Samar Qureshi

1009149583

1. Extract the two channels and compute $P_1$ and $P_2$

P1 (Channel 1):           -13.79 dB
P2 (Channel 2):           -13.79 dB
P2 relative to P1:        0.00 dB -> same average power 

The powers are negative because for audio signals normalized between -1 and 1, the average power is less than 1, leading to a negative dB value. My original .wav file was mono, so I added code to duplicate it into two identical channels, so the power sare identical, and they sound exactly the same

2. Form the sum of the channels & compute power

P(sum = ch1+ch2):         -7.77 dB
P(sum) relative to P1:    6.02 dB

Since the channels are identical, the observed +6.02dB difference is correct for summing two identical signals.

Sum becomes 2x mono signal, so it is visibly larger in amplitude.

3. Form the difference of the channels and compute power

P(diff = ch1-ch2):        -Inf dB
P(diff) relative to P1:   -Inf dB

Since both channels are identical, 10log(0) is infinity. The difference signal has zero power, since all samples are zero. The difference channel normally contains stereo information, but here it is exacly zero

4. FFT and 99% Power Bandwidth

--- 99% power bandwidth ---
99% power bandwidth ≈ 0.59 kHz = 590Hz

• 99% of the total signal power is contained between 0 and 590Hz
• That means the audio is heavily dominaed by low frequency content (i.e., bass and lower harmonics)

Spectrum is strong near low frequencies, then gradually rolls off. 99% of the energy is below ~600Hz.

5. Filter the signal to 10kHz, 5kHz, 3kHz, 1kHz

Filter(1:N) = 1;
Filter(Nb+2:end-Nb) = 0;