Why does my windowed-sinc function have non-linear phase?

Question

Similar to many tutorials on the web, I've tried implementing a windowed-sinc lowpass filter using the following python functions:

def black_wind(w):
''' blackman window of width w'''
    samps = np.arange(w)
    return (0.42  - 0.5 * np.cos(2 * np.pi * samps/ (w-1)) + 0.08 * np.cos(4 * np.pi * samps/ (w-1)))

def lp_win_sinc(tw, fc, n):
''' lowpass sinc impulse response 
Parameters:
    tw = approximate transition width [fraction of nyquist freq]
    fc = cutoff freq [fraction of nyquest freq]
    n = length of output. 
Returns:
    s = impulse response of windowed-sinc filter appended zero-padding
    to make len(s) = n
'''
    m = int(np.ceil( 4./tw / 2) * 2) 
    samps = np.arange(m+1)
    shift = samps - m/2
    shift[m/2] = 1
    h = np.sin(2 * np.pi * fc * shift)/shift
    h[m/2] = 2 * np.pi * fc
    h = h * black_wind(m+1)
    h = h / h.sum()
    s = np.zeros(n)
    s[:len(h)] = h
    return s

For input: 'tw = 0.05', 'fc = 0.2', 'n = 6000', the magnitude of the fft seems reasonable.

tw = 0.05
fc = 0.2
n = 6000
lp = lp_win_sinc(tw, fc, n)
f_lp = np.fft.rfft(lp)
plt.figure()
x = np.linspace(0, 0.5, len(f_lp))
plt.plot(x, np.abs(f_lp))

magnitude of lowpass filter response

however, the phase is non-linear above ~fc.

plt.figure()
x = np.linspace(0, 0.5, len(f_lp))
plt.plot(x, np.unwrap(np.angle(f_lp)))

phase of lowpass filter response

Given the symmetry of the non-zero-padded portion of the impulse response, I would expect the resulting phase to be linear. Can someone explain what is going on? Perhaps I'm using a numpy function incorrectly, or maybe my expectations are incorrect. I'm very grateful for any help.

***********************EDIT***********************

based on some of the helpful comments to this question and some more work, I wrote a function that produces zero phase delay and is therefore a bit easier to interpret the np.angle() results.

def lp_win_sinc(tw, fc, n):
    m = int(np.ceil( 2./tw) * 2) 
    samps = np.arange(m+1)
    shift = samps - m/2
    shift[m/2] = 1
    h = np.sin(2 * np.pi * fc * shift)/shift
    h[m/2] = 2 * np.pi * fc
    h = h * np.blackman(m+1)
    h = h / h.sum()
    s = np.zeros(n)
    s[:len(h)] = h
    return np.roll(s, -m/2)

The main change here is using np.roll() to place the line of symmetry at t=0.

Answer 1

The magnitudes in the stop band are crossing zero. The phase of the coefficient after a zero crossing will jump by 180 degrees, which is confusing np.angle()/np.unwrap(). -1*180° = 1*0°

Answer 2

The phase as shown in your graph is in fact linear. It's a constant slope in the passband, corresponding to a constant delay in the time domain. It's a much steeper slope, which renders as wrapping around at 2pi boundaries, in the stopband. But the value of the phase in the stopband is not particularly important since those frequencies aren't going to come through the filter anyway.