# A voice signal sample is stored as one byte. Frequency range is 16 Hz to 20 Hz. What is the memory size required to store 4 minutes voice signal?

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#### rakhel(mnnit)

• 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

#### vignyan

• Aug 22nd, 2005

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

#### raj

• Feb 28th, 2006

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

#### prem kiran

• 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 samples

and each sample has 8 bits ,so min required memory is 8*8*240=1920 bytes

#### Sundeep Borra

• 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 Seconds

1sample->1 Byte

1Second->8 Samples

240 Seconds->(240*8) = 1920 Samples

#### krishna

• 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

#### janani

• 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.

#### Faheem

• Jul 12th, 2007

Nyquist–Shannon sampling theorem states that perfect reconstruction of a signal is possible when the sampling frequency is greater than twice the maximum frequency of the signal being sampled...so imagine sampling a 16Hz signal with 8hz...there's gonna be a sample every 2 cycles!..it has to be 40

#### bexdeep Profile Answers by bexdeep

• Aug 17th, 2008

Voice sample = 8 bit
Smaple frequency = 1/20 s = 0.05s
for 4 minutes = 4*60(minto sec)/0.05 = 4800 sample
s0 total memory = 8 * 4800 = 38400 bits = 4800 Bytes = 4.69 KB

#### geekijk Profile Answers by geekijk

• Aug 28th, 2008

banwidth of signal=4Hz
according to niquist theorem sampling frequency>=2*bandwidth
so sampling freq=2*4Hz(8 samples /sec)
1 samples is stored as 1 byte
so 8*8 bits/sec
so for 4 mins bits are = (4*60)*(64)=15KB
so asn is 15 KB .

#### adityajain Profile Answers by adityajain

• Nov 21st, 2009

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.

• Sep 30th, 2012

bt practically i required 1.07 mb to record 1.31 min speech,
any1 kno its calculation?  