Can i plot spectrum of a signal in Matlab

I want to plot spectrum diagram which gives the frequency content of a signal x(t) for example if i draw spectrum of x(t) = 14 cos(200*pi*t -pi/3)

by hand it will be two arrows at -100 and +100 with values 7e^(jpi/3) and 7e^(-jpi/3)

Can some software or matlab can do this for me

Answer 1

[Wayne King]

Fs = 1000;
t = 0:1/Fs:1-(1/Fs);
x = 14*cos(200*pi*t-pi/3);
xdft = (1/length(x))*fft(x);
freq = -500:(Fs/length(x)):500-(Fs/length(x));
plot(freq,abs(fftshift(xdft)));

[Wayne King]

Your frequency above is actually 100 Hz (2*100*pi*t). You have not given the sampling rate, so I just assumed 1000 Hz here.

[Amna]

Can anyone plz explain what will be the spectrum of j*cos(2*pi*10*t) , j*exp(-j*2*pi*10*t),

-j*exp(-j*2*pi*10*t) and -j*cos(2*pi*10*t) ?? please plot them i'm not asking for magnitude plot show me actual plots please help.

Answer 2

[moonman]

Wonderful Wayne King. u again rocked

Can u tell me what will be the minimum sampling rate that will be required to sample the signal without aliasing. Since the frequency is 100 Hz, so i think 200 Hz is minimum. am i right

Secondly i want to ask that if the signal is like this cos(1000*pi*t) +sin(300*pi*t) so we have to see the largest frequency component in the equation and according to that we will select minimum freq to avoid aliasing. So in this example it will be 1000 Hz (500*2 as per nyqust).* Am i right*

[Wayne King]

Yes, 200 is right, but I would sample at least a bit more than that, 200 would put the 100 Hz component right at the Nyquist. I would avoid that.

Not sure exactly what you're asking in the 2nd, you should sample based on the bandwidth of your signal, which is 2000 Hz here (you have a component at -1000 and 1000) so 1000-(-1000)=2000. You have to sample at least at the bandwidth (2 kHz).

[Wayne King]

sorry I misread above-- your bandwidth is 1 kHz (500-(-500))=1000. I misread the frequency as (2*1000*pi) instead of (1000*pi). you have to sample at least at the bandwidth. (1 kHz).

Answer 3

[moonman]

In this signal

x(t) = 14 cos(200*pi*t -pi/3)

we have to write it like

x(t) = 14 cos(2*pi*100*t -pi/3) ----> cos(2*pi*f*t) form

So freq is 100 Hz ---*Am i right*

Similarly in equation

cos(1000*pi*t) +sin(300*pi*t)

the max frequency is 500 Hz and other is 150 Hz.

So minimum sampling rate for this signal as per nyquist will be 500*2=1000Hz

am i right

[Wayne King]

Oh sorry! yes, for some reason, I read 2*pi*1000 in your first post, you are right. it's 500 Hz so it has to be sampled at a minimum of 1 kHz.

Answer 4

[moonman]

One last query regarding plot

It is nice to see the spectrum through code but it does not show the amplitude values i mean e^(j*pi) what so ever they are how can i come to know abt those or i have to do calcualtions by hand for those

[moonman]

Furthermore code is not working when i have to multiply two waveform

x = cos(1000*pi*t+pi/3)*(sin(500*pi*t+pi/4))

Code is giving error

??? Error using ==> mtimes

Inner matrix dimensions must agree.

Error in ==> spectrum at 3

x = cos(1000*pi*t+pi/3)*(sin(500*pi*t+pi/4));

[Wayne King]

The factor your refer to is the phase, you can get that by plotting the angle() (but that tends to be messy unless you unwrap it which is another matter), or you can query it directly. For example, xdft(101) corresponds to the DFT bin for 100 Hz in the example I gave you. Because the bins are spaced at Fs/length(x) with the first bin corresponding to zero frequency, DC. If you query

angle(xdft(101))

you get -1.0472, which is -pi/3

Answer 5

[Wayne King]

as far as your multiplication problem, use the .* operator

x = cos(1000*pi*t+pi/3).*(sin(500*pi*t+pi/4));

[moonman]

ok i m also getting error with dot operator

Fs = 1000;

t = 0:1/Fs:1-(1/Fs);

x = cos(1000*pi*t+pi/3).*(sin(500*pi*t+pi/4));

xdft = (1/length(x))*fft(x);

freq = -2000:(Fs/length(x)):2000-(Fs/length(x));

plot(freq,abs(fftshift(xdft)))

[moonman]

Sorry Problem is resolved

I was not putting one bracket

Answer 6

[Wayne King]

That is because you changed the sampling frequency, in your frequency vector

freq = -2000:(Fs/length(x)):2000-(Fs/length(x));

Your frequency vector is set up for a 4000 Hz sampling rate. However you left

Fs = 1000;

So your sizes are not correct.

Also you are creating a modulated signal above, so you have to think about what the spectrum of that signal is. That is different than just adding components.

Fs = 4000; 
t = 0:1/Fs:1-(1/Fs); 
x = cos(1000*pi*t+pi/3).*(sin(500*pi*t+pi/4)); 
xdft = (1/length(x))*fft(x); 
freq = -2000:(Fs/length(x)):2000-(Fs/length(x)); 
plot(freq,abs(fftshift(xdft)));

Answer 7

[moonman]

Great Answer Bundles of Thanks again