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Unformatted text preview: = m =n n ∑ x[m] m =−∞ Homogeneity:
Let x1[ n ] = g[ n ] . Then y1[ n ] = n ∑ g[m] m =−∞ Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] = n n m =−∞ m =−∞ ∑ K g[m] = K ∑ g[m] = K y [n] . Homogeneous.
Let x1[ n ] = g[ n ] . Then y1[ n ] = n ∑ g[m] m =−∞ Solutions 36 1 M. J. Roberts  8/16/04 Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] =
Let x 3 [ n ] = g[ n ] + h[ n ] .
Then y 3 [ n ] = n ∑ m =−∞ n ∑ h[m] m =−∞ (g[m] + h[m]) = n ∑ m =−∞ g[ m] + n ∑ h[m] = y [n] + y [n] .
1 m =−∞ 2 Additive.
Since the system is homogeneous and additive it is also linear.
The system is also incrementally linear because it is linear.
The system is not statically nonlinear because it is linear.
Time Invariance:
Let x1[ n ] = g[ n ] . Then y1[ n ] = n ∑ g[m] . m =−∞ Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] = n ∑ g[m − n ] .
0 m =−∞ The first equation can be rewritten as
y1[ n − n 0 ] =
Let m = q − n 0 . Then
y1[ n − n 0 ] = n − n0 ∑ g[m] m =−∞ n ∑ g[q − n ] = y [n] q =−∞ 0 2 Time invariant
Stability:
If the excitation is a constant, the response increases without bound.
n
Also the solution of the homogeneous difference equation is y h [ n ] = K (1) = K . Therefore
the eigenvalue is 1 whose magnitude is not less than 1 and the system must be BIBO
unstable.
Unstable
Causality:
At any discrete time, n = n 0 , the response depends only on the excitation at that discrete time
and previous discrete times.
Causal.
Memory:
At any discrete time, n = n 0 , the response depends on the excitation at that discrete time and
previous discrete times.
System has memory.
Invertibility:
Inverting the functional relationship, Solutions 37 M. J. Roberts  8/16/04 y[ n ] = n ∑ x[m] . m =−∞ Invertible.
Taking the first backward difference of both sides of the original system equation,
y[ n ] − y[ n − 1] = n n −1 m =−∞ m =−∞ ∑ x[m] − ∑ x[m] x[ n ] = y[ n ] − y[ n − 1]
The excitation is uniquely determined by the response.
Invertible.
10. Show that a system with excitation, x[ n ] , and response, y[ n ] , described by
y[ n ] = rect ( x[ n ]) ,
is nonlinear, time invariant and noninvertible.
11. Show that the system of Figure E11 is nonlinear, timeinvariant, static and invertible. 5
x[n] y[n] 10
Figure E11 A DT system
y[ n ] = 10 x[ n ] − 5 , The system is incrementally linear because the only deviation from linearity is caused by the
presence of the nonzero, zeroexcitation response.
Invertibility:
Solving the system equation for the excitation as a function of the response,
x[ n ] =
Invertible. y[ n ] + 5
10 12. Show that the system of Figure E12 is timeinvariant, BIBO stable, and causal. Solutions 38 M. J. Roberts  8/16/04 x[n] y[n] 1
4 D D y[n2] 1
4
1
2
Figure E12 A DT system
Homogeneity:
Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ]
Let x 2 [ n ] = K g[ n ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = K g[ n ]
Multiply the first equation by K. 4...
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 Spring '09
 ATOUSA

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