Fourier transforms
So far we have seen how to write periodic signals as sums of scaled complex exponentials at different frequencies (Topic 5). This representation allows us to predict how linear time- invariant systems transform periodic input signals, by using the frequency response (Topics 6 and 7). What happens if the input to an LTI system is not periodic? Is there a frequency- domain representation of aperiodic signals that is useful for understanding the response of LTI systems to general input signals?
In this topic we discuss the discrete-time Fourier transform of a DT aperiodic signal, and the continuous-time Fourier transform of a CT aperiodic signal. We then use the frequency response of an LTI system to predict how LTI systems transform general input signals. The Fourier transforms (and their Fourier series counterparts) have a number of useful properties. The last part of the topic summarizes these.
8.1 Discrete-time Fourier transform (DTFT)
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We have seen that a periodic DT signal has a decomposition as a sum of scaled complex expo- nentials at integer multiples of the fundamental frequency. For many aperiodic DT signals, we will be able to write them as an integral of scaled complex exponentials at all possible discrete frequencies.
We can’t do this for all DT signals, though. One condition that ensures that an aperiodic DT signal x has a frequency-domain decomposition is that it is absolutely summable.
Absolutely summable
A DT signal x is absolutely summable if
|x[n]|<∞. n=−∞
Discrete-time Fourier transform (DTFT)
If x is an absolutely summable DT signal then the discrete-time Fourier transform of x at
frequency ω is the complex number
X(ω) = x[k]e−jωk. (8.1)
The function X is called the discrete-time Fourier transform (DTFT) of x.
This definition is useful, because it gives us a way to express the signal x as an integral of scaled complex exponentials (at different frequencies). This ‘reconstruction’ formula is called the inverse discrete-time Fourier transform.
To see why this holds, we substitute the expression for X(ω) from (8.1) into (8.2). Then the right-hand side becomes
Inverse DTFT
If x is an absolutely summable DT signal and X is its discrete-time Fourier transform then 1 2π jωn
X(ω)e dω. (8.2)
x[k]e−jωkejωn dω = x[k]
2πδ[l] (check this directly!) the right-hand side simplifies to ∞
x[k]δ[n−k]=x[n]. k=−∞
2π 0 k=−∞ k=−∞
where we have exchanged the order of integration and summation. Using the fact that 2π ejωl dω =
• The condition that x is absolutely summable ensures that the infinite sum in the definition
of X(ω) converges (uniformly) for each ω, and so that X is a function.
• The DTFT X(ω) is periodic with period 2π, even when x is an aperiodic signal. This
reflects the fact that there is a largest frequency in the discrete-time setting.
• We can compute the integral in the definition of the inverse DTFT (8.2) over any interval
of length 2π. This doesn’t change the result because X(ω) is periodic with period 2π. 8.1.1 Examples
ejω(n−k) dω 2π 0
Example 8.1: DTFT of unit impulse
The discrete-time Fourier transform of the DT unit impulse δ is
δ[k]e−jωk = 1 for all ω.
The representation of δ given by the inverse DTFT is
1 2π jωn
We can check this formula directly by evaluating the integral. If n = 0 then
If n ̸= 0 then since ej2πn = 1.
1 2π jωn 1 1 j2πn
e dω=2πjn e
1 2π jωn 1 2π
2π e dω = 2π dω = 1.
Example 8.2: DTFT of finite duration signal
Consider the finite duration DT signal
The DTFT is
1 ifn=0
2 if n = 1
ifn=2 4 if n = 3
0 otherwise.
X(ω) = x[n]e−jωn = x[0]+x[1]e−jω+x[2]e−j2ω+x[3]e−j3ω = 1+2e−jω−e−j2ω+4e−j3ω.
Example 8.3: DTFT of decaying exponential
Let x[n] = 2−nu[n] be the signal shown below. x
··· n −2 −1 0 1 2 3 4
The discrete-time Fourier transform is
Here we have used the fact that
−jωn x[n]e =
−n −jωn 2 e
= 1 − e−jω/2.
αn=1−α wheneverαisacomplexnumberwith|α|<1. n=0
We used this result with α = e−jω/2. The formula is valid in this case because |e−jω/2| = 1/2 < 1 for all ω.
The magnitude of the DTFT X, shown over one period, is |X(ω)|
The phase of the DTFT X, shown over one period, is ∠X(ω)
8.1.2 Properties of discrete-time Fourier transforms
The discrete-time Fourier transform has a number of properties. These help us compute more complicated Fourier transforms from simpler ones. Many of these are very similar to the prop- erties of Fourier series we discussed in topic 5.
Conjugate symmetry
If x is a real DT signal (i.e., x[n] is a real number for all times n) then its DTFT X satisfies X(−ω) = X(ω)∗. This implies that
|X(−ω)| = |X(ω)| the magnitude of the DTFT is even ∠X(−ω) = −∠X(ω) the phase of the DTFT is odd
To see why this holds, suppose x is a real DT signal. Then x[k]∗ = x[k] for all k and so
X(−ω) = x[k]ejωk = x[k]e−jωk∗ = x[k]e−jωk = X(ω)∗.
k=−∞ k=−∞ k=−∞
Time-shifting in DT
If x is a DT signal with DTFT X and y = DelayN(x) (i.e., y[n] = x[n−N] for all n) then Y (ω) = e−jωN X(ω)
for all ω.
So, when we shift a signal in time, the Fourier transform is multiplied by a complex exponential (in frequency). To see why this is true, we could use the definition of the DTFT
Y (ω) = y[k]e−jωk = x[k − N]e−jωk
k=−∞ k=−∞ ∞
x[k′]e−jω(k′+N) (putting k′ = k − N) ∞
= e−jωN x[k′]e−jωk′ = e−jωN X(ω). k=−∞
Frequency-shifting and modulation in DT
If x is a DT signal with DTFT X and y[n] = ejω0nx[n] for all n then Y (ω) = X(ω − ω0)
for all ω.
So, if we multiply a signal by a complex exponential at some frequency ω0, we shift the Fourier transform by ω0.
To see why this is true, we use the definition of the DTFT
Y (ω) = y[k]e−jωk = x[k]ejω0ke−jωk
k=−∞ k=−∞ ∞
= x[k]e−j(ω−ω0)k = X(ω − ω0). k=−∞
For Fourier series, we saw that multiplication in time corresponded to an appropriate notion of convolution in frequency. The same holds for the DTFT.
Suppose x1 and x2 are DT signals with DTFTs X1 and X2 respectively. For any complex numbers α1 and α2, the DTFT of α1x1 + α2x2 is α1X1 + α2X2.
Multiplication
Suppose x1 and x2 are DT signals with DTFTs X1 and X2 respectively. If y[n] = x1[n]x2[n] for all n is the product of x1 and x2 then
Y (ω) = 2π
is the periodic convolution of X1 and X2.
X1(Ω)X2(ω − Ω)dΩ for all ω
To see why this is true, we first note that the inverse DTFT of X2(ω − Ω) is, by the modulation property, ejΩnx2[n] for all n. Then, because the inverse DTFT is linear, we can take the inverse DTFT of both sides to obtain
1 2π jΩn
X1(Ω) e X1(Ω)e
For Fourier series, we saw that an appropriate notion of convolution in time corresponded to multiplication in frequency. The same holds for the DTFT.
Computing convolutions directly from the definition, as we did in topic 3 was quite tedious. This gives us an alternative approach to computing the convolution of two signals. First we take the Fourier transform of both signals, then we multiply the Fourier transforms, then we compute the inverse Fourier transform of the result.
To see why the convolution property holds, we can use the time-shifting property of the
x2[n] dΩ jΩn
for all n for all n
= x1[n]x2[n] for all n.
Convolution
Suppose x1 and x2 are DT signals with DTFTs X1 and X2 respectively. If y = x1 ∗ x2 is the DT convolution of x1 and x2 then Y (ω) = X1(ω)X2(ω) for all ω.
DTFT. First note that
y[n] = x1[k]x2[n − k]. k=−∞
By the time-shifting property, we know that the DTFT of the signal defined by x2[n−k] for all n, is given by e−jkωX2(ω) for all ω. By linearity of the DTFT we can then see that
Y(ω)= x1[k]e−jkωX2(ω) forallω
= x1[k]e−jkω X2(ω) for all ω k=−∞
= X1(ω)X2(ω) for all ω.
Parseval’s relation
The total energy of a DT signal x is
|x[n]|2.
Parseval’s relation tells us that the total energy of the DTFT is related to the total energy
1 2 π ∞
2π 0 |X(ω)|2 dω =
As usual, since the DTFT is periodic with period 2π, the integral on the left can be taken
over any interval of length 2π.
To see why Parseval’s relation holds, we can use a similar argument to the one we used for Fourier series. If X is the DTFT of x then the DTFT of the signal defined by z[n] = x∗[n] for all n, is Z(ω) = X(−ω)∗ (by the conjugation property). If y[n] = x[n]z[n] = |x[n]|2 then by the multiplication property, we know that
= 2π In particular, setting ω = 0 we see that
Y(0)=2π 0 |X(Ω)| dΩ.
On the other hand, from the definition of the DTFT we know that
Y (0) = y[n]e−j0n = |x[n]|2. n=−∞ n=−∞
X(Ω)Z(ω−Ω)dΩ ∗
X(Ω)X(Ω − ω) dΩ. 1 2π 2
Summary of Section 8.1
If x is an absolutely summable discrete-time signal then it has a frequency domain de- scription as an integral of scaled unit complex exponentials at different frequencies. The scalings in the decomposition are called the discrete-time Fourier transform (DTFT) of x. Various operations on signals (scaling, addition, convolution, multiplication) give rise to corresponding operations on the Fourier transforms of those signals. By thinking of a signal as being built out of ‘simple’ signals, we can use these properties to see how the Fourier transform of a signal can be built out of a few basic Fourier transform of ‘simple’ signals.
8.2 Continuous-time Fourier transform (CTFT)
We have seen that a periodic CT signal has a decomposition as a sum of scaled complex exponen- tials at integer multiples of the fundamental frequency. For many aperiodic CT signals, we will be able to write them as an integral of scaled complex exponentials at all possible frequencies.
We can’t do this for all CT signals, though. If a CT signal is absolutely integrable (see below), and has some other regularity properties, then it has a frequency-domain decomposi- tion.
Absolutely integrable
A CT signal x is absolutely integrable if
−∞ |x(t)| dt < ∞.
Continuous-time Fourier transform (CTFT)
If x is an absolutely integrable CT signal then the continuous-time Fourier transform of x at frequency ω is the complex number
The function X is called the continuous-time Fourier transform (CTFT) of x.
∞ −jωt x(t)e dt.
This definition is useful, because it gives us a way to express the signal x as an integral of scaled complex exponentials (at different frequencies). This ‘reconstruction’ formula is called the inverse continuous-time Fourier transform.
Inverse CTFT
If x is a CT signal and X is its continuous-time Fourier transform then, under appropriate technical assumptions on x,
X(ω)e dω. (8.3)
• The condition that x is absolutely integrable ensures that the integral (over an infinite
interval) in the definition of X(ω) converges for each ω, and so that X is a function.
• The CTFT X(ω) is, in general, aperiodic.
• For a periodic CT signal x, there were technical assumptions (e.g., the Dirichlet condi- tions) needed to ensure we could reconstruct x (except at discontinuities) from its Fourier coefficients Xk. For an aperiodic CT signal x, similar technical assumptions are needed to ensure that the inverse CTFT formula allows us to correctly reconstruct x from its Fourier transform X. In particular, if
– x is absolutely integrable;
– x has finitely many maxima and minima in any finite interval; and
– x has finitely many discontinuities in any finite interval (and these discontinuous ‘jumps’ are of finite size)
then the right-hand side of (8.3) converges to x(t) for all t except where x is discontinuous.
• In topic 10 we will see a generalization of the CTFT (the Laplace transform) that is valid
for all CT signals, not just absolutely integrable signals.
8.2.1 Examples
Example 8.4: CTFT of decaying exponential
Consider the signal
x(t) = e−atu(t)
where a is a positive real number. The CTFT of this signal is
∞ −at −jωt
e u(t)e dt
∞ −(a+jω)t
−1 −(a+jω)t∞
= a+jωe =1.
To compute the magnitude and phase of X(ω) it is useful to rewrite it in rectangular form
X(ω)= 1 =a−jω 1 = a −j ω . a+jω a−jωa+jω a2 +ω2 a2 +ω2
Then the magnitude of X(ω) is
|X(ω)|= The phase of X(ω) is
∠X(ω) = tan−1
a 2 ω 2
= √a2 +ω2.
a2 +ω2 + −ω/(a2 + ω2)
a/(a2 + ω2)
= tan−1(−ω/a) = − tan−1(ω/a).
Example 8.5: CTFT of two-sided decaying exponential
Consider the signal
where a is a positive real number. The CTFT of this signal is
x(t) = e−a|t| = eatu(−t) + e−atu(t) ∞ −a|t| −jωt
X(ω)= e e dt −∞
∞ −(a+jω)t 0 −(−a+jω)t
= e dt+ e dt
−1 −(a+jω)t∞ −1 −(−a+jω)t0
= a+jωe =1+1
a+jω a−jω = 2a .
Note that this CTFT is real-valued and positive for all ω. Hence |X(ω)| = X(ω) for all ω and ∠X(ω) = 0 for all ω.
Example 8.6
The continuous-time Fourier transform of the CT unit impulse δ is ∞ −jωt
δ(t)e dt = 1 for all ω. −∞
The representation of δ given by the inverse CTFT is 1∞ jωt
Note that the CTFT X(ω) = 1 for all ω is not absolutely integrable, and δ itself is not a function in the usual sense. This is a sign that care needs to be taken to interpret this representation rigorously.
8.2.2 Properties of continuous-time Fourier transforms
The Fourier transform has a number of properties. These help us compute more complicated Fourier transforms from simpler ones. Many of these are very similar to the properties of Fourier series, and the properties of discrete-time Fourier transforms. As such, we will be fairly brief in this section.
Conjugate symmetry in CT
If x is a real CT signal (i.e., x(t) is a real number for all t) then its CTFT X satisfies X(−ω) = X(ω)∗. This implies that
|X(−ω)| = |X(ω)| the magnitude of the DTFT is even ∠X(−ω) = −∠X(ω) the phase of the DTFT is odd
Time-shifting in CT
IfxisaCTsignalwithCTFTX andy=Delayτ(x)(i.e.,y(t)=x(t−τ)forallt)then Y (ω) = e−jωτ X(ω)
for all ω.
Time-scaling in CT
If x is a CT signal with CTFT X and y(t) = x(at) for all t (where a is a non-zero real
number) then
Y(ω)= 1Xω forallω. |a| a
We will just show this in the case a > 0. This is true because
where the third equality holds because integral.
e−j(ω/a)t′ x(t′) a dt′
we have made the change of variable t′ = at in the
Frequency-shifting and modulation in CT
If x is a CT signal with CTFT X and y(t) = ejω0tx(t) for all t then Y (ω) = X(ω − ω0)
for all ω.
Suppose x1 and x2 are CT signals with CTFTs X1 and X2 respectively. For any complex numbers α1 and α2, the CTFT of α1×1 + α2×2 is α1X1 + α2X2.
Convolution
Suppose x1 and x2 are CT signals with CTFTs X1 and X2 respectively. If y = x1 ∗ x2 is the CT convolution of x1 and x2 then Y (ω) = X1(ω)X2(ω) for all ω.
Parseval’s relation
The total energy of a CT signal is
∞2 −∞ |x(t)| dt.
Parseval’s relation tells us that the total energy of the CT Fourier transform is very closely related to the total energy of x. If X is the CTFT of x then
1∞2∞2 2π |X(ω)| dω = |x(t)|
Summary of Section 8.2
If x is an absolutely integrable continuous-time signal then it has a frequency domain description as an integral of scaled unit complex exponentials at different frequencies. The scalings in the decomposition are called the continuous-time Fourier transform (CTFT) of
x. Various operations on signals (scaling, addition, convolution, multiplication) give rise to corresponding operations on the Fourier transforms of those signals. By thinking of a signal as being built out of ‘simple’ signals, we can use these properties to see how the Fourier transform of a signal can be built out of a few basic Fourier transform of ‘simple’ signals.
8.3 LTI systems and Fourier transforms
8.3.1 Response of a discrete-time LTI system to a general input
In Topic 7 on filtering, we saw how to use a Fourier series decomposition of a periodic signal, together with the frequency response of an LTI system, to find the system output when the input is periodic. In particular, the Fourier series coefficients of the output were the Fourier series coefficients of the input, scaled by the frequency response (at an appropriate frequency).
We now use the Fourier transform to relate the Fourier transform of the system output to the Fourier transform of the system input and the frequency response of the system.
Suppose x is the input and y is the output of a DT LTI system with frequency response H. Using the inverse DTFT (8.2) we can write x as an integral of scaled complex exponentials:
the output is
1 2π jωn
X(ω)H(ω)e dω.
1 2π jωn
where X is the DTFT of x. Then (by linearity and the definition of the frequency response)
We can also state this relationship purely in the frequency domain.
8.3.2 Frequency response and impulse response for DT LTI systems
In topic 6 we saw a formula relating the impulse response and the frequency response of a DT LTI system. Now we have defined the DTFT we can summarize this relationship concisely.
If X is the DTFT of the input and Y is the DTFT of the output of a discrete-time LTI system with frequency response H then
Y (ω) = H(ω)X(ω) for all frequencies ω.
Frequency response and impulse response of DT LTI systems
The frequency response of a DT LTI system is the discrete-time Fourier transform of the impulse response.
Example 8.7: Impulse response of a DT ideal low-pass filter
Let HLP be the frequency response of an ideal low-pass filter with cutoff frequency ωc. Note that HLP is periodic with period 2π (since it is the frequency response of a DT LTI
system). Over one period −π < ω < π it is defined by 1 if−ωc ≤ω≤ωc HLP. Doing so we obtain HLP(ω)= 0 if−π<ω<−ωc andωc <ω<π. We can find the impulse response hLP[n] of this system by taking the inverse DTFT of hLP[n] = 2π HLP(ω)e dω =2π e dω = 1 ejωnωc 2πjn −ωc 1 ejωcn−e−jωcn = πn 2j = sin(ωcn). Note that we computed the integral in the formula for the inverse DTFT over the period −π < ω < π rather than the period 0 < ω < 2π because it was more convenient to do so. Note, also, that hLP[n] is non-zero for both positive and negative values of n. This shows that the ideal low-pass filter is a non-causal system. Sinc function The function sin(t) if t ̸= 0 sinc(t) = t 1 if t = 0 appears quite often when dealing with Fourier transforms. Although it is fairly unnecessary to define a new function name, it is often used. The sinc function looks like t −8π π −π0π 4π 8π We define sinc(0) = 1 because this makes the function continuous. Its main features are that • sinc(0) = 1 • sinc(kπ) = 0 for any integer k • sinc(t) oscillates just like sin(t) but also decays like 1/t. As an example of expressing things in terms of the sinc function, in the previous example, we can write hLP in terms of the sinc function as hLP[n] = ωπ sinc(ωcn). c This makes it clear that hLP[0] = π , for instance. ωc 8.3.3 Response of a continuous-time LTI system to a general input The Fourier transform gives us an alternative representation of a CT signal. This representation is very useful when we want to understand how an LTI system transforms a signal. Suppose x is the input and y is the output of a CT LTI system with frequency response H. Using the inverse CTFT (8.3) we can write x as an integral of scaled complex exponentials: where X is the CTFT of x. Then (by linearity and the definition of the frequeny response) the output is X(ω)H(ω)e dω. We can also state this relationship purely in the frequency domain. 8.3.4 Frequency response and impulse response for CT LTI systems In topic 6 we saw a formula relating the impulse response and the frequency response of a CT LTI system. Now we have defined the CTFT we can summarize this relationship 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com