A3. Frequency Representation of Continuous Time and Discrete Time Signals

Transcription

1 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 bacground 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) We have seen that a sinusoidal signal with amplitude A, frequency F and phase α is represented in terms of complex exponentials as A jα j2π F t A jα j2πft x( 2πF t + α) = e e + e e 2 2 As we loo at this signal, we see that it is the superposition of two complex exponentials with frequencies F and F. Each complex exponential is multiplied by a complex jα number A e ± depending on magnitude and phase only. This leads to a graphical 2 representation of the signal in terms of magnitude A / 2 and phase ± α versus frequency, as shown in the figure below.

2 Frequency domain representation of the signal x ( 2π Ft + α) in terms of magnitude and phase versus frequency. Let s see a numerical example. Example. Consider the signal x( = 1 cos(2π t +.15). Clearly it has an amplitude A = 1 (Volts, Amps or whatever physical quantity the signal is associated to), a frequency of F 1 Hz (don t forget the factor 2 as in 2 π ) and phase α =. 15 = radians. This signal can be represented in terms of complex exponentials as j.15 j2π t j.15 j2π t x( = ( 5e ) e + ( 5e ) e which shows the two frequency components at ± 1 Hz each one with magnitude 5 and phase ±. 15 radians. This is shown as a plot of magnitude and phase vs frequency in the figure below. Frequency Domain Representation of the signal x( = 1 cos(2π t +.15) in the example. Notice two facts. First the definition of a negative frequency. When we tal about frequency (say in your radio), you always assume a positive number. Radio station (say) KAZU 9.3 implies a signal broadcast around a carrier with frequency 9.3MHz. It is sort of understood that the frequency is a positive number. The negative frequency ( F = 1. Hz in the example) is a mathematical artifact, since we are expressing sinusoids in terms of complex exponentials. Since each complex

3 j exponential 2 πf t e = cos( 2πF + j sin(2πf t ) has an imaginary part (multiplied by j = 1, imaginary because it does not exist in nature) we need to add its own j complex conjugate 2 πf t e = cos( 2πF j sin(2πf t ) so that we cancel it out. For the same reason, the complex coefficient of the negative exponential 5e j.15 in the example, is the complex conjugate of the coefficient 5e j.15 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 F, 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 A / 2 = 2, which implies A = 4, the phase is α =. 3 radians and the frequency F = 5Hz. Then the continuous time signal is given by x( = 4cos(1π t +.3), and no other sgnal has the same representation. There is no amguity!

4 Magnitude and Phase for the Example. 3. Frequency Representation of Sinusoids: Discrete Time VIDEO: Frequency Representation of Discrete Time Sinusoids (8:33) In the discrete time domain, we have exactly the same representation in terms of the digital frequency ω = 2πF / FS. 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 F. S Then a sinusoid x [ n] ω n + α) in discrete time, with amplitude A, phase α and digital frequency ω can be expressed in terms of complex exponentials as A jα jω n A jα jωn x[ n] = e e + e e 2 2 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.

5 Frequency Domain representation of the discrete time sinusoid x [ n] ω n + α). In order to illustrate this concept let us see an example. Example. Consider the discrete time sinusoid x[ n] = 8cos(.45π n.35). It has an amplitude of A = 8, phase α =. 35 radians and digital frequency.45π radians. This signal can be expanded in terms of complex exponentials as j.35 j.45π n j.35 j. 45π n x[ n] = ( 4e ) e + ( 4e ) e This shows two frequency components at ω = ±. 45π radians with magnitudes 4 and phase. 35 radians. The frequency domain representation of this signal is shown in the figure below. Frequency Domain Representation of the signal x[ n] = 8cos(.45π n.35) in the example. Again we see that the magnitude and phase of the two components (at ω = ±. 45π radians) are related having the same magnitude and opposite phase.

6 4. Discrete Time Sinusoids and Frequency Domain Ambiguity VIDEO: Frequency Domain Ambiguity in Discrete Time (13:1) 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 x [n]. In order to see this let s start with two discrete time sinusoids x [ n] ω n + α) x n] ω n + ) 1[ 1 α Let the two frequencies ω,ω 1 be related as ω ω 2 1 = + π, with an integer. Then we can write x [ ] as 1 n x 1[ n] ω n + 2π n + α) ωn + α) where the rightmost equality comes from the fact that both, n are integer by assumption, and an integer multiple of 2 π does not affect the angle. It is easy to verify, with this choice of frequencies x 1[ n] = x[ n] for all n. That is to say that they are not distinguishable from each other, even though they have different frequencies. Tae another signal x 2[ n] ω2n α) Let the two frequencies ω,ω 2 be related as ω ω 2 2 = + π, with an integer. Then we can write x [ ] as 2 n x 2[ n] ω n + 2π n α) ωn + α) So we can see that even this signal is not distuinguishable from x [ n ], since x n] = x [ ] for all n. 2[ 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

7 ω = ω + 2π 1 ω2 = 2π ω with any arbitray integer, yield the same signal in the time domain. Let s see this with an example. Example. Consider the discrete time sinusoid x[ n] = 5cos(.1π n +.2) and let s try to find other sinusoids with different frequencies but the same sample values in the time domain. From the expression ω1 = ω + 2π, let (say) = 1 and we obtain ω1 =.1π + 2π = 2. 1π. If we choose = 2 we have the digital frequency ω2 =.1π + 4π = 4. 1π and so on. Therefore the sinusoids x[ n] = 5cos(.1π n +.2) = 5cos(2.1π n +.2) = 5cos(4.1π n +.2) =... and many more are indistinguishable from each other. Similarly, using the expression ω2 = 2π ω, let (say) = 1,2,3,... and we obtain the digital frequencies 1.9π, 3.9π, 5.9π,... 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 (9:27)

8 Now we want to try to resolve this ambiguity so there is no confusion when we loo at the frequency domain. In order to do so we mae the following observation. Consider all the digital frequencies in the interval π ω < π 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 ω + 2π > π ω + 2π > π In other words they are outside the interval π, + π as shown in the figure below. Frequences of the Aliases For a digital frequency in the interval π ω < π all aliases are outside this interval. In fact, notice from the last example, that the digital frequency ω =. 1π radians has all aliases at frequencies ω = 2.1π, 4.1π,...,1.9π,3.9π,... all with values larger than π. So, if we now in advance that the signal has frequency between π and + π there is only one choice, in this case ω =. 1π and none else. This is shown in the figure below. For all sinusoids with digital frequency π ω < π 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 π ω < π and the time domain signals.

9 As a consequence of this, we loo at the digital frequency spectrum of discrete time signals only within the interval π ω < π radians. Everything else is just an alias and it can be represented by a frequency within the main interval π ω < π. 5. Sampling of Continuous Time Sinusoids VIDEO: Sampled Sinusoids (2:3) 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 π ω < π. This has a very significant consequence on the choice of the sampling frequency for a given signal and it leads to the well nown sampling theorem which we are going to explain next. Consider a sinusoid with frequency F Hz, sampled at F S Hz (or samples per second). As shown in the figure below, in order for the the two digital frequencies ± ω = ± F / to be within the required interval π ω < π we need 2π F S F < F S 2 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 F and F S and we obtain the two conditions on the frequencies as F F Fs + 2π + 2π = 2π F F s s

10 2π F F Fs F + 2π = 2π s This shows that a sinusoid with frequency F, after sampling is not distuinguishable from other sinusoids with frequencies F + FS and F S F, with an arbitrary integer. In other words, as shown in the figure below, after sampling all sinusoids with frequencies F + FS, F + 2F, and S F S F, 2F S F, all lead to the same sampled sinusoid with frequency ω = F / F. 2π The Frequency of a Sinuosoid and its Aliases S F s All Sinusoids with Frequencies F, F + FS, F + 2F, S F S F, 2F S F, give all the same sampled signal. Example. Consider a signal with frequency F = 2Hz sampled at F S = 1Hz. Then, after sampling, it would not be distinguishable from its aliases at frequencies F + F = S 12Hz, 22Hz,... and F 8, 12 S F = Hz Hz,... All these sinusoids here, after sampling, map into the same sinusoid with digital frequency ω = 2 / 5 radians. π The importance of this fact is that there is no loss of information in sampling a sinusoid with frequency F < F / 2. In other words, as shown in the figure below, we can S exactly reconstruct it from its samples.

11 Sampling Thoerem for One Sinusoid There is no loss of information in sampling a sinusoid with frequency F < F / 2. S 6. Extension to General Signals: the Sampling Theorem VIDEO: Fourier Series Expansion (2:15) All these considerations can be extended to more general signals. It can be shown that any periodic signal with period T can be expanded as a Fourier Series of the form + j2πft x( = a e = where F = 1/ T is the fundamental frequency, and a are the Fourier Coefficients, in general complex. Example. Tae a continuous time sinusoid x( = 5cos(2π t +.1) with frequency F = 1Hz and period 1/ 1 3 T = F = sec. Then, as seen above, we can expand it as j.1 j2π t j.1 j2π t the sum of two complex exponentials x( = ( 2.5e ) e + ( 2.5e ) e. This is a particular case of a Fourier Series with fundamental frequency F = 1Hz and j.1 j.1 coefficients a 1 = 2. 5e, a 1 = 2. 5e and a = for all 1. 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

12 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 x (, assumed periodic, and its Fourier Series coefficients a. In order to do so, first notice a fundamental property of complex expoentials, called orthogonality. This property states that, if we tae two complex esponentials of the form j2πmft e with, m any two integers, then we can write j2πft e and + T / 2 1 j2πf t j2πmft e e dt [ m ] T = d T / 2 where δ [ m ] = 1 if m = and δ [ m ] = if m, and again T = 1/ F. Applying this to the expression of the Fourier Series, we obtain a way of computing the Fourier Series coefficients, as 1 + T / T / 2 j2πf t ( ) = 1 j2 m Ft x( e dt a m e π dt T T m= T / 2 T / 2 The rightmost term is δ [ m ] which is always zero, apart from when = m. Therefore we obtain a = T We can illustrate this with an example. + / 2 1 T T / 2 x( e j2πft dt Periodic Signal for the Example. Example. Consider the periodic signal shown above. It is a square wave, with period 3 3 T = 4 1 sec. Each pulse has a width of 2 1 sec. From the period we determine the fundamental frequency coefficients as F = / T 25Hz and the new compute the Fourier Series 1 =

13 a 3 1 = 1 2e j2π 25t 1 if and, for =, a = 2 = dt 3 1 dt = sin 2 ( π / 2) π. A plot of the Fourier Coefficients (magnitude only) is shown in the figure below. Notice that they decay to zero at the index 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 T / 2 1 T T / 2 x( 2 dt = + = a 2 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 + = a 2 and therefore, for all signals lim 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 (3:39) Sampling Theorem. A signal with bandwidth (ie maximum frequency) B can be sampled at a sampling rate F S > 2B without any loss of information.

14 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 F S > 2B without loss of information. Let s see a couple of tpical examples. Example. A telephone quality signal has a maximum bandwidth of sampled at a sampling frequency of at least F S = 8Hz. B = 4Hz. It can be Example. The human hearing goes up to a maximum frequency Compact Dis the music is sampled at F S = 44. 1Hz. B = 22Hz. In a