
π∞ −
−∞
⎛ ⎞=⎜ ⎟
⎝ ⎠∫ .
02
* *
0
( ) .nj tTnG g t e dt
T
π∞
−∞
⎛ ⎞=⎜ ⎟
⎝ ⎠∫
2
*2 2
0 0 0 0 0
1 1n n
n n nG G GT T T T T
∞ ∞
=−∞ =−∞
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑ ∑
This relation is known as Parseval’s power Theorem. It states that the average power of a periodic signal ( )pg t is equal to the sum of the squared amplitudes of all the harmonic components of the signal ( )pg t .
10. SPECTRAL DENSITY : (Energy or power per unit Area). The Spectral Density of the periodic or non periodic signal ( )g t represents the distribution of Power or Energy in the frequency domain. (or) The total area under the spectral density as a function of frequency is equal to total energy or Average power of the signal. (i) Energy Spectral Density (esd) (ii) Power Spectral Density (psd) (i) Energy Spectral Density : This spectral density gives the distribution of energy of the signal in the frequency domain . we know that
2 2( ) ( )E g t dt G f df∞ ∞
−∞ −∞
= =∫ ∫
( )G f is the amplitude spectrum. Let’s denote the squared amplitude spectrum 2( )G f of the signal ( )g t by ( )g fΨ 2( ) ( )g f G fΨ = ∴ The Energy of the signal ( )g t in terms of ( )g fΨ is
( )gE f df∞
−∞
= Ψ∫ .
The above equation shows that the total energy of the signal is given by total area under curve ( )g fΨ ∴ ( )g fΨ represents the energy spectral density of the signal ( )g t in joules per Hertz. We know that amplitude spectrum of real valued signal is an even function of f. ∴ Energy Spectral Density of such signal is symmetrical about the vertical axis at frequency f = 0.
(ii) Power Spectral Density : This spectral density function gives the distribution of power of the signal in the frequency domain .
0
0
22 2
20 0 0
2
1 1( )
T
pnT
nP g t dt GT T T
∞
=−∞−
⎛ ⎞= = ⎜ ⎟
⎝ ⎠∑∫
Let ( )
pgS f = power spectral density
( )gpP S f df∞
−∞
∴ = ∫
2
20 0 0
1( )gpn
n nS f G fT T T
δ∞
=−∞
⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∑
2
20 0 0
1n
n nP G f dfT T T
δ∞ ∞
=−∞−∞
⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∑∫
2
20 0 0
1n
n nG f dfT T T
δ∞∞
=−∞ −∞
⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∑ ∫
=2
20 0
1n
nGT T
∞
=−∞
⎛ ⎞⎜ ⎟⎝ ⎠
∑
( )gpP S f df∞
−∞
∴ = ∫
∴ The average power is the area under the power spectral density ( )gpS f .
