how can plot this Periodic Signal with E Amplitude using Matalb and calculate FFT for it.

Question

abdo daood 2015-7-17

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/230291-how-can-plot-this-periodic-signal-with-e-amplitude-using-matalb-and-calculate-fft-for-it

编辑： Star Strider 2015-7-23

采纳的回答： Star Strider

2 个评论
显示无隐藏无

Walter Roberson 2015-7-17

The value is ambiguous at t = T/2, being both +E and -E. Your ranges are inclusive so the boundary conditions are part of two different ranges, but it only makes a difference at t = T/2

Walter Roberson 2015-7-18

You need to rewrite this in periodic form, such as by transforming t to mod(t,T).

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2015-7-18

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/230291-how-can-plot-this-periodic-signal-with-e-amplitude-using-matalb-and-calculate-fft-for-it#answer_186442

编辑：Star Strider 2015-7-23

在 MATLAB Online 中打开

You’re not calculating the fft but the analytic Fourier transform. To do this, use the definition of the Fourier transform of x(t):

X(w) = int(x(t) * exp(j*w*t), t, T1, T2)

defining w=2*pi*f.

Specifically, with the Symbolic Math Toolbox:

syms E T t w
S1 = int(4*E/T*t * exp(j*w*t), t, 0, T/4);
S2 = int(E * exp(j*w*t), t, T/4, T/2);
S3 = int(-E * exp(j*w*t), t, T/2, 3*T/4);
S4 = int(4*E/T*(t-T) * exp(j*w*t), t, 3*T/4, T);
X(w) = S1 + S2 + S3 + S4;
X(w) = simplify(collect(X(w)))                      % Exponential Expression
X(w) = simplify(collect(rewrite(X(w), 'sincos')))   % Trigonometric Expression
X(w) =
-(4*E - 4*E*exp(T*w*1i) - 4*E*exp((T*w*1i)/4) + 4*E*exp((T*w*3i)/4) + E*T*w*exp((T*w*1i)/2)*2i - E*T*w*exp((T*w*3i)/2)*1i + E*T*w*exp((T*w*3i)/4)*1i)/(T*w^2)
X(w) =
((E*T*cos((3*T*w)/2)*1i + 2*E*T*sin((T*w)/2) - E*T*sin((3*T*w)/2) + E*T*sin((3*T*w)/4) - E*T*cos((3*T*w)/4)*1i - E*T*cos((T*w)/2)*2i)*w + 4*E*cos(T*w) - 4*E + 4*E*cos((T*w)/4) - 4*E*cos((3*T*w)/4) + E*sin(T*w)*4i + E*sin((T*w)/4)*4i - E*sin((3*T*w)/4)*4i)/(T*w^2)

7 个评论
显示 5更早的评论隐藏 5更早的评论

abdo daood 2015-7-21

在 MATLAB Online 中打开

Thanks for your helping, but after I understood your code and tried to show the signal on matlab using my won code after writing your code but the signal does not shown.

Calculate fourier coefficients

 clear 
 clc
 syms E T t  k
C0   = ( int(4*E/T*t, t, 0, T/4)+ int(E, t, T/4, T/2)+ int(-E, t, T/2, 3*T/2)+int(4*E/T*(t-T), t, 3*T/4, T))/T; % For k equal zero.
S1   = int(4*E/T*t * exp(-j*2*pi*k/T*t), t, 0, T/4);        % k diffiernt from zero.
S2   = int(E * exp(-j*2*pi*k/T*t), t, T/4, T/2);            % k diffiernt from zero.
S3   = int(-E * exp(-j*2*pi*k/T*t), t, T/2, 3*T/2);         % k diffiernt from zero.
S4   = int(4*E/T*(t-T) * exp(-j*2*pi*k/T*t), t, 3*T/4, T);  % k diffiernt from zero.
C(k) = (S1 + S2 + S3 + S4)/T;
C(k) = simplify(collect(C(k)));                             % Exponential Expression
C(k) = simplify(collect(rewrite(C(k), 'sincos')));          % Trigonometric Expression.
clear
clc
E=5;
k=1:1:50;
C =(- 2.*E + 2.*E.*exp(-pi.*k.*2i) + 2.*E.*exp(-(pi.*k.*1i)/2) - 2.*E.*exp(-(pi.*k.*3i)/2) + E.*pi.*k.*exp(-pi.*k*3i).*(exp(pi.*k.*2i).*2i + exp((pi.*k.*3i)/2).*1i - 1i))/(2.*pi.^2.*k.^2);

plot(k,C)

but the signal does not appear .. it should be like the picture.

are there any mistake .. please ???

Star Strider 2015-7-21

在 MATLAB Online 中打开

The mistake may be mine in not including the plots earlier. Here is the full code with all the possible plots:

syms E T t w
S1 = int(4*E/T*t * exp(j*w*t), t, 0, T/4);
S2 = int(E * exp(j*w*t), t, T/4, T/2);
S3 = int(-E * exp(j*w*t), t, T/2, 3*T/2);
S4 = int(4*E/T*(t-T) * exp(j*w*t), t, 3*T/4, T);
X(w) = S1 + S2 + S3 + S4;
X(w) = simplify(collect(X(w)))                      % Exponential Expression
X(w) = simplify(collect(rewrite(X(w), 'sincos')))   % Trigonometric Expression
Xw = matlabFunction(X(w))
E = 5.00;
T = 0.02;
x1 = @(t) 4*E/T*t;
x4 = @(t)4*E/T*(t-T);
figure(1)
plot([0, 0.005], x1([0, 0.005]), 'b', 'LineWidth',1)
hold on
plot([0.005, 0.01],  E*[1, 1], 'b', 'LineWidth',1)
plot([0.01, 0.015], -E*[1, 1], 'b', 'LineWidth',1)
plot([0.015, 0.02], x4([0.015, 0.02]), 'b', 'LineWidth',1)
plot([0.01 0.01], [-5 5], 'b', 'LineWidth',1)
hold off
grid
figure(2)
ezplot(@(w) real(Xw(E,T,w)), [-20*pi, 20*pi])
hold on
ezplot(@(w) imag(Xw(E,T,w)), [-20*pi, 20*pi])
hold off
grid
legend('\itRe\rm\{X(\omega)\}', '\itIm\rm\{X(\omega)\}')
title('Real and Imaginary Components of X(\omega)')
figure(3)
ezplot(@(w) abs(Xw(E,T,w)), [-20*pi, 20*pi])
grid
title('|X(\omega)|')

abdo daood 2015-7-22

编辑：abdo daood 2015-7-22

sorry , but does X(w) is same C(k) , right? or X(w) is fourier series of the signal .. I think its same of C(K) .. and about your answer and your code which is :

syms E T t w

S1 = int(4*E/T*t * exp(j*w*t), t, 0, T/4);

S2 = int(E * exp(j*w*t), t, T/4, T/2);

S3 = int(-E * exp(j*w*t), t, T/2, 3*T/2);

S4 = int(4*E/T*(t-T) * exp(j*w*t), t, 3*T/4, T);

X(w) = S1 + S2 + S3 + S4;

but inside the integral should be (-) because the rule is the new picture below and the should divide the all integral on the period also in the picture down .. and the signal is odd so the real term of every value of C(k) should be zero .. but it'not on matlab here .. why .. this is my new code ..please do it on your lab-top and where are the mistakes...

My code is

clear

clc

syms t k

E=5

T=0.02

evalin(symengine,'assume(k,Type::Integer)'); % tell matlab that k is %integer always .

C0 = ( int(4*E/T*t, t, 0, T/4)+ int(E, t, T/4, T/2)+ int(-E, t, T/2, 3*T/2)+int(4*E/T*(t-T), t, 3*T/4, T))/T;

S1= int(4*E/T*t*exp(-j*2*pi*k/T*t), t, 0, T/4); % k not zero.

S2 = int(E * exp(-j*2*pi*k/T*t), t, T/4, T/2);% k is not zero.

S3= int(-E * exp(-j*2*pi*k/T*t), t, T/2, 3*T/2); % k is not zero.

S4 = int(4*E/T*(t-T)* exp(-j*2*pi*k/T*t), t, 3*T/4, T);% k isn't zero.

C(k) = (S1 + S2 + S3 + S4)/T;

------------------- and the rules of fourier series is ..

abdo daood 2015-7-22

在 MATLAB Online 中打开

>> >> C(1)

ans =

- (((pi*1i)/2 - 1)*5i)/pi^2 + ((pi*1i + 2 - 2i)*5i)/(2*pi^2) + (- 5/2 + 5i/2)/pi - 5/pi^2

>> ans =

-0.7958 + 1.8090i

so the real term is not zero ???? are there problem in my code ??

Star Strider 2015-7-22

编辑：Star Strider 2015-7-23

在 MATLAB Online 中打开

The constant term is X(0) because it is the direct-current term where omega is zero. When I evaluate it as my X(w)/T, it equals 1.25. That may be an artifact of the way the waveform was defined. (The X(0) value should never be complex.) I calculated it exactly the way the problem defined it, and calculated the Fourier transform, not the Fourier series, so there was no ‘k’ term because no sampling interval or sampling frequency was stated in the problem you posted. I trust the maths.

The use of -j or +j in the Fourier transform is a matter of convention. (I followed the convention in the MATLAB documentation.) Both are correct, so long as the opposite signs are used in the Fourier and inverse Fourier transforms.

The integral from -Inf to +Inf is not appropriate to this waveform because it is not defined as a periodic function, at least as the problem was posted. It was defined as a single pulse on the interval (0,T).

EDIT — There was a typographical error in transcribing ‘S3’. It should be:

S3 = int(-E * exp(j*w*t), t, T/2, 3*T/4);

With that correction, everything works.

请先登录，再进行评论。