Fourier Transform | 10 | Signals and Systems

ABSTRACT

Control systems are an area of electrical engineering where signals and systems are used A control system is designed to regulate the behavior of one or more variables in some desired manner Control systems play major roles in our everyday life Household appliances such as heating and air-conditioning systems, switch-controlled thermostats, washers and dryers, cruise control in automobiles, elevators, traffic lights, manufacturing plants, navigation systems-all utilize control systems In the aerospace field, precision guidance of space probes, the wide range of operational modes of the space shuttle, and the ability to maneuver space vehicles remotely from earth all require knowledge of control systems In the manufacturing sector, repetitive production line operations are increasingly performed by robots, which are programmable control systems designed to operate for many hours without fatigue

Control engineering applies signals and systems theory It is not limited to any specific engineering discipline but may involve environmental, chemical, aeronautical, mechanical, civil, and electrical engineering For example, a typical task for a

control engineer might be to design a speed regulator for a disk drive head A thorough understanding of control systems techniques is essential to the electrical engineer and is of great value for designing control systems to perform the desired task

In Chapter 4, we saw that Fourier series enabled us to represent a periodic signal as a linear combination of infinite sinusoids The Fourier series expansion reveals the frequency content of the periodic signal It is also possible to analyze the frequency content of nonperiodic signals The tool that enables us to do this is the Fourier transform (also known as Fourier integral) The transform assumes that a nonperiodic (or aperiodic) signal is a periodic signal with an infinite period The Fourier transform is important in analyzing and processing signals in science and engineering, especially in medical images, computerized axis tomography (CAT), and magnetic resonance imaging (MRI)

The Fourier transform is similar to Laplace transform Both are integral transforms The Fourier transform may be regarded a special case of Laplace transform with s = jω, when the Laplace transform exists Just like the bilateral Laplace transform, Fourier transform can deal with systems having inputs for t < 0 as well as those for t >0 Fourier transform is very useful in linear systems, filtering, communication systems, and digital signal processing, in situations where Laplace transform does not exist

The chapter begins by defining the Fourier transform from Fourier series Then we present and prove the properties of Fourier transform, which are useful in deriving Fourier transform pairs We discuss Parseval’s theorem, compare the Laplace and Fourier transforms, and see how the Fourier transform is applied in circuit analysis, amplitude modulation, and sampling We finally learn how to use MATLAB® in Fourier analysis

There is a close connection between the definition of Fourier series and the Fourier transform for functions x(t) that are zero outside of an interval For such a function, we can calculate its Fourier series on any interval that includes the points where x(t) is not identically zero The Fourier transform is also defined for such a function As we increase the length of the interval on which we calculate the Fourier series, the Fourier series coefficients begin to look like the Fourier transform and the sum of the Fourier series of x(t) begins to look like the inverse Fourier transform To explain this more precisely, suppose T is large enough so that the interval [−T/2, T/2] contains the interval on which x is not identically zero

Recall that if x(t) is a periodic function with period T, then we can expand it in a complex Fourier Series,

x t c e n

n jn t( ) =

where

c T

x t e dtn jn t

= -

( ) w (52)

The spacing between adjacent harmonics is

Dw w w w p= +( ) - = =n n

T 1 20 0 0 (53)

We substitute Equation 52 into Equation 51 and putting 1 2

T = w

p we obtain

x t T

x t e dt e n

jn t( ) = ( ) é

ê ê ê ê

ú ú ú ú

jn tx t e dt e =-¥

å ò ( ) é

ê ê ê ê

ú ú ú ú

Dw p

1 2

jn tx t e dt e =-¥

å ò ( ) é

ê ê ê ê

ú ú ú ú

(54)

In view of the previous discussion, as T → ∞, we can put ω0 as ∆ω and replace the sum over the discrete frequencies nω0 by an integral over all frequencies We replace nω0 by a general frequency variable ω We then obtain the double integral representation:

x t x t e dt e dj t j t( ) = ( ) é

ë ê ê

û ú ú

òò12p ww w (55) The inner integral is known as the Fourier transform of x(t) and is represented by X(ω), that is,

X x t x t e dtj tw w( ) = ( )éë ùû = ( ) - -¥

òF (56) where F is the Fourier transform operator

The Fourier transform of a signal x(t) is the integration of the product of x(t) and e−jωt over the interval from −∞ to +∞

X(ω) is an integral transformation of x(t) from the time-domain to the frequencydomain and is generally a complex function X(ω) is known as the spectrum of x(t) The plot of its magnitude |X(ω)| versus ω is called the amplitude spectrum, while that of its phase ∠X(ω) versus ω is called the phase spectrum If a signal is real-valued, its magnitude spectrum is even, that is, |X(ω)| = |X(−ω)| and its phase spectrum is odd, that is, ∠X(ω) = −∠X(−ω) Both the amplitude and phase spectra provide the physical interpretation of the Fourier transform

We can write Equation 55 in terms of X(ω) and we obtain the inverse Fourier transform as

x t X X e dj t( ) = ( )éë ùû = ( ) -¥

òF w p w ww 1

2 (57)

We say that the signal x(t) and its transform X(ω) form a Fourier transform pair and denote their relationship by

x t X( ) ( )Û w (58)

We will denote signals by lowercase letters and their transforms by uppercase letters Not all functions have Fourier transforms The necessary conditions required for

a function to be transformable are called the Dirichlet conditions similar to the Fourier series These are listed as:

1 x(t) is bounded 2 x(t) has a finite number of maxima and minima within any finite interval 3 x(t) has a finite number of discontinuities within any finite interval

Furthermore, each of these discontinuities must be finite 4 x(t) is absolutely integrable, that is,

x t dt( ) <¥ -¥

ò (59) Therefore, absolutely integrable signals that are continuous or that have a finite number of discontinuities have Fourier transforms

Example 5.1

Example 5.2

Example 5.3

The properties of the Fourier transform are summarized as follows These properties have important interpretations and meanings in the context of signals and signal processing The properties of the Fourier transform provides valuable insight into how signals operating in the time-domain are described in the frequency-domain Many properties like linearity, time scaling and shifting, frequency shifting, time differentiation and integration, frequency differentiation, duality, convolution, and Parseval’s Theorem are discussed as follows

The Fourier transform is a linear transform That is, suppose we have two functions x1(t) and x2(t), with Fourier transform given by X1(ω) and X2(ω), respectively, then the Fourier transform of x1(t) and x2(t) can be easily found as

F a x t a x t a X a X1 1 2 2 1 1 2 2( ) + ( )éë ùû = ( ) + ( )w w (510)

where a1 and a2 are constants (real or complex numbers) Equation 510 can be easily shown to be true via using the definition of the Fourier transform

F a x t a x t a x t a x t e dt

( ) + ( )éë ùû = ( ) + ( )

ò [ ] w

x t e dt a x t e dt

a X a X

( ) + ( )

= ( ) + ( )

w w

(511)

This can be extended to a linear combination of an arbitrary number of signals

The scaling property of the Fourier transformation is as follows If F w( ) = ( )éë ùûx t and a is a real constant Then

F x at a

X a

( )éë ùû = æ è ç

ö ø ÷

1 w (512)

resulting in a new frequency ω/a Equation 512 can be found using the definition

F x at x at e dtj t( )éë ùû = ( ) - -¥

Letting λ = at, dλ = adt so that

F x at x e d a a

X a

j a( )éë ùû = ( ) =

æ è ç

ö ø ÷

ò l l w wl 1

(513)

Now if a is positive, (a > 0),

F x at x e d a a

X a a

X a

j a( )éë ùû = ( ) =

æ è ç

ö ø ÷ =

æ è ç

ö ø ÷

l l w w wl 1 1

If a is negative (a > 0),

F x at x e d a

x e d a a

j a( )éë ùû = ( ) = - ( ) = -

ò l l l l wl wl 1 w

a X

æ è ç

ö ø ÷

= æ è ç

ö ø ÷

Therefore, we can see that for the general case of scaling with a real number a we get Equation 512 The scaling property is sometimes also called the reciprocal spreading property

Another simple property of the Fourier transform is the time shifting If

X x tw( ) = ( )éë ùûF and t0 is a constant, then

F x t t e Xj t-( )éë ùû = ( )-0 0w w (514)

If the original function x(t) is shifted in time by a constant amount, it should have the same magnitude of the spectrum, X(ω) That is, a time delay does not cause the frequency content of X(ω) to change at all Since the complex exponential always has a magnitude of 1, we see that time delay alters the phase of X(ω) but not its magnitude

By definition,

F x t t x t t e dtj t-( )éë ùû = -( )éë ùû - -¥

Letting λ = t − t0, dλ = dt, and t = λ + t0, so that

F x t t x e d e x e dj t j t j-( )éë ùû = ( ) = ( ) - +( ) -

ò e Xj t0

By following similar steps,

F x t t e Xj t+( )éë ùû = ( )0 0w w (516)

This property forms a basis for every radio and TV transmitter This property states

that if X x tw( ) = ( )éë ùûF and ω0 is constant, then

F x t e Xj t( )éë ùû = -( )

w w w0 0 (517)

Note that this is a dual of the time shifting property The previous equation means that a shifting in the frequency-domain is equivalent to a phase shift in the timedomain By definition,

F x t e x t e e dt

x t e dt

( )éë ùû = ( )

( ) ==

0 X w w-( ) -¥

(518)

If X x tw( ) = ( )éë ùûF , then the Fourier transform of the derivative of x(t) is given by

F ¢( )éë ùû = ( )x t j Xw w (519)

This states that the transform of the derivative of x(t) is obtained by multiplying its transform X(ω) by jω By definition,

x t X X e dj t( ) = ( )éë ùû = ( )- -¥

òF 1 12w p w ww (520)

Taking the derivative of both sides with respect to t gives

dx t dt

j X e d j X

x t j X

j t( ) = ( ) = ( )éë ùû

¢( )éë ùû =

òwp w w w w

w w

2 1

( ) (521)

The nth derivative of the Equation 519 is

F x t j Xn n( ) ( )éë ùû = ( ) ( )w w (522)

This property states that if X x tw( ) = ( )éë ùûF then

F -( ) ( )éëê

ù ûú

= ( )jt x t dd X n

nw w (523)

By definition,

d d

X d d

x t e dt x t d d

e n

j w

w w w

w w( ) = ( ) æ

ç ç

÷ ÷

= ( )- -¥

x t jt e dt jt x t e dt

ò= ( ) -( ) = -( ) ( )w w ¥

ò = -F (( ) )( )jt x tn

(524)

If X x tw( ) = ( )éë ùûF then, the integration of function x(t) is given by

F -¥ ò ( )

ë ê ê

û ú ú

= ( )

+ ( ) ( ) t

x t dt X

j X w w

p d w0 (525)

Integrating x(t) in the time-domain leads to division of X(ω) by jω plus an additive term πX(0)δ(ω) to account for the possibility of the dc component that may appear in the integral of x(t) If we replace ω by 0 in Equation 56

X x t dt0( ) = ( ) -¥

ò (526) indicating that the dc component is zero when the integral of x(t) over all time vanishes

Duality between the time and frequency domains is another important property of Fourier transforms This property relates to the fact that the analysis equation and synthesis equation look almost identical except for a factor of 1/2π and the difference of a minus sign in the exponential in the integral The duality property states that if X(ω) is the Fourier transform of x(t), then the Fourier transform of X(t) is 2πx(−ω); that is,

F Fx t X X t x( )éë ùû = ( ) Þ ( )éë ùû = -( )w p w2 (527)

This expresses the fact that the Fourier transform pairs are symmetric To derive the property, from Equation 57

x t X X e dj t( ) = ( )éë ùû = ( )- -¥

òF 1 12w p w ww (528)

2p w wwx t X e dj t( ) = ( ) -¥

Replacing t with −t, in Equation 529, we obtain

2p w wwx t X e dj t-( ) = ( ) - -¥

By interchanging t and ω, and comparing the result with the Equation 528

2p w wx X t e dt X tj t-( ) = ( ) = ( )éë ùû--¥

ò F (531)

According to the convolution property, the Fourier transform of the convolution of two time functions is the product of their corresponding Fourier transforms If x(t) and h(t) are two signals, their convolution y(t) is given by the convolution integral

y t h t x t h x t d( ) = ( )* ( ) = ( ) -( ) -¥

ò t t t (532)

If X(ω), H(ω), and Y(ω) are the Fourier transforms of x(t), h(t), and y(t) respectively, then

Y h t x t H X( ) [ ( )* ( )] ( ) ( )w w w= =F (533)

which states that the Fourier transform of the convolution of two-time functions is the product of their corresponding Fourier transforms

To derive the convolution property, we take the Fourier transform on both sides of Equation 532 we get

Y h x t d e dtj tw t t t w( ) = ( ) -( ) é

ë ê ê

û ú ú

Y h x t e dt dj tw t t tw( ) = ( ) -( ) é

ë ê ê

û ú ú

By taking λ= t − τ so that t = λ + τ and dt = dλ

Y h x e d d

Y h e

w t l l t

w t

w l t( ) = ( ) ( ) é

ë ê ê

û ú ú

( ) = ( )

wt wlt l l w wd x e d H Xj

ò ò ( ) = ( ) ( )

(535)

Table 51 presents these and other properties of Fourier transform, while Table 52 lists the transform pairs of some common functions Note the similarities between these tables on Fourier transform and Tables 31 and 32 on Laplace transform

Example 5.4

TABLE 5.1 Properties of the Fourier Transform

TABLE 5.2 Fourier Transform Pairs

TABLE 5.2 (Continued) Fourier Transform Pairs

Example 5.5

In general, we would use Equation 57 to find the inverse Fourier transform x(t) of X(ω) Quite often, the Fourier transform is in the form of a rational function In this case, we perform a partial fraction expansion of X(ω) and finding the corresponding time-domain signals using the Fourier transform pairs in Table 52 The procedure is similar to what we did in Section 34 for Laplace transform

Example 5.6

The Fourier transform is used extensively in a variety of fields such as optics, spectroscopy, acoustics, computer science, and electrical engineering In electrical engineering, it is applied in circuit analysis, communications systems, and signal processing Here we consider three application areas: circuit analysis, amplitude modulation (AM), and sampling Filtering is another area of application, but we have already considered this in Section 463

We use Fourier transform to analyze circuits with nonsinusoidal excitations exactly in the same way we use phasor techniques to analyze circuits with sinusoidal excitations For example, Ohm’s law is written as

V Z Iw w w( ) = ( ) ( ) (536)

where V(ω) and I(ω) are the Fourier transforms of the voltage and current respectively Z(ω) is the impedance

The impedances of resistors, inductors, and capacitors are the same as in phasor analysis, that is,

R R L j L C j C

Þ Þ

w 1

(537)

The Fourier approach to circuit analysis generalizes the phasor technique It involves three steps First, we transform the circuit elements into frequency-domain as in Equation 557 and take the Fourier transform of the excitations Second, we apply circuit techniques such as Kirchhoff’s voltage law, Kirchhoff’s current law, voltage division, current division, source transformation, node or mesh analysis to find the unknown response (current or voltage) Third, we finally take the inverse Fourier transform to get the response in the time-domain (We should note that Fourier analysis cannot handle circuits with initial conditions)

As usual, the transfer function H(ω) is the ratio of the output response Y(ω) to the input excitation X(ω):

H Y

X ( ) ( )( )w

w w

= (538)

Y H X( ) ( ) ( )w w w= (539)

Equation 539 states that if we know the transfer function H(ω) and the input X(ω), we can find the output Y(ω)

Example 5.7

Example 5.8

AM is one of the applications of the Fourier transform Transmission of information through space has become part of modern society, especially with the wireless telephony However, transmitting intelligent signals-such as for speech and music-is not practical: it requires a lot of power and large antennas One way of transmitting low-frequency audio information (50 Hz to 20 kHz) is to transmit it along with a high-frequency signal, called a carrier Any of the three characteristics

(amplitude, frequency, or phase) of a carrier can be controlled, to be able to carry the intelligent signal, called the modulating signal. Here, we are interested in the control of the carrier’s amplitude This is known as AM

Amplitude modulation (AM) is a process whereby we let the modulating signal control the amplitude of the carrier

Consider the signal multiplier shown in Figure 513 We focus only on the case when the modulating signal m(t) = cosωot The output of the multiplier is

y t x t to( ) ( )cos= w (540) Since the multiplication in the time-domain always results in convolution in the frequency-domain,

Y X

X X

( ) ( )* ( ) ( )

( ) ( )

w p

w p d w w d w w

w w w w

= - + +éë ùû

1 2

(541)

Thus, modulation results in shifting the spectrum of the signal as illustrated in Figure 514 The signal with frequency ωc−ωo is known as the lower sideband, while the one with frequency ωc + ωo is known as the upper sideband

At the receiving end of the transmission, the audio information is received from the modulated carrier by a process known as demodulation. This shifts back the message spectrum to its original low frequency location A synchronous demodulation multiplies the modulated signal y(t) with cosωot as shown in Figure 515

z t y t to( ) ( )cos= w (542) Taking the Fourier transform as we did before,

Z Y Y

X X X

( )

w w w w w

= -( ) + +( )éë ùû

= + -( ) + +( )

1 2

1 4

2 1 4

(543)

This is illustrated in Figure 516 The signal z(t) is passed through a low-pass filter (see Figure 517a), which passes only the low-frequency components Thus, x(t) is extracted as shown in Figure 517b

Example 5.9

Sampling is an important operation in signal processing It may be regarded as a way of reducing analog signals to discrete signals Thus sampling is the bridge from

continuous to discrete signals In analog systems we process entire signals, but in digital systems, only samples of signals are necessary Sampling can be done by using a train of pulses or impulses We will use impulse sampling here

Consider the continuous-time signal x(t) depicted in Figure 518a We multiply by a train of impulses δ(t−nTs) as shown in Figure 518b, where Ts is the sampling interval and fs = 1/Ts is the sampling frequency (or rate) The value of x(t) at point nTs is

x t t nT x nT t nTs s s( ) ( ) ( ) ( )d d-= - (544)

The sampled signal xs(t) in Figure 518c is, therefore,

x t x t t nTs s n

( ) ( ) ( )= - =-¥

åd (545)

We take the Fourier transform of this and apply the frequency convolution property

X X n

T X n

( ) ( ) * ( )

( )* ( )

w p

w w d w w

w d w w

= éë ùû -

= -

1 2

1 T

X n s

( )w w-=-¥

(546)

where ωs = 2π/Ts We have used item 23 in Table 52 Equation 546 states that the Fourier transform Xs(ω) of the sampled signal is a sum of translates of the Fourier transform of the original signal X(ω) at a rate of 1/Ts

To ensure that the sampling process is successful, we must carefully select the sampling interval This is stated in the following sampling theorem:

A band-limited signal, with bandwidth W hertz, may be completely recovered from its samples if taken at a frequency at least twice as high as 2W samples per second

(In practice, a signal is made band-limited by passing it through a low-pass filter before sampling it) For a signal, with no frequency component higher than W hertz, there is no loss of information if the sampling frequency is at least twice the highest frequency in the signal, that is,

1 2 T

f W s

s= ³ (547)

The minimum sampling frequency fs = 2W is known as the Nyquist frequency or rate, and the maximum spacing 1/fs is the Nyquist interval, named after Harry Nyquist (1889-1976) for his paper published in 1928 The Nyquist rate is the minimum sampling rate necessary to preserve the information x(t) A signal is said to be oversampled if it is sampled at a rate greater than its Nyquist rate It is undersampled if it is sampled at less than the Nyquist rate

Example 5.10

Parseval’s theorem demonstrates another application of the Fourier transform It states the relationship between energy in the time and frequency domains The theorem is important in communications and signal processing

If p(t) is the power associated with the signal, the energy carried by the signal is

E p t dt= -¥

ò ( ) (548)

It is convenient to use a 1 − Ω resistor as the base for energy calculation For a 1 − Ω resistor, p(t) = v2(t) = i2(t) = x2(t), where x(t) stands for either voltage or current Thus Equation 548 becomes

E x t dt1 2W -¥

= ò ( ) (549)

Parseval’s theorem states that this same energy can be calculated in the frequencydomain as

E x t dt X d1 2 2 1

= =ò ò( ) | ( ) |p w w (550)

indicating that the Fourier transform is an energy-conserving relation

Parseval’s theorem states that the total energy delivered to a 1 − Ω resistor equals the total area under the square x2(t) or 1/2π times the total area under the magnitude of the Fourier transform squared |X(ω)|2

Parseval’s theorem is named after the French mathematician Marc-Antoine Parseval des Chênes (1755-1836) It relates energy between the time-domain and the frequency-domain It provides the physical significance of X(ω), namely, that |X(ω)|2 is the energy density (in joules per hertz) corresponding to x(t) The plot of |X(ω)|2 versus ω is the energy spectrum of the signal x(t)

To derive Equation 550, we begin with Equation 549 and substitute Equation 57 for one of the x(t)′s

E x t dt x t X e d dtj t1 2 1

= = é

ë ê ê

û ú úò òò( ) ( ) ( )p w w

w (551)

The function x(t) can be moved inside the inner integral, since the integral does not involve time:

E x t X e d dtj t1 1

= òòp w ww( ) ( ) (552)

Reversing the order of integration,

E X x t e dt d

X X d

1 2

= é

ë ê ê

û ú ú

= - =

òòp w w

p w w w

w( ) ( )

( ) ( ) -¥

ò ò X X d( ) * ( )w w w

(553)

But X(ω)X* (ω) = |X(ω)|2 Hence,

E x t dt X d1 2 2 1

= =ò ò( ) | ( ) |p w w (554)

as expected Equation 554 shows the energy of x(t) can be computed directly in the time-domain or indirectly from X(ω) in the frequency-domain

Since |X(ω)|2 is an even function, we may integrate from 0 to ∞ and double the result; that is,

E x t dt X d1 2 2

1 W

= =ò ò( ) | ( ) |p w w (555)

We may calculate the energy in any frequency band ω1 < ω.< ω2 as

E X d1 2 1

W = òp w w w

| ( ) | (556)

We should note that Parseval’s theorem, as stated here, applies to nonperiodic signals Parseval’s theorem for periodic signals was covered in Section 446

Example 5.11

The Fourier and Laplace transforms are closely related, as evident in their definitions, which are repeated here for convenience The general (bilateral) Laplace transform of x(t) is

X s x t e dtst( ) ( )= - -¥

where s = σ + jω If we let s = jω, Equation 557 becomes

X x t e dtj t( ) ( )w w= - -¥

which is the Fourier transform of x(t) For a causal signal x(t) (ie, positive-time sig-

nal, x(t) = 0, t < 0) and | ( ) |x t dt < ¥ -¥

X X s

s j( ) ( )w w= = (559)

showing that the Fourier transform is a special case of the Laplace transform with s = jω We can use Equation 559 to obtain the Fourier transform of signals for which the Fourier integral converges This integral will converge when all the poles of F ( )s lie in the left half of the s-plane and not on the jω axis

The Laplace transform defined and used in Chapter 4 is one-sided (or unilateral) in that the integral is over 0 < t < ∞, restricting its usefulness to only positive-time signals, x(t), t > 0 The Fourier transform accommodates signals existing for all time

Although the Laplace transform has some advantages over the Fourier transform in circuit analysis, being an extension of Fourier analysis to nonperiodic signals, Fourier transform uses the same frequency-domain representation as phasor analysis Fourier transform shares some operational properties with the Laplace transform but has some unique properties For circuit analysis, Laplace transform is more widely used because the Laplace integral converges for a wider range of signals and automatically incorporates initial conditions

MATLAB® function fourier can be used to find the Fourier transform of a signal x(t), while the function ifourier can be used to find the inverse Fourier transform of X(ω) How to use them can be displayed with the help Fourier and help ifourier commands Since transfer functions are already functions of frequency, it is straightforward and easy to plot them using MATLAB

Example 5.12

Example 5.13

Example 5.14

1 The Fourier transform is defined as

X x t x t e dtj tw w( ) = ( )éë ùû = ( ) -¥

It converts a nonperiodic signal x(t) into a transform X(ω) 2 The inverse Fourier transform of X(ω) is defined as

x t X X e dj t( ) [ ( )] ( )= =- -¥

òF 1 12w p w ww

3 The magnitude |X(ω)| plotted against ω is known as the magnitude spectrum of x(t), while the plot of |X(ω)|2 against ω is known as the energy spectrum The angle of X(ω) plotted against ω is known as the phase spectrum

4 Common Fourier transform properties and pairs are in Tables 51 and 52 respectively

5 If H(ω) is the transfer function of a system, the output Y(ω) can be obtained from the input X(ω) using

Y H X( ) ( ) ( )w w w=

6 Parseval’s theorem provides the energy relationship between a signal x(t) and its Fourier transform X(ω) The 1 − Ω energy is

E x t dt X d1 2 2 1

= =ò ò( ) | ( ) |p w w

7 Applications of the Fourier transform include circuit analysis, AM, and sampling Applying the Fourier transform approach to circuit analysis involves transforming the excitation and circuit elements into the frequency-domain, solving for the unknown response, and transforming the response to the timedomain using the inverse Fourier transform For AM, the modulating signal is used to control the amplitude of the carrier For sampling application, we found that no information is lost in sampling if the sampling frequency is equal to at least twice the Nyquist rate

8 If x(t) = 0 for t ≤ 0, we can obtain the Fourier transform of signal x(t) from the one-sided Laplace transform X(s) of x(t) by replacing s with jω

9 MATLAB commands can be used to find the Fourier transform of a signal or its inverse

5.1 The Fourier transform of x(t) is called the spectrum of x(t) (a) True, (b) false 5.2 The Fourier transform and the Laplace transform of δ(t) are the same (a) True, (b) false 5.3 Which of these functions is not Fourier transformable? (a) etu(−t), (b) te−3tu(t), (c) 1/t, (d) |t|u(t) 5.4 If x(t) = 2δ(t) and y(t) = dx/dt, then Y(ω) is (a) 2δ′(ω), (b) 2δ(ω), (c) 2jω, (d) 2/jω 5.5 The inverse Fourier transform of e

j j-

w3 is

(a) e−3t, (b) e−3t u(t − 1), (c) e−3 (t − 1), (d) e−3(t − 1)u(t − 1)

5.6 The property that relates e X t Xj t ow w w0 ( ) ( )Û - is (a) Time shift, (b) frequency shift, (c) time differentiation, (d) frequency

differentiation

5.7 Evaluating the integral 8 2

4 2 d w

w w( )-

+ -¥

ò d gives (a) 0, (b) 1 (c) 2, (d) ∞ 5.8 A unit step current is applied through a 1 H inductor with no initial current

The voltage across the inductor is (a) u(t), (b) sgn(t), (c) e−tu(t), (d) δ(t) 5.9 The total energy of a signal is independent of its phase spectrum (a) True, (b) false

5.10 The MATLAB command for finding the inverse Fourier transform is (a) invfourier, (b) ifourier, (c) fourierinv, (d) fourier

Answers: 51a, 52a, 53c,54 c,55d, 56b, 57b, 58d, 59b, 510b

5.1 Find the Fourier transform of

x t

t ( )

< < < <

- < <

í ï ï

î ï ï

1 0 1 0 1 2 1 2 3 0 otherwise

5.2 Obtain the Fourier transforms of the following signals:

(a) x t

t 1

1 2 1 0 1 1 1 1 2 0

( )

- < < - - < < < <

í ï ï

î ï ï otherwise

(b) x t

t 2

1 2 1 0 1 1 1 1 2 0

( )

- - < < - - < <

< <

í ï ï

î ï ï otherwise

5.3 Calculate the Fourier transforms of the signals in Figure 521 5.4 Compute the Fourier transform of the signal in Figure 522 5.5 Find the Fourier transform of the signal shown in Figure 523

5.6 Using the properties of the Fourier transform, determine the Fourier transform of these signals:

(a) x(t) = sin(2t)u(t) (b) y(t) = e−tcos(3t)u(t) (c) z(t) = te−2tu(t) 5.7 Verify that

(a) cos ( )w w pdtd t= -¥

(b) j e d to o j t o2 d w w d w w w w w( ) ( ) sin( )+ - -éë ùû =

ò 5.8 Find the Fourier transform of the following signals: (a) x1(t) = e−3tsin10tu(t) (b) x2(t) = e−4tcos10tu(t) 5.9 (a) Find the Fourier transform of x(t) = 2cos(ωot + θ) (b) Use the result in part (a) to derive the Fourier transform of y(t) =

2cos(100πt−π/4) 5.10 Given that F F( ) [ ( )]w = f t , prove the following results using the definition of

Fourier transform

(a) F Ff t t eo j to-( )éë ùû = ( )- w w

(b) F F df t

dt j( )é

ë ê ê

û ú ú

= ( )w w

(c)F F[ ( )] ( )f t-= -w (d) F Ftf t j d

d( )éë ùû = ( )w w 5.11 A signal has Fourier transform

X jj( )w

w w w

= + - + +

5 2 12

Determine the Fourier transform of these signals: (a) y(t) = x(2t − 5) (b) y(t) = tx(t) (c) y(t) = x(t)sin2t

(d) y t d x t

dt ( ) ( )=

5.12 A signal f(t) has a Fourier transform given by

F ( ) ( )w w

w w = +

- + 5 1

8 62 j

Without finding f(t), find the Fourier transform of the following: (a) f(t − 3) (b) f(4t) (c) e−j2tf(t) (d) f(−2t) 5.13 The Fourier transform of a signal x(t) is

X j( ) ( )w w pd w= +

Derive the Fourier transform of the following signals: (a) x1(t) = 4x(2t−1) (b) x2(t) = x(t−2)sint (c) x t

d dt

x t3( ) ( )= (d) x4(t) = tx(t) 5.14 Obtain the Fourier transform of each of the following signals: (a) x(t) = 4cos2t + 2δ(t) (b) y(t) = 4sin2t−2sin2(t−1)−2sin2(t + 1) 5.15 If X(ω) is the Fourier transform of x(t), show that

x t a x t a X a( ) ( ) ( )cos+ + - Û 2 w w

5.16 Use the convolution property to derive the Fourier transform of the timedomain product y(t) = sin2t ⋅.u(t)

5.17 Let a signal x(t) have the Fourier transformX(ω) shown in Figure 524 Without computing x(t), sketch Y(ω) when

(a) y(t) = x(2t) (b) y(t) = x(t/4) (c) y(t) = x(t)cos2t (d) y(t) = x(t)sin2t 5.18 Find the Fourier transform of the following signals: (a) x(t) = δ(t−2) + δ(t−1) + δ(t + 1) + δ(t + 2) (b) y(t) = 4te−tu(t) (c) z(t) = 2e−tsin(4t)u(t) 5.19 Verify the Fourier transform pairs:

te u t

a j a at-Þ

+ >( ) ( ) ,

1 02w

5.20 Verify the following identities:

(a) X x t dt( ) ( )0 = -¥

ò (b) x X d( ) ( )0 1

2 =

òp w w 5.21 (a) A signal x(t) has the equivalent bandwidth B and equivalent duration T

defined as

x t dt

x B

X d

X = =-¥

ò ò( ) ( ) ,

( ) ( )0 0 w w

Show that BT = 2π (b) Determine the equivalent bandwidth of (i) Π(t/τ), (ii) Λ(t/τ)

5.22 If X x t( ) [ ( )]w = F and a and b are constants, prove the following identities:

(a) F x t a

b a e Xjab+æ è ç

ö ø ÷

ë ê

û ú = ( )w w

(b) F[ ( ) ( )] ( )x t t a X a e j ad w-= -

b a e Xjab( ) -æ è ç

ö ø ÷

ë ê

û ú = ( )-* d ww

5.23 Let x(t) be given by x t t t( ) , ,

= £ £ì

í î

0 1 0 otherwise

(a) Sketch x(t) (b) Obtain X(ω) by using differentiation property 5.24 (a) If x(t) is real and even, show that X(ω) is real and even (b) Give an example for part (a) 5.25 If x(t) and y(t) are two signals with Fourier transforms X(ω) and Y(ω), show that

x t y t dt X Y d( ) * ( ) ( ) * ( ) -¥

ò ò= 12p w w w

5.26 Consider X(ω) as the Fourier transform of x(t) in Figure 525 Evaluate (a) X(0)

(b) X d( )w w -¥

x t

t t( ) sin , ,

= < <ì

í î

0 2 0

p otherwise

528 Find x(t) for each of the following:

(a) X j( ) ( )w pd w

w = -

+ 4 1

(b) X j( ) ( )w pd w w= -4 2

(a) X jj( )w w w

= + 2

1 (b) Y(ω) = 10e−|ω|

w w =

- + +2 5 6 5.30 Determine the inverse Fourier transform of the signal in Figure 526

5.31 The transfer function of a system is

H j( )w w= +

10 2

Determine the response y(t) for the following inputs: (a) x(t) = 5cos(2t) (b) x(t) = δ(t) (c) x(t) = e−4tu(t) 5.32 A signal x(t) is passed through a system as shown in Figure 527 (a) Find the transfer function H(ω) (b) If x(t) = cos(t), find y(t) (c) If x(t) = e−3tu(t), find y(t)

5.33 The transfer function of a circuit is

H j( )w w= +

2 2

If the input signal to the circuit is vs(t) = e−4tu(t), find the output signal Assume all initial conditions are zero

5.34 Suppose vs(t) = u(t) for t > 0 Determine i(t) in the circuit of Figure 528 using Fourier transform

5.35 Use Fourier transform to find i(t) in the circuit of Figure 529 if vs(t) = 10e−2tu(t) 5.36 Find the voltage vo(t) in the circuit of Figure 530 Let is(t) = 8e−tu(t) A 5.37 For the circuit in Figure 531, find the transfer function H(ω) = Vo(ω)/Vi(ω) 5.38 Obtain the response vo(t) of the circuit in Figure 532, where vs(t) = e−2tu(t) V 5.39 Refer to the filter shown in Figure 533 Find the transfer function and deter-

mine the type of filter 5.40 A continuous-time linear system is described by

dy t dt

y t x t( ) ( ) ( )+ =3

Find the output y(t) due to the input x(t) = e−2tu(t)

5.41 Determine the transfer function of the linear system characterized by

d y t dt

dy t dt

y t dx t dt

x t 2

3 5 6 ( ) ( ) ( ) ( ) ( )+ + = +

and find its impulse response 5.42 Find the response of a system described by

dy t dt

y t x t x t e u tt( ) ( ) ( ), ( ) ( )+ = = -2

using the Fourier transform 5.43 In a system, the input signal x(t) is amplitude modulated by m(t) = 2 + cosωot

The response y(t) = m(t)x(t) Find Y(ω) in terms of X(ω) 5.44 In a communication system, two signals

x t t x t t1 2100 2 400( ) cos( ) ( ) cos( )= =p pand

are multiplied to give x3(t) = x1(t)x2(t) Sketch the frequency spectra of x1(t), x2(t), and x3(t)

5.45 Determine the Nyquist rates for these signals: (a) x(t) = 10cos(60πt) (b) x(t) = 3cos(40πt) + 2sin(50πt) 5.46 A signal x(t) is passed through a low-pass filter If the filter blocks frequencies

above 6 Hz, find the output y(t) for the following cases: (a) x(t) = 2−3cos(2πt) + 4sin(13πt) + 10cos(15πt) (b) x(t) = 4−2sin2(4πt)

5.47 Find the energy dissipated by the resistor in the circuit of Figure 534 5.48 Calculate the total energy of the signal whose magnitude spectrum is shown in

Figure 535

5.49 If x(t) and X(ω) are Fourier transform pair and

X( ) , | |

, | |w w w

= < >

ì í î

10 2 0 2

Let y t d x t dt

( ) ( )= 2

2 Find | ( ) |y t dt2 -¥

5.50 Use MATLAB to obtain the Fourier transform of the following signals: (a) x(t) = texp(−t)u(t) (b) y(t) = (e−2t−e−t)u(t) (c) z(t) = δ(t + 1/2) + e−1u(t) 5.51 Use MATLAB to computer the Fourier transform of

x t e t u t t( ) ( ) ( )= -10 42

Plot the magnitude |X(ω)|

5.52 Use MATLAB to find the inverse Fourier transform of

(a) X e j j

1 1 ( )w

= -

(b) X j2 1 2( ) ( )w d w

w =

( )( )w pd w w w

= + +

5.53 A system has the transfer function

H s s s

s s s s j( ) = + +

+ + + =

2 10 16 8 w

Use MATLAB to plot the frequency response 5.54 Use MATLAB to plot the transfer function

H jj j( )

( . ) ( . )( . )w

w w w

= + + +

100 1 0 01 1 0 02 1 0 05

5.55 The three-pole Butterworth filter has the transfer function

H s

s s s s jc

( ) ,= + + +

=w w w w

w 3

Use MATLAB to sketch H(s) for ωc = 2π 5.56 Use MATLAB to plot the transfer function

H j( ) . .w w w= + -

1 1 0 2 0 05 2