I'm trying to plot the phase of Fourier transform of the function below:  \begin{gather*} x[n] = 2e^{-0.9|n|} \: \: n \in [-5,5]\\ x[n] = 0 \:\: n \notin [-5,5] \end{gather*}

which is equal to the function below:

\begin{gather*} x[n] = 2e^{-0.9|n|} \times \mathrm{rect}(5) \end{gather*} so I calculated Fourier transform like below:  \begin{gather*} x[n] = 2e^{-0.9|n|} \times \mathrm{rect}(5)\\ y[n] = \alpha^{|n|} \:\: |\alpha |< 1 \\ Y[e^{jw}] = \frac{1-\alpha^2}{1+\alpha^2-2\alpha \cos(w)}\\ \longrightarrow x[n] = 2(\frac{1}{e^{0.9}})^{|n|} \longrightarrow\\ X[e^{jw}] = \frac{1}{2\pi} 2 \frac{1-(\frac{1}{e^{0.9}})^2}{1+(\frac{1}{e^{0.9}})^2-2(\frac{1}{e^{0.9}}) \cos(w)} * \frac{\sin\left(w(\frac{1}{2}+5\right)}{\sin\left(\frac{w}{2}\right)}\\ X[e^{jw}] = \frac{1}{\pi} \frac{1-(\frac{1}{e^{0.9}})^2}{1+(\frac{1}{e^{0.9}})^2-2(\frac{1}{e^{0.9}}) \cos(w)} * \frac{\sin\left(\frac{11w}{2}\right)}{\sin\left(\frac{w}{2}\right)} \end{gather*} in which * represents convolution so as you see I shouldn't get linear phase because both sides of the convolution doesn't even have phase. However, when I draw it in Matlab I get linear phase and I'm stuck. I don't know why.

x = @(t) 2*exp(-0.9 .* abs(t)).* ((t<= 5) & (t>= -5));
time_step =1;
t = -5:time_step:5;
yfft = fftshift(fft(x(t)));
f = linspace(-pi, pi, numel(yfft));
plot(f,unwrap(angle(yfft)));
xlim([-pi,pi]);

I also added N number to fft so it pad the function by zero but result is same as before.  

The plot is like above and phase is linear which it shouldn't be.

I'm trying to plot the phase of Fourier transform of the function below: 

\begin{gather*} x[n] = 2e^{-0.9|n|} \: \: n \in [-5,5]\\ x[n] = 0 \:\: n \notin [-5,5] \end{gather*}

which is equal to the function below:

\begin{gather*} x[n] = 2e^{-0.9|n|} \times \mathrm{rect}(5) \end{gather*}

so I calculated the Fourier transform like below: 

\begin{gather*} x[n] = 2e^{-0.9|n|} \times \mathrm{rect}(5)\\ y[n] = \alpha^{|n|} \:\: |\alpha |< 1 \\ Y[e^{jw}] = \frac{1-\alpha^2}{1+\alpha^2-2\alpha \cos(w)}\\ \longrightarrow x[n] = 2(\frac{1}{e^{0.9}})^{|n|} \longrightarrow\\ X[e^{jw}] = \frac{1}{2\pi} 2 \frac{1-(\frac{1}{e^{0.9}})^2}{1+(\frac{1}{e^{0.9}})^2-2(\frac{1}{e^{0.9}}) \cos(w)} * \frac{\sin\left(w(\frac{1}{2}+5\right)}{\sin\left(\frac{w}{2}\right)}\\ X[e^{jw}] = \frac{1}{\pi} \frac{1-(\frac{1}{e^{0.9}})^2}{1+(\frac{1}{e^{0.9}})^2-2(\frac{1}{e^{0.9}}) \cos(w)} * \frac{\sin\left(\frac{11w}{2}\right)}{\sin\left(\frac{w}{2}\right)} \end{gather*}

in which * represents a convolution. As you see I shouldn't get linear phase because both sides of the convolution don't even have phase. However, when I draw it in Matlab I get linear phase and I'm stuck. I don't know why.

x = @(t) 2*exp(-0.9 .* abs(t)).* ((t<= 5) & (t>= -5));
time_step =1;
t = -5:time_step:5;
yfft = fftshift(fft(x(t)));
f = linspace(-pi, pi, numel(yfft));
plot(f,unwrap(angle(yfft)));
xlim([-pi,pi]);

I also added N number to FFT so it pads the function with zero, but the result is same as before. 

The plot is like above and phase is linear which it shouldn't be.

I'm trying to plot Phase of Fourier transform of function bellow : \begin{gather*} x[n] = 2e^{-0.9|n|} \: \: n \in [-5,5]\\ x[n] = 0 \:\: n \notin [-5,5] \end{gather*} which is equal to function bellow:

\begin{gather*} x[n] = 2e^{-0.9|n|} \times rect(5) \end{gather*} so I calculated Fourier transform like below: \begin{gather*} x[n] = 2e^{-0.9|n|} \times rect(5)\\ y[n] = \alpha^{|n|} \:\: |\alpha |< 1 \\ Y[e^{jw}] = \frac{1-\alpha^2}{1+\alpha^2-2\alpha cos(w)}\\ \longrightarrow x[n] = 2(\frac{1}{e^{0.9}})^{|n|} \longrightarrow\\ X[e^{jw}] = \frac{1}{2\pi} 2 \frac{1-(\frac{1}{e^{0.9}})^2}{1+(\frac{1}{e^{0.9}})^2-2(\frac{1}{e^{0.9}}) cos(w)} * \frac{sin(w(\frac{1}{2}+5)}{sin(\frac{w}{2})}\\ X[e^{jw}] = \frac{1}{\pi} \frac{1-(\frac{1}{e^{0.9}})^2}{1+(\frac{1}{e^{0.9}})^2-2(\frac{1}{e^{0.9}}) cos(w)} * \frac{sin(\frac{11w}{2})}{sin(\frac{w}{2})} \end{gather*} in which * represents convolution so as you see I shouldn't get linear phase because both side of convolution doesn't have even phase but as i draw it in matlab i get linear phase and I'm stuck. I don't know why? $$ $$ can any one help me?

x = @(t) 2*exp(-0.9 .* abs(t)).* ((t<= 5) & (t>= -5));
time_step =1;
t = -5:time_step:5;
yfft = fftshift(fft(x(t)));
f = linspace(-pi, pi, numel(yfft));
plot(f,unwrap(angle(yfft)));
xlim([-pi,pi]);

I also added N number to fft so it pad the function by zero but result is same as before. the plot is like above and phase is linear which it shouldn't be.

I'm trying to plot the phase of Fourier transform of the function below: \begin{gather*} x[n] = 2e^{-0.9|n|} \: \: n \in [-5,5]\\ x[n] = 0 \:\: n \notin [-5,5] \end{gather*}

which is equal to the function below:

\begin{gather*} x[n] = 2e^{-0.9|n|} \times \mathrm{rect}(5) \end{gather*} so I calculated Fourier transform like below: \begin{gather*} x[n] = 2e^{-0.9|n|} \times \mathrm{rect}(5)\\ y[n] = \alpha^{|n|} \:\: |\alpha |< 1 \\ Y[e^{jw}] = \frac{1-\alpha^2}{1+\alpha^2-2\alpha \cos(w)}\\ \longrightarrow x[n] = 2(\frac{1}{e^{0.9}})^{|n|} \longrightarrow\\ X[e^{jw}] = \frac{1}{2\pi} 2 \frac{1-(\frac{1}{e^{0.9}})^2}{1+(\frac{1}{e^{0.9}})^2-2(\frac{1}{e^{0.9}}) \cos(w)} * \frac{\sin\left(w(\frac{1}{2}+5\right)}{\sin\left(\frac{w}{2}\right)}\\ X[e^{jw}] = \frac{1}{\pi} \frac{1-(\frac{1}{e^{0.9}})^2}{1+(\frac{1}{e^{0.9}})^2-2(\frac{1}{e^{0.9}}) \cos(w)} * \frac{\sin\left(\frac{11w}{2}\right)}{\sin\left(\frac{w}{2}\right)} \end{gather*} in which * represents convolution so as you see I shouldn't get linear phase because both sides of the convolution doesn't even have phase. However, when I draw it in Matlab I get linear phase and I'm stuck. I don't know why.

x = @(t) 2*exp(-0.9 .* abs(t)).* ((t<= 5) & (t>= -5));
time_step =1;
t = -5:time_step:5;
yfft = fftshift(fft(x(t)));
f = linspace(-pi, pi, numel(yfft));
plot(f,unwrap(angle(yfft)));
xlim([-pi,pi]);

I also added N number to fft so it pad the function by zero but result is same as before.

The plot is like above and phase is linear which it shouldn't be.