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posted by  RainyHeaven on 10/4/2008 4:10:26 PM  |  status: Closed  

Questions on general properties of systems.

Course Textbook Chapter Problem
Signal Theory Signals and System 2nd Edition N/A N/A
Question Details:
i have to determine which properties hold and which do not hold for each of the following continuous-time systems. and i've to justify my answer...
please help me.
 
properties: memoryless, time invariant, linear, causal, stable.
 
a) y(t) = x(t-2) + x(2-t)
b) y(t) = [cos(3t)]x(t)
 
 
 
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posted by Avana(EE-SME) on 10/6/2008 5:55:06 AM  |  status: Live
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RainyHeaven's comment:
"Ty so much~!"
Response Details:
Memoryless :-
   If the output at any time depends only on the input at that same time then the system is memoryless.
a)      Given continuous-time system is y(t) = x(t-2)+x(2-t)
   Therefore, the given system with MEMORYLESS as it at any time depends only on input.
b)      Given continuous-time system is y(t) = cos(3t)x(t)
   Therefore, the given system with MEMORY.
Causal :-
   
The output of a causal system at the present time depends on only present and/or past values of input, not on its future values. 
a)      Given continuous-time system is y(t) = x(t-2)+x(2-t)
         
         
   Thus , the given system is NON-CAUSAL it depends on future values.
b)      Given continuous-time system is y(t) = cos(3t)x(t)
         
         
   Thus , the given system is CAUSAL as it depends on present values.  
Linear :-
    condition for linear.
a)      Given continuous-time system is y(t) = x(t-2)+x(2-t)
      
      
      
   Hence,
   Therefore , the given system is LINEAR.
b)      Given continuous-time system is y(t) = cos(3t)x(t)
      
      
      
   Hence,
   Therefore , the given system is LINEAR.
Stable :-
   If a system is bounded-input/bounded output is stable.
a)      Given continuous-time system is y(t) = x(t-2)+x(2-t)
      y(t), x(t) are bounded  between -2 and 2.
      Therefore , given system is STABLE.
b)      Given continuous-time system is y(t) = cos(3t)x(t)
      y(t) and x(t) are not bounded.
   Therefore , the given system is UN STABLE. 
 
 
Hope this helps you..............
 
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