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The system is not time invariant
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(a)The system is time invariant
  
 
Let time delay = t
 
Let time delay = t
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equals to Yk[n]=<math>(k+1)^2</math> &delta;[n-(k+1)]->time delay
 
equals to Yk[n]=<math>(k+1)^2</math> &delta;[n-(k+1)]->time delay
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 +
equals to Y(k+t)[n]=<math>(k+t+1)^2</math> &delta;[n-(k+t+1)]
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yields the same result
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----
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(b)
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According to row 1 of the table, input x[n]=u[n] yields the output Y[n]=u[n-1].

Latest revision as of 15:42, 10 September 2008

(a)The system is time invariant

Let time delay = t

then

Xk[n]=δ[n-k] -> time delay -> system

equals to X(k+t)[n]=δ[n-(k+t)]->system

equals to X(k+t)[n]=δ[n-k-t]->system

equals to Y(k+t)[n]=$ (k+t+1)^2 $ δ[n-(k+t+1)]

while

Xk[n]=δ[n-k] -> system -> time delay

equals to Yk[n]=$ (k+1)^2 $ δ[n-(k+1)]->time delay

equals to Y(k+t)[n]=$ (k+t+1)^2 $ δ[n-(k+t+1)]

yields the same result


(b)

According to row 1 of the table, input x[n]=u[n] yields the output Y[n]=u[n-1].

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang