The Fourier Transform

As we have seen, any (suﬃciently smooth) function f(t) that is periodic can be built out of sin’s and cos’s. We have also seen that complex exponentials may be used in place of sin’s and cos’s. We shall now use complex exponentials because they lead to less writing and simpler computations, but yet can easily be converted into sin’s and cos’s. If f(t) has period 2ℓ, its (complex) Fourier series expansion is

ℓ ∞ π π ik ℓ t 1 ik ℓ t f(t)= cke with ck = 2ℓ f(t)e− dt (1) k= ℓ X−∞ Z− ik π t ik π (t+2ℓ) Not surprisingly, each term cke ℓ in this expansion also has period 2ℓ, because cke ℓ = ik π t i2kπ ik π t cke ℓ e = cke ℓ . We now develop an expansion for non-periodic functions, by allowing complex exponentials (or equivalently sin’s and cos’s) of all possible periods, not just 2ℓ, for some fixed ℓ. So, from now on, do not assume that f(t) is periodic. For simplicity we’ll only develop the expansions for functions that are zero for all sufficiently large t . | | With a little more work, one can show that our conclusions apply to a much broader class of functions. Let L > 0 be sufficiently large that f(t) = 0 for all t L. We can get a Fourier series expansion for the part | | ≥ of f(t) with Lintegral. For each integer k, th π π define the k frequency to be ωk = k L and denote by ∆ω = L the spacing, ωk+1 ωk, between any two iωt − successive frequencies. Also define fˆ(ω)= ∞ f(t)e− dt. Since f(t) = 0 for all t L −∞ | |≥ L π ∞ R π ∞ 1 ik L t 1 i(k L )t 1 iωk t 1 ˆ 1 ˆ ck(L)= 2L f(t)e− dt = 2L f(t)e− dt = 2L f(t)e− dt = 2L f(ωk)= 2π f(ωk) ∆ω L Z− Z−∞ Z−∞ In this notation, ∞ 1 ˆ iωkt f(t)= FL(t)= 2π f(ωk)e ∆ω (3) k= X−∞ for any L

1 ˆ iωt y = 2π f(ω)e

ω 0 ω1 ω2 ω3 ∆ω

small slices of width ∆ω and approximate, as in the above ﬁgure, the area under the part of the graph of 1 ˆ iωt 1 ˆ iωk t 2π f(ω)e with ω between ωk and ωk + ∆ω by the area of a rectangle of base ∆ω and height 2π f(ωk)e . 1 iωt 1 iωk t ∞ ˆ ∞ ˆ This approximates the integral 2π f(ω)e dω by the sum 2π k= f(ωk)e ∆ω. As L the −∞ → ∞ approximation gets better and better−∞ so that the sum approaches the integral. So taking the limit of (3) as R P L gives → ∞ 1 ∞ ˆ iωt ˆ ∞ iωt f(t)= 2π f(ω)e dω where f(ω)= f(t)e− dt (4) Z−∞ Z−∞ The function fˆ is called the Fourier transform of f. It is to be thought of as the frequency proﬁle of the signal f(t).

Example 1 Suppose that a signal gets turned on at t = 0 and then decays exponentially, so that

at f(t)= e− if t 0 0 if t<≥ 0 for some a> 0. The Fourier transform of this signal is

t(a+iω) ∞ iωt ∞ at iωt ∞ t(a+iω) e− ∞ 1 fˆ(ω)= f(t)e− dt = e− e− dt = e− dt = = 0 0 (a + iω) 0 a + iω Z−∞ Z Z −

√ 2 2 1 ω ˆ iφ(ω) Since a + iω has modulus a + ω and phase tan− a , we have that f(ω) = A(ω)e with A(ω) = 1 = 1 and φ(ω)= tan 1 ω . The amplitude A(ω) and phase φ(ω) are plotted below. a+iω √a2+ω2 − a | | − A(ω)

a a ω ϕ(ω)

a a ω

Example 2 Suppose that a signal consists of a single rectangular pulse of width 1 and height 1. Let’s say that it gets turned on at t = 1 and turned oﬀ at t = 1 . The standard name for this “normalized” − 2 2 rectangular pulse is 1 1 if 1

1/2 when ω = 0. When ω = 0, rect(0) = 1/2 dt = 1. By l’Hˆopital’s rule 6 − R d sin ω 1 cos ω lim rect(ω) = lim 2 2 = lim 2 2 2 =1= rect(0) ω 0 ω 0 ω 0 → → ω → 1 d d so rect(ω) is continuous at ω = 0. There is a standard function called “sinc” that is deﬁned(1) by sinc ω = sin ω ω ω . In this notation rect(ω) = sinc 2 . Here is a graph of rect(ω). d d 1 d rect(ω)

d ω 2π 2π −

Properties of the Fourier Transform

Linearity

If α and β are any constants and we build a new function h(t)= αf(t)+ βg(t) as a linear combination of two old functions f(t) and g(t), then the Fourier transform of h is

∞ iωt ∞ iωt ∞ iωt ∞ iωt hˆ(ω)= h(t)e− dt = αf(t)+ βg(t) e− dt = α f(t)e− dt + β g(t)e− dt (L) Z−∞ Z−∞ Z−∞ Z−∞ = αfˆ(ω)+ βgˆ(ω)

Time Shifting

Suppose that we build a new function h(t)= f(t t ) by time shifting a function f(t) by t . The easy − 0 0 way to check the direction of the shift is to note that if the original signal f(t) has a jump when its argument t = a, then the new signal h(t) = f(t t ) has a jump when t t = a, which is at t = a + t , t units to − 0 − 0 0 0 the right of the original jump.

t0 f(t) h(t)= f(t t ) − 0

t a a + t0 The Fourier transform of h is

′ ∞ iωt ∞ iωt ∞ iω(t +t0) hˆ(ω)= h(t)e− dt = f(t t )e− dt = f(t′)e− dt′ where t = t′ + t , dt = dt′ − 0 0 Z−∞ Z−∞ Z−∞ (T) ′ iωt0 ∞ iωt iωt0 = e− f(t′)e− dt′ = e− fˆ(ω) Z−∞

(1) sin ω There are actually two different commonly used definitions. The first, which we shall use, is sinc ω = ω . The second is sin πω sinc ω = πω . It is sometimes called the normalized sinc function.

Scaling

t If we build a new function h(t)= f α by scaling time by a factor of α> 0, then the Fourier transform of h is

′ ∞ iωt ∞ t iωt ∞ iωαt hˆ(ω)= h(t)e− dt = f e− dt = α f(t′)e− dt′ where t = αt′, dt = αdt′ α (S) Z−∞ Z−∞ Z−∞ = αfˆ(αω)

Example 3 Now consider a signal that consists of a single rectangular pulse of height H, width W and centre C. y H y = rHWC (t)

t C W C + W − 2 2 t C The function rHWC (t) = H rect −W (with rect the normalized rectangular pulse of Example 2) has height H and jumps when t C = 1 , i.e. t = C 1 W and so is the speciﬁed signal. By combining properties (L), −W 2 2 ± ± t C (T) and (S), we can determine the Fourier transform of rHWC (t) = H rect −W for any H, C and W . We build it up in three steps. First we consider the special case in which H = 1 and C = 0. Then we have R (t) = rect t . So, ◦ 1 W according to the scaling property (S) with

t t t f(t) = rect(t), h(t)= R1(t) = rect W = f W = f α with α = W we have ˆ ˆ ˆ 2 Wω 2 Wω R1(ω)= h(ω)= αf(αω)= W rect(Wω)= W Wω sin 2 = ω sin 2

t C Next we allow a nonzero C too and consider R d(t) = rect − = R (t C). By (T), with ◦ 2 W 1 − f(t)= R (t), h(t)= R (t)= R (t C)= f(t t ), with t = C 1 2 1 − − 0 0

the Fourier transform is

iωt0 iωC iωC Rˆ2(ω)= hˆ(ω)= e− fˆ(ω)= e− Rˆ1(ω)= We− rect Wω d t C Finally, by (L), with α = H and β = 0, the Fourier transform of rHWC = H rect − = HR (t) is ◦ W 2 ˆ iωC iωC 2 Wω iωC 2H Wω rˆHWC (ω)= HR2(ω)= HWe− rect(Wω)= HWe− Wω sin 2 = e− ω sin 2

d Example 4 Now suppose that we have a signal that consists of a number of rectangular pulses of various heights and widths. Here is an example

y 2

y = s(t) 1

t 2 10 1 2 3 4 − − t C We can write this signal as a sum of rectangular pulses. As we saw in the last example, rHWC (t)= H rect −W is a single rectangular signal with height H, centre C and width W . So H =2, W =1, C = 1.5 for n =1 − s(t)= s1(t)+ s2(t)+ s3(t) where sn(t)= rHWC (t) with H =1, W =2, C =1 for n =2 ( H =0.5, W =2, C =3 for n =3

iωC 2H Wω So, using (L) andr ˆHWC (ω)= e− ω sin 2 , which we derived in Example 3,

3 4 i 2 ω ω 2 iω 1 i3ω sˆ(ω)=ˆs1(ω)+ˆs2(ω)+ˆs3(ω)= ω e sin 2 + ω e− sin ω + ω e− sin ω

Diﬀerentiation

If we build a new function h(t)= f ′(t) by diﬀerentiating an old function f(t), then the Fourier transform of h is ∞ iωt ∞ iωt hˆ(ω)= h(t)e− dt = f ′(t)e− dt Z−∞ Z−∞ iωt iωt Now integrate by parts with u = e− and dv = f ′(t) dt so that du = iωe− dt and v = f(t). Assuming − that f( ) = 0, this gives ±∞

∞ ∞ ∞ ∞ iωt hˆ(ω)= u dv = uv v du = f(t) ( iω)e− dt = iωfˆ(ω) (D) − − − Z−∞ −∞ Z−∞ Z−∞

Example 5 The diﬀerentiation property is going to be useful when we use the Fourier transform to solve diﬀerential equations. As an example, let’s take another look at the RLC circuit

RL + + x(t) y(t) C − −

thinking of the voltage x(t) as an input signal and the voltage y(t) as an output signal. If we feed any input signal x(t) into an RLC circuit, we get an output y(t) which obeys the diﬀerential equation

LCy′′(t)+ RCy′(t)+ y(t)= x(t) (5)

You should be able to quickly derive this equation, which is also (16) in the notes “Fourier Series”, on your own. Take the Fourier transform of this whole equation and use that

the Fourier transform of y′(t) is iωyˆ(ω) and ◦ 2 the Fourier transform of y′′(t) is iω times the Fourier transform of y′(t) and so is ω yˆ(ω) ◦ − So the Fourier transform of (5) is

LCω2yˆ(ω)+ iRCωyˆ(ω)+ˆy(ω)=ˆx(ω) − This is trivially solved for xˆ(ω) yˆ(ω)= LCω2+iRCω+1 (6) − which exhibits classic resonant behaviour. The circuit acts independently on each frequency ω component of the signal. The amplitude yˆ(ω) of the frequency ω part of the output signal is the amplitude xˆ(ω) of | | 1 1 | | 2 the frequency ω part of the input signal multiplied by A(ω)= LCω +iRCω+1 = 2 2 2 2 2 . |− | √(1 LCω ) +R C ω A −

ω We shall shortly learn how to convert (6) into an equation that expresses the time domain output signal y(t) in terms of the time domain input signal x(t).

Example 6 The Fourier transform, rect(ω), of the rectangular pulse function of Example 2 decays rather 1 slowly, like ω for large ω. We can try suppressing large frequencies by eliminating the discontinuities at t = 1 in rect(t). For example d ± 2 5 0 if t 8 y ≤− y = g(t) 4(t + 5 ) if 5 t 3 1  8 − 8 ≤ ≤− 8  3 3 g(t)=  1 if 8 t 8  − ≤ ≤  4( 5 t) if 3 t 5 8 − 8 ≤ ≤ 8 t 5 5 3 3 5 0 if t 8 8 8 8  8 − −  ≥  It is not very diﬃcult to evaluate the Fourier transformg ˆ(ω) directly. But it easier to use properties (L)–(D), since y 0 if t 5 ≤− 8 4 if 5 t 3 3 5  − 8 ≤ ≤− 8  8 8  3 3  g′(t)=  0 if 8 t 8  = s4(t)+ s5(t) 5 3 t  − ≤ ≤   4 if 3 t 5  − 8 − 8 − 8 ≤ ≤ 8 5  0 if t  y = g′(t)  ≥ 8    1 1   H =4, W = 4 , C = 2 for n =4 where sn(t)= rHWC (t) with − H = 4, W = 1 , C = 1 for n =5 ( − 4 2 By Example 3, the Fourier transform of

8 iω/2 ω 8 iω/2 ω 8 ω ω sˆ (ω)+ˆs (ω)= e sin e− sin =2i sin sin 4 5 ω 8 − ω 8 ω 2 8

1 By Property (D), the Fourier transform of g′(t) is iω timesg ˆ(ω). So the Fourier transform of g(t) is iω times the Fourier transform of g′(t):

1 8 ω ω 16 ω ω gˆ(ω)= iω 2i ω sin 2 sin 8 = ω2 sin 2 sin 8 This looks somewhat like the the Fourier transform in Example 2 but exhibits faster decay for large ω.

1 gˆ(ω)

ω 2π 2π −

Parseval’s Relation

The energy carried by a signal f(t) is deﬁned(2) to be

∞ ∞ f(t) 2 dt = f(t)f(t) dt | | Z−∞ Z−∞ We would like to express this energy in terms of the Fourier transform fˆ(ω). To do so, substitute in that

1 ∞ ˆ iωt 1 ∞ ˆ iωt 1 ∞ ˆ iωt f(t)= 2π f(ω)e dω = 2π f(ω) e dω = 2π f(ω)e− dω Z−∞ Z−∞ Z−∞ This gives

∞ 2 1 ∞ ∞ iωt 1 ∞ ∞ iωt f(t) dt = f(t)fˆ(ω)e− dω dt = fˆ(ω) f(t)e− dt dω | | 2π 2π Z−∞ Z−∞ Z−∞ Z−∞ Z−∞ ∞ ∞ = 1 fˆ(ω) fˆ(ω) dω = 1 fˆ(ω) 2 dω 2π 2π | | Z−∞ Z−∞ ∞ 2 1 ∞ ˆ 2 This formula, f(t) dt = 2π f(ω) dω, is called Parseval’s relation. −∞ | | −∞ | | R R Duality

The duality property says that if we build a new time–domain function g(t)= fˆ(t) by exchanging the roles of time and frequency, then the Fourier transform of g is

gˆ(ω)=2πf( ω) (Du) − To verify this, just write down just the deﬁnition ofg ˆ(ω) and the Fourier inversion formula (4) for f(t) and, in both integrals, make a change of variables so that the integration variable is s:

∞ iωt ∞ iωt t=s ∞ iωs gˆ(ω)= g(t)e− dt = fˆ(t)e− dt = fˆ(s) e− ds Z−∞ Z−∞ Z−∞ 1 ∞ ˆ iωt ω=s 1 ∞ ˆ ist f(t)= 2π f(ω)e dω = 2π f(s) e ds Z−∞ Z−∞ Sog ˆ(ω), which is given by the last integral of the first line is exactly 2π times the last integral of the second line with t replaced by ω, which is 2πf( ω). − − (2) By way of motivation for this definition, imagine that f(t) is the voltage v(t) at time t across a resistor of resistance R. 1 2 Then the power (i.e. energy per unit time) dissipated by the resistor is v(t)i(t) = R v(t) . So the total energy dissipated 1 2 by the resistor will be R v(t) dt, which, up to a constant, agrees with our definition. R c Joel Feldman. 2007. All rights reserved. March 1, 2007 The Fourier Transform 7

sin t Example 7 In this example, we shall compute the Fourier transform of sinc(t)= t . Our starting point is Example 2, where we saw that the Fourier transform of the rectangular pulse rect(t) of height one and 2 ω ω ˆ width one is rect(ω)= ω sin 2 = sinc 2 . So, by duality (Du), with f(t) = rect(t) and g(t)= f(t)= rect(t), t the Fourier transform of g(t) = sinc 2 isg ˆ(ω)=2πf( ω)=2π rect( ω)=2π rect(ω), since rect(t) is d − t − d even. So we now know that the Fourier transform of sinc 2 is 2π rect(t). To ﬁnd the Fourier transform 1 t t of sinc(t), we just need to scale the 2 out of 2 . So we apply the scaling property (S) with f(t) = sinc 2 t 1 and h(t) = sinc(t) = f(2t) = f α where α = 2 . By (S), the Fourier transform of h(t) = sinc(t) is αfˆ(αω)= α 2π rect(αω)= π rect ω . 2 Multiplication and Convolution

A very common operation in signal processing is that of filtering. It is used to eliminate high frequency noise and also to eliminate sixty cycle hum, arising from the ordinary household AC current. It is also used to extract the signal from any one desired radio or TV station. In general, the filter is described by a function Hˆ (ω) of frequency. For example you might have Hˆ (ω) = 1 for desired frequencies and Hˆ (ω) = 0 for undesirable frequencies. When a signal f(t) is fed into the filter, an output signal g(t), whose Fourier transform is fˆ(ω)Hˆ (ω) is produced. For example, the RLC circuit of Example 5 is a filter with ˆ 1 H(ω)= LCω2+iRCω+1 . The− question “what is g(t)?” remains. Of course it is the inverse Fourier transform

1 ∞ ˆ ˆ iωt g(t)= 2π f(ω)H(ω)e dω Z−∞ of fˆ(ω)Hˆ (ω), but we would like to express it more directly in terms of the original time–domain signal f(t). iωτ So let’s substitute fˆ(ω)= ∞ f(τ)e− dτ (which expresses fˆ(ω) in terms f(t)) and see if we can simplify the result. −∞ R 1 ∞ ∞ iωτ ˆ iωt g(t)= 2π f(τ)e− H(ω)e dτdω Z−∞ Z−∞ ∞ 1 ∞ ˆ iω(t τ) = f(τ) 2π H(ω)e − dω dτ Z−∞ Z−∞ ∞ = f(τ)H(t τ) dτ − Z−∞ The last integral is called a convolution integral and is denoted

∞ (f H)(t)= f(τ)H(t τ) dτ ∗ − Z−∞

If we make the change of variables τ = t τ ′, dτ = dτ ′ we see that we can also express − −

−∞ ∞ (f H)(t)= f(t τ ′)H(τ ′) ( dτ ′)= f(t τ ′)H(τ ′) dτ ′ ∗ − − − Z∞ Z−∞ We conclude that gˆ(ω)= fˆ(ω)Hˆ (ω)= g(t) = (f H)(t) (C) ⇒ ∗

Example 8 The convolution integral gives a sort of moving weighted average, with the weighting determined by H. Its value at time t involves the values of f for times near t. I say “sort of” because there is no requirement that ∞ H(t) dt = 1 or even that H(t) 0. −∞ ≥ R c Joel Feldman. 2007. All rights reserved. March 1, 2007 The Fourier Transform 8

Suppose for example, that we use the ﬁlter with

ˆ 2 ω H(ω)= ω sin 2

The graph of this function was given in Example 2. It is a “low pass filter” in the sense that it lets through most small frequencies ω and suppresses high frequencies. It is not really a practical filter, but I have chosen it anyway because it is relatively easy to see the effect of this filter in t–space. From Example 2 we know that 1 1 H(t)= 1 if 2

1 t 1 ∞ 2 − 2 (f H)(t)= f(t τ)H(τ) dτ = f(t τ) dτ = f(τ ′) dτ ′ where τ ′ = t τ, dτ ′ = dτ ∗ − 1 − − t+ 1 − − Z−∞ Z− 2 Z 2 1 t+ 2 = f(τ ′) dτ ′ t 1 Z − 2

1 1 which is the area under the part of the graph of f(τ ′) from τ ′ = t to τ ′ = t + . To be concrete, suppose − 2 2 that the input signal is y

1 if τ ′ 0 f(τ ′)= ≥ 0 if τ ′ < 0 n τ ′ 1 1 Then the output signal at time t is the area under the part of this graph from τ ′ = t 2 to τ ′ = t + 2 . 1 1 1 1 − if t + < 0, that is t < , then f(τ ′) = 0 for all t < τ ′ < t + (see the ﬁgure below) so that ◦ 2 − 2 − 2 2 (f H)(t)=0 ∗ y y = f(τ ′)

1 1 t 2 t+ 2 τ − ′

1 1 1 1 1 if t + 2 0 but t 2 0, that is 2 t 2 , then f(τ ′) = 0 for t 2 < τ ′ < 0 and f(τ ′) = 1 for ◦ ≥ − ≤ − ≤ ≤ 1 − 1 1 t+ 2 t+ 2 1 0 < τ ′

1 1 t 2 t+ 2 τ − ′ 1 1 1 1 1 t+ 2 if t > 0, that is t> , then f(τ ′) = 1 for all t < τ ′

1 1 t 2 t+ 2 τ − ′ All together, the output signal is

y y = (f H)(t) ∗ 0 if t< 1 − 2 (f H)(t)= t + 1 if 1 t 1 ∗  2 − 2 ≤ ≤ 2  1 if t> 1 2 1 1 t 2 2  − Example 9 For a similar, but more complicated example, we can through the procedure of Example 8 with H(t) still being rect(t) but with f(t) replaced by the more complicated signal s(t) of Example 4. This time the output signal at time t is

1 t+ 1 ∞ 2 2 (s H)(t)= s(t τ)H(τ) dτ = s(t τ) dτ = s(τ ′) dτ ′ where τ ′ = t τ ∗ − 1 − t 1 − Z−∞ Z− 2 Z − 2

1 1 which is the area under the part of the graph of s(τ ′) from τ ′ = t 2 to τ ′ = t + 2 . We can read off this − 1 area from the figures below. In each figure, the left hand vertical dotted line is τ ′ = t 2 and the right hand 1 − 1 1 vertical dotted line is τ ′ = t+ . The first figure has t so small that the entire interval t < τ ′

y = s(τ ) y = s(τ ) 1 ′ 1 ′

1 1 t 2 t+ 2 2 10 1 2 3 4 τ ′ 2 10 1 2 3 4 τ ′ − − − − − Case: t + 1 < 2, that is t< 5 Case: 2 t + 1 1, that is 5 t 3 2 − − 2 − ≤ 2 ≤− − 2 ≤ ≤− 2 area = 2((t + 1 ) ( 2)) = 2t +5 area = 0 2 − − y y 2 2

y = s(τ ) y = s(τ ) 1 ′ 1 ′

2 10 1 2 3 4 τ ′ 2 10 1 2 3 4 τ ′ − − − − Case: 1 t + 1 0, that is 3 t 1 Case: 0 t + 1 1, that is 1 t 1 − ≤ 2 ≤ − 2 ≤ ≤− 2 ≤ 2 ≤ − 2 ≤ ≤ 2 area = 2(( 1) (t 1 )) = 1 2t area = t + 1 − − − 2 − − 2 y y 2 2

y = s(τ ) 1 ′ 1

2 10 1 2 3 4 τ ′ 2 10 1 2 3 4 τ ′ − − − − Case: 1 t + 1 2, that is 1 t 3 Case: 2 t + 1 3, that is 3 t 5 ≤ 2 ≤ 2 ≤ ≤ 2 ≤ 2 ≤ 2 ≤ ≤ 2 area = [2 (t 1 )+ 1 [(t + 1 ) 2] = 7 1 t area = 1 − − 2 2 2 − 4 − 2

y y 2 2

1 1

2 10 1 2 3 4 τ ′ 2 10 1 2 3 4 τ ′ − − − − Case: 3 t + 1 4, that is 5 t 7 Case: 4 t + 1 5, that is 7 t 9 ≤ 2 ≤ 2 ≤ ≤ 2 ≤ 2 ≤ 2 ≤ ≤ 2 area = 1 area = 1 [4 (t 1 )] = 9 1 t 2 2 − − 2 4 − 2 y 2

y = s(τ ) 1 ′

1 1 2 10 1 2 3 4 t 2 t+ 2 − − − Case: t 1 > 4, that is t> 9 − 2 2 area = 0

All together, the graph of the output signal is

(s H)(t) ∗ 2

t 2 10 1 2 3 4 − −

Impulses

The inverse Fourier transform H(t) of Hˆ (ω) is called the impulse response function of the ﬁlter, because it is the output generated when the input is an impulse at time 0. An impulse, usually denoted δ(t) (and called a “delta function”) takes the value 0 for all times t = 0 and the value at time t = 0. In fact it is 6 ∞ so inﬁnite at time 0 that the area under its graph is exactly 1. Of course there isn’t any such function, in y ∞ y = δ(t)

the usual sense of the word. But it is possible to generalize the concept of a function (to something called a distribution or a generalized function) so as to accommodate delta functions. One generalization involves

taking the limit as ε 0 of “approximate delta functions” like y → 1 2ε 1 2ε if εcontinuous function f(t) and that is easy. Because δ(t) = 0 for all t = 0, δ(t)f(t) is the−∞ same as δ(t)f(0). (Both are zero for t = 0.) 6 R 6 Because f(0) is a constant, the area under δ(t)f(0) is f(0) times the area under δ(t), which we already said is 1. So ∞ δ(t)f(t) dt = f(0) Z−∞ iωt In particular, choosing f(t)= e− gives the Fourier transform of δ(t):

ˆ ∞ iωt iωt δ(ω)= δ(t)e− dt = e− t=0 =1 Z−∞

iωt0 By the time shifting property (T), the Fourier transform of δ(t t ) (where t is a constant) is e− . We − 0 0 may come to the same conclusion by ﬁrst making the change of variables τ = t t to give − 0

∞ iωt τ=t t0 ∞ iω(τ+t0) δ(t t )e− dt =− δ(τ)e− dτ − 0 Z−∞ Z−∞

iω(τ+t0) and then applying ∞ δ(τ)f(τ) dτ = f(0) with f(τ)= e− . Returning to the−∞ impulse response function, we can now verify that the output generated by a ﬁlter R Hˆ (ω) in response to the impulse input signal δ(t) is indeed

∞ (δ H)(t)= δ(τ)H(t τ) dτ = H(t) ∗ − Z−∞ where we have applied ∞ δ(τ)f(τ) dτ = f(0) with f(τ)= H(t τ). −∞ − R Example 10 We saw in (6) that for an RLC circuit

ˆ 1 H(ω)= LCω2+iRCω+1 − To make the numbers work out cleanly, let’s choose R = 1, L = 6 and C = 5. Then

ˆ 1 1 H(ω)= 6ω2+i5ω+1 = (3iω+1)(2iω+1) − Recall from Example 1 that

at f(t)= e− if t 0 fˆ(ω)= 1 (7) 0 if t<≥ 0 ⇒ a+iω So we can determine the impulse response function H(t) for the RLC ﬁlter just by using partial fractions to 1 write H(ω) as a linear combination of a+iω ’s. Since

1 a b a(2iω +1)+ b(3iω + 1) (2a +3b)iω + (a + b) Hˆ (ω)= = + = = (3iω + 1)(2iω + 1) 3iω +1 2iω +1 (3iω + 1)(2iω + 1) (3iω + 1)(2iω + 1) 2a +3b =0, a + b =1 b =1 a, 2a + 3(1 a)=0 a =3, b = 2 ⇐⇒ ⇐⇒ − − ⇐⇒ −

1 1 we have, using (7) with a = 3 and a = 2 ,

3 2 1 1 t/3 t/2 Hˆ (ω)= = H(t)= e− e− if t 0 3iω +1 − 2iω +1 1 + iω − 1 + iω ⇒ 0− if t<≥ 0 3 2 which has graph

y = H(t)

Example 10 (again) Here is a sneakier way to do the partial fraction expansion of Example 10. We know that Hˆ (ω) has a partial fraction expansion of the form

1 a b Hˆ (ω)= = + (3iω + 1)(2iω + 1) 3iω +1 2iω +1

The fast way to determine a is to multiply both sides of this equation by (3iω + 1)

1 b(3iω + 1) = a + 2iω +1 2iω +1 and then evaluate both sides at 3iω +1 = 0. Then the b term becomes zero, because of the factor (3iω + 1) in the numerator and we get 1 1 a = = 1 =3 2iω +1 1 iω= 3 3 −

Similarly, multiplying by (2iω + 1) rather than (3iω + 1),

1 1 1 b = = = 1 = 2 3iω +1 3iω +1 1 − 2iω+1=0 iω= 2 2 − −

Properties of the Fourier Transform

Property Signal Fourier Transform

1 ∞ iωt ∞ iωt x(t)= 2π xˆ(ω)e dω xˆ(ω)= x(t)e− dt −∞ −∞ 1 ∞ iωt ∞ iωt y(t)= 2π R yˆ(ω)e dω yˆ(ω)= R y(t)e− dt −∞ −∞ Linearity Ax(t)+ ByR (t) Axˆ(ω)+R Byˆ(ω)

iωt0 Time shifting x(t t ) e− xˆ(ω) − 0 Frequency shifting eiω0tx(t) xˆ(ω ω ) − 0 Scaling x t α xˆ(αω) α | | t t0 iωt0 Time shift & scaling x − ) α e− xˆ(αω) α | | iω0t ω ω0 Frequency shift & scaling α e x(αt) xˆ − | | α Conjugation x(t) xˆ( ω) − Time reversal x( t) xˆ( ω) − − t–Diﬀerentiation x′(t) iωxˆ(ω) x(n)(t) (iω)nxˆ(ω) d ω–Diﬀerentiation tx(t) i dω xˆ(ω) n d n t x(t) i dω xˆ(ω) Convolution ∞ x(τ)y(t τ) dτ xˆ(ω)ˆy(ω) −∞ − 1 ∞ Multiplication Rx(t)y(t) 2π xˆ(θ)ˆy(ω θ) dθ −∞ − Duality xˆ(t) 2πxR( ω) − ∞ 2 1 ∞ 2 Parseval x(t) dt = 2π xˆ(ω) dω −∞ | | −∞ | | R R at 0 if t< 0, 1 e− u(t)= at (a constant, Re a> 0) e− if t> 0 a + iω n at at e− if t< 0, 1 e− u( t)= (a constant, Re a< 0) − 0 if t> 0 −a + iω 2a e a t (a constant, Re a> 0) − | | a2 + ω2

1 if t < 1 | | 2 ω 2 ω Boxcar in time rect(t)= 1 sinc 2 = ω sin 2 ( 0 if t > 2 | | H if t C < W 2 iωC Wω iωC 2H Wω General boxcar rHWC (t)= | − | HWe sinc = e sin W − 2 − ω 2 ( 0 if t C > 2 | − | 1 if ω < 1/2 Boxcar in frequency 1 sinc t = 1 sin t rect(ω)= 2π 2 πt 2 0 if |ω| > 1/2 | | iωt0 Impulse in time δ(t t ) e− − 0 iωt0 δ(t t ) x(t) e− x(t ) − 0 0 Single frequency eiω0t 2πδ(ω ω ) − 0