SOLUTIONS

INEL 5309- 036 – DSP I – QUIZ ONE
Prof. Domingo Rodríguez – Thursday, September 18, 2014
SOLUTIONS
Problem 1a (40 Points): Linear Convolution
The output signal of the FIR filter shown in Figure 1 below is given by the
formula y[n ]  h[0]x[n ]  h[1]x[n  1]  h[2]x[n  2] . Compute the output signal
using LINEAR convolution if the input is x[n ]   1,1.
Figure 1
Problem 1a (40 Points): Solution
The equation of the system is described by the formula above:
y[n ]  h[0] x[n ]  h[1] x[n  1]  h[2] x[n  2]
If the input
x[n ] is set to  [n ] , then the output y[n ] becomes h[n ] :
h[n ]  h[0] [n ]  h[1] [n  1]  h[2] [n  2]
From Figure 1 given above, the values of h[n ] are obtained:
h[0]  1, h[1]  1, h[2]  2
INEL 5309- 036 – DSP I – QUIZ ONE
Prof. Domingo Rodríguez – Thursday, September 18, 2014
SOLUTIONS
A first, efficient, approach is to use a matrix-vector computation:
 y[0]  h[0] 0 
  1
1 0 
 y[1]   h[1] h[0] x[0]   1  1  1  2
 



    

 y[2] h[2] h[1]   x[1]    2  1    1   3
 y[3]  0


  2
h[2]

 
 0  2
 
A second approach is to use direct linear convolution operation:
1
y[n ]   x[k ]h[n  k ]  x[0]h[n ]  x[1]h[n  1]; n  Z 4
k 0
y[0]  x[0]h[0]  x[1]h[ 1]  x[0]h[0]
y[1]  x[0]h[1]  x[1]h[0]
y[2]  x[0]h[2]  x[1]h[1]
y[3]  x[0]h[3]  x[1]h[2]  x[1]h[2]
y[0]  x[0]h[0]  ( 1)( 1)  1
y[1]  x[0]h[1]  x[1]h[0]  ( 1)( 1)  ( 1)( 1)  2
y[2]  x[0]h[2]  x[1]h[1]  ( 1)( 2)  ( 1)( 1)  3
y[3]  x[0]h[3]  x[1]h[2]  x[1]h[2]  ( 1)( 2)  2
INEL 5309- 036 – DSP I – QUIZ ONE
Prof. Domingo Rodríguez – Thursday, September 18, 2014
SOLUTIONS
Problem 1a (60 Points): Linear & TI Properties
Determine whether or not the system T x[n ]  n 2 x[n  nd ] is linear and/or TI.


A system y[n ]  T x[n ] is said to be a linear system if it satisfies the
following condition or identity known as the linearity property:
T ax1 [n ]  bx 2 [n ]  aT x1 [n ]  bT x 2 [n ]
The equation of the system is given by
y[n ]  T x[n ]  n 2 x[n  nd ]
Computing the r.h.s.:
aT x1 [n ]  an 2 x1 [n  nd ] ;
bT x2 [n ]  bn 2 x 2 [n  nd ]
aT x1 [n ]  bT x2 [n ]  an 2 x1 [n  nd ]  bn 2 x2 [n  nd ]
Computing the l.h.s.:
Using an intermediate step:
Let
s[n ]  ax1 [n ]  bx 2 [n ] .
Think of this signal as a new signal.
T s[n ]  n 2 s[n  nd ]
Substituting for the original sum
s[n ]  ax1 [n ]  bx 2 [n ] , we get:
T ax1 [n ]  bx 2 [n ]  n 2 ax1 [n  nd ]  bx 2 [n  nd ]
T ax1 [n ]  bx2 [n ]  n 2 ax1 [n  nd ]  n 2 bx2 [n  nd ]
Comparing the r.h.s. and the l.h.s., we conclude that the system is linear.
INEL 5309- 036 – DSP I – QUIZ ONE
Prof. Domingo Rodríguez – Thursday, September 18, 2014

SOLUTIONS

A system y[n ]  T x[n ] is said to be time invariant if it satisfies the
following condition or identity known as the time invariant property:
T x[n  n0 ]  y[n  n0 ]
Computing the r.h.s.:
y[n  n0 ]  ( n  n0 ) 2 x[n  n0  nd ] ;
Computing the l.h.s.:
Let
s[n ]  x[n  n0 ] ;
T s[n ]  n 2 s[n  nd ]
Substituting for the original delay
s[n ]  x[n  n0 ] , we get:
T x[n  n0 ]]  n 2 x[n  nd  n0 ]
Comparing the r.h.s. and the l.h.s., we conclude that the system IS NOT TI.

Download Report