EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005) SOLUTIONS

Solutions to Assignment 5

1. Consider the following discrete-time periodic signal:

x[n] = 1 + sin

(

2π

5n

)

+√

3 cos

(

2π

5n

)

+ cos

(

4π

5n +

π

2

)

(10)

[2] (a) The fundamental period of x[n] is N = 5 and the fundamental frequency of x[n]is ω0 = 2π

N= 2π

5.

[4] (b) To find the Fourier series coefficients for x[n], simply express x[n] as follows:

x[n] = 1 +1

2j

[

ej(2π/5)n − e−j(2π/5)n]

+

√3

2

[

ej(2π/5)n + e−j(2π/5)n]

+1

2

[

ej[(4π/5)n+π/2] − e−j[(4π/5)n+π/2]]

= 1 +

(√3

2+

1

2j

)

ej(2π/5)n +

(√3

2− 1

2j

)

e−j(2π/5)n

+

(

1

2ejπ/2

)

ej2(2π/5)n +

(

1

2e−jπ/2

)

e−j2(2π/5)n (11)

Thus, the FS coefficients are:

a0 = 1→ |a0| = 1; ∠a0 = 0

a1 =

√3

2+

1

2j=

√3

2− 1

2j → |a1| = 1; ∠a1 = tan−1

(

− 1√3

)

= −π

6

a−1 =

√3

2− 1

2j=

√3

2+

1

2j → |a−1| = 1; ∠a−1 = tan−1

(

1√3

)

=π

6

a2 =1

2ejπ/2 → |a2| =

1

2; ∠a2 =

π

2

a−2 =1

2e−jπ/2 → |a−2| =

1

2; ∠a−2 = −π

2a3 = a−2; a4 = a−1

[4] (c) Plots the magnitude and phase spectra over two periods are shown in Figure 18.

2. Let x[n] be a periodic signal with fundamental period N and FS coefficients ak.

[4] (a) First Difference: Let y[n] = x[n− 1]. The Fourier series coefficients bk of y[n] canbe computed as follows:

bk =1

N

∑

<N>

y[n]e−jkω0n =1

N

∑

<N>

x[n− 1]e−jkω0n

=1

N

∑

<N>

x[n]e−jkω0(n+1) =1

N

∑

<N>

(

x[n]e−jkω0ne−jkω0

)

=

(

1

N

∑

<N>

x[n]e−jkω0n

)

e−jkω0 = ake−jkω0 = ake

−jk(2π/N)

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EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005) SOLUTIONS

� � � � � � � � � � �

|| ka

k

� � � � �

��� � �����

� �

��� � ��� �

�

�

�

�

� �

�

�

�

� �k

ka∠

6/π−

2/π

2/π−

6/π

6/π−

2/π

2/π−

6/π

Figure 18: Magnitude and phase spectra of x[n] (plotted over two periods).

Hence x[n− 1]FS←→ ake

−jk(2π/N) and it follows that:

x[n]− x[n− 1]FS←→ ak

[

1− e−jk(2π/N)]

[3] (b) Now let

x[n] =

{

1 0 ≤ n ≤ 70 8 ≤ n ≤ 9

(12)

be a periodic signal with fundamental period N = 10 and Fourier series coefficientsak. Then the fundamental frequency of x[n] is ω0 = 2π/N = 2π/10. Also, let

g[n] = x[n]− x[n− 1] (13)

be a periodic signal with fundamental period N = 10 and Fourier series coefficientsak. Also, let

g[n] = x[n]− x[n− 1]

Clearly, g[n] has the same fundamental period N = 10. The function g[n] can bewritten as

g[n] =

1, n = 00, 1 ≤ n ≤ 7−1, n = 8

0, n = 9

(14)

Plot of g[n] in one period is shown below.

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EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005) SOLUTIONS

0 1 2 3 4 5 6 7 8 9−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1g[n]

n

Figure 19: Plot of g[n] over two periods.

[3] (c) The Fourier series coefficients of g[n] can be computed as follows:

bk =1

N

∑

<N>

g[n]e−jkω0n =1

10

9∑

n=0

g[n]e−jkω0n

=1

10

[

e−jkω00 − e−jkω08]

=1

10

[

1− e−jk(2π/10)8]

(d) Since g[n] = x[n]−x[n− 1], using the first-difference property established earlier,the Fourier series coefficients ak and bk are related as follows:

[2]bk = ak − e−jk(2π/10)ak (15)

Hence,

ak =bk

1− e−jk(2π/10)=

(1/10)[

1− e−jk(2π/10)8]

ak − e−jk(2π/10)ak

(16)

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EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005) SOLUTIONS

3. (MATLAB)(a)

[5]function a=dtfs(x);

N=length(x); % The fundamental period of x

a=zeros(1,N); % Initializing the vector of FS coefficients

for k=1:N

for n=0:N-1

a(k)=a(k)+x(n+1)*exp(-j*(k-1)*(2*pi/N)*n);

% Note how the indexes k and n are entered in the exponent of exp(.)

end

a(k)=a(k)/N;

end

(b)

[7]% Computing the FS coefficients using dtfs

Nvec=[32,64,128,256,512];

for h=1:length(Nvec)

N=Nvec(h);

x=0.9.^[0:N-1];% create one period of x[n], a decaying exponential

t0=clock; % set t0 to the current time

X=dtfs(x); % store the DTFS of x[n] in X

dtfstime(h)=etime(clock,t0); %store the elapsed time in vector dtfstime;

end

% Repeat using fft

for h=1:length(Nvec)

N=Nvec(h);

x=0.9.^[0:N-1];% create one period of x[n], a decaying exponential

t0=clock; % set t0 to the current time

X=fft(x)/N; % store the DTFS of x[n] in X

ffttime(h)=etime(clock,t0); %store the elapsed time in vector ffttime;

end

p1=loglog(Nvec,dtfstime,’marker’,’none’,’color’,’k’,’linewidth’,1.0);

hold on;

p2=loglog(Nvec,ffttime,’marker’,’none’,’color’,’k’,’linewidth’,1.0,’linestyle’,’--’);

xlabel(’{\itN}’,’FontName’,’Times New Roman’,’FontSize’,16);

ylabel(’Elapsed Time (sec)’,’FontName’,’Times New Roman’,’FontSize’,16);

set(gca,’FontSize’,16,’FontName’,’Times New Roman’);

The results of dtfstime and ffttime obtained in my computer are listed below:

dtfstime =

1.6000e-002 4.7000e-002 1.8700e-001 7.3400e-001 2.9540e+000

ffttime =

0 0 0 0 0

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EE351–Spectrum Analysis and Discrete Time Systems (Fall 2005) SOLUTIONS

It appears that fft is too fast for timing. Finally, Figure 20 plots dtfstime usingloglog.

101

102

103

10−2

10−1

100

101

N

Ela

psed

Tim

e (s

ec)

Figure 20: Plot of dtfstime.

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