Fourier Transform of Time Functions (DC Signal, Periodic Signals, and Pulsed Cosine)

Example Find the Fourier Transform of the constant 1. Use duality and the fact that the transform of δ(t) is 1

We can also define a Fourier Transform for periodic signals. If a signal has both periodic and aperiodic components, then this will enable us to use one transform to deal with both the periodic and aperiodic components.

Can we take the Fourier Transform of a periodic signal? For example, we can write:

but we can not directly calculate this integral because it does not converge. So we will generalize the Fourier Transform to include impulses in the frequency domain.

We can use either the Duality or Modulation Properties to show that:

We can check this by taking Inverse Fourier Transform (and using the sifting property):

Now let's consider a general periodic signal x(t) that we can represent as a Fourier Series:

Since we know that:

by linearity of the Fourier Transform, we get that:

Thus we see that a periodic signal has a Fourier Transform that is an infinite impulse train at discrete frequencies kω0 with weights of 2πCk (many or most of the weights are usually 0).

Example Find the Fourier Transform of cos(ω0t).

Example Find the Fourier Transform of the pulsed cosine:

You can do it directly and also using the multiplication property.

Let's show this graphically:

In this section, we will discuss sampling of continuous time signals. Sampling a continuous time signal is used, for example, in A/D conversion, such as would be done in digitizing music for storage on a CD, digitizing a movie for storage on a DVD, or taking a digital picture.

Fourier Transforms of Sampled Signals

To start, we define the continuous time impulse train as:

As you just saw, p(t) is an infinite train of continuous time impulse functions, spaced Ts seconds apart.

Now, x(t) is the continuous time signal we wish to sample. We will model sampling as multiplying a signal by p(t). Let xs(t) = x(t)p(t) be the sampled signal. Then

Now, we will derive the Sampling Theorem . To do this, we will examine our signals in the frequency domain. To start, let p(t) have a Fourier Transform P(ω), x(t) have a Fourier Transform X(ω), and xs(t) have a Fourier Transform Xs(ω).

Then, because xs(t) = x(t)p(t), by the Multiplication Property,

Now let's find the Fourier Transform of p(t). Because the infinite impulse train is periodic, we will use the Fourier Transform of periodic signals:

where Ck are the Fourier Series coefficients of the periodic signal.

Let's find the Fourier Series coefficients Ck for the periodic impulse train p(t):

by the sifting property. Therefore

Thus, an impulse train in time has a Fourier Transform that is a impulse train in frequency. The spacing between impulses in time is Ts, and the spacing between impulses in frequency is ω0 = 2π/Ts. We see that if we increase the spacing in time between impulses, this will decrease the spacing between impulses in frequency, and vice versa. This is a very important result!

Now, to finish our derivation of the Sampling Theorem, we will go back and determine Xs(ω).

We saw that:

Therefore,

or we get replicated, scaled versions of X(ω), spaced every ω0 apart in frequency:

From this development and observing the above figure, we can derive our Sampling Theorem. This is one of the most important results you will see this quarter.

As you can see from the figure if ω0 - ωc < ωc, we would get overlap of the replicates of X(ω) in frequency. This is known as "aliasing." Therefore, to avoid aliasing, we require ω0 - ωc > ωc or ω0 > 2ωc. If we avoid aliasing, we can recover x(t) from its samples. (Usually, we choose a sampling rate a bit higher than twice the highest frequency since filters are not ideal.)

ω0 - ωc < ωc , thus aliasing occured

We hear music up to 20 kHz and CD sampling rate is 44.1 kHz. (Dogs would need a higher quality CD since they hear higher frequencies than humans.)

We can recover x(t) from its sampled version xs (t) by using a low pass filter to recover the center island:

output

input to a filter

Example 1 Given a signal x(t) with Fourier Transform with cutoff frequency ωc as shown:

you are given three different pulse trains with periods

, , and ,

draw the sampled spectrum in each case. Which case(s) experiences aliasing?

Example 2 The inverse Fourier Transform of the signal in the previous example is

Draw the sampled signals using the sampling trains of the previous example

( , , and ).

Notice how aliasing looks in the time domain.

Sampled signal

Sampled signal

Sampled signal

Given that an LTI system has the relationship y(t) = x(t) * h(t) where x(t) is the input, y(t) is the output, and h(t) is the impulse response of the system, we know that Y(ω) = X(ω) H(ω). H(ω) is a filter to eliminate unwanted parts of the frequency spectrum. There are a number of different types of filters including ideal low pass, high pass, band pass, and band stop, which are plotted here. These ideal filters have Fourier Transforms of the form rect(ω/2ωc), where ωc is the cutoff frequency. The corresponding time domain functions are sincs.

Applications of the Fourier Transform of Ideal Filters (Sinusoidal Amplitude Modulation)

Example 1 Given two functions x1(t) = cos(500πt) and x2(t) = cos(1000πt), form a new signal x3(t) = x1(t) x2(t).

Draw the frequency response of a low pass filter that passes the low frequency component of the signal and blocks the high frequency component of the signal.

Modulation is multiplying a signal in time by complex exponentials (such as sine waves and cosine waves) to shift the signal to a desired frequency band. This is done, for example, by radio (or television) to assign different parts of the frequency spectrum to different radio (or television) stations. Remember that the Modulation Theorem told us that:

We will use the Modulation Theorem in an example of AM radio.

Example 2 AM radio - Double-sideband, suppressed carrier, amplitude modulation

Let x(t) be a music signal with Fourier Transform X(ω).

To modulate this signal, we'll form and transmity(t) = [x(t) + B] cos(ωct + φ)

where ωc is the carrier frequency of the radio station (e.g. 770 kHz), and B is the carrier. Let B=0 which means that we have a suppressed carrier and let φ = 0 for simplification. In this case, y(t) = x(t) cos(ωct).

Then

Now, let's look at the Fourier Transform of the modulated signal y(t). Y(ω) = ?

Your radio demodulates the signal by multiplying the received signal y(t) by another cosine wave to form a third signal z(t):

To recover the original music signal x(t) from the demodulated signal z(t), you filter z(t) with a low-pass filter by multiplying by a rect function in frequency. This corresponds to convolving with a sinc function in time.