# Is ifft(fft(x).*fft(h)) faster or conv(x,h) ?

53 vues (au cours des 30 derniers jours)
Alim Huseynov le 1 Déc 2017
Commenté : Rik le 11 Mar 2020
Dear All,
I need to find out which one is faster to obtain convolution? - Linear convolution using just 'conv'? - or Obtaining Convolution from ifft(fft)? As far as I have coded and identified using tic/toc, that Matlab performs linear convolution faster than ifft(fft). But textbooks say fft is faster.
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Alim Huseynov le 11 Mar 2020
Modifié(e) : Rik le 11 Mar 2020
Section 1. (for Problem 1)
format LONGENG; %increased precision
x=x'; %Column vector was transposed for making it row vector.
yy37=[1:37]; % 37 place line vector created for further storage
h=ones(1,18); % this line vector will accept given values
h(1)=-0.0000000000000020464110886333966;
h(2)=0.020185108207322531;
h(3)=-0.0000000000000032166461686436967;
h(4)=0.0063932717248746463;
h(5)=-0.0000000000000042068450824985665;
h(6)=-0.030584021511458139;
h(7)=-0.0000000000000037927619003410752;
h(8)=0.0092608976386982338;
h(9)=-0.0000000000000034686968012612998;
h(10)=0.055319499353126252;
h(11)=-0.0000000000000029165858917179788;
h(12)=-0.062431158774528053;
h(13)=-0.0000000000000030126051803342088;
h(14)=-0.079105057829546827;
h(15)=-0.0000000000000030966220578734095;
h(16)=0.3013776867026185;
h(17)=-0.0000000000000028385702197172922;
h(18)=0.58901882620106349;
yy37(1)=0.0053712613864042745; %the value manually given because there is no y(0)
yy37(2:19)=h; %line vector copied to filter coefficients
for i=1:18 % for loop used for the rest of values.
yy37(19+i)=yy37(19-i);
end
t = zeros(1,100); % Vector for storing time values
for n = 1:100 %Looping repeats 100 times
tic; % Timer started
x_c=conv(x,yy37); %Operation of convolution
t(n)=toc; %During each loop relevant value of timer will be copied and timer will be reset
end
t_aver=mean(t); %Average time spent for operation
********************************************************************************
Section 2A (Problem 2)
format LONGENG; %increased precision
x=x'; %Column vector was transposed for making it row vector.
yy37=[1:37]; % 37 place line vector created for further storage
h=ones(1,18); % this line vector will accept given values
h(1)=-0.0000000000000020464110886333966;
h(2)=0.020185108207322531;
h(3)=-0.0000000000000032166461686436967;
h(4)=0.0063932717248746463;
h(5)=-0.0000000000000042068450824985665;
h(6)=-0.030584021511458139;
h(7)=-0.0000000000000037927619003410752;
h(8)=0.0092608976386982338;
h(9)=-0.0000000000000034686968012612998;
h(10)=0.055319499353126252;
h(11)=-0.0000000000000029165858917179788;
h(12)=-0.062431158774528053;
h(13)=-0.0000000000000030126051803342088;
h(14)=-0.079105057829546827;
h(15)=-0.0000000000000030966220578734095;
h(16)=0.3013776867026185;
h(17)=-0.0000000000000028385702197172922;
h(18)=0.58901882620106349;
yy37(1)=0.0053712613864042745; %the value manually given because there is no y(0)
yy37(2:19)=h; %line vector copied to filter coefficients
for i=1:18 % for loop used for the rest of values.
yy37(19+i)=yy37(19-i);
end
t = zeros(1,100); % Vector for storing time values
for n = 1:100 %Looping repeats 100 times
tic; % Timer started
t(n)=toc;
end
t_aver=mean(t); %Average time spent for operation
*******************************************************************************
Section 2B (Problem 2)
x = randi([1 1000],1,10000);
y = randi([1 1000],1,10000);
time1=zeros(1,20);
time2=zeros(1,20);
n = length(x) + length(y) - 1;
for i=1:10
tic
z1 = conv(x,y);
time1(i)=toc;
end
for i=1:10
tic
z2 = ifft(fft(x,n) .* fft(y,n));
time2(i)=toc;
end
t1_normal=mean(time1)
t2_fft=mean(time2)
*******************************************************************************
Section 2C (Problem 2)
x = randi([1 100],1,10000);
y = randi([1 100],1,10000);
time1=zeros(1,5);
time2=zeros(1,5);
time_l=zeros(1,1000);
time_2=zeros(1,1000);
for b=1:2000
y1=y(1:b*5);
for i=1:5
tic
z1 = conv(x,y1);
time1(i)=toc;
end
n = length(x) + length(y1) - 1;
for i=1:5
tic
z2 = ifft(fft(x,n) .* fft(y1,n));
time2(i)=toc;
end
time_1(b)=mean(time1);
time_2(b)=mean(time2);
end
figure(1)
plot(5*[1:2000],time_1,'r')
hold on
plot(5*[1:2000],time_2,'g')
legend('normal','fft')
xlabel('Length of second signal')
ylabel('Time spent')
set(findall(gcf,'-property','FontSize'),'FontSize',14); %size
title('Plot Showing how time spent on convolution differs with technique applied')
************************************************************************************************************************
Section 3
format LONGENG; %increased precision
x=x'; % Column vector transposed
yy37=[1:37]; % 37 place line vector created for further storage
h=ones(1,18); % this line vector will accept given values
h(1)=-0.0000000000000020464110886333966;
h(2)=0.020185108207322531;
h(3)=-0.0000000000000032166461686436967;
h(4)=0.0063932717248746463;
h(5)=-0.0000000000000042068450824985665;
h(6)=-0.030584021511458139;
h(7)=-0.0000000000000037927619003410752;
h(8)=0.0092608976386982338;
h(9)=-0.0000000000000034686968012612998;
h(10)=0.055319499353126252;
h(11)=-0.0000000000000029165858917179788;
h(12)=-0.062431158774528053;
h(13)=-0.0000000000000030126051803342088;
h(14)=-0.079105057829546827;
h(15)=-0.0000000000000030966220578734095;
h(16)=0.3013776867026185;
h(17)=-0.0000000000000028385702197172922;
h(18)=0.58901882620106349;
yy37(1)=0.0053712613864042745; %the value manually given because there is no y(0)
yy37(2:19)=h; %line vector copied to filter coefficients
for i=1:18 % for loop used for the rest of values.
yy37(19+i)=yy37(19-i);
end
% The lenght of x will be step by step increased (65 steps) and during
% each step both Linear and FFT willl be used for obtaining convolution.
% The mean time of each operation on total 3 will be stored for further
% discussion
M1_time=zeros(65,1); % time variable vector for obtaining mean time during each cycle from Linear
M2_time=zeros(65,1); % time variable vector for obtaining mean time during each cycle from FFT
for k=1:65 % Loop will be repeated 65 times, each time value of k will be incremented by +1 starting from 1.
x_new=x(1:100000*k); % During Each loop the lenght of x_new signal will be increased by 100000*k samples.
t1 = zeros(1,3); % Vector for storing time values
for n = 1:3 %Looping repeats 3 times
tic; % Timer started
x_c=conv(x_new,yy37); %Operation of convolution
t1(n)=toc; %During each loop relevant value of timer will be copied and timer will be reset
end
t2 = zeros(1,3); % Vector for storing time values
for n = 1:3 %Looping repeats 3 times
tic; % Timer started
t2(n)=toc;
end
%Average value of time spent for operation obtained and stored in respective element of row vector.
M1_time(k)=mean(t1);
M2_time(k)=mean(t2);
end
figure(1)
plot(M1_time)
hold on
plot(M2_time)
xlabel('Number of elements, in 10^5')
ylabel('Time, sec')
legend ('Linear Conv','FFT')
Rik le 11 Mar 2020
Just a note: format LONGENG will not increase calculation precision, it will only affect how values are shown in the command window.

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### Réponses (2)

Rik le 11 Mar 2020
Why would a textbook say ifft(fft()) is faster? That doens't make sense. If that was the case, Mathworks would have implemented conv a bit like this:
function out=conv(x,h)
out=ifft(fft(x).*fft(h));
end
The mere fact that they didn't should tell you the real conv function is faster than the one I put here.
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Honglei Chen le 11 Mar 2020
Time is not the best way to compare the two approaches. Rather, the best approach to describe is the computation complexity, i.e., when the size of the signal increases, how does the time of computation increases accordingly. It is in this comparison that the FFt approach shows the advantage. The theory should be available in any standard DSP text book and here is a webpage for a summary
HTH
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Rik le 11 Mar 2020
I would expect the implementation in Matlab to be a black box: it is not relevant if internally there is an FFT or direct approach, as long as it gives the correct output. That is true in general for internal functions. Your comment about computational complexity still holds, but I expect the Mathworks employees that engineered the conv function to know about this.

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