A3. Frequency Representation of Continuous Time and Discrete Time Signals Objectives

• Define the magnitude and phase plots of continuous time sinusoidal signals • Extend the magnitude and phase plots to discrete time signals • Define aliasing for sampled sinusoids • State the sampling theorem for sinusoidal signals • Extend the sampling theorem to general signals of interest

1. Introduction We begin this section with a frequency representation of sinusoids both in continuous time and discrete time, to extend it to more general signals. The goals of this development are twofold: a. provide an early introduction to the frequency spectrum with minimal mathematical background and b. introduce the sampling theorem, at the basis of any Digital Signal Processing implementation.

2. Frequency Representation of Sinusoids: Continuous Time

VIDEO: Frequency Representation of Continuous Time Sinusoids (18:13)

http://faculty.nps.edu/rcristi/eo3404/a-signals/videos/section3-seg1_media/section3-seg1-0.wmv

We have seen that a sinusoidal signal with amplitude A , frequency 0F and phase α is represented in terms of complex exponentials as

tFjjtFjj eeAeeAtFAtx 00 220 22

)2cos()( παπααπ −−

+

=+=

As we look at this signal, we see that it is the superposition of two complex exponentials with frequencies 0F and 0F− . Each complex exponential is multiplied by a complex

number αjA e±2 depending on magnitude and phase only. This leads to a graphical

representation of the signal in terms of magnitude 2/A and phase α± versus frequency, as shown in the figure below.

Frequency domain representation of the signal )2cos()( 0 απ += tFAtx in

terms of magnitude and phase versus frequency. Let’s see a numerical example. Example. Consider the signal )15.02000cos(10)( += ttx π . Clearly it has an amplitude

10=A (Volts, Amps or whatever physical quantity the signal is associated to), a frequency of 10000 =F Hz (don’t forget the factor “2” as in π2 ) and phase 15.0=α radians. This signal can be represented in terms of complex exponentials as

( ) ( ) tjjtjj eeeetx ππ 200015.0200015.0 55)( −−+= which shows the two frequency components at 1000± Hz each one with magnitude 5 and phase 15.0± radians. This is shown as a plot

of magnitude and phase vs frequency in the figure below.

Frequency Domain Representation of the signal )15.02000cos(10)( += ttx π in

the example. Notice two facts. First the definition of a “negative frequency”. When we talk about frequency (say in your radio), you always assume a positive number. Radio station (say) KAZU 90.3 implies a signal broadcast around a carrier with frequency 90.3MHz. It is sort of understood that the frequency is a positive number. The negative frequency ( kHzF 0.10 −=− in the example) is a mathematical artifact, since we are expressing sinusoids in terms of complex exponentials. Since each complex

exponential )2sin()2cos( 002 0 tFjtFe tFj πππ += has an imaginary part (multiplied by

1−=j , “imaginary” because it does not exist in nature) we need to add its own complex conjugate )2sin()2cos( 00

2 0 tFjtFe tFj πππ −=− so that we cancel it out. For the

same reason, the complex coefficient of the negative exponential 15.05 je− in the example, is the complex conjugate of the coefficient 15.05 je multiplying the exponential with positive frequency. Another important observation is that there is a one-to-one correspondence between a continuous time sinusoid and its frequency domain representation. Referring to the figure below, given a sinusoid with given Amplitude A , phase α and frequency 0F , we can represent it (as we have seen ) in the frequency domain. On the other hand, given the frequency representation on the right, showing amplitude phase and frequency, there is only sinusoid in continuous time with these parameters. In other words, there is no ambiguity in the time and frequency domain representations.

In Continuous Time there is no ambiguity between Time and Frequency

Domain representation of a sinusoid. Since there are no ambiguities, at least theoretically there are no limits in value of amplitude, phase and frequency. In particular this means that the frequency F can assume any value in the whole range from ∞− to ∞+ . Example. Suppose the magnitude and phase are as shown in the figure below. First, we verify that it is symmetrical (same magnitude and opposite phase for positive and negative frequencies), which is necessary for a real signal. Then we can see that the magnitude is 22/ =A , which implies 4=A , the phase is 3.0=α radians and the frequency HzF 5000 = . Then the continuous time signal is given by

)3.01000cos(4)( += ttx π , and no other sgnal has the same representation. There is no amguity!

Magnitude and Phase for the Example.

3. Frequency Representation of Sinusoids: Discrete Time

VIDEO: Frequency Representation of Discrete Time Sinusoids (08:33)

http://faculty.nps.edu/rcristi/eo3404/a-signals/videos/section3-seg1_media/section3-seg1-1.wmv

In the discrete time domain, we have exactly the same representation in terms of the digital frequency SFF /2πω = . Recall that this is expressed in “radians” (NOT radians per second) and it has no dimensions since it is a “relative frequency”, relative to the sampling frequency SF . Then a sinusoid )cos(][ 0 αω += nAnx in discrete time, with amplitude A , phase α and digital frequency 0ω can be expressed in terms of complex exponentials as

njjnjj eeAeeAnx 00

22][ ωαωα −−

+

=

This is the same form as in the continuous time domain. We have magnitude and phase versus frequency ω as shown in the figure below.

Frequency Domain representation of the discrete time sinusoid

)cos(][ 0 αω += nAnx . In order to illustrate this concept let us see an example. Example. Consider the discrete time sinusoid )35.045.0cos(8][ −= nnx π . It has an amplitude of 8=A , phase 35.0−=α radians and digital frequency π45.0 radians. This signal can be expanded in terms of complex exponentials as

( ) ( ) njjnjj eeeenx ππ 45.035.045.035.0 44][ −− += This shows two frequency components at πω 45.0±= radians with magnitudes 4 and phase 35.0 radians. The frequency

domain representation of this signal is shown in the figure below.

Frequency Domain Representation of the signal )35.045.0cos(8][ −= nnx π in

the example. Again we see that the magnitude and phase of the two components (at πω 45.0±= radians) are related having the same magnitude and opposite phase.

4. Discrete Time Sinusoids and Frequency Domain Ambiguity

VIDEO: Frequency Domain Ambiguity in Discrete Time (13:01)

http://faculty.nps.edu/rcristi/eo3404/a-signals/videos/section3-seg1_media/section3-seg1-2.wmv In continuous time we have seen that there is a one to one correspondence between a continuous time sinusoid and its frequency domain representation. In other words we can go from the time domain to the frequency domain and viceversa without ambiguity. In discrete time this is not so, in the sense that there exists an infinite number of sinusoids having different digital frequencies but with the same samples ][nx . In order to see this let’s start with two discrete time sinusoids

)cos(][ 00 αω += nAnx )cos(][ 11 αω += nAnx

Let the two frequencies 10 ,ωω be related as πωω 201 k+= , with k an integer. Then we can write ][1 nx as

)cos()2cos(][ 001 αωαπω +=++= nAnknAnx

where the rightmost equality comes from the fact that both nk, are integer by assumption, and an integer multiple of π2 does not affect the angle. It is easy to verify, with this choice of frequencies ][][ 01 nxnx = for all n . That is to say that they are not distinguishable from each other, even though they have different frequencies. Take another signal

)cos(][ 22 αω −= nAnx Let the two frequencies 20 ,ωω be related as πωω 202 k+−= , with k an integer. Then we can write ][2 nx as

)cos()2cos(][ 002 αωαπω +=−+−= nAnknAnx So we can see that even this signal is not distuinguishable from ][0 nx , since

][][ 02 nxnx = for all n . From these two examples we can see that, in discrete time, there is a lot of ambiguity in the frequency domain representation, since the frequencies

πωω 201 k+=

02 2 ωπω −= k with k any arbitray integer, yield the same signal in the time domain. Let’s see this with an example. Example. Consider the discrete time sinusoid )2.01.0cos(5][ += nnx π and let’s try to find other sinusoids with different frequencies but the same sample values in the time domain. From the expression πωω 201 k+= , let (say) 1=k and we obtain

πππω 1.221.01 =+= . If we choose 2=k we have the digital frequency πππω 1.441.02 =+= and so on. Therefore the sinusoids

...)2.01.4cos(5)2.01.2cos(5)2.01.0cos(5][ =+=+=+= nnnnx πππ and many more are indistinguishable from each other. Similarly, using the expression 02 2 ωπω −= k , let (say) ,...3,2,1=k and we obtain the digital frequencies ,...9.5,9.3,9.1 πππ and so on, all having the same sample values in the time domain. The figure below, show this ambiguity. For simplicity we show only the magnitude plot, but it is understood that there is a phase plot as well.

In discrete time, different frequencies can give the same time domain

signal.

VIDEO: Aliased Frequencies (09:27)

http://faculty.nps.edu/rcristi/eo3404/a-signals/videos/section3-seg2_media/section3-seg2-0.wmv

Now we want to try to resolve this ambiguity so there is no confusion when we look at the frequency domain. In order to do so we make the following observation. Consider all the digital frequencies in the interval

πωπ <≤− 0 Then it is easy to verify (simply try to believe!) that all other frequencies with the same samples (which we call aliases) are such that, for any integer 0≠k

ππω >+ 20 k

ππω >+− 20 k In other words they are outside the interval ππ +− , as shown in the figure below. Frequences of the Aliases

For a digital frequency in the interval πωπ <≤− 0 all aliases are outside this

interval. In fact, notice from the last example, that the digital frequency πω 1.00 = radians has all aliases at frequencies ,...9.3,9.1,...,1.4,1.2 ππππω = all with values larger than π . So, if we know in advance that the signal has frequency between π− and π+ there is only one choice, in this case πω 1.00 = and none else. This is shown in the figure below. For all sinusoids with digital frequency πωπ <≤− 0 there is a one to one correspondence (ie, no ambiguity, no aliases) between time domain signal and frequency domain representation.

In discrete time, there is a one to one correspondence between the

frequency interval πωπ <≤− 0 and the time domain signals.

As a consequence of this, we look at the digital frequency spectrum of discrete time signals only within the interval πωπ <≤− 0 radians. Everything else is just an alias and it can be represented by a frequency within the main interval πωπ <≤− 0 .

5. Sampling of Continuous Time Sinusoids

VIDEO: Sampled Sinusoids (20:03)

http://faculty.nps.edu/rcristi/eo3404/a-signals/videos/section3-seg2_media/section3-seg2-1.wmv In the past section we saw that, in order to have a one to one correspondence between frequency representation and time signal, we need the digital frequency of a sinusoid to be within the interval πωπ <≤− 0 . This has a very significant consequence on the choice of the sampling frequency for a given signal and it leads to the well known sampling theorem which we are going to explain next. Consider a sinusoid with frequency 0F Hz, sampled at SF Hz (or samples per second). As shown in the figure below, in order for the the two digital frequencies

SFF /2 00 πω ±=± to be within the required interval πωπ <≤− 0 we need

20 0

SFF <≤

Sampling a Continuous Time Sinusoid.

Now we want to see which other frequencies would give aliases, ie the same sampled sinusoids. Rewrite the aliasing conditions in terms 0F and SF and we obtain the two conditions on the frequencies as

s

s

s FkFFk

FF +

=+ 00 222 πππ

s

s

s FFkFk

FF 00 222 −

=+− πππ

This shows that a sinusoid with frequency 0F , after sampling is not distuinguishable from other sinusoids with frequencies SkFF +0 and 0FkFS − , with 0≠k an arbitrary integer. In other words, as shown in the figure below, after sampling all sinusoids with frequencies SFF +0 , SFF 20 + ,… and 0FFS − , 02 FFS − , … all lead to the same sampled sinusoid with frequency SFF /2 00 πω = . The Frequency of a Sinuosoid and its Aliases

All Sinusoids with Frequencies 0F , SFF +0 , SFF 20 + , … 0FFS − , 02 FFS − ,…

give all the same sampled signal. Example. Consider a signal with frequency kHzF 20 = sampled at kHzFS 10= . Then, after sampling, it would not be distinguishable from its aliases at frequencies

,...22,120 kHzkHzkFF S =+ and ,...12,80 kHzkHzFkFS =− All these sinusoids here, after sampling, map into the same sinusoid with digital frequency 5/20 πω = radians. The importance of this fact is that there is no loss of information in sampling a sinusoid with frequency 2/0 0 SFF <≤ . In other words, as shown in the figure below, we can exactly reconstruct it from its samples.

Sampling Thoerem for One Sinusoid

There is no loss of information in sampling a sinusoid with frequency 2/0 0 SFF <≤ .

6. Extension to General Signals: the Sampling Theorem

VIDEO: Fourier Series Expansion (20:15)

http://faculty.nps.edu/rcristi/eo3404/a-signals/videos/section3-seg2_media/section3-seg2-2.wmv

All these considerations can be extended to more general signals. It can be shown that any periodic signal with period 0T can be expanded as a Fourier Series of the form

∑+∞

−∞=

=k

tkFjkeatx 02)( π

where 00 /1 TF = is the fundamental frequency, and ka are the Fourier Coefficients, in general complex. Example. Take a continuous time sinusoid )1.02000cos(5)( += ttx π with frequency

HzF 10000 = and period sec10/1 300

−== FT . Then, as seen above, we can expand it as

the sum of two complex exponentials ( ) ( ) tjjtjj eeeetx ππ 20001.020001.0 5.25.2)( −−+= . This is a particular case of a Fourier Series with fundamental frequency HzF 10000 = and coefficients 1.0

1 5.2 jea = , 1.01 5.2 jea −− = and 0=ka for all 1−≠k .

Although the treatment of the Fourier Series will be subject of a later module, in this section we just want to emphasize the concepts of frequency spectrum of a signal and

the distribution of its power in the frequency domain. This will help understanding how to choose the sampling frequency of a signal so to minimize any loss of information due to the sampling process. In order to do so, we need to understand the relation between the signal )(tx , assumed periodic, and its Fourier Series coefficients ka . In order to do so, first notice a fundamental property of complex expoentials, called “orthogonality”. This property states that, if we take two complex esponentials of the form tkFje 02π and

tmFje 02π with mk, any two integers, then we can write

][1 2/

2/

22

0

0

0

00 kmdteeT

T

T

tmFjtkFj −=∫+

−

− dππ

where 1][ =− kmd if km = and 0][ =− kmd if km ≠ , and again 00 /1 FT = . Applying this to the expression of the Fourier Series, we obtain a way of computing the Fourier Series coefficients, as

( )∑ ∫∫∞+

−∞=

+

−

−+

−

−

=

m

T

T

tFkmjm

T

T

tkFj dteT

adtetxT

2/

2/

2

0

2/

2/

2

0

0

0

0

0

0

01)(1 ππ

The rightmost term is ][ km −d which is always zero, apart from when mk = . Therefore we obtain

∫+

−

−=2/

2/

2

0

0

0

0)(1 T

T

tkFjk dtetx

Ta π

We can illustrate this with an example.

Periodic Signal for the Example.

Example. Consider the periodic signal shown above. It is a square wave, with period

sec104 30

−×=T . Each pulse has a width of sec102 3−× . From the period we determine the fundamental frequency HzTF 250/1 00 == and the new compute the Fourier Series coefficients as

( ) 2/sin221041

3

3

10

10

25023 k

kdtea tkjk π

ππ =×

= ∫−

−−

−−

if 0≠k and, for 0=k , 121041

3

3

10

1030 =

×= ∫

−

−−− dta . A plot of the Fourier Coefficients

(magnitude only) is shown in the figure below. Notice that they decay to zero at the index k goes to infinity.

Magnitude of Fourier Series Coefficients. An important property called Parseval’s Theorem, relates the magnitude of the Fourier Coefficients to the Average Power of a signal, as

∑∫+∞

−∞=−

=k

k

T

T

adttxT

22/

2/

2

0

|||)(|1 0

0

In other words, the sum of the magnitude square of the Fourier Series Coefficients yields the average Power of the signal. Since all signals in nature have finite power, then

∞<∑+∞

−∞=kka 2||

and therefore, for all signals 0lim =±∞→ kk

a . This means that most of the energy of a signal

is within frequency range called the Bandwidth of the signal. Then we can state the sampling theorem as follows.

VIDEO: Sampling Theorem (03:39)

http://faculty.nps.edu/rcristi/eo3404/a-signals/videos/section3-seg2_media/section3-seg2-3.wmv Sampling Theorem. A signal with bandwidth (ie maximum frequency) B can be sampled at a sampling rate

BFS 2> without any loss of information.

This is hown in the fiure below. Sampling of a Bandlimited Signal

A bandlimited signal with maximum frequency B can be sampled at a rate

BFS 2> without loss of information. Let’s see a couple of tpical examples. Example. A telephone quality signal has a maximum bandwidth of kHzB 4= . It can be sampled at a sampling frequency of at least kHzFS 8= . Example. The human hearing goes up to a maximum frequency kHzB 22= . In a Compact Disk the music is sampled at kHzFS 1.44= .