22842 1 frequency analysis using dtft

INTRODUCTION TO DISCRETE-TIME FOURIER TRANSFORM (DTFT)

In this chapter, we will discuss about the Fourier transform of discrete-time

signals, i.e. discrete-time Fourier transform (DTFT). There are many similarities in the

analysis of continuous – time and discrete – time signals using Fourier series and there

are also important differences between continuous-time Fourier series (CTFS) and

discrete – time Fourier series (DTFS). For example, the Fourier series representation of

a discrete – time periodic signal is finite series but the Fourier series representation of

continuous – time periodic signal is infinite series. Also we will see in this chapter, there

are corresponding differences between continuous – time Fourier transform (CTFT) and

discrete – time Fourier transform (DTFT).

Here we will extend the Fourier series description of discrete – time periodic

signals in order to develop a Fourier transform representation for discrete – time non-

periodic signals.

url: development-of-discrete-time-fourier-transform.aspx

Title: Development Of Discrete Time Fourier Transform, DTFT, Digital Signal

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Description: Discrete – time Fourier transform, DTFT is a transformation tool that transforms discrete – time signal from time-domain to frequency-domain. In this section, we will develop an expression for discrete – time Fourier transform...

Keywords: DTFT, discrete time fourier transform, continuous time period, periodic signal, non periodic signal, digital signal processing assignment help, homework help, sample homework

DEVELOPMENT OF THE DISCRETE – TIME FOURIER TRANSFORM (DTFT)

Discrete – time Fourier transform (DTFT) is a transformation tool that transforms

discrete – time signal from time-domain to frequency-domain. In this section, we will

develop an expression for discrete – time Fourier transform (DTFT) for discrete – time

signals. We have already learnt that the Fourier series coefficient for a continuous –

time periodic square wave can be viewed as samples of an envelope function. As the

fundamental period of the continuous – time periodic square wave increases, its

samples become more and more finely spaced. Here a non-periodic signal s(t) is used

to construct a periodic signal s(t) that equals to S(t) over one fundamental period. As the

period approaches infinity, s(t) will be equal to s(t) over larger and larger intervals of

time and the Fourier series representation for s(t). In this section, we will develop an

analogous procedure for deriving an expression for discrete – time Fourier transform

(DTFT) for discrete – time non – periodic signals.

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url: expression-for-discrete-time-fourier-transform.aspx

Title: Expression For Discrete Time Fourier Transform, Digital Signal Processing

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Description: Expression for discrete time fourier transform. Discrete – time Fourier transform, DTFT is a transformation tool that transforms discrete – time signal from time-domain to frequency-domain...

Keywords: fourier transform, discrete tme sequence, fourier series, digital signal processing assignment help, homework help, sample homework

Expression For Discrete – time Fourier Transform

Consider a general discrete – time sequence s(n) of finite duration. That is,

s (n )≠0 , for−N1≤n≤N 2

= 0, otherwise

From this non – periodic signal, we can construct a periodic signal or sequence ~s (n) for which s(n) is of one period. A discrete – time non-periodic signal s(n) is shown

in Fig. 4.1(α) and a periodic signal ~s (n) which is constructed form s(n) for which s(n) is

of one fundamental period is shown in Fig. 4.1 (b).

If fundamental period N0 approaches infinity then ~s (n )=s (n ) for any finite value of

n.

Fourier series representation of ~s (n) can be expressed as

~s (n) = ∑

k=¿N0>¿A k ejk ω0 n¿

¿

where Ak is the Fourier series coefficient and ω0= 2π/N0 is the fundamental frequency.

N0 is the fundamental period.

Fourier series coefficients Ak are given by

Ak = 1N 0

∑n=¿N0>¿

~s (n)e− jkω0n¿

¿

Since s(n) = ~s (n) over one period that includes the interval -N1 ≤ n ≤ N2.

Therefore replacing ~s (n) by s(n) in Eq. we get

Ak=1N 0

∑n=n1

N 2

s (n ) e− j 󠏉k ω0n

But s (n ) is zero outside the interval – N 1≤n≤N2

S (e jω )= ∑n=−∞

∞

s (n ) e jωn

We see that the Fourier series coefficients Ak are directly proportional to samples

of S(e jkω0)

Ak=1N 0

S (e jk ω0 )

Now combining Eqsand we get

~s (n)= ∑

k=¿N0>¿ A k ejk ω0 n= ∑

k=¿N0 >¿ 1N 0

S(e jk ω0)e jk ω0n

¿¿¿

¿

Since ω0 ¿2πN0

∨ 1N0

=ω0

2 π, Eq. can be rewritten as

~s (n )= ∑

k=¿N 0>¿ω0

2πS (e jk ω0n )e jkω0 n¿

¿

= 1

2π ∑k=¿N 0>¿ S (e jk ω0)e jkω0 nω ¿

¿

As fundamental period N0 increases, fundamental frequency ω0 decreases. As

N0 approaches infinity, the Eq. passes to an integral.

Eq. can be graphically interpreted in Fig.

From Eq., S (e jω )is periodic∈ωwith period 2π and therefore e jωn is also periodic in ω

with same period. Thus, the product S (e jω )e jωn in Eq. 4.8 represents the area of

rectangle of height S (e jj ω0 )e jω0n and width ω0. As fundamental frequency ω0 approaches zero, the summation in Eqbecomes an integral. The summation is carried out over N0 at consecutive intervals of width ω0 but the total interval of integration will always be of width 2π.

Therefore, as N0 → ∞ or ω0 → 0.~s (n) = s(n), the Eq. becomes

S (n )= 12π∫2π

S (e jω)e jωndω

Where S (e jω )= ∑n=−∞

∞s (n ) e jωn

¿¿

The function S (e jω ) is called the Discrete – time Fourier Transform (DTFT). Eq. is

called DTFT synthesis equation. Eq. is referred to as DTFT analysis equation. These

equations show how a non-periodic sequence can be thought of as a linear combination

of complex exponential functions.

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url: dtft-ctft-differences-and-similarities.aspx

Title: Transform & Logarithms, Comparison, Digital Signal Processing Assignment

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Description: The discrete – time Fourier transform (DTFT) has many similarities with the continuous – time Fourier transform (CTFT) such as linearity, convolution property...

Keywords: DTFT. CTFT, discrete time fourier transform, continous time fourier transform, periodic, digital signal processing assignment help, homework help, sample homework

DTFT And CTFT

The discrete – time Fourier transform (DTFT) has many similarities with the continuous

– time Fourier transform (CTFT) such as linearity, convolution property etc. The major

differences between the DTFT and CTFT are :

(i) DTFT is periodic in ω with period 2 π but CTFT is not periodic.

(ii) DTFT has finite interval of integration in the synthesis equation but CTFT has

infinite interval of integration in the synthesis equation.

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url: convergence-of-dtft.aspx

Title: Convergence Of DTFT, Digital Signal Processing Assignment Help, Homework Help

Description: Here we will discuss about the convergence of the infinite summation in the DTFT analysis equation given as...

Keywords: convergence of DTFT, digital signal processing assignment help, homework help, sample homework

CONVERGENCE OF THE DTFT

Here we will discuss about the convergence of the infinite summation in the

DTFT analysis equation given as

S (e jω )= ∑n=−∞

∞

s (n ) e− jωn ….(4.11)

The conditions on s(n) that guarantee the convergence of this sum are direct

counterparts of the convergence conditions for the CTFT Eq. 4.11 will converge either if

signal s(n) is absolutely sum able i.e.,

∑n=−∞

∞

|s (n )|<∞ ….(4.12)

or, if the sequence has finite energy, i.e.,

∑−∞

∞

|s (n )|2<∞ ….(4.13)

But, there are no issues associated with DTFT synthesis equations given by

s (n )= 12π∫2π

S (e jω)e jωndω ….(4.14)

as the integral in this synthesis equation is over a finite interval of integration. This is a

similar situation as for the discrete – time Fourier series (DTFS) synthesis equation

which involves a finite sum. Consequently, there is no issue of convergence associated

with DTFT synthesis equation.

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url: dtft-solved-examples1.aspx

Title: Determine DTFT, Solved Examples, Digital Signal Processing Assignment Help, Homework Help

Description: Solved examples based on discrete time fourier transform. Determine DTFT of a discrete – time signal...

Keywords: DTFT. Discrete time fourier transform, solved examples, digital signal processing assignment help, homework help, sample homework

Solved Examples Of DTFT 1

Solved Example : Determine DTFT of a discrete – time signal s ( t )=Anu (n ) ,|A|<1.

Solution. DTFT of s(n) is given by

S (e jω )=DTFT {s (n ) }= ∑n=−∞

∞

s (n ) e− jωn

= ∑n=−∞

∞

s (n ) e− jωn …..(1)

But u(n) is the unit – step sequence and it is defined as

u (n )={1 , n≥00 , n<0 …..(2)

Substituting Eq. 2 in Eq. 1, we obtain

S (e jω )=∑n=0

∞

An1e jωn=∑n=0

∞

(A e− jω )n

(This is a geometric progression of infinite number of terms.)

= 1

1−Ae− jω

Or S(e jω) = 1

1−Ae− jω

Note that DTFT of any discrete – time sequence s(n) is periodic function in ω

with period 2π.


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url: dtft-solved-examples2.aspx

Title: Determine DTFT, Discrete Signal, Solved Examples, Digital Signal Processing


Description: Solved examples based on discrete time fourier transform. Determine DTFT of a discrete – time signal...

Keywords: DTFT. Discrete time fourier transform, solved examples, digital signal processing assignment help, homework help, sample homework

Solved Examples Of DTFT 2

Solved Example Determine the DTFT of the discrete-time signal s (n )=A|n|,|A|<1.

Solution. DTFT of the discrete-time signal is determined;

S (e jω )=DTFT {s (n ) }= ∑n=−∞

∞

s (n ) e jωn

= ∑n=−∞

∞

A¿n∨¿ e−jωn=∑

n=0

∞

Ane− jωn+ ∑n=−∞

−1

A−ne jωn

¿

Substituting n = -m in the second summation, we obtain

S (e jω )=∑n=0

∞

Ane− jωn+∑m=1

∞

Am e jωm …(1)

Both of the summations given inEq. 1 are infinite geometric progressions that we

can evaluate in closed form, producing

S (e jω )=∑n=0

∞

(A e jω)m

¿ 11−Ae− jω+ A e jω

1−A eω

= 1−Ae jω+A e jω (1−A e jω)

(1−Ae− jω) (1−A e jω)

= 1−A e jω+Ae jω−A2

1−2 A ( eω+e− jω

2 )+A2

or S (e jω )= 1−A2

1−2 Acosω+A2 .....(2)

The discrete – time signal s(n) is shown in Fig. 4.3 and its DTFT is shown in Fig.

4.4. Here S (e jω )= 1−A2

1−2 A cosω+A2 is a real valued function of ω.



url: determine-dtft-of-discrete-time-rectangular -pulse.aspx

Title: DTFT, Discrete Time Rectangular Pulse, Comparison, Digital Signal Processing


Description: Solved examples based on discrete time fourier transform. Determine the DTFT of the discrete – time rectangular pulse given as ...

Keywords: DTFT, discrete time rectangular pulse, digital signal processing assignment help, homework help, sample homework

Determine DTFT Of Discrete – Time Rectangular Pulse

Example: Determine the DTFT of the discrete – time rectangular pulse given as

s (n )={1 ,|n|≤N 1

0 ,|n|>N1

DTFT of this rectangular pulse is determined as

S (e jω )=DTFT {s (n ) }= ∑n=−∞

∞

s (n ) e jωn

= ∑n=N1

N1

1 e− jωn= ∑n=−N1

N 1

e− jωn ..(1)

This is a geometric series or progression of finite number of terms.

Eq. 1 can be rewritten as

S (e jω )= ∑n=−N1

N1

e jωn=sinω [N 1+( 1

2 )]sin(ω

2 ¿)¿

The discrete – time Fourier transform of rectangular pulse is shown in Fig. 4.6.

DTFT of a discrete-time rectangular pulse is periodic with period 2 π but the

CTFT of the continuous – time rectangular pulse is not periodic.



url: fourier-transform-of-discrete-time-periodic-signal.aspx

Title: Fourier Transform Of Discrete Time Periodic Signal, Digital Signal Processing


Description: Fourier transform for discrete – time periodic signals is interpreted as the Fourier transform of a periodic signal as an impulse train in the frequency domain...

Keywords: fourier transform of discrete time periodic signal, fourier transform, digital signal processing assignment help, homework help, sample homework

FOURIER TRANSFORM OF DISCRETE – TIME PERIODIC SIGNALS

In this section, we will study about the Fourier transform for discrete-time periodic

signals. Fourier transform for discrete – time periodic signals is interpreted as the

Fourier transform of a periodic signal as an impulse train in the frequency domain.

To derive an expression for Fourier transform for discrete – time periodic signals,

we consider the signal

S(n) = ejωn ….(4.15)

We know that the CTFT of e jω0 tcan be interpreted as an impulse at ω = ω0.

Therefore, we expect the same type of Fourier transform that results for the discrete –

time signal of Eq. 4.15. However, the DTFT must be periodic in ω with period 2π.

Fourier transform of s(n) = ejωn should have impulses at ω0 ,ω0±2π ,ω0 ,±4 π and so on.

Fourier transform of s(n) is the impulse train. It is given by Eq. 4.16 and illustrated in

Fig. 4.7.

S (e jω )= ∑m=−∞

∞

2π δ (ω−ω0−2πm ) ….(4.16)

For checking the validity of expression given in Eq. 4.16, we will determine the

inverse Fourier transform.

Substituting the Eq. 4.16 into the DTFT synthesis equation given as

12π∫2 π

S (e jω )e jω󐏉󐏉dω

= 12π∫2 π [ ∑m=−∞

∞

2 πδ (ω−ω0−2 πm)]e jωndω ….(4.17)

We know that any interval of length 2π includes exactly one impulse in the

summation given by Eq. 4.16. Therefore, if the interval of integration is so chosen that it

includes the impulse located at ω0 + 2πℓ, then Eq. 4.17 becomes

12π∫2 π

S (e jω )e jω ndω=e j (ω0+2π l)n=e jωn󐏉 ….(4.18)

and the DTFT is given by

S (e jω )= ∑k=−∞

∞

2π A kδ (ω−2πkN0 ) ….(4.19)

From the above analysis, we conclude that the Fourier transform of a discrete –

time periodic signal can be directly constructed from its Fourier series coefficients Ak.

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url: determine-dtft-example.aspx

Title: Determine DTFT, examples, Comparison, Digital Signal Processing Assignment Help, Homework Help

Description: Determine the DTFT of the discrete – time periodic signal. Transtutors is the best place to get answers to all your doubts regarding discrete time fourier transform...

Keywords: DTFT, discrete time periodic signal, eulers relation, digital signal processing assignment help, homework help, sample homework

Determine DTFT

Example: Determine the DTFT of the discrete – time periodic signal

s(n) = cos ω0n

with fundamental frequency ω0 = 2π/5

Solution. Using Euler’s relation, signal s(n) = cos ω0n can be written as

s (n )=cosω0n=e jω0n+e

− jω0 n

2 …(1)

= 12e jω0n+ 1

2e− jω0n

But we know that DTFT of a period signal s(n) = e jω0n is given by

S (e jω )=DTFT {s (n ) }=DTFT {e jω0n }

= ∑m=−∞

∞

2πδ (ω−ω0−2πm ) ….(2)

Similarly, DTFT of ?Eq. 1 can be determined as

S (e jω )=DTFT {s (n ) }=DTFT {12e jω0n+ 1

2e− j ω0n}

= 12DTFT {e jω0n }+ 1

2DTFT {e− jω0n }

= 12 ∑

m=−∞

∞

2πδ (ω−ω0−2 πm)+ 12 ∑

m=−∞

∞

2πδ (ω+ω0−2 πm)

(Substitutingω0=2π5 )

or Se jω¿= ∑m=−∞

∞

2πδ(ω−2 π5

−2πm)

+ ∑m=−∞

∞

πδ (ω+2 π3

−2πm) ,,,,(3)

S (e jπ )= ∑m=−∞

∞

πδ (ω−2π5

−2 πm)+ ∑m=−∞

∞

π δ(ω+2 π5

−2πm)

= {…+πδ(ω−2π5

+2 π)+πδ (ω−2 π5 )+πδ (ω−2π

5−2π )+…}

+{….+πδ (ω+ 2π5

+2π )+πδ(ω+2 π3 )+πδ (ω+ 2π

5−2 π)+….}

or S (e jω )=πδ (ω−2 π5 )+π δ(ω+ 2π

3 ),−π ≤ω≤π …(4)

Here S (e jω ) repeats periodically witha period of 2π . DTFT of s (n )=cosω0n is shown∈fig



url: determine-dtft-of-discrete-time-periodic-impulse-train.aspx

Title: DTFT Of Discrete Time Periodic Impulse Train, Comparison, Digital Signal


Description: Explanation regarding DTFT of the discrete time periodic impulse train. Determine the DTFT of the discrete-time periodic impulse train given by...

Keywords: logarithm, transform, comparison, laplace transform, fourier transform, time signal, z transform, digital signal processing assignment help, homework help, sample homework

Determine the DTFT of the discrete-time periodic impulse train

Example: Determine the DTFT of the discrete-time periodic impulse train given by

s (n )=∑k=∞

∞

δ (n−k N 0)

Solution. Discrete – time periodic impulse train s (n )= ∑k=−∞

∞

δ (n−k N0) is sketched in Fig.

4.9.

Now, w can calculate the Fourier series coefficients for this discrete-time periodic

impulse train s(n) = ∑k=−∞

∞

δ (n−k N0 ) directly from equation given below

Ak = 1N 0

∑n=¿N0>¿¿

s (n ) e− jk ω0n

Substituting s(n) = ∑k=−∞

∞

δ (n−k N0 ) and choosing the interval of summation as 0

≤n≤N0 – 1 in Eq. 1, we obtain

Ak = 1N 0

∑n=0

N 0−1

∑k=−∞

∞

δ (n−k N0 )e− jk ω0n=

1N 0 …(2)

DTFT of s(n) is determined as

S (e jω )=DTFT {s (n ) }=DTFT ¿

= ∑k=−∞

∞

2π Ak δ (ω−2πkN0 ) …(3)


S (e jω )=2πN 0

∑k=−∞

∞

δ (ω−2πkN0 ) ….(4)

DTFT of discrete – time periodic impulse train s(n) = ∑k=−∞

∞

δ (n−k N0 ) is shown∈Fig .4 .10 .

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url: properties-of-dtft.aspx

Title: Properties Of DTFT, Digital Signal Processing Assignment Help, Homework Help

Description: this section, we will study about the various properties of Fourier transform of discrete – time signals. Fourier transform of discrete – time signals is also referred to as discrete – time Fourier transform...

Keywords: properties of DTFT, fourier transform, CTFT, digital signal processing assignment help, homework help, sample homework

PROPERTIES OF THE DTFT

In this section, we will study about the various properties of Fourier transform of

discrete – time signals. Fourier transform of discrete – time signals is also referred to

as discrete – time Fourier transform (DTFT). These properties are often useful in

reducing the complexity in evaluation of the Fourier transform and inverse Fourier

transform. Also, we will see some of the similarities and differences between CTFT and

DTFT. The derivation of DTFT properties is essentially identical to its continuous – time

counterpart, i.e. CTFT. Also, because of the close relationship between the Fourier

series and Fourier transform, many of the DTFT properties translate directly into

corresponding DTFS properties.

Here we will use one notation similar to that used for CTFT to indicate the pairing

of a signal and its Fourier transform. That is

S (e jω )=DTFT {s (n ) }

s (n )=Inverse DTFT {S(e jω)}

s (n )=DTFT↔

S (e jω )

In this section, we will discuss following properties of the DTFT:

1. Periodicity of the DTFT

2. Linearity of the DTFT

3. Time shifting and frequency shifting

4. Complex-conjugation and conjugate symmetry

5. Differencing and accumulation in time-domain

6. Time reversal of a discrete – time sequence

7. Time – expansion

8. Differentiation in frequency – domain

9. Parseval’ relation for DTFT

10.The convolution property for DTFT

11.The multiplication property for DTFT.

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url: periodicity-of-dtft.aspx

Title: Periodicity Of DTFT, Digital Signal Processing Assignment Help, Homework Help

Description: We have already discussed that discrete – time Fourier transform (DTFT) of a discrete – time signal is always periodic in ω with period 2π. A periodic function is a function which repeats its value after some fixed value of independent variable...

Keywords: periodicity of DTFT, discrete time fourier transform, digital signal processing assignment help, homework help, sample homework

Periodicity of the DTFT

We have already discussed that discrete – time Fourier transform (DTFT) of a

discrete – time signal is always periodic in ω with period 2π. A periodic function is a

function which repeats its value after some fixed value of independent variable. This

fixed value of independent variable is called the period of the periodic function.

The DTFT of s(n) is always periodic in ω with period 2π, i.e.

s (n ) DTFT↔

S (e jω )

S (e j (ω+2π ) )=S (e jω)

This is in contrast of CTFT, which in general is not a periodic function.

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url: linearity-of-dtft.aspx

Title: Linearity Of DTFT, Digital Signal Processing Assignment Help, Homework Help

Description: Properties of discrete time fourier transform (DTFT) includes linearity of DTFT. The DTFT is also a linear transformation tool as CTFT...

Keywords: DTFT, CTFT, linear transformation, digital signal processing assignment help, homework help, sample homework

Linearity of DTFT

The DTFT is also a linear transformation tool as CTFT.

s1 (n )DTFT↔

S1 (e jω)¿

s2 (n ) DTFT↔

S2 (e jω)

where s1(n) and s2(n) are discrete – time sequences whose DTFs are

S1 (e jω )∧S2 (e jω ) ,respectively .

From the property of linearity, it is true for above two discrete – time sequences

s1(n) and s2(n) given as

A1 s1 (n )+A2 s2 (n ) DTFT↔

A1S1 (e jω)+A2S2(ejω)

where A1 and A2 are constants.

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url: time-shifting-and-frequency-shifting.aspx

Title: Time Shifting & Frequency Shifting, Digital Signal Processing Assignment Help, Homework Help

Description: Here we will first study the effect of time shifting of a discrete-time sequence s(n) on its DTFT and then the effect of frequency shifting...

Keywords: DTFT, frequency shifting, time shifting, digital signal processing assignment help, homework help, sample homework

Time-Shifting and Frequency Shifting

Here we will first study the effect of time shifting of a discrete-time sequence s(n)

on its DTFT S (e jω )and then the effect of frequency shifting.

If s (n ) DTFT↔

S (e jω )

then s (n−n0 )DTFT↔

e− jω0S(e jω) ….(4.23)

where s (n−n0 ) is the time-shifted version of s(n).

Now will see the effect of frequency shifting.

If s (n ) DTFT↔

S (e jω )

then e j ω0n s (n )DTFT↔

S ¿

where S(e j(ω−ω0 )) is the frequency – shifted version of S(e jω).

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url: complex-and-conjugate-symmetry.aspx

Title: Complex & Conjugate Symmetry, Digital Signal Processing Assignment Help, Homework Help

Description: We can obtain complex conjugation of a complex discrete-time signal s(n) by reversing the sign of the imaginary part of the complex signal...

Keywords: DTFT, complex conjugation, conjugate symmetry, discrete time signal, digital signal processing assignment help, homework help, sample homework

Complex Conjugation And Conjugate Symmetry

We can obtain complex conjugation of a complex discrete-time signal s(n) by

reversing the sign of the imaginary part of the complex signal s(n).

Let s(n) be the complex discrete – time signal s (n )=sR (n )+ j sI (n) and its DTFT is

also complex, i.e.,

S (e jω )=SR (e jω )+ j S I (ejω)

where sR(n) and sr (n) are real and imaginary parts of s(n), and SR (ew ) and S I (ejω)

are also real and imaginary parts of S (e jω ) ,respectively.

Now complex-conjugation of s(n) is given by

S¿ (n )=sR (n )− j sr (n )

and complex conjugation of S(e jω) is given by

S¿ (e jω)=SR (e jω)− j S I (e jω )

If s (n ) DTFT↔

s (EJω )

then s¿ (n )DTFT↔

S¿(e− jω) …(4.25)

Also, if signal s(n) is a real-valued function, then its DTFT S(e jω) will be conjugate

symmetric. That is

S (e jω )=S¿ (e jω) for s (n ) real ….(4.26)

From Eq. 4.26, we can say that Re {S (e jω)} is an even function of ω and I m

{S (e jω)} is an odd function of ω. Similarly, the magnitude of S(e jω) is an even function

and the phase angle S (e jω ) is an odd S(e jω) is an odd function.

Also, we can decompose sequence s(n) into even and odd parts. Furthermore,

Ev {s (n ) }=sE (n )DTFT↔

ReS (e jω)=SR (e jω)

and Od {s (n ) }=s0 (n )DTFT↔

j I m {S (e jω )}= j S I (e jω )

where Ev {s(n)} and Od {s(n)} are the even and odd parts, respectively of s(n).

Specifically, if s(n) is real and even, its DTFT is also rean and even.

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url: differencing-and-accumulation-in-time.aspx

Title: Differencing & Accumulation In Time, Digital Signal Processing Assignment Help, Homework Help

Description: Accumulation is the discrete-time counterpart of integration. Integration is used for continuous-time signals. Inverse of accumulation is referred to as first differencing...

Keywords: differencing and accumulation in time, DTFT, discrete time signal, digital signal processing assignment help, homework help, sample homework

Differencing and Accumulation in Time

Accumulation is the discrete-time counterpart of integration. Integration is used

for continuous-time signals. Inverse of accumulation is referred to as first differencing.

Let s(n) be a discrete-time signal with DTFT S (e jω ) . The first differencing of signal

s(n) is given by s(n) is given by s(n) – s (n−1 ) . DTFT of first differencing of signal s(n)

can be determined by using the properties of linearity and time-shifting as

s (n )−s (n−1 )DTFT↔

S (e jω )

= (1−e− jω )S (e jω) …..(4.27)

The accumulation of signal s(n) is given by

Y(n) = ∑m=−∞

n

s (m) …..(4.28)

Eq. 4.28 can also be expressed as

y (n )− y (n−1 )=s(n) ….(4.29)

Thus, we con conclude that DTFT of y (n) should be related to the DTFT of s(n)

division by (1-e-jω). This is not perfectly correct but the precise relation is given by

∑m=−∞

∞

s (m )DTFT↔

11−e jω S (e jω)+πS (e j0 ) ∑

k=−∞

∞

δ (ω−2πk ) ….(4.30)

The impulse train on the RHS of Eq. 4.30 shows the dc or average value that can

result from summation.

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url: accumulation-property-of-dtft-examples.aspx

Title: Accumulation Property Of DTFT, Solved Examples, Digital Signal Processing


Description: Solved examples based on accumulation property of the DTFT. Accumulation is the discrete-time counterpart of integration…

Keywords: differencing and accumulation in time, DTFT, discrete time signal, digital signal processing assignment help, homework help, sample homework

Solved Examples Of Differencing and Accumulation in Time

Example: Determine the DTFT of the unit step function s (n )=u (n) using the

accumulation property of DTFT.

Solution. We know that

g (n )=δ (n ) DTFT↔

G (e jω )=1 …..(1)

Also we have studied that the unit-step function u(n) is the running um of the unit

– impulse function δ(n). This relation is given as

s (n )= ∑m=−∞

∞

g(m)

Now, taking the DTFT of both sides of Eq. 2 and using accumulation property of

DTFT, we obtain

S (e jω )=DTFT { ∑m=−∞

∞

g (m )}or S (e jω )= 1

(1−e− jω )G (e jω)+πG (e j 0) ∑

k=−∞

∞

δ (ω−2πk ) …(3)

But from Eq. 1, we have

G (e jω )=1∧G (e j0 )=1

Substituting these values in Eq. 3, we obtain

S (e jω )= 1(1−e− jω )

+π ∑k=−∞

∞

㲴(ω−2 πk¿)¿ ….(4)

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url: time-reversal-of-discrete-time-signal.aspx

Title: Time Reversal Of Discrete-Time Signals, Digital Signal Processing Assignment Help, Homework Help

Description: Taking mirror image or folding of a discrete-time sequence is called time reversal. Consider a discrete-time signal or sequence s(n) whose mirror image is given by...

Keywords: time reversal of discrete time signal, DTFT, digital signal processing assignment help, homework help, sample homework

Time Reversal of a Discrete-Time Signals

Taking mirror image or folding of a discrete-time sequence is called time

reversal. Consider a discrete-time signal or sequence s(n) whose mirror image is given

by s (−n ) .

If s (n ) DTFT↔

S (e jω )

then s (−n ) DTFT↔

S(e− jω) …..(4.31)

Now we can prove this property as

y (n )=s (−n) …..(4.32)

Taking DTFT of both sides of Eq. 4.32, we obtain

Y (e jω )=DTFT { y (n)}= ∑n=−∞

∞

s (−n ) e− jωn …..(4.33)

Now substituting m=−n in Eq. 4.33, we obtain

Y (e jω )= ∑m=−∞

221 e

s (m ) e− jω(−m)=S (e− jω) ….(4.34)

That is,

s (−n ) DTFT↔

S(e− jω) …..(4.35)

Hence , it is to be proved.

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url: time-expansion-of-discrete-time-signals.aspx

Title: Time Expansion Of Discrete Time Signals, Digital Signal Processing Assignment Help, Homework Help

Description: Time expansion of a discrete – time signal is equivalent to multiplication of independent variable, i.e., time of a signal by an integer scalar quantity A, which is greater than unity...

Keywords: discrete time signal, time expansion, DTFT, CTFT, digital signal processing assignment help, homework help, sample homework

Time Expansion of a Discrete – Time Signals

Time expansion of a discrete – time signal is equivalent to multiplication of

independent variable, i.e., time of a signal by an integer scalar quantity A, which is

greater than unity.

We have already derived time expansion property of CTFT which is given by

s (At )CTFT↔

1|a|

S ( jωA ) …..(4.36)

We cannot slow down the signal by choosing A <1. On the other hand, if we let

A be an integer other than ±1. For example A = 2 then s(2n), we cannot merely speed

up the original signal. That is, since n can take on only integer values the signal s(2n)

consists of the even samples of original signal s(n) alone.

Time – expanded signal is given by

s( k ) (n )={s( nk ), if n is amultiple of k0 , if n is not amultiple of k

…..(4.37)

Where k is a positive integer.

For k = 2, the sequence s( k )(n) is obtained from original sequence s(n) by placing

k – 1 zeros between successive values of the original sequence. Now we can say that

s( k ) (n ) is a slowed-down version of s(n).

One sequence s(n) and its slowed down version s( k ) (n ) for k = 2 are shown in Fig.

4.11.

Since s( k )(n) equals zero unless n is multiple of k.

For n = mk, DTFT of s( k ) (n ) is given by

S(k)¿ ejω¿=DTFT {s( k )(n)}= ∑

n=−∞

∞

s (k ) (n ) e− jωn

= ∑n=−∞

∞

s (k ) (mk ) e− jωmk

Furthermore, since s( k ) (mk ) , we find that

S( k ) (ejω )= ∑

m=−∞

∞

s (m )e− j (kω )m=S (e jkω )

That is s( k ) (n )DTFT↔

S (e jkω)

As the discrete – time signal is spread out and slowed down in time by taking k >

1, its DTFT is compressed. For example, since S(e jω) is periodic in ω with period 2π

and S (e jω ) is also periodic in ω with period 2π/k.

Now we can conclude that there is inverse relationship between time and

frequency domains. In other words, we can say that as value of k increases, s( k )(n)

spreads out while its discrete – time Fourier transform (DTFT) is compressed.

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url: time-expansion-of-discrete-time-signals-solved-examples.aspx

Title: Solved Examples Of Time Expansion Of Discrete Time Signals, Digital Signal


Description: Determine the DTFT of the discrete – time sequence s(n) shown in Fig. using time-expansion property of DTFT. Time expansion of a discrete – time signal is equivalent to...

Keywords: discrete time signal, time expansion, DTFT, CTFT, digital signal processing assignment help, homework help, sample homework

Solved Examples Of Time Expansion of a Discrete – Time Signals

Example: Determine the DTFT of the discrete – time sequence s(n) shown in Fig. using

time-expansion property of DTFT.

Solution. This sequence s(n) can be related to the simpler sequence g(n) shown in

Fig. (a).

The discrete – time sequence s(n) is related to the discrete – time sequence g(n)

as

s (n )=g2 (n )+2 g( 2)(n−1) …..(1)

where g2 (n )={g( n2 ) , if n is even0 , if n is odd

…….(2)

and g2(n-1) is the shifted version of g2(n) by one unit to the right. The signals

g( 2)(n−1) and 2g (2 )(n−1) are depicted in Fig. 4.13(b) and Fig. 4.13 (c) respectively.

Here, we relate given sequence s(n) in terms of a simpler sequence g(n) which is

a discrete-time rectangular pulse.

DTFT of g(n) = G (e jω )= ∑n=−∞

−∞

g (n ) e− jωn

= ∑n=0

4

1e− jωn=1[ 1−(e− jω)5 ]

[1−e− jω ]

=1−e− j5ω

1−e− jω = e− j 󐏉󐏉 5ω

2 [ e j 5ω /2−e−5 jω /2

2 j ]

e− jω /2[ e jω

2−e−ω /2

2 j ]

or G (e jω )=e− j2ω{sin (󐏉5ω2 )

sin(ω2

) } …(3)

g (n ) DTFT↔

G (e jω )=e− j2ω{sin (5ω2 )

sin (ω2 ) }then from time-expansion property

g( k ) (n )DTFT↔

G ¿

For k = 2, g( 2) (n )DTFT↔

G (e j 2ω )=e− j2× 2ω {sin( 5×22 )

sin( 2ω2 ) }

= e− j4ω{sin 󠏉(5ω)

sin 󠏉(ω) } ….(4)

Now using the linearity and time-shifting properties, we get

2g (2 ) (n ) DTFT↔

e− jω {2G (e j2ω )}

= 2e− jω [ sin (5ω )sin (ω ) ]

= 2e j5ω[ sin (5ω)sin (ω) ] ….(5)

Combining Eqs. 4 and 5, we have

s (n )=g (2) (n )+2g (2 ) (n−1 )DTFT↔

e j4ω[ sin (5ω )sin (ω) ]

+2e− j5ω[ sin (5ω)sin (ω) ]

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url: differentiation-in-frequency-domain.aspx

Title: Differentiation In Frequency-Domain, Digital Signal Processing Assignment Help, Homework Help

Description: Explanations regarding differentiation in frequency domain. Here we will study the differentiation of DTFT using the DTFT analysis equation...

Keywords: DTFT, differentiation in frequency domain, digital signal processing assignment help, homework help, sample homework

Differentiation in Frequency-Domain

Here we will study the differentiation of DTFT of s(n), S (e jω )withrespect ¿ω.Let

s (n ) DTFT↔

S (e jω )

Using the DTFT analysis equation S (e jω )= ∑n=−∞

∞

s (n ) e− jωn and differentiating both

sides with respect to ω, we obtain

dS (e jω )d ω

= dd ω [ ∑n=−∞

∞

s (n ) e− jωn ]= ∑

n=−∞

∞

s (n ) [− jne− jωn]

= − j ∑n=−∞

∞

ns (n ) e jωn

or jdS(e jω)d ω

=− j2 ∑n=−∞

∞

n s (n ) e− jωn

or jdS(e jω)d ω

= ∑n=−∞

∞

(ns (n ) )e− jωn=DTFT {ns (n)}

or n s (n ) DTFT↔

jdS (e jω)d ω

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url: parsevals-relation-for-dtft.aspx

Title: Parsevals Relation Of DTFT, Comparison, Digital Signal Processing Assignment Help, Homework Help

Description: Parsevals relation for DTFT states that the total energy in a discrete – time signal s(n) may be determined either by computing the energy per unit time and summing over all time or by computing the energy per unit frequency...

Keywords: parsevals relation for DTFT, energy-density spectrum, digital signal processing assignment help, homework help, sample homework

Parseval’s Relation for DTFT

Parseval’s relation for DTFT states that the total energy in a discrete – time

signal s(n) may be determined either by computing the energy per unit time |s(n)|2 and

summing over all time or by computing the energy per unit frequency |S (e jω )2/2π| and

integrating over a full 2π interval of distinct discrete – time frequencies. In analogous

with the continuous – time signal ¿is called the energy – density spectrum of the signal

s (n ) . Parseval’s relation for DTFT is given as

∑n=−∞

∞

|s (10 )|2= 12π ∫2π

|S (e jω)|2d ω

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url: convolution-property.aspx

Title: Convolution Property, Digital Signal Processing Assignment Help, Homework Help

Description: Here we will discuss the importance of the DTFT with regard to its effect on the convolution operation and analysis of discrete – time LTI systems. By using DTFT, convolution of two discrete – time signals...

Keywords: convolution property, DTFT, discrete time signal, discrete – time LTI systems, digital signal processing assignment help, homework help, sample homework

The Convolution Property

Here we will discuss the importance of the DTFT with regard to its effect on the

convolution operation and analysis of discrete – time LTI systems. By using DTFT,

convolution of two discrete – time signals s1 (n )∧s2(n) is converted into multiplication of

DTFT of individual discrete – time signals.

If s (n ) DTFT↔

S1 (e jω )

and s2 (n ) DTFT↔

S1(ejω)

then from convolution property of DTFT

s (n )=s1 (n )∗s2 (n )DTFT↔

S (e jω)=S1 (e jω )S2(ejω) …(4.41)

Now we are going to apply DTFT for representing and analyzing discrete-time

LTI systems. Specifically, if s(n), h (n )∧ y (n) are the input, impulse response and output,

respectively of a discrete-time LTI system.

Output y (n) is determined by convolving s(n) and h(n) given as

y (n )=s (n )∗h(n) ….(4.42)

By using DTFT, Eq. 4.42 can be expressed as

Y (e− jω )=S (e jω)H (e jω) ….(4.43)

Where S (e jω ) ,H (e jω )∧Y (e jω) are the DTFTs pf s (n ) , h (n )∧ y (n ) , respectively.

Now, combining Eqs. 4.42 and 4.43, we get

y (n )=s (n )∗h (n ) DTFT↔

y (e jω )=S (e jω)H (e jω ) …..(4.44)

Where H (e jω) is the discrete – time Fourier transform (DTFT) of the impulse

response h(n) of the discrete – time LTI system. It is also called frequency response of

discrete – time LTI system.

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url: determine-frequency-response-of-discrete-time-lti-system.aspx

Title: Determine Frequency Response Of Discrete Time LTI System, Digital Signal


Description: The frequency response of a discrete – time LTI system is equal to the DTFT of the impulse response of the system. The frequency response is determined as...

Keywords: frequency response of discrete time LTI system, impulse response, frequency response, convolution property of DTFT, digital signal processing assignment help, homework help, sample homework

Determine The Frequency Response Of A Discrete – Time LTI System

Example: Determine the frequency response of a discrete – time LTI system with

impulse response h (n )=δ (n−n0 ) . Also determine output for this system.

Solution. The frequency response of a discrete – time LTI system is equal to the DTFT

of the impulse response h(n) of the system. The frequency response is determined as

H (e jω )=DTFT {h(n)}= ∑n=−∞

∞

h (n )e jωn

∑n=−∞

∞

δ (n−n0 )e jωn=e− jωn0 …(1)

We know from convolution property of DTFT

y (n )=h (n )∗s (n ) DTFT↔

Y (e jω )=H (e jω)S (e jω )

or Y (e jω )=H (e jω )S (e jω) …(2)


Y (e jω )=e− jωn0S (e jω ) …(3)

Output y (n) of above discrete – time LTI system in determined by taking the

inverse DTFT of Eq. 3. Taking inverse DTFT, we get

y (n )=Inverse DTFT ¿

= Inverse DTFT {e− jω n0S(e jω)}=s (n−n0) …(4)

Note in this example output y(n) is equal to shifted version of input s(n) by a

constant time n0. The frequency response H (e jω )=e− jωn0 is purely time-shifted and has

unity magnitude at all frequencies. Its phase characteristics are equal to – ωn0, i.e., it is

linear with frequency.

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url: determine-impulse-response.aspx

Title: Determine Impulse Response, Comparison, Digital Signal Processing


Description: Solved Examples based on impulse response. determine the impulse response h(n) of a discrete-time ideal low pass filter whose frequency response...

Keywords: impulse response, discrete time signal, DTFT, digital signal processing assignment help, homework help, sample homework

Determine The Impulse Response

Example: determine the impulse response h(n) of a discrete-time ideal low pass filter

whose frequency response H (e jω) is shown in Fig

Solution. Impulse response h(n) of discrete – time ideal low pass filter is equal to

inverse DTFT of the frequency response H (e jω).

H(n) = Inverse DTFT {H (e jω)}

= 12π∫2 π

H (e− jω)e jωnd ω .....(1)

In particular, using –π ≤ ω ≤ π as the interval of integration in Eq. 1, we obtain

h (n )= 12 π∫−π

π

H (e jω )e jωnd ω

= 12π ∫

–ωc

ω

1e jωnd ω= 12 π [ 1

jne jωn ]

–ωc

ω0

= 1nπ [ e jωn−e− jωn

2 j ]= 1nπ

sinωcn

or h (n )=sinωc nπn

…(2)

This impulse response h(n) is shown in Fig.

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url: discrete-time-lti-system-solved-problems.aspx

Title: Determine Output Of Discrete Time LTI System, Digital Signal Processing


Description: Solved examples to determine the output of discrete time LTI system with impulse response...

Keywords: discrete time LTI system, impulse response, DTFT, digital signal processing assignment help, homework help, sample homework

Discrete – Time LTI System –Solved Problems

Example: Determine the output y(n) of a discrete – time LTI system with impulse

response h (n )=Anu (n )with|A|<1to an input s(n) = Bnu (n )with |B| < 1.

Solution. Output y(n) is determined by using convolution property of DTFT as

y (n )=h (n )∗s (n ) DTFT↔

y (e jω )=H (e jω)S (e jω )

or Y ¿) = H (e jω )S (e jω) …(1)

Now we will determine H(ejω) and S(ejω) as

H (e jω )=DTFT {h (n ) }= ∑n=−∞

∞

h (n )e− jωn

¿ ∑n=−∞

∞

Anu (n ) e jωn

[But u (n ) is defined asu (n )={1 , n≥00 ,n<0 ]

= ∑n=0

∞

An e− jωn=∑n=0

∞

( Ae− jω )n

or H (e jω )= 11−Ae− jω ….(2)

Similarly, we can determine S(ejω) as

S (e jω )=DTFT {s (n ) }= ∑n=−∞

∞

s (n ) e− jωn

= ∑n=−∞

∞

Bnu (n ) e jωn=∑n=0

∞

Bn e− jωn

= ∑n=0

∞

(Be jω )n= 11−Be− jω

or S (e jω )= 11−Be jω …(3)

Substituting Eqs. 2 and 3 in Eq. 1, we get

Y (e jω )=H (e jω )S (e jω)

= ( 11−Ae− jω)+( 1

1−B e− jω )+¿

(using partial fraction expansion method)

or Y (e jω )=α1

1−Ae− jω+α 2

1−Be− jω …..(4)

Determination of values of α1 and α2 :

1−A e− jω=0∨e jω= 1A,

α 1=1

(1−Be− jω)= 1

(1−B 1A )

= A(A−B)

1-Be− jω=0∨e− jω= 1B,

α 2=1

(1−Ae jω )= 1

(1−A 1B )

= B(B−A )

= −B(A−B)

Substituting the values of α 1∧α 2 in Eq. 4, we get

Y (e jω )=( AA−B )

1−Ae− jω+( −BA−B )

1−Be jω ….(5)

For determining output y(n), we take inverse DTFT of both sides of Eq. 5, we get

y (n )=Inverse DTFT {Y (e jω )}

¿ Inverse DTFT { ( AA−B )

1−A e− jω+( −BA−B )

1−Be jω }=( A

A−B )Invese DTFT { 11−A e jω }−( B

A−B ) InverseDTFT { 11−Be− jω}

= ( AA−B )Anu (n )−( B

A−B )Bnu (n )

or y (n )= (A−B ) {An+1u (n )−Bn+1u (n ) }

Note that for α1 = α2 , the partial fraction expansion of Eq. 4 is not valid.

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url: multiplication-property.aspx

Title: Multiplication Property, Digital Signal Processing Assignment Help, Homework Help

Description: Explanation regarding multiplication property of DTFT with examples. The multiplication property of DTFT is used in the sampling and communication theory....

Keywords: multiplication property, DTFT, inverse DTFT, digital signal processing assignment help, homework help, sample homework

The Multiplication Property

The multiplication property of DTFT is used in the sampling and communication

theory.

Consider s(n) is a sequence which is a product of two sequence s1(n) and s2(n).

s (n )=s1 (n ) s2(n) …..(4.45)

Let s1 (n )DTFT↔

S1 (e jω)

s2(n) DTFT↔

S2(ejω)

s (n ) DTFT↔

Se (e jω)

Taking the DTFT of both sides of Eq. 4.45, we get

S (e jω )=DTFT {s (n)}

¿ ∑n=−∞

∞

s (n ) e− jωn= ∑n=−∞

∞

s1 (n ) s2 (n ) e− jωn ….(4.46)

Since from inverse DTFT

S1(n) = 12π∫2 π

S1 (e j0 )e j0nd 0 ….(4.47)

Substituting Eq. 4.47 in Eq. 4.46, we obtain

S (e jω )= ∑n=−∞

∞

s1 (n ) s2 (n ) e− jωn

= ∑n=−∞

∞

s2 (n ){ 12π∫2π

S1 (e j0 )e j0nd0}e− jωn ….(4.48)

Interchanging the order of summation and integration in Eq. 4.48, we get

S (e jω )= 12π∫2π

S1 (e j0 ) [ ∑n=−∞

∞

s2 (n ) e− j (ω−0) n]d0 …(4.49)

But S2 ¿ …(4.50)

Substituting the Eq. 4.50 in Eq. 4.49, we get

S (e jω )= 12π∫2π

S1 (e j0 ) S2 (e j (ω−0) )d 0 ….(4.51)

Eq. 4.51 corresponds to a periodic convolution of S1(ejω).

Integral in Eq. 4.51 can be evaluated over an interval of length 2π. The usual form of

convolution is often referred to as non-periodic as the integral ranges from -∞ to ∞.

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url: multiplication-property-solved-examples.aspx

Title: Multiplication Property, Solved Examples, Digital Signal Processing Assignment Help, Homework Help

Description: Solved examples based on the multiplication property of DTFT signal. etermine the DTFT of the multiplication of two discrete – time signals given as....

Keywords: multiplication property, solved examples, DTFT, inverse DTFT, digital signal processing assignment help, homework help, sample homework

The Multiplication Property – Solved Examples

Example: Determine the DTFT of the multiplication of two discrete – time signals given

as s(n) = s1 (n ) , s2 (n ) .

where s1(n) = sin 󠏉( 3πn

4)

πn∧s2 (n )=

sin 󠏉( πn2

)

πn

Solution. From the multiplication property of DTFT , we know that DTFT of s(n), i.e.,

S(e jω) is the periodic convolution of S1(e jω )∧S2 (e jω) . Here integration is done over any

interval of length 2π. In this periodic convolution, we have chosen the interval -π < 0 π.

The multiplication property of DTFT is given as

s (n )=s1 (n ) s2 (n )DTFT↔

S (e jω )= 12π∫2 π

S1 (e j0 ) S2 (e j (ω−0 ) )d0

or S (e jω )= 12π∫2π

S1 (e j0 ) S2 (e j (ω−0) )d 0 …(1)

Eq. 1 represents a periodic convolution, it resembles non-periodic convolution

except for that the integration is limited to the interval –π <0 ≤ π.

Eq. 1 can be converted into an ordinary convolution by defining

´S1(ejω )={S1 (e jω) , for−π<ω≤ π

0 , otherwise….(2)

Substituting Eq. 2 in Eq. 1, we get

S (e jω )= 12π∫−π

πS1 (e j0 )S2 (e j (ω−0 ))d 0

¿¿

= 12π ∫

−∞

∞

S1 (e j0 )S2 (e j (ω−0 ))d 0

Thus, S (e jω )is 12π the non-periodic convolution of the rectangular pulse S1 (e jω)

and periodic square wave S2 (e jω) .Both S1(ejω ) and S2 (e jω )are shown in Fig. 4.17. the result

of this convolution is the DTFT of s(n), i.e., S(e jω) shown in Fig. 4.17(c).

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url: tabulation-of-properties-of-dtft.aspx

Title: Tabulation Of Properties Of DTFT, Digital Signal Processing Assignment Help, Homework Help

Description: Explanations regarding tabulation properties of DTFT with examples...

Keywords: tabulation properties of DTFT, Time-shifting, Frequency shifting, Time

reversal, digital signal processing assignment help, homework help, sample homework

TABULATION OF PROPERTIES OF DTFT

A number of important properties of the DTFT are summarized in Table

Table Summary of Important Properties of DTFT

DTFT property Non-periodic signal DTFT

s (n )

s1 (n )

s2(n)

S (e jω ) , S1 (e jω )

These are periodic with period

2 π

Linearity

Time-shifting

Frequency shifting

Complex

conjugation

Time reversal

A1 s1 (n )+A2 s2(n)

s (n−n0 )

e jω0n s (n )

s∗(n )

s( k ) (n )

= {s ( nk ) , if n is multiple of

k

= 0, if n is not multiple of

A1S1 (e jω)+A2S2(ejω)

where A1∧A2are scalar

constant

e− jωn0S (e jω)

where n0 is constant time

S (e j (ω−ω0) )

S∗(e− jω)

S (e jkω )

Convolution of two

discrete-time signal

Multiplication of two

discrete-time signal

First Difference of a

discrete – time signal in

time

Accumulation of a

discrete – time signal

Differentiation in

frequency

Conjugate Symmetry for

real signals

k

s1 (n )∗s2 (n )

s1 (n ) s2 (n )

s (n )−s (n−1 )

∑k=−∞

n

s (k )

Ns(n)

s(n)

S1 (e jω )S2 (e jω )

12π∫2 π

S1 (e j0 )S2 (e j (ω−0 ) )d0

(1−e jω) S (e jω )

11−e− jω S (e jω )+π S (e j 0 )

× ∑k=−∞

∞

δ (ω−π k )

jds(e jω)d ω

{ S (e jω)=S∗(e jω )ℜ {S (e jω )}=ℜ {S (e− jω) }I m {S (e jω )}=−ℑ {S (e− jω) }

|S (e jω )=|S (e− jω)||

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url: duality-between-ctft-analysis-and-ctft-sysnthesis-equation.aspx

Title: Duality Between CTFT Analysis & CTFF Synthesis Equation, Digital Signal


Description: There is a duality between the CTFT analysis equation and CTFF synthesis equation. Both equation are given as...

Keywords: duality, DTFT, CTFT, DTFS, digital signal processing assignment help, homework help, sample homework

DUALITY Between The CTFT Analysis Equation And CTFF Synthesis Equation

There is a duality between the CTFT analysis equation and CTFF synthesis

equation. Both equation are given as

CTFT analysis equation,

S ( jω )=∫−∞

∞

s (t )e− jωtdt

and CTFT synthesis equation

s(t) = 12π ∫

−ȡ∞

∞

S ( jω )e jωt 󐏉dω

But there is no corresponding duality between the DTFT analysis equation and

DTFT synthesis equation. These equations are given as

DTFT analysis equation

S (e jω )=∑n−∞

∞

s (n ) e− jωn ..(4.54)

and DTFT synthesis equation

s (n )= 12π∫2π

S (e jω)e jωndω ….(4.55)

However, there is a duality between the DTFS equations. These equations are

given as

DTFS analysis equation,

Ak = 1N 0

∑n=¿N0>¿¿

s (n ) e− jk ω0n …(4.56)

and DTFS synthesis equation,

s(n) = ∑k=¿N0>¿¿

A k ejkω0n

….(4.57)

where Ak is the Fourier series coefficients of a discrete-time periodic signal s(n).

Also, there is a duality relationship between the DTFT and CTFS.

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url: duality-in-dtfs.aspx

Title: duality in dtfs, Comparison, Digital Signal Processing Assignment Help, Homework Help

Description: Since the Fourier series coefficients Ak of the periodic signal s(n) are themselves a periodic sequence, we can expand the sequence Ak in a Fourier series. The duality property for discrete – time Fourier series...

Keywords: duality in DTFS, fourier series, digital signal processing assignment help, homework help, sample homework

Duality in the DTFS

Since the Fourier series coefficients Ak of the periodic signal s(n) are themselves

a periodic sequence, we can expand the sequence Ak in a Fourier series. The duality

property for discrete – time Fourier series (DTFS) implies that the Fourier series

coefficients for the periodic sequence Ak are the values of (1/N0)s(-n).

In other words, the duality property for DTFS implies that the Fourier series

coefficients for the periodic sequence Ak is proportional to the values of the original

sequence which is reversed or folded in time.

Proof :

To prove the duality property for DTFS, we consider two periodic sequences with

period N0. These two sequences are related through the summation given as

s1 (m )= 1N 0

∑l=¿N0>¿ s2 (l ) e− jlω0 m¿

¿ ….(4.58)

Where ω0 = 2πN 0

If we let m = k and l=n, Eq . (4.58 )becomes

s1 (k )= ∑n=N0>¿s2 (n) e− jnω0k ¿

¿

= ∑n=N 0>¿ s2 (n ) e−jk ω 0k ¿

¿ …..(4.59)

Comparing Eq. (4.59) with DTFS analysis equation given as

Ak=1

N 0n=¿N 0>¿∑ s (n ) e− jkω0n ¿ ….(4.60)

We see that the sequence

s1 (k ) corresponds¿ theℱ series coefficients of the signal s2 (n ) .

Eq. 4.60 can be represented as

s (n ) DTFS↔

Ak

Similarly, Eq. 4.59 can be represented as

s2 (n ) DTFS↔

s1 ( k ) …..(4.61)

Alternatively, if we let m = n and l=−k, Eq. 4.58 becomes

s1 (n )= 1N0

∑k=¿N0>¿ s2 (– k ) e jk ω0n¿

¿

¿ ∑k=¿N0 >¿

1N0

s2 (−k ) e jk ω0n ¿

¿ …..(4.62)

Now, comparing Eq. 4.62 with DTFS synthesis equation given as

s (n )= ∑k=¿N 0>¿ Ak e

jk ω0n¿

¿ …..(4.63)

We obtain, s1 (n )DTFS↔

Ak e− jk ω0n0 …..(4.64)

Now, we can conclude that 1N 0

s2(−k) …(4.65)

and ejmω0n s (n )DTFS

↔Ak−m .(4.66)

Duality is often useful in reducing the complexity of the calculations involved in

determining Fourier series representations.

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url: verify-duality-of-periodic.aspx

Title: Verify Duality Of Periodic Signal, Solved Examples, Digital Signal Processing


Description: Solved examples based on the duality. Verify the duality of the following periodic signal with a period...

Keywords: verify the duality of periodic signal, Fourier series,

digital signal processing assignment help, homework help, sample homework

Verify the duality of the following periodic signal

Example: Verify the duality of the following periodic signal with a period of N0=9

s (n )={19

sin(󐏉5πn9 )

sin( 5 πn9 )

, n≠multiple of 9 ……….(1)

Solution. We have already studied that the Fourier series coefficients of a discrete-

time square (rectangular) wave is given as

Ak={ 1N0

sin[2πk (N 1+12 )

N 0]

sin ( πkN 0 ), k ≠0 ,± N0±2N 0 ,…

…..(2)

{2N1+1N 0

, k=0±N0±2N 0,…

Duality, then suggests that the Fourier series coefficients for s(n) must be in the

form of a rectangular square wave. The rectangular (square) wave with period N0=9 is

such that

g (n )={ 1 ,|n|≤20 ,2<|n|≤ 4 ….(3)

The Fourier series coefficients Bk for discrete-time square wave g(n) is

determined as

Bk={19

sin( 5πn9 )

sin( π n9 )

{59, for k=multiple of 9

The discrete-time Fourier series (DTFS) analysis equation for sequence g(n) is

given as

Bk=1N0

∑n=¿N0>¿ g (n )e−jkω0 n¿

¿

¿ 19 ∑

n=−2

2

1 (e )− j k( 2π

9 )n

¿ 19 ∑

n=−2

2

(1 )e(− jk2 πn)/9 ..(5)

Interchanging the names of the variables k and n and putting Bn=s (n ) , we obtain

Bn=19 ∑

k=−2

2

e− jn 2πk

9

or s (n )=19 ∑

k=−2

2

e−( j2πnk )/9 …(6)

Also letting k=−k ' in the sum of RHS of Eq. 6, we get

1

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22842 1 frequency analysis using dtft

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