RE: A voice signal sample is stored as one byte. Frequency range is 16 Hz to 20 Hz. What is the memory s...
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
RE: A voice signal sample is stored as one byte. Frequ...
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.
RE: A voice signal sample is stored as one byte. Frequ...
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.
RE: A voice signal sample is stored as one byte. Frequ...
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
RE: 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?
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 .
