**rakhel(mnnit)**
Answered On : Jul 16th, 2005

Its 9600 bytes. as the voice frequency irange is in between 16-20,so sampling frequency must be grater than 40hz,so number of samples in 4min=40*240=9600 samples hemce we need a memory of 9600 bytes to store the voice signal

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**vignyan**
Answered On : Aug 22nd, 2005

Its just 1440 bytes. Bandpass sampling frequency is just 6Hz. so just 6*240 bytes.

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**raj**
Answered On : Feb 28th, 2006

Would u plz explain how it is 1440 bytes??????????????

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**prem kiran**
Answered On : Aug 1st, 2006

since the signal is in the band 16-20 HZ . bandwidth is 4HZ and so 8HZ sampling rate enough..i.e 8 s/sec. 4min=240sec.4min duration has 8*240 samplesand each sample has 8 bits ,so min required memory is 8*8*240=1920 bytes

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**Sundeep Borra**
Answered On : Aug 8th, 2006

It's 1920Bytes. As the bandwidth (f2-f1) is 20-16=4Hz. The sampling frequency should be twice the bandwidth (2B). the sampling frequency is 8Hz= 8 samples= 8 cycles/second. 4 minutes=(4*60) =240 Seconds1sample->1 Byte1Second->8 Samples240 Seconds->(240*8) = 1920 Samples

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**krishna**
Answered On : Dec 30th, 2006

Band width : 4Hzsampling freq : 2*4(=8)Hz or above4 minutes = 4*60secs = 240 secsnumber of byte required minimum = 240*8=1920 bytes

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**janani**
Answered On : May 16th, 2007

Two things i wish to point here.

1) They have said that they are representing each sample by 8 bits. which implies that the sampling frequency is quite high. and that for one cycle of the minimum freq we get 256 samples.

2) Regarding the criterion that the sampling frequency should be twice of the bandwidth. this is applicable only when we are dealing with base band signal. whereas here it is pass band (since 16 to 20 is significant in our case).

So the sampling frequency should be quite high considering the two facts.

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**Faheem**
Answered On : Jul 12th, 2007

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Voice sample = 8 bitSmaple frequency = 1/20 s = 0.05sfor 4 minutes = 4*60(minto sec)/0.05 = 4800 samples0 total memory = 8 * 4800 = 38400 bits = 4800 Bytes = 4.69 KB

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Banwidth of signal=4Hzaccording to niquist theorem sampling frequency>=2*bandwidthso sampling freq=2*4Hz(8 samples /sec)1 samples is stored as 1 byteso 8*8 bits/secso for 4 mins bits are = (4*60)*(64)=15KBso asn is 15 KB .

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Voice signal is sampled at 8000 samples per second. (Human voice is in the range of 300Hz-3400Hz, and according to nyquist sampling rate we take it more than double the rate).For a 4 min. voice signal, we have Memory = 4*60*8000 Bytes = 1875KB = 1.83 MB.

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**adi**
Answered On : Sep 30th, 2012

Bt practically i required 1.07 mb to record 1.31 min speech,

any1 kno its calculation?

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