Hi Munish
Can you explain the modified code for this lines?
>St(s,N)*(2/Fs);
>w=((0:((N/2)-1))/(N/2))*(Fs/2);
Thanks in advance
>
>
>
> Hi Sonia
>
>I modified your code a bit. I think it will solve your problem. Try it.
>
>____________________________________________________
>
>clear all;
>clc;
>Fs0;
>t=[0:100]/Fs;
>s1=5*sin(2*pi*5*t);
>s2*sin(2*pi*10*t);
>s=(s1+s2);
>figure(1);
>plot(t,s)
>N%60;
>St(s,N)*(2/Fs);
>w=((0:((N/2)-1))/(N/2))*(Fs/2);
>figure(2);
>plot(w,abs([S(1:(N/2))']))
>xlabel('FREQUENCY');
>Ylabel('Magnitude');
>____________________________________________
>
>Regards
>Munish
>GEMS IT
>
>
>
>-----Original Message-----
>From: navaneetha krishnan [mailto:]
>Sent: Thursday, May 29, 2003 12:47 AM
>To: ;
>Subject: Re: [matlab] FFT Magnitude doubt
>
>
>The FFT output depends on the length of the FFT as
>well as the input amplitude. It also depends on the
>DFT definition used.
>
>For a good text on DFT see
> http://www-ccrma.stanford.edu/~jos/mdft/
>
>Navan
>
>--- Sonia Ramwaswamy <>
>wrote:
>> Hi
>>
>> Sorry there was some problem with previous mail. Its
>> not ==== but == only. I dont know hot it got
>> changed. Any how find below the new piece of code.
>>
>>
>*********************************************************
>> clear all;
>> clc;
>> Fs=0;
>> t==[0:100]/Fs;
>> s1==sin(2*pi*5*t);
>> s2==sin(2*pi*10*t);
>> s==(s1+s2);
>> figure(1);
>> plot(t,s)
>> N=24;
>> S=yt(s,N);
>> w==((0:((N/2)-1))/(N/2))*(Fs/2);
>> figure(2);
>> plot(w,abs([S(1:(N/2))']))
>> xlabel('FREQUENCY');
>> Ylabel('Magnitude');
>>
>*******************************************************
>>
>> Now the magnitude of the FFT curve is 50. But the
>> input signal is having amplitude of 1 in both the
>> cases. Then how come during FFT this thing got
>> magnified to 50.
>> Can anybody explain this thing?
>>
>> Sonia
>
>
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