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    Two signals need to be time division multiplexed. 
1) cos(2pi 10^6 t) 
2) cos(2pi 4000t) 
Both are flat top sampled with a sampling freq of "double the Nyquist rate". 

Then they are encoded with 4 bit codes. The bit dureation is 1microsecond. 
For synchronisation 1 bit (1 microsecond) is added. 

The time duration between two bits is 
a) 9s b) 9 microsecond c) 0.9 microsecond d) none.. 

The bit rate of the TDM is 
a) 96 kbps b) 9.6 kbps c) 9 kbps d) 1000 bps.. 

jus not able to understand the problem..  

Author	Message
manju Senior Member Joined: 20Feb2007 Online Status: Offline Posts: 221	Topic: TDM Posted: 22Feb2007 at 2:37pm
	Two signals need to be time division multiplexed. 1) cos(2pi 10^6 t) 2) cos(2pi 4000t) Both are flat top sampled with a sampling freq of "double the Nyquist rate". Then they are encoded with 4 bit codes. The bit dureation is 1microsecond. For synchronisation 1 bit (1 microsecond) is added. The time duration between two bits is a) 9s b) 9 microsecond c) 0.9 microsecond d) none.. The bit rate of the TDM is a) 96 kbps b) 9.6 kbps c) 9 kbps d) 1000 bps.. jus not able to understand the problem.. Post Resume: Click here to Upload your Resume & Apply for Jobs

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