This preview shows page 1. Sign up to view the full content.
Unformatted text preview: K y1[ n ] − K y1[ n − 1] + 2K y1[ n − 2] = K g[ n ]
Then, equating results,
4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = 4 K y1[ n ] − K y1[ n − 1] + 2K y1[ n − 2]
If this equation is to be satisfied for all n,
y 2 [ n ] = K y1[ n ] .
Homogeneous.
Additivity:
Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ]
Let x 2 [ n ] = h[ n ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = h[ n ]
Let x 3 [ n ] = g[ n ] + h[ n ] . Then 4 y 3 [ n ] − y 3 [ n − 1] + 2 y 3 [ n − 2] = g[ n ] + h[ n ]
Add the two first two equations.
4 ( y1[ n ] + y 2 [ n ]) − ( y1[ n − 1] + y 2 [ n − 1]) + 2( y1[ n − 2] + y 2 [ n − 2]) = g[ n ] + h[ n ]
Then, equating results,
4 y 3 [ n ] − y 3 [ n − 1] + 2 y 3 [ n − 2] = 4 ( y1[ n ] + y 2 [ n ]) − ( y1[ n − 1] + y 2 [ n − 1]) + 2( y1[ n − 2] + y 2 [ n − 2]) If this equation is to be satisfied for any arbitrary excitation for all n,
Additive. y 3 [ n ] = y1[ n ] + y 2 [ n ]. Since the system is both homogeneous and additive, it is linear.
Since the system is linear it is also incrementally linear.
Since the system is linear, it is not statically nonlinear. Solutions 39 M. J. Roberts  8/16/04 Time Invariance:
Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ]
Let x 2 [ n ] = g[ n − n 0 ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = g[ n − n 0 ]
We can rewrite the first equation as
4 y1[ n − n 0 ] − y1[ n − n 0 − 1] + 2 y1[ n − n 0 − 2] = g[ n − n 0 ]
Then, equating results,
4 y1[ n − n 0 ] − y1[ n − n 0 − 1] + 2 y1[ n − n 0 − 2] = 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2]
If this equation is to be satisfied for any arbitrary excitation for all n, then
y 2 [ n ] = y1[ n − n 0 ] . Time Invariant. Stability:
The eigenvalues of the system homogeneous solution are found from the characteristic
equation,
4α 2 − α + 2 = 0 .
They are
α1,2 = 0.125 ± j 0.696 or α1,2 = 0.7071e ± j1.3931
Therefore the homogeneous solution is of the form,
y h [ n ] = K h1 ( 0.7071) e+ j1.3931n + K h 2 ( 0.7071) e− j1.3931n
n n and, as n approaches infinity the homogeneous solution approaches zero and the total
solution approaches the particular solution. The particular solution is bounded because it
consists of functions of the same form as the excitation and all its unique differences and the
excitation is bounded in the BIBO stability test. Therefore if x is bounded, so is y.
Stable.
Causality:
We can rearrange the system equation into
y[ n ] = 1
(x[n] + y[n − 1] + 2 y[n − 2])
4 showing that the response at time, n, depends on the excitation at time, n, and the response at
previous times. It does not depend on any future values of the excitation.
Causal.
Memory:
The response depends on past values of the response.
The system has memory.
Invertibility: Solutions 310 M. J. Roberts  8/16/04 The original system equation, 4 y[ n ] − y[ n − 1] + 2 y[ n − 2] = x[ n ], expresses the excitation in
terms of the response.
Invertible.
13. Find the impulse responses of these systems.
(a) y[ n ] = x[ n ] − x[ n − 1]
The impulse response is very easily found by direct iteration to be
h[ n ] = δ [ n ] − δ [ n − 1] .
Also, using linearity and superposition...
View
Full
Document
This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.
 Spring '09
 ATOUSA

Click to edit the document details