EE 413 : Digital Signal Processing
SOLUTION TO HOMEWORK # 1
Problem 1.(a) T x n( ( ))= −x n( )
Linear: T ax n( 1( )+bx n2( ))=ax1(− +n) bx2(− =n) aT x n( ( ))1 +bT bx n( 2( ))
Not Shift invariant: T x n n( ( − 0))= − −x n n( 0) y n n( − 0)= − +x n n( 0)
Stable: For x n( )M, n
( ( ))
( )
T x n
=
x n
−
M
Not Causal: For n < 0, it depends on the future values of x(n).
(b) T x n( ( )) ln( (= x n−2)) Not Linear: 1 2 1 2 1 2 1 2
(
( )
( )) ln(
(
2)
(
2))
( ( ))
(
( ))
ln( (
2))
ln( (
2))
T ax n
bx n
ax n
bx n
aT x n
bT bx n
a
x n
b
x n
+
=
− +
−
+
=
−
+
−
Shift invariant: T x n n( ( − 0)) ln( (= x n− −2 n0))=y n n( − 0)Not Stable: For x n( ) 0= M, n
( ( ))
T x n =
Causal: It does not depend on the future values.
(c) 0 ( ( )) k ( ) ( ) k T x n a x n k u n = =
− + Not Linear:
1 2 1 2 0 1 2 1 2 0 0 1 2 0 ( ( ) ( )) ( ) ( ) ( ) ( ( )) ( ( )) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) k k k k k k k k T ax n bx n a ax n k bx n k u n aT x n bT x n a a x n k au n b a x n k bu n a ax n k bx n k a b u n = = = = + = − + − + + = − + + − + = − + − + +
Not Shift invariant:
0 0 0 0 0 0 0 ( ( )) ( ) ( ) ( ) ( ) ( ) k k k k T x n n a x n n k u n y n n a x n n k u n n = = − = − − + − = − − + −
Stable: Forx n
( )
M
,
n
0 0 0 ( ) k k k k k k a x n k a M M a = = = − =
Therefore, the system is stable if 0 k k a =
.(d) ( ( )) ( ) ( ) m T x n x m n m =− =
− Linear:
1 2 1 2 1 2 1 2 1 2 1 2 ( ( ) ( )) ( ) ( ) ( ) ( ) ( ) ( ( )) ( ( )) ( ) ( ) ( ) ( ) ( ) ( ) m m m T ax n bx n ax m bx m n m ax n bx n aT x n bT x n a x m n m b x m n m ax n bx n =− =− =− + = + − = + = + = − + − = +
Shift invariant: ( ( 0)) ( ) ( ) ( 0) m T x n n x m n m y n n =− − =
− = − Stable: Forx n
( )
M
,
n
( ) ( ) m x m n m M =− −
Causal: It does not depend on the future values.
Problem 2.
Using Fourier series, we can find the Fourier series coefficients by 1 (2 / ) 0 1 ( ) N jk N n k n a x n e N − − = =
In this case, N = 6. We also have 5 0 0 1 ( ). 6n a x n =
=
But from Fact#2, 5 0 ( ) 2 n x n = =
0 1 3 a = When k=3, 3 5 5 0 0 1 1 ( ) ( )( 1) 6 6 j n n n n a x n e− x n = ==
=
− . But from Fact#3, 7 2 ( )( 1)n 1 n x n = − =
3 1 6 a = From Parseval’s theorem, the average power of x(n) is 5 20 . x k n P a =
=
To get the minimum power x(n), we set a1=a2=a4=a5= . 0Therefore, we have 0 1, 1 2 0, 3 1, 4 5 0
3 6
a = a =a = a = a =a = . Using Fourier series,
1 (2 / ) 0 ( ) N jk N n k n x n − a e = =
In this case, 3(2 / 6) 0 3 1 1 1 1 ( ) ( 1) 3 6 3 6 j n j n n x n =a +a e = + e = + − Problem 3. (a) [1 0 0 0] [1 2 3 4] [1 2 3 4] = (b) [1 1 1 1] [1 1 1 1] [1 2 3 4 3 2 1] = (c) [1 2 3 4] [0.25 0.25 0.25 0.25] [0.25 0.75 1.50 2.50 2.25 1.75 1.00] = Problem 4.• Solve the difference equation using direct approach ( ) 4 ( 1) ( )
y n + y n− =x n
(0) 4 ( 1) y = − y − = (1) 4 (0) 3 y = − y = − (2) 4 (1) 4( 3 ) 13 y = − y = − − = (3) 4 (2) 4(13 ) 51 y = − y = − = − (4) 4 (3) 4( 51 ) 205 y = − y = − − = … 0 ( ) n ( 4) ( )k p k y n u n = =
−Consider homogeneous equation
1 0 ( ) 4 ( 1) 0 1 1 4 0 4 1 ( ) 4 n y n y n z y n C − + − = + = = =
Therefore, the solution of the D.E. is
0 1 ( ) ( 4) ( ) 4 n n k k y n C u n = = + −
The term 1 4 n C
diverges when n → −
. That means C = 0 and the solution becomes 0 ( ) n ( 4) ( )k k y n u n = =
− .When n → , y(n) → that means this system is not BIBO stable.
Or• Solve the difference equation using Z-transform 1 1 ( ) 4 ( 1) ( ) ( ) 4 ( ) ( ) ( ) 1 ( ) ( ) 1 4 y n y n x n Y z z Y z X z Y z H z X z z − − + − = + = = = +
There is one pole at z = −4. Because the system is causal, the ROC is the area outside the circle with radius of 4. This ROC does not include the unit circle. Therefore, this system is not BIBO stable, i.e., y(n) is not bounded.
Problem 5. Given Fs = 40,000 (a) ( ) ( ) ( ) ( ) ( ) ( ) s s y n x n ax n TF h n n a n TF = + − = + − (b) 0 0 ( ) ( ) ( ) ( ) m s m m s m y n a x n m TF h n a n m TF = = = − = −
Problem 6.
( ) ( ) ( 1) ( 1) ( ) z n =x n +ay n− −ay n− =x n
If H1 represents the echo on a telephone line, then H2 is an echo canceller. In practice, the echo is a more
complex system and we have to model H2 as an adaptive system.
Problem 7.
(a) h n( )=(n+ +1) (n−1)
(b) Not causal because impulse response is not zero for all n < 0 (has a value at n = −1). (c) Linear. If x n1( ) is input, then y n1( )=x n1( + +1) x n1( −1) If x n2( ) is input, then y n2( )=x n2( + +1) x n2( −1) If a x n1 1( )+a x n2 2( ) is input, then
1 1 2 2 1 1 2 2 1 1 1 2 2 2 1 1 2 2 ( ) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( ) ( ) y n a x n a x n a x n a x n a x n x n a x n x n a y n a y n = + + + + − + − = + + − + + + − = +(d) FIR system because its impulse response has nonzero values for only a finite number of n. (e) H e( j)=DTFT h n( ( ))=DTFT( ( n+ +1) (n−1))=ej+e−j =2cos
(f) Yes, this system BIBO stable since
( ) ( 1) ( 1) 2 n n h n n n =− =− = + + − =
. Problem 8. ( ) 0.5 ( 1) ( ) 0.5 ( 1), 0 y n + y n− =x n − x n− n(a) Taking DTFT on both sides Y e( j) 1 0.5 + e−j=X e( j) 1 0.5 − e−j 1 0.5 ( ) ( ) ( ) 1 0.5 j j j j j e Y e H e X e e − − − = = + (b) 1/ 2 1/ 2 2 2 1/ 2 1/ 2 2 2 1 0.5 1 0.5cos 0.5sin 1 0.5 ( ) 1 0.5cos 0.5sin 1 0.5 1 0.5
(1 0.5cos ) (0.5sin ) 1.25 cos 5 4cos
5 4cos 1.25 cos (1 0.5cos ) (0.5sin ) j j j j j e j e H e j e e − − − − − − + − = = = + − + + − + − − = = = + + + +
(c) arg ( ) tan 1 0.5sin tan 1 0.5sin
1 0.5cos 1 0.5cos j H e = − − − − − +
H
1H
2x(n)
y(n)
z(n)
Problem 9. (a) 1 3 / 4 3 / 4 ( 1) ( 1) 3 / 4 3 / 4 1 ( ) { ( )} ( ) 2 1 4 1 4 2 2 3 sin ( 1) 4 4 ( 1) j j j n j j n j n h n DTFT H e H e e d e e d e d n n − − − − − − = = = = − = −
(b) This impulse response has values for an infinite number of indices (n) therefore it represents an IIR filter. Problem 10. (a) { ( 2)} 2 { ( )} 2 1 1 ( 1) z Z u n z Z u n z z z z − − − = = =
− − . Same ROC asZ u n{ ( )}, i.e., z 1
(b) 1 2 2 1 { ( 1)} { ( 1)} , 1 1 ( 1) ( 1) d d z z Z nu n z Z u n z z z z dz dz z z z − − = − − = − = − − = − − − . (c) 3 3 3( 1) 3 1 3 3 1 3 3 3 3 { n ( 1)} { n ( 1)} { n ( )} e z z e , Z e u n Z e e u n e z Z e u n z e z e z e − − − − − − − − − − − − − − = − = = = − − . (d) 2 2 1 ( 1) 1 1 1 1 1 1 1 1 1 1 1 1 2 {( 1) ( 1)} {( 1) ( 1)} { ( )} { ( )} { ( )} , ( ) n n n n n Z n e u n Z n e e u n e z Z ne nu n d d d e z z Z ne u n e z z z Z e u n dz dz dz d d z e z z e dz dz z e d z e z e z dz z e − − − − − − − − − − − − − − − − − − − − − = − − = = − = − − = − − − = − 2 1 1 2 1 2 1 2 2 1 1 4 1 4 1 2 1 3 , ( ) ( ) 2 ( ) 1 2 , ( ) ( ) ( ) ( ) d z e z e dz z e z e z z e z e e z e z e z e z e z e − − − − − − − − − − − − = − − − = − = − − − − − . Problem 11. 3 2 2 23 3 ( ) 3 16 32 3 1 4 3 1 4 8 4 8 z X z z z z z z z z z − + = = + + − + − +
Consider the term
2 23 3 16 32 3 1 4 8 z z z − − + , 2 23 3 16 32 3 1 1 1 4 8 2 4 z A B z z z z − = + − + − − ,
1/ 2 23 3 5 16 32 1 2 4 z z A z = − = = − , 1/ 4 23 3 17 16 32 1 16 2 z z B z = − = = − − , 17 5 ( ) 3 2 16 1 1 4 2 4 X z z z = + +z− −z− 2 17 5 3 2 16 ( ) 1 1 4 2 4 z z X z z z z z = + + − − − Because ROC is 0.5<|z|<, 3 5 1 17 1 ( ) ( 2) ( 1) ( ) 4 2 2 16 4 n n x n = n+ + n+ + − u n Problem 12. 1 2 2 1 2 1 2 1 2 ( ) ( 1) ( ) ( ) 2 3 1 2 3 ( )(1 2 3 ) ( )( ) ( ) 2 ( 1) 3 ( 2) ( 1) ( 2) ( ) 2 ( 1) 3 ( 2) ( 1) ( 2) Y z z z z H z X z z z z z Y z z z X z z z y n y n y n x n x n y n y n y n x n x n − − − − − − − − + + = = = − + − + − + = + − − + − = − + − = − − − + − + −
The frequency response can be found by substitute z by ej under the condition that the system is stable.
However, given that this system is causal and there are 2 poles at 1 j 2, the ROC will be z 1 j 2 . This ROC does not cover the unit circle. Therefore, this system is not stable and we cannot determine its frequency response.
