1In the previous three lessons, we discussed the Fourier Series, which is for periodic signals. This lesson will cover the Fourier Transformwhich can be used to analyze aperiodic signals. (Later on, we'll see how we can also use it for periodic signals.) The Fourier Transform is another method for representing signals and systems in the frequency domain

Chapter-5 Fourier Transform

2On the -axis, distance between two consecutive aks is now 0=2/T, the fundamental frequency.

3Bridge Between Fourier Series and Transform

Consider the periodic signal x(t) below:

We know that the Fourier coefficients for x(t) will be:

==

00

)sin(

,2

10

1

kk

kTk

TT

ak

x(t)

t-T1 0 T1 T-T

Bridge Between Fourier Series and Transform (contd)

Now, sketch ak on the -axis:

On the -axis, distance between two consecutive aks is now 0=2/T, the fundamental frequency.

ak

-20 -0 0 0 20

2T1/T

Bridge Between Fourier Series and Transform (contd)

As the period T, the fundamental frequency 00.So, the distance between the two consecutive aksbecomes zero, and the sketch of ak becomes continuous, what is called as Fourier Transform.

At the other side, as T, the signal x(t) becomes aperiodic and takes the form:

This means the Fourier Transform can represent an aperiodic signal on the frequency-domain.

x(t)

t-T1 0 T1

4Bridge to Fourier Transform Periodic Signals Fourier Series Aperiodic Signals Fourier Transform Consider an aperiodic signal x(t). Now, consider

periodic version of x(t), xp(t), where x(t) is repeated over a period T.

We can say that xp(t) is the limiting case of x(t) as T approaches to infinity, i.e.

Since xp(t) is periodic, we can write Fourier series representation for it:

where 0=2/T.

)(lim)( txtx pT =

===

2/

2/

00 )(1)(T

T

tjkpk

k

tjkkp dtetxT

aeatx

Bridge to Fourier Transform (contd)

Since x(t)=0 outside of the interval (-T/2,T/2):

Define an envelope function:

So,

Now, substitute that in the Fourier series representation of xp(t):

= dtetxjX tj )()(

= dtetxT

a tjkk 0)(1

)(1 0jXTak =

=

=k

tjkp ejXTtx 0)(1)( 0

Bridge to Fourier Transform (contd)

Since 2/T= 0:

As T, this sum will approach to an integral because there will a continuum of Fourier terms in the Fourier series.

Finally, the Fourier Transform of x(t):

And, the Inverse Fourier Transform of x(t):

= dtetxjX tj )()(

= dejXtx tj)(

21)(

=

=k

tjkp ejXtx 00 0)(2

1)(

5Convergence of Fourier Transform

Similar to Fourier Series, convergence conditions apply for Fourier Transform: The signal must be absolutely integrable

Over a finite interval of time, the signal must have finite number of maxima and minima (or variations)

Over a finite interval of time, the signal must have finite number of discontinuities. Also, those discontinuities must be finite.

6Inverse Fourier Transform:It is always possible to move back from the frequency-domain to time-domain, by either summing the terms of the Fourier Series or by Inverse Fourier Transform

7Properties of the Fourier Transform

The point of this lesson is that knowledge of the properties of the Fourier Transform can save you a lot of work. We will cover some of the important Fourier Transform properties here

Linearity

Because the Fourier Transform is linear, we can write:

F[a x1(t) + bx2(t)] = aX1() + bX2()

where X1() is the Fourier Transform of x1(t)and X2() is the Fourier Transform of x2(t).

Time Scaling

8Time Shifting

Duality

In general, the Duality property is very useful because it can enable to solve Fourier Transforms that would be difficult to compute directly (such as taking the Fourier Transform of a sincfunction). The Duality Property tells us that if x(t) has a Fourier Transform X(), then if we form a new function of time that has the functional form of the transform, X(t), it will have a Fourier Transform x() that has the functional form of the original time function (but is a function of frequency). Mathematically, we can write:

9Notice that the second term in the last line is simply the Fourier Transform integral of the function X(t), i.e.

Example 1 Using the Fourier Transform integral equation, directly find the Fourier Transform of

x(t) = e-at u(t), a> 0

Example 2 Using the results of Example 1 and the Duality

Property, find the Fourier Transform ofThe Convolution Property

It tells us that convolution in time corresponds to multiplication in the frequency domain. Therefore, we can avoid doing convolution by taking Fourier Transforms! In many cases, this will be much more convenient than directly performing the convolution.

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The convolution property states that:

Let us show that. To start, let

The Convolution Property

Then we can take the Fourier Transform of y(t) and plug in the convolution integral for y(t) (notice how we've marked the integrals with dt and d to keep track of them):

Therefore,

We've just shown that the Fourier Transform of the convolution of two functions is simply the product of the Fourier Transforms of the functions. This means that for linear, time-invariant systems, where the input/output relationship is described by a convolution, you can avoid convolution by using Fourier Transforms. This is a very powerful result.

Multiplication of Signals

It states that the Fourier Transform of the product of two signals in time is the convolution of the two Fourier Transforms.

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Proof

Therefore,

Example 2 Find the Fourier Transform of x(t) = sinc2(t) (Hint: use the Multiplication Property).

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Example 1 Find the inverse Fourier Transform of

Here is a plot of this function:

This tells us that modulation (such as multiplication in time by a complex exponential, cosine wave, or sine wave) corresponds to a frequency shift in the frequency domain.

DC LevelExample Find the Fourier Transform of the constant 1. Use duality and the fact that the transform of (t) is 1

We know ejt2(-0)=0ee

Differentiation in timeTransform of derivativesSuppose that f(n) is piecewise continuous, and absolutely integrable on R. Then

In particular

and

)()()}({ )( wFiwtfF nn =

)()}({ ' wiwFtfF = )()}({ 2'' wFwtfF =

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but we can not directly calculate this integral because it does not converge. For many periodic signals, the integral for Fourier transform may not work. Luckily, there is an easier way of determining Fourier transforms of periodic signalsSo we will generalize the Fourier Transform to include impulses in the frequency domain. We can use either the Duality or Modulation Properties to show that. So we will generalize the Fourier Transform to include impulses in the frequency domain. We can use either the Duality or Modulation Properties to show that:

We can check this by taking Inverse Fourier Transform (and using the sifting property)

Now let's consider a general periodic signal x(t) that we can represent as a Fourier Series:

Since we know that:

by linearity of the Fourier Transform, we get that:

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Example Find the Fourier Transform of cos(0t)Pulsed Cosine

=

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Fourier Transforms of Sampled Signals

In this lesson, we will discuss sampling of continuous time signals. Sampling a continuous time signal is used, for example, in A/D conversion, such as would be done in digitizing music for storage on a CD, digitizing a movie for storage on a DVD, or taking a digital picture.

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To start, we define the continuous time impulse train as:

As you just saw, p(t) is an infinite train of continuous time impulse functions, spaced Ts seconds apart.

Now, x(t) is the continuous time signal we wish to sample. We will model sampling as multiplying a signal by p(t

Let xs(t) = x(t)p(t) be the sampled signal. Then

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Now, we will derive the Sampling Theorem . To do this, we will examine our signals in the frequency domain.To start, let p(t) have a Fourier Transform P(), x(t) have a Fourier Transform X(), and xs(t) have a Fourier Transform Xs().

Then, because xs(t) = x(t)p(t), by the Multiplication Property,

Now let's find the Fourier Transform of p(t). Because the infinite impulse train is periodic, we will use the Fourier Transform of periodic signals:

where Ck are the Fourier Series coefficients of the periodic signal.

Let's find the Fourier Series coefficients Ck for the periodic impulse train p(t):

Therefore

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Now, to finish our derivation of the Sampling Theorem, we will go back and determine Xs().

We saw that:

Therefore,

or we get replicated, scaled versions of X(), spaced every 0 apart in frequency:

From this development and observing the above figure, we can derive our Sampling Theorem. This is one of the most important results.

As you can see from the figure if 0 - c < c, we would get overlap of the replicates of X() in frequency. This is known as "aliasing." Therefore, to avoid aliasing, we require 0 - c > c or 0 > 2c. If we avoid aliasing, we can recover x(t) from its samples. (Usually, we choose a sampling rate a bit higher than twice the highest frequency since filters are not ideal.)

We hear music up to 20 kHz and CD sampling rate is 44.1 kHz. (Dogs would need a higher quality CD since they hear higher frequencies than humans.)

We can recover x(t) from its sampled version xs (t) by using a low pass filter to recover the center island:

After low pass filtering

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Sampling Theorem Conditions

Sampling Theorem: A continuous-time signal x(t) can be uniquely reconstructed from its samples xs(t) with two conditions:

x(t) must be band-limited with a maximum frequency M Sampling frequency s of xs(t) must be greater than 2M, i.e. s>2M.

The second condition is also known as Nyquist Criterion. s is referred as Nyquist Frequency, i.e. the smallest

possible sampling frequency in order to recover the original analog signal from its samples.

Aliasing (Under-sampling)

What happens when sampling frequency is less than Nyquist Frequency, i.e. s

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Example 1 Given a signal x(t) with Fourier Transform with cutoff frequency c as shown:

you are given three different pulse trains with periods

Draw the sampled spectrum in each case. Which case(s) experiences aliasing?

Example 2 The inverse Fourier Transform of the signal in the previous example is

Draw the sampled signals using the sampling trains of the previous example

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Applications of the Fourier Transform:Ideal Filters

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Example 1 Given two functions x1(t) = cos(500t) and x2(t) = cos(1000t), form a new signal x3(t) = x1(t) x2(t).Draw the frequency response of a low pass filter that passes the low frequency component of the signal and blocks the high frequency component of the signal.

Sinusoidal Amplitude Modulation Modulation is multiplying a signal in time by

complex exponentials (such as sine waves and cosine waves) to shift the signal to a desired frequency band. This is done, for example, by radio (or television) to assign different parts of the frequency spectrum to different radio (or television) stations. Remember that the Modulation Theorem told us that:

We will use the Modulation Theorem in an example of AM radio.

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Example 2 AM radio - Double-sideband, suppressed carrier, amplitude modulation

Let x(t) be a music signal with Fourier Transform X().

To modulate this signal, we'll form and transmit

Your radio demodulates the signal by multiplying the received signal y(t) by another cosine wave to form a third signal z(t):

To recover the original music signal x(t) from the demodulated signal z(t), you filter z(t) with a low-pass filter by multiplying by a rect function in frequency. This corresponds to convolving with a sinc function in time.