Aliasing

CHAPTER 5 Signals, Systems, and Transforms

5.10 The Discrete Fourier Transform and Fast Fourier Transform

5.10.3 Aliasing

The sampled version of a signal may not be unique. Two non-identical signals whose sampled versions are identical are said to be aliases of each other. We now study the conditions under which aliasing is possible.

To understand the problem, we need to re-examine the nature of an impulse train, s_T. Recall that this is a periodic function with a period of T. Therefore, its fundamental angular frequency

(EQ 53)

Now, any periodic signal can be expressed as a Fourier series using Equation 32, restated here:

(EQ 54)

The kth term of this series is given by

(EQ 55)

where, in the second step, we expand s_T using Equation 45. Note that the integration limits are in the range [-T/2, T/2]. In this range, the only delta signal in the summation that is non zero is . So, the infinite sum reduces to a single term,

. Moreover, because is zero valued other than at the origin, we can expand the integral to cover the entire real line, to get

aⁿu nT( )e^–jωnT

n=–∞

∞

∑

^(ae–jωT)n

n=0

∞

∑

= ae^–jωT e^–jωT

1 1–ae^–jωT ---

x nT[ ] 1 ω_s

--- X˜( )ejω ^jωnTdω

0 ω_s

∫

=

ω_s 2π ---T

=

ω_s 2π ---T

=

s_T c_ke^jkωst

k=–∞

∞

∑

=

c_k 1

T--- s_Te^–jkωstdt

T 2--- –

T ---2

∫

=

1

T--- δ(t–nT)

n=–∞

∞

∑

^e–jkωstdt

T ---2 –

T 2---

∫

=

δ( )t δ( )et ^–jkωst δ( )t

DRAFT - Version 2 - The Discrete Fourier Transform and Fast Fourier Transform

(EQ 56)

But, from Equation 15, the integral is simply the value of at t = 0, which is 1, so and we can rewrite Equa- tion 54 as

(EQ 57)

Let us use this to compute the Fourier transform of the impulse train. From the linearity property, we only need to compute the transform of . From Row 3 of Table 2, this is given by . Therefore, we have

(EQ 58)

(EQ 59)

This is an alternative form of the transform we already saw in Equation 46. However, this form gives us additional insight.

Notice that the transform is an infinite series of impulses in the frequency domain that are separated by the fundamental fre- quency . In other words, it is an impulse train in the transform domain. Therefore, we have the beautiful result that the discrete-time Fourier transform of an impulse train in the time domain is an impulse train in the transform domain!

We can use this result to explain the need for bandlimiting x(t). From Equation 49, we see that the discrete (sampled) signal x[nT] is the product of x(t) and s_T. Therefore from the convolution property of the Fourier transform, , the transform of x[nT] (which we previously computed in Equation 51) can also be written as

(EQ 60)

Carefully examine the expression within the summation, which is the convolution of with a frequency-shifted impulse. From Equation 10, we see that this reduces to , which is the (continuous) Fourier transform of x(t) shifted in frequency by . Therefore, the summation represents the addition of scaled and frequency-shifted replicas of

(see Figure 12). That is, sampling a signal x(t) with a sampler of period T to produce x[nT] causes the transform of the sampled signal to infinitely replicate the transform of the original signal with a frequency spacing of .

Suppose that the support of (the range of values for which it is non-zero) is smaller than . Then, the shift-and-add operation will result in creating multiple replicas of that do not touch each other. We can pick any one copy (by mul-

c_k 1

T--- δ( )et ^–jkωstdt

∞ –

∞

∫

=

e^–jkωst c_k 1

T---

=

s_T 1

T---e^jkωst

k=–∞

∞

∑

=

e^jkωst 2πδ ω( –kω_s)

s_T 2π

---δ ωT ( –kω_s)

k=–∞

∞

∑

↔

s_T ω_s δ ω( –kω_s)

k=–∞

∞

∑

↔

ω_s

X˜( )jω

X˜( )jω 1

2π---X j( )ω ω_s δ ω( –kω_s)

k=–∞

∞

∑

⊗

= ω_s 2π

--- X jω( )⊗δ ω( –kω_s)

k=–∞

∞

∑

=

1

T--- X jω( )⊗δ ω( –kω_s)

k=–∞

∞

∑

=

X jω( ) X jω( –kω_s)

ω_s X jω( )

ω_s

X jω( ) ω_s

X jω( )

DRAFT - Version 2 -The Discrete-Time-and-Frequency Fourier Transform and the Fast Fourier Transform (FFT)

157

tiplying it with a pulse of width in the frequency domain, also called a band-pass filter) and take the inverse transform to recover x(t). In other words, the original continuous signal can be recovered despite digital sampling! This is quite a neat trick, for we have gone from the continuous domain to a discrete domain with no loss of information. The corresponding condition is called the Nyquist criterion.

To get some insight into this condition, recall that the Fourier transform of a cosine signal of frequency (Row 4 of Table 2) results in the creation of two symmetric impulses at . Because any signal can be represented as an integral of sinusoids using the Fourier transform, intuitively, if the signal has a highest inherent frequency component of , then its Fourier transform will have a bandwidth of (see Figure 12). If we wish to sample such a signal in a way that preserves all the information of the original signal, we require . This is an alternative statement of the Nyquist criterion and can be remembered as: to prevent aliasing, the sampling function should have a frequency that is at least twice that of the highest frequency component of a signal.

FIGURE 12. The effect of sampling on the Fourier transform of a signal

This result is widely applicable to computer networking, where we are nearly always dealing with sampled signals. It is important to ensure that, given as estimate of highest estimated frequency of the underlying signal, the sample rate is at least twice as fast. Recall that any signal with a sharp transition (such as a pulse signal) has a Fourier transform that has an infinite support, so that it is not bandlimited. Sampling any such function is guaranteed to introduce aliasing. Therefore, it is impor- tant to ensure, in practice, that most of the signal energy lies within .

5.10.4 The Discrete-Time-and-Frequency Fourier Transform and the Fast Fourier Transform (FFT)

So far, we have placed no restrictions on the form of the transformed signal. In the discrete-time Fourier transform, for exam- ple, the transformed signal extends over all frequencies. Suppose we introduce the restriction that the transformed signal must be represented by a finite set of discrete frequencies. In other words, the transformed signal must modulate a finite impulse train in the frequency domain. In this case, we call the transform the discrete-time-and-frequency Fourier transform, that is usually abbreviated as the ‘discrete Fourier transform’ or DFT. The Fast Fourier Transform (FFT) is a clever tech- nique for rapidly computing the DFT that we will study later in this section.

It can be shown that the DFT of a discrete-time function cannot be uniquely defined unless the function is either time-limited or, if eternal, then periodic with a finite period. Moreover, the duration of the function (or its period) must be an integer mul- tiple of the sampling period T. We have already seen that a discrete-time function can be represented as a modulated impulse train, whose discrete-time Fourier transform is a modulated impulse train in the frequency domain. This fact can be used to show that a discrete-time function that is either time-limited or periodic with N samples per period is discrete and periodic in the transform domain with N component frequencies.

We denote the discrete-time signal, which needs to be specified at N instants, as x[0], x[T], x[2T],..., x[(N-1)T]. We denote the period of the signal as T₀, with T₀= NT. Then, the sampling frequency is given by

ω_s

ω_L ω_L

±

ω_L 2ω_L

ω_s>2ω_L

X(jω)

ω

s_T(jω)

1

ω ω_s -ω_s

X(jω)

ω -ω_s

~

ω_L

-ω_L 0 -ω_L ω_L

ω_s 0 0

ω_s

±

DRAFT - Version 2 - The Discrete Fourier Transform and Fast Fourier Transform

(EQ 61)

and the signal frequency (corresponding to the period over which the signal repeats) is given by

(EQ 62)

The kth frequency of the DFT of this signal is denoted where the term indicates that the transform domain is complex and discrete, with a fundamental frequency of . This term of the DFT is given by

(EQ 63)

The corresponding inverse transform is given by

(EQ 64)

EXAMPLE 25: DFT

Compute the DFT of the function shown in Figure 13. (Assume T = 1, T₀ = 9).

Solution: We have x[0] = 0, x[1] = 1, x[2] = 2, x[3] = 3, x[4] = 4, x[5] = 3, x[6] = 2, x[7] = 1, x[8] = 0. Also, N = = 9, and .

FIGURE 13. Figure for Example 23

The first Fourier value, is given by = = .

This is also the arithmetic mean of the input signal (the first DFT coefficient is the always the average of the signal values because this represents the DC value of the signal).

The second Fourier value, with , is is given by ω_s 2π

---T

=

ω₀ 2π T₀ ---

=

X jkω[ ₀] jkω₀

ω₀

X jkω[ ₀] 1

NT--- x nT[ ]e^–jkω0nT

n=0 N–1

∑

=

x nT[ ] T X jkω[ ₀]e^jkω0nT

k=0 N–1

∑

=

T₀ ---T ω₀ 2π

---9

=

x[nT]

t 0 1 2 3 4 5 6 7 1

2 3 4

8

X[ ]0 1

NT--- x nT[ ]e^–j0.ω0nT

n=0 N–1

∑

_N^{----1 x nT[ ]}

n=0 N–1

∑

¹₉^{--- 1( +2+3+4+3+2+1) = 16---}₉

k = 1 X j2π ---9

DRAFT - Version 2 -The Fast Fourier Transform

159

= .

Recall that this can be reduced to a complex quantity of the form a + jb by expanding each term using Euler’s formula.

The third Fourier value, with , is given by

= .

The remaining values are computed similarly.

„