Finding Fourier Transform of tri(\frac{t}{2\pi })Cos(2\pi (\frac{5}{\pi })t)

[frenzal_dude]

I want to input the following function so I can find the Fourier Transform of it:

$$tri(\frac{t}{2\pi })Cos(2\pi (\frac{5}{\pi })t)$$

I couldn't find a simple way of doing a tri function so this is what I inputted in matlab:

$$a(t_{1}) = (\frac{t_{1}}{\pi }+1)Cos(2\pi (\frac{5}{\pi })t_{1})$$ where -π < t1 < 0

$$b(t_{2}) = (-\frac{t_{2}}{\pi }+1)Cos(2\pi (\frac{5}{\pi })t_{2})$$ where 0 < t2 < π

$$g(t) = a(t_{1}) + b(t_{2})$$

Here is what I typed into matlab:
>> t1 = -pi:0.01:0;
>> t2 = 0:0.01:pi;
>> g = ((t1/pi) + 1)*cos(2*pi*(5/pi)*t1) + ((-t2/pi) + 1)*cos(2*pi*(5/pi)*t2);

But I get this error:
? Error using ==> mtimes
Inner matrix dimensions must agree.

I looked in the workspace and t1 and t2 both have the same dimensions of 1x315 (just different min anad max values).

Thanks for your help.

 

[frenzal_dude]

OK I managed to get this to work:

g=((t1/pi) + 1).*cos(2*pi*(5/pi)*t1) + ((-t2/pi) + 1).*cos(2*pi*(5/pi)*t2);

But how can I plot this? I tried creating a new time vector t=-pi:0.001:pi;

But then I get this error:
? Error using ==> plot
Vectors must be the same lengths.

 

[frenzal_dude]

OK worked it out! :)

To do a tri function multipled with another function, you need to break it up into two components and think of the tri as two straight lines, and then use a rect function for each of the two components, as shown below:

t = -pi:0.1:pi;
y1 = ( (-pi <= t) .* (t < 0 ) ); %rect[(t-pi/2)/pi]
g1 = y1.*((t/pi)+1).*cos(2*pi*(5/pi)*t);
y2 = ( (0 <= t) .* (t < pi ) ); %rect[(t+pi/2)/pi]
g2 = y2.*((-t/pi)+1).*cos(2*pi*(5/pi)*t);
plot(t,g1+g2);