Fourier representation of a periodic pulse-train

Lesson Title: Fourier representation of a periodic pulse-train

Lesson Number: 05

Background:

In Chapter 3, the Fourier representation (analysis) of a periodic signal was developed. The basis functions were defined to be the complex exponential basis functions

, k=[0 ,1 , , ¥]

which form an orthogonal set. The synthesis formula (Eq. 3.27) is given by:

where the analysis formula satisfies:

which defines the amplitude of the k^th harmonic which is physically at frequency f_k = k f₀, where f₀= 1/T₀ with T₀ being the fundamental period. Using Eulers equation, that states that

e^jf(t) = cos(f(t)) + j sin(f(t))

the Fourier series can, if desired, be expressed in terms of cosines and series.

Challenge:

In Section 3.6.1, the Fourier series of a periodic pulse was derived. Specifically, for a continuous-time signal x(t) having the waveform (called 50% duty cycle):

1.0

-2T₀ -T₀ 0 0.5T₀ T₀ 2T₀

Figure 1: Periodic pulse train.

The Fourier coefficients are computed as follows for k¹0:

(imaginary)

since e^-j2p=1 and e^-jp=-1, a_k = 0 if k is even (k¹0) and a_k = 1/jpk if k is odd. For k=0, however

The magnitude spectrum is given in Figure 3-16 and interpreted below, assuming T₀=1/25.

-175 -75 -25 0 25 75 175

Figure 2: Magnitude spectrum of the signal shown in Figure 1.

What if T₀ ¥. What is the resulting Fourier series and magnitude spectrum. Does it make sense?

Consider the slightly modified version of x(t) shown below. Predict the resulting spectral representation.

0.5

-0.5

-2T₀ -T₀ 0 0.5T₀ T₀ 2T₀

Figure 3: Modified 50% duty cycle periodic pulse train.

Consider another modified version of x(t) shown below. Predict the resulting spectral representation and validate with formal computation?

1.0

-2T₀ -T₀ 0 T₀ 2T₀

Figure 4: Modified 50% duty cycle periodic pulse train.

The analysis requested in Question 3 relates to the analysis equation. How could you demonstrate that the answer makes sense in the context of the synthesis equation?

Response:

Q. What if T₀ ¥. What is the resulting Fourier series and magnitude spectrum. Does it make sense?

A. Based on the previous analysis, in the limit a₀=1 and a_k 0 for all k¹0. The first harmonic is located at f₁=¥. However, even though the signal x(t) has a value of unity for tI[0, ¥), the DC value of the signal remains one-half. This can be justified on the basis that x(t) also has a value of zero for tI(-¥, 0), that is half-on, half-off.

Q. Consider the slightly modified version of x(t) shown below. Predict the resulting spectral representation.

0.5

-0.5

-2T₀ -T₀ 0 0.5T₀ T₀ 2T₀

A. The essential difference is the signal DC bias of ½. The Fourier transform of the modified signal is therefore given by a_k = 0 for k is even, k¹0, and a_k = 1/jpk if k is odd, and a₀ = 0 (DC).

Q. Consider another modified version of x(t) shown below. Predict the resulting spectral representation and validate with formal computation?

1.0

-2T₀ -T₀ 0 T₀ 2T₀

A.      The modified signal is equal to the original x(t) shift in time by T₀/4 seconds. The Fourier coefficients can be computed as shown below. For k¹0:

which also introduces the classic sin(x)/x or sinc(x) function. The DC term is computed to be:

as before. The coefficients a_k can then be computed to be a_k = with all the even coefficients (except a₀) having a value of zero.

The synthesis equation, therefore, is given by

which simplifies to

The magnitude response is sketched below (based on the exponential form):

-175 -75 -25 0 25 75 175

which is identical to that shown in Figure 2. The signals associated with these two graphs therefore identical. Well, no. While agreeing in magnitude response, the phase responses are 90 apart. The 90 difference is due to the T₀/4 temporal delay between the two signals.

Q. The analysis requested in Question 3 relates to the analysis equation. How could you demonstrate that the answer makes sense in the context of the synthesis equation?

A. Using a general purpose computer, the signal x(t) described by the synthesis equation can be generated and displayed. The following MatLab program was created and executed. The results are shown below.

% cosine coefficients

a0 = 0.5 * ones(1,100);

a1 = (2/pi) * ones(1,100);

a3 = -(2/(3*pi)) * ones(1,100);

a5 = (2/(5*pi)) * ones(1,100);

% pure sinusoids

x0 = ones(1,100);

n = 0:99;

x1 = cos(2*pi*0.01*n);

x3 = cos(2*pi*0.03*n);

x5 = cos(2*pi*0.05*n);

figure

plot(n,x0,n,x1,n,x3,n,x5);

figure;

subplot(1,4,1)

% construct time-domain signal

y0 = a0.*x0;

plot(n,y0);

y1=y0+a1.*x1;

subplot(1,4,2)

plot(n,y1);

y3=y1+a3.*x3;

subplot(1,4,3)

plot(n,y3);

y5=y3+a5.*x5;

subplot(1,4,4)

plot(n,y5);

x5(t)

x3(t)

x1(t)

x0(t)

f1=0.5, f3=1.5, f5=2.5

(Left) First three non-zero harmonics (f₁=0.5, f₃=1.5, f₅=2.5), (Right) Constructed cosine series reconstruction starting with x₀(t)=a₀cos(0t)= a₀, x₁(t)= x₀(t)+a₁cos(0.5t), x₃(t)= x₁(t)+a₃cos(1.5t), x₅(t)= x₃(t)+a₅cos(2.5t). The data suggested that x_i(t) is converging to a periodic pulse train with period T₀=2. The ringing appearing on the rightmost graph is due to what is called Gibbs phenomenon.

References:

Principles of Signals and System, F. Taylor, McGraw Hill, 1994