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Lesson Title: Fourier representation of a periodic pulsetrain
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 f_{0}, where f_{0}= 1/T_{0} with T_{0 }being the fundamental period. Using Eulers equation, that states that
e^{j}^{f(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 continuoustime signal x(t) having the waveform (called 50% duty cycle):
1.0
2T_{0} T_{0} 0 0.5T_{0} T_{0} 2T_{0}
Figure 1: Periodic pulse train.
The Fourier coefficients are computed as follows for k¹0:
_{} (imaginary)
since e^{j2}^{p}=1 and e^{j}^{p}=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 316 and interpreted below, assuming T_{0}=1/25.
175 75 25 0 25 75 175
Figure 2: Magnitude spectrum of the signal shown in Figure 1.
What if T_{0} ¥. 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
2T_{0} T_{0} 0 0.5T_{0} T_{0} 2T_{0}
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
2T_{0} T_{0} 0 T_{0} 2T_{0}
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 T_{0} ¥. What is the resulting Fourier series and magnitude spectrum. Does it make sense?
A. Based on the previous analysis, in the limit a_{0}=1 and a_{k} 0 for all k¹0. The first harmonic is located at f_{1}=¥. However, even though the signal x(t) has a value of unity for tI[0, ¥), the DC value of the signal remains onehalf. This can be justified on the basis that x(t) also has a value of zero for tI(¥, 0), that is halfon, halfoff.
Q. Consider the slightly modified version of x(t) shown below. Predict the resulting spectral representation.
0.5
0.5
2T_{0} T_{0} 0 0.5T_{0} T_{0} 2T_{0}
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 a_{0} = 0 (DC).
Q. Consider another modified version of x(t) shown below. Predict the resulting spectral representation and validate with formal computation?
1.0
2T_{0} T_{0} 0 T_{0} 2T_{0}
A. The modified signal is equal to the original x(t) shift in time by T_{0}/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 a_{0}) 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 T_{0}/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 timedomain 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);





(Left) First three nonzero harmonics (f_{1}=0.5, f_{3}=1.5, f_{5}=2.5), (Right) Constructed cosine series reconstruction starting with x_{0}(t)=a_{0}cos(0t)= a_{0}, x_{1}(t)= x_{0}(t)+a_{1}cos(0.5t), x_{3}(t)= x_{1}(t)+a_{3}cos(1.5t), x_{5}(t)= x_{3}(t)+a_{5}cos(2.5t). The data suggested that x_{i}(t) is converging to a periodic pulse train with period T_{0}=2. The ringing appearing on the rightmost graph is due to what is called Gibbs phenomenon.
Motivation: The production of frequencydomain representation of periodic continuoustime signals has been presented and exemplified. However, because the Fourier coefficient production rules are based on calculus (integration), the analysis technique does not easily lend itself to analysis by a generalpurpose digital computer. Discretetime signals, however, do admit digital computer analysis. The theory upon which the discretetime signals are created, called the Sampling Theorem, is the logically the next topic of study.
References:
Principles of Signals and System, F. Taylor, McGraw Hill, 1994
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