Fourier Transforms and Sampling

Samantha R Summerson

19 October, 2009

1 Fourier Transform

Recall the formulas for the Fourier transform:

F {s(t)} = S(f) =∫ ∞−∞

s(t)e−j2πftdt,

F−1 {S(f)} = s(t) =∫ ∞−∞

S(f)ej2πftdf.

Suppose s(t) is periodic. Then we can write it using the Fourier series,

s(t) =∞∑

l=−∞

clej2πltT .

We can compute the Fourier transform of the signal using its Fourier series representation.

S(f) =∫ ∞−∞

( ∞∑l=−∞

clej2πltT

)e−j2πftdt,

=∞∑

l=−∞

cl

∫ ∞−∞

ej2πltT e−j2πftdt,

=∞∑

l=−∞

clδ

(f − l

T

).

The function δ(t) is the Dirac delta function:

δ(t) ={

1 t = 00 t 6= 0 .

This means that in order to find the Fourier transform of a periodic signal, we only need to find the Fourierseries coefficients.

Example 1. Find the Fourier transform of

s(t) = cos(2πf0t).

We can re-write the signal using Euler’s formula:

s(t) =12ej2πf0t +

12e−j2πf0t.

1

Thus, the Fourier series coefficients are

ak ={

12 k = −1, 10 o/w

,

and so the Fourier transform is

S(f) = a−1δ

(f +

1T

)+ a1δ

(f − 1

T

),

=12δ (f + f0) +

12δ (f − f0) .

In general, for non-periodic signals, the Fourier transform has many nice properties. I recommendlooking at CTFT tables online or in the course book. Two nice properties to highlight are the operations ofdifferentiation and integration in the time domain. Consider a signal s(t) and take the derivative.

y(t) =d

dts(t),

=d

dt

∫ ∞−∞

S(f)ej2πftdf,

=∫ ∞−∞

S(f)d

dt

{ej2πft

}df,

=∫ ∞−∞

S(f)j2πfej2πftdf,

=∫ ∞−∞

Y (f)ej2πftdf

Thus,d

dts(t)↔ j2πfS(f).

Differentiating in the time domain corresponds to mutliplying by j2πf in the frequency domain. Similarly,we can show that integrating in the time domain corresponds to dividing by j2πf in the frequency domain(if S(0) = 0). ∫ t

−∞s(τ)dτ ↔ 1

j2πfS(f)

2 Sampling

We can create discrete-time signals by sampling continuous-time signals at regular intervals of length Ts. Ifwe multiply a continuous-time signal s(t) with a train of Dirac delta functions, we have a sampled signal:

s(t)×∞∑

n=−∞δ(t− nTs) =

∞∑n=−∞

s(t)δ(t− nTs),

=∞∑

n=−∞s(nTs)δ(t− nTs).

The sampling period, or interval, is Ts, and the sampling frequency, or rate, is fs = 1Ts

. If s(t) is band-limited,we can prevent aliasing (overlap in the frequency domain) by selecting Ts such that

Ts <1

2W,

2

where W is the bandwidth of the signal. This is the as the Nyquist-Shannon Sampling theorem. We refer tof = 1

2Tsas the Nyquist frequency since it is the highest frequency at which a signal can contain energy and

remain compatible with the sampling theorem.If we wish to filter a discrete-time signal that originates from a continuous-time signal, does it matter

the order in which we perform the operations, i.e. if we sample and then filter versus filter and then sample?Consider the two systems shown below:

s(t) Samplers(n)

H(f) y1(n)

Figure 1: System 1: Sampling and then filtering.

s(t) H(f)x(t)

Sampler y2(n)

Figure 2: System 2: Filtering and then sampling.

Does y1(n) = y2(n) for these systems? In class we showed that

F {s(n)} =∑

akS

(f − k

Ts

).

We will use this result in order to show that, in fact, the two signals are not equal. In the first system, theFourier transform for s(n), the output of the sampler, is exactly the formula we have above. If we put thissignal through a LTI filter, the Fourier transform of the output is

Y1(f) = F{s(n)}H(f),

=∑

akS

(f − k

Ts

)H(f).

For the second system, we know thatX(f) = S(f)H(f),

and then we can use the same formula above for the spectrum of the sampled signal:

Y2(f) = F {y2(n)} ,

=∑

bkXf f −k

T

),

=∑

bkS f −k

T

)H

(f − k

T

).

We can see in general the two formulas are not equal. Let’s consider the above problem in pictures.Sampling the signal creates multiples copies of the spectrum of the signal centered at different frequencies.If we low-pass filter this sampled signal using a filter with passband size WHz, as in the first system,

then we will get the original signal back and the spectrum Y1(f) is the same as S(f). If we use the LPF firston s(t), the spectrum is unchanged since it falls within the passband. If we then sample, as in the secondsystem, the spectrum of the output, Y2(f), is the spectrum of s(nT ). Thus, the two signals are differentsince their frequency content is different.

3

f

|S(f)|

W−W

Figure 3: Magnitude of the spectrum of s(t).

f

|S(f)|

W−W

. . .. . .

Figure 4: Magnitude of the spectrum of s(nT ).

Example 2. What should the sampling period be for the sinc function,

s(t) = sinc(πt)?

Recall that the sinc function is defined as

sinc(πt) =sin(πt)πt

.

We know thatsin(2πWt)

πt↔ S(f) =

{1 −W < f < W0 o/w

.

Thus, our signal has bandwidth W = 12 . By the sampling theorem, we require

T <1

2(

12

) = 1.

Any sampling interval less than one will suffice (equivalently, any sampling frequency greater than one). Ifwe select T = 1

2 , then

s(n) =sin(π 1

2n)

π 12n

,

=

1 n = 00 n even2πn n = 1, 5, 9, ...−2πn n = 3, 7, 11, ...

.

4