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Waveforms [4 marks] Given that the sampling rate is 44100 Hz, what is the Nyquist frequency? (2 marks) Answer: 22050 Hz A tone at 12000 Hz is above the Nyquist frequency but below the sampling rate. It aliases to sound like 6000 Hz. What is the sampling rate? (2 marks) Answer: sample rate is 18000 Hz 3. Pitch & Tempo [12 marks] Given that the frequency of the note E4 is 329 Hz, what are the frequencies of E5 and E2? (4 marks) Answer: E5 = 659 Hz (within 2 Hz is acceptable) E2 = 82 Hz Given a Csound score consisting of 80 beats, please write a tempo statement that delivers a constant tempo and the resulting audio signal should last for exactly 50 seconds. (2 marks) Answer: t0 96 (zero marks for non-constant tempo) Given the following tempo statement, t0 60 8 120 what are the instantaneous tempos at the 5th and 10th beat? (4 marks) Answer: 5th beat = 97.5 (60 + (120 - 60) * (5 / 8), error of 2 Hz is acceptable) 10th beat = 120 Write down a tempo statement which sustains a tempo of 60 bpm for 5 seconds and suddenly changes to 96 bpm at the 5th second. (2 marks) Answer: t0 60 5 60 5 96 (deduce 1 mark for 1 mistake) 4. Additive Synthesis [15 marks] If the fundamental frequency is 300 Hz, what is the frequency of the 3rd harmonic? (2 marks) Answer: 900 Hz For an instrument tone, we know that 2 harmonic frequencies are 300 Hz and 500 Hz, but we do not know their harmonic numbers. What is the greatest possible fundamental frequency? (2 marks) Answer: 100 Hz (GCD of 300 and 500) Fill in the blanks in the following Csound orchestra file which generate a sine wave with simple attack-decay amplitude envelope. (6 marks) ; sinewave.orc in just intonation ; 1st instrument sine wave sr = 22050 kr = 2205 ksmps = 10 nchnls = 1 ;----------------------------------------------------- instr 1 ; sine wave idur = p3 iamp = p4 ; p4 controls the amplitude ifreq = p5 ; set tuning ratio in Hertz iattack = p6 ; attack time idecay = p7 ; decay time isus = idur - iattack - idecay iwave = 1 aenv linseg ___________________________________ asig oscili iamp,ifreq,iwave ; signal out ____________________________ ; output endin Answers: 0,iattack,1,isus,1,idecay,0 (4 marks) asig * aenv (2 marks)  The following orchestra (add.orc) and score (add.sco) files synthesize a sound with fundamental frequency at 200 Hz using additive synthesis of 5 harmonics. Look at the following spectrum for the individual amplitudes of each harmonic. Fill in the missing fields in the score file (within an accuracy of 0.3). (5 marks, 0.5 marks for each answer)  ; add.orc sr = 22050 kr = 2205 ksmps = 10 nchnls = 1 ;------------------------------------------------------ instr 1 ; sine wave idur = p3 iamp = p4 ; p4 controls the amplitude ifreq = p5 * 200 ; set tuning ratio in Hertz iwave = 1 asig oscili iamp, ifreq, iwave ; signal out asig ; output endin ;------------------------------------------------------  ; add.sco f1 0 16385 10 1 ; start dur amp pitch i1 1 2 ____ _____ ;fundamental i1 1 2 ____ _____ ;2nd harmonic i1 1 2 ____ _____ ;3rd harmonic i1 1 2 ____ _____ ;4th harmonic i1 1 2 ____ _____ ;5th harmonic end Answer: (0.5 marks for each answer, amp corr. 0.3) i1 1 2 3.8 1 i1 1 2 1.5 2 i1 1 2 4.7 3 i1 1 2 3.8 4 i1 1 2 5 5  5. Wavetable Synthesis [14 marks] Write down the wavetable statement for a sine wave of table length being 16385 starting at time 0. (2 marks) f1 Answer: f1 0 16385 10 1 (0.5 mark for each, -10 or 10 is both OK) The following orchestra (wav.orc) and score (wav.sco) files synthesize a sound of 500 Hz with 4 harmonics using 2 wavetables together. The table size is 4096. We only have the following data for the amplitudes for the two wavetables, but they should sound similarly loud in the synthesis. Fill in the missing values in the wavetable statements in the score file. (12 marks, 1 mark for each answer) Harmonic No.Amplitude for f1Amplitude for f2153110.9522231338860.6449610.44 ; wav.orc sr = 22050 kr = 2205 ksmps = 10 nchnls = 1 ;------------------------------------------------------ instr 1 idur = p3 iamp = p4 ; p4 controls the amplitude ifreq = p5 ; set tuning ratio in Hertz iattack = p6 ; attack time idecay = p7 ; decay time isus = idur - iattack idecay aenv linseg 0,iattack,iamp,isus,iamp,idecay,0 asig1 oscili aenv, ifreq, 1 ; signal 1 asig2 oscili aenv, ifreq, 2 ; signal 2 aout = asig1 + asig2 abal balance aout, asig1 out abal endin ;------------------------------------------------------  ; wav.sco ; ****** define 2 wavetables here ****** ____ 0 4096 _______ _______ _______ _______ _______ ____ 0 4096 _______ _______ _______ _______ _______ ; start dur amp freq attack decay i1 1 2 10000 500 .02 .05 end Answer: (1 mark for each answer) f1 0 4096 10 5311 223 3886 961 (must be normalized) (f1 0 4096 10 1 0.04 0.73 0.18) f2 0 4096 10 0.95 1 0.64 0.44 (as long as ratio is correct is OK)  6. Vibrato & FM Synthesis [12 marks] What vibrato rate would make a vibrato become a FM synthesis? (2 marks) Answer: 20 Hz Present the modulation index in terms of vibrato width and modulation frequency. (2 marks) Answer: modulation index = vibrato width / modulation frequency Given that the carrier frequency is 400 Hz and the modulation frequency is 600 Hz, what is the fundamental frequency? (2 marks) Answer: 200 Hz (GCD of 400 and 600) The following orchestra file generates FM instrument tones. Please fill in the blanks to complete the orchestra file. (6 marks) instr 1 ; clarinet and brass tones idur = p3 iamp = p4 icarfr = p5 ; carrier frequency imodfr = p6 ; modulator frequency index = p7 ; modulation index iwave = 1 ; wavetable 1 (sine wave) aenv linseg 0, .1, iamp, idur-.3, iamp, .2, 0 amod oscili _____________, ____________, iwave acar oscili _____________, ____________, iwave out _____________ endin Answers: amod oscili index*imodfr, imodfr,(2 marks) iwave acar oscili aenv, icarfr+amod,(2 marks) iwave out acar (2 marks)  7. FM Spectra [6 marks] In real life, negative amplitudes or negative frequencies are not possible for a sound. However, negative values often arise in FM spectra. What do negative amplitudes and negative frequencies mean respectively? (4 marks) Answer: Negative amplitudes 180 degree phase shift, signal weaker Negative frequencies folds up to corresponding positive frequencies For a simple formant FM instrument with carrier frequency set to the 6th harmonic, which sidebands contribute to the amplitude of the 10th harmonic? (2 marks) Answer: ak = J(k-nc)(I) - J-(k+nc)(I) k = 10, nc = 6 k-nc = 4 k+nc = 16 Sidebands 4 and -16 contribute to the amplitude of the 10th harmonic. 8. Voices [8 marks] What is the decibel level of a sound with an amplitude of 10000? Given that the reference amplitude is 1000. (2 marks) Answer: 20 dB Given that the amplitude of a sound is 20000 and the decibel level is 40 dB. What is the reference amplitude? (2 marks) Answer: 200 If we increase the decibel level of a sound by 1 dB, how many times will be the amplitude increased? (2 marks) Answer: 1.12 times (10 ^ (1/20)) If we increase the amplitude of a sound by 2 times, how many decibels will be the decibel level increased? (2 marks) Answer: 6.02 dB (20log(2)) 9. Effects & Filters [16 marks] What type of filter would be used for each of the following cases? (4 marks) Brightening the signal Separating the harmonics Removing hissing noise Removing the electric hum at 50Hz Answer: High-pass filter Band-pass filter Low-pass filter Notch/band-stop filter Assuming that the full amplitude is 1, what is the amplitude at the cut-off frequency after applying a low-pass filter? Show clearly how you arrive at your answer. (2 marks) Answer: At the cut-off frequency the power is halved. power = amp2; amp = sqrt(power) 1/2 power = sqrt(1/2) amp = 70.7% amp The following orchestra (filter.orc) and score (filter.sco) files take a 12-second long wave file (Rondeau.wav) as input. Originally a band-pass filter with dynamically changing center frequency and bandwidth was applied, but now we want to implement an equivalent low-pass filter followed by a high-pass filter. Fill in the blanks in the following orchestra file to implement the two filters which should result in a similar resultant signal as the band-pass filter. (10 marks, 1 mark for each answer) ;filter.orc sr = 22050 kr = 2205 ksmps = 10 nchnls = 1 ;------------------------------------------------------ instr 1 idur = p3 asig soundin "Rondeau.wav" ;*************************************************** ; Deleted lines of band-pass filter ; kbw linseg 100, idur, 1900 ; kfiltfr linseg 1950, idur, 1050 ; afilt reson asig, kfiltfr, kbw ;*************************************************** kfiltfr linseg _______, _______, _______ afilt1 _______ _______, _______ ;low-pass filter afilt2 _______ _______, _______ ;high-pass filter abal balance _______, asig out abal ; output endin ;------------------------------------------------------ Answer: kfiltfr linseg 1900, idur, 100 afilt1 tone asig, 2000 (no marks if filter is wrong) afilt2 atone afilt1, kfiltfr (no marks if filter is wrong) abal balance afilt2, asig  ;filter.sco ; start dur i1 0 12 end 10. Comb Filters & All-Pass Filters [4 marks] The following is a Csound statement applying a comb filter, acomb comb gacomb, iring, iloop Write down an equivalent Csound statement using an all-pass filter. (2 marks) Answer: acomb alpass gacomb, iring, iloop (zero marks for wrong opcode) For applying a comb filter using the statement in (a), we need to extend the duration (p3) of the note. Which variable (gacomb, iring or iloop) is related to the extension of the duration of the note? (2 marks) Answer: iring 11. Reverb [4 marks] In a room of length 15m (ignore the width), how long a distance will a sound wave travel between its being produced and absorbed if the reverberation time is 1.3s? (Given the sound speed in the air is 350m/s) How many times approximately does the sound bounce before being absorbed? (2 marks) Answer: Distance: 350 * 1.3 = 455m Times: 455 / 15 = ~30 times Which two types of filters does the Schroeder reverberator include? 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