CEDC606: Digital Signal Processing

Lecture Notes 2: The z-transform

Ramez Koudsieh, Ph.D.

Faculty of Engineering
Department of Robotics and Intelligent Systems
Manara University
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 Chapter 2
The z-transform
1. Introduction
2. The z-transform
3. Transfer function of LTI systems
4. Linear constant-coefficient difference equations
5. Connections between pole-zero locations and time-domain behavior
6. The one-sided z-transform

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 1. Introduction
▪ The z-transform plays the same role in the analysis of DTLTI systems as the
Laplace transform does in the analysis of CTLTI systems.
▪ The z-transform is an extension of the DTFT to address two problems:
• First, there are many useful signals in practice, such as nu[n], for which the
DT Fourier transform does not exist.
• Second, the transient response of a system due to initial conditions or due
to changing inputs cannot be computed using the DTFT approach.
▪ The decomposition of an arbitrary sequence into a linear combination
k =− x [k ] [n − k ] shows that every LTI

of scaled and shifted impulses, x [n ] =
system can be represented by the convolution sum:
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 k =− x [k ]h[n − k ] = k =− h[k ]x [n − k ]
 
y[n ] =
▪ The impulse response sequence h[n] specifies completely the behavior and
the properties of the associated LTI system.
▪ In general, any sequence that passes through a LTI system changes shape.
We now ask: is there any sequence that retains its shape when it passes
through an LTI system?
Let us consider the complex exponential sequence: x[n] = zn, for all n
where z is a complex variable defined everywhere on the complex plane
y[n ] = k =− h[k ]z
 n −k
= ( k =−

h [k ]z −k
z n
)=H (z )z n
, for all n

▪ Thus, the output sequence is the same complex exponential as the input
sequence, multiplied by a constant H(z) that depends on the value of z.
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 ▪ The quantity H(z), as a function the complex variable z, is known as the
system function or transfer function of the system.
▪ The complex exponential sequences are eigenfunctions of LTI systems.
▪ The constant H(z), for a specified value of the complex variable z, is the
eigenvalue associated with the eigenfunction zn.
2. The z-transform 
X (z ) = Z{x [n ]} =  x [n ]z −n two-sided or bilateral z-transform
n =−
▪ Since the z-transform is an infinite power series, it exists only for those values
of z for which this series converges.
▪ The region of convergence (ROC) of X(z) is the set of all values of z for which
X(z) attains a finite value.
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 Let us express the complex variable z in polar form as: z = rej

n =−
 −n − j n
X (z ) z =re j = x [n ]r e
  
 X (z ) =  x [n ]r e−n − j n
  x [n ]r e− n − j n
=  x [n ]r −n
n =− n =− n =−

−1   
x [n ] x [n ]
X (z )   x [n ]r −n
+  r n
=  x [ − n ]r +  n
n

n =− n =0 n =1 n =0 r

 n =1
 n
x [ − n ]r converges for all points inside a circle of radius r1


x [n ]
 r n
converges for all points outside a circle of radius r2
n =0

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 X(z) converges for all points within an annular region of the form r2  r  r1
▪ Note: The ROC depends only on r and not on .

n =−
j  − j n
X (z ) z =e j = X (e ) = x [n ]e = F {x [n ]}

▪ Note: The discrete-time Fourier transform X(ej) may be

viewed as a special case of the z-transform X(z).
▪ The values of z for which X(z) = 0 are called zeros of X(z), and
the values of z for which X(z) is infinite are known as poles.
▪ Note: The ROC cannot include any poles.
▪ For finite duration sequences the ROC is the entire z-plane, with the possible
exception of z = 0 or z = ∞.
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 ▪ For infinite duration sequences the ROC have one of the following shapes:
• Right-sided (x[n] = 0, n  n0) ⇒ ROC: |z|  r.
• Left-sided (x[n] = 0, n  n0) ⇒ ROC: |z|  r.
• Two-sided ⇒ ROC: r2  |z|  r1.

▪ Example 1: z-Transform of the unit-impulse

n =−
 −n
Z{ [n ]} = x [n ]z = x [ 0]z 0
= 1 Converges at every point in the z-plane

▪ Example 2: z-Transform of a causal exponential signal x[n] = an u[n]

  
1 z
X (z ) =  a u[n ]z =  a z =  (az ) =
n −n n −n −1 n
−1
=
n =− n =0 n =0 1 − az z −a
converge if: |az−1|  1 ⇒ |z|  |a|
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 ▪ Note: The z-Transform of an anticausal exponential signal x[n] = −an u[−n − ] is
 −1 
1
X (z ) =  −a u[ − n − 1]z =  −a z = −  (a z ) =
n −n n −n −1 n
−1
, |z|  |a|
n =− n =− n =1 1 − az
n n 1
Z{a u[n ]} = Z{ − a u[ − n − 1]} = −1
 ROC must be specified with X(z)
1 − az
The inverse z-transform
▪ The recovery of a sequence x[n] from its z-transform (X(z) and ROC) can be
done using the formula (inverse z-transform ):
x [n ] = Z −1 {X (z )} = 1
2 j C X (z )z n −1dz

where C is a counterclockwise closed circular contour centered at the origin

and with radius r such that  is in the ROC of X.
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 1
▪ Example 3: Finding the inverse z-transform of −1
n
1 − az
z
x [n ] = 21 j  dz C is a circle at radius greater than |a|. f(z) = zn.
C z −a

If n ≥ 0, f(z) has no poles inside C. The only is z = a ⇒ x[n] = f(a) = an, n ≥ 0.

If n  0, f(z) = zn has an nth-order pole at z = 0, which is also inside C. Thus
there are contributions from both poles. For n = −1 we have
1 1 1
x [ − 1] = 2 j 
1
dz = + =0
C z (z − a ) z − a z = 0 z z =a
1 d  1  1
For n = −2 we have x [ − 2] = 1
2 j C 2
z (z − a )
dz =   + 2
dz  z − a  z = 0 z z =a
=0

Continuing in the same way we can show that x[n] = 0, n  0 ⇒ x[n] = anu[n].
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 ▪ For rational functions, the inverse z-transform can be more easily computed
using partial fraction expansions (PFE).
▪ Example 4: Finding the inverse z-transform using partial fractions
1 + z −1
X (z ) =
(1 − z −1 )(1 − 0.5z −1 )
4 3
X (z ) = −1
−
1−z 1 − 0.5z −1
ROC: |z|  1, both fractions are the z-transform of causal sequences. Hence
x [n ] = 4u[n ] − 3 ( )
n
1
2
u[n ]
x [n ] = −4u[ − n − 1] + 3 ( ) u[ − n − 1]
n
ROC: |z|  ½, 1
2
anticausal
x [n ] = −4u[ − n − 1] − 3 ( 2 ) u[n ]
n
ROC: ½  |z|  1, 1 noncausal
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 Properties of z-Transform

Property x[n] X(z) ROC

Linearity ax1[n] + bx2[n] aX1(z) + bX2(z) ⊃ (R 1 ∩ R 2)
Time shifting x[n − k] X(z)z −k R ± {0 or ∞}
Time reversal x[−n] X(z−1) R −1
Multiply by exp. an x[n] X(z/a) |a|R
Differentiate in z nx[n] − z dX(z)/dz R
Convolution x1[n] x2[n] X1(z) X2(z) ⊃ (R 1 ∩ R 2)
Correlation rx1x2[l] = x1[l] x2[−l] X1(z) X2(z−1) ⊃ (R 1 ∩ R 2)
z
k =− x [k ]
n
Summation X (z ) ⊃ (R ∩ (z  1))
z −1
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 ▪ Example 5: z-Transform of a cosine signal x [n ] = cos(0n )u[n ]
cos(0n )u[n ] = 21 e j 0n u[n ] + 21 e − j 0n u[n ]
Z{cos(0n )u[n ]} = 21 Z{e j 0n u[n ]} + 21 Z{e − j 0n u[n ]}
1/2 1/2 1 − cos(0 )z −1
= j 0 −1
+ − j 0 −1
= ROC is |z|  1
1 −e z 1 −e z 1 − 2cos(0 )z −1 + z −2

▪ Example 6: z-Transform of a sine signal x [n ] = sin(0n )u[n ]

sin(0n )u[n ] = 1
2j
e j 0n u[n ] − 1
2j
e − j 0n u[n ]
Z{sin(0n )u[n ]} = 1
2j
Z{e j 0n u[n ]} − 1
2j
Z{e − j 0n u[n ]}
1/2 j 1/2 j sin(0 )z −1
= j 0 −1
− − j 0 −1
= ROC is |z|  1
1 −e z 1 −e z 1 − 2cos(0 )z −1 + z −2
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 ▪ Example 7: autocorrelation sequence of the signal x[n] = anu[n], −1  a  1
Rxx (z ) = Z{rxx [n ]} = X (z )X (z −1 )
1 −1 1
X (z ) = −1
, |z|  |a| X (z ) = , |z|  1/|a|
1 − az 1 − az
1 1 1 1
Rxx (z ) = = , a  z 
1 − az −1 1 − az 1 − a (z + z −1 ) + a 2 a

1 n
rxx [n ] = 2
a , −n  Since the ROC of Rxx(z) is a ring,
1−a
rxx[n] is a two-sided signal, even if x[n] is causal.

▪ Initial and final value properties of the z-transform applies to causal signals
only. Initial value: x [0] = lim X (z ) Final value: lim x [n ] = lim (z − 1)X (z )
z → n → z →1
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 3. Transfer function of LTI systems
For relaxed LTI system
k =− x [k ]h[n − k ]

x[n] h[n] y[n] y[n ] = x [n ]  h[n ] =
X(z) H(z) Y(z) Y (z ) = X (z )H (z )

▪ Example 8: Determine the response of a system with impulse response

h[n] = anu[n], |a|  1 to the input x[n] = u[n] using the convolution theorem.
 
1 1
H (z ) = a n −n
z =
1 − az −1
, z a and X (z ) = z −n
=
1−z −1
, z 1
n =0 n =0

1
Y (z ) = −1 −1
, z  max{ a , 1} = 1
(1 − az )(1 − z )
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 1  1 a 
Y (z ) =  −1
− −1 
, z 1
1 −a 1 − z 1 − az 
n +1
1 1 − a
y[n ] = (1 − a n +1)u[n ] = u[n ]
1−a 1−a
Causality and stability
▪ A TF H(z) with the ROC that is the exterior of a circle, including ∞, is a
necessary and sufficient condition for DTLTI system to be causal.
▪ An LTI system with transfer function H(z) = N(z)/D(z) is causal if and only if:
1. the ROC is |z|  |p|, where p is the outermost pole and 2. deg N  deg D.
▪ An LTI system is stable if and only if the ROC of H(z) includes the unit circle
|z| = 1.
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 ▪ A causal LTI system with rational transfer function H(z) is stable if and only if
all poles of H(z) are inside the unit circle.
▪ The conditions for causality and stability are different and that one does not
imply the other.
▪ For example, a causal system may be stable or unstable, just as a noncausal
system may be stable or unstable.
▪ Similarly, an unstable system may be either causal or noncausal, just as a
stable system may be causal or noncausal.
▪ Example 9: A linear time-invariant system is characterized by the transfer
function: 1 z z 
H (z ) =  −
2  z − 23 z + 21 
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 If ROC: |z|  3/2, the system is causal and unstable
h [n ] = 2 ( 2 ) u [n ] − 2 ( − 2 ) u [n ]
1 3 n 1 1 n

If ROC: 1/2  |z|  3/2, the system is noncausal and stable

h [n ] = ( )
1 3 n
−2 2 u [ − n − 1] − 1
2 ( )
1 n
−2 u [n ]
If ROC: |z|  1/2, the system is anticausal and unstable
h [n ] = − 21 ( 23 ) u [ − n − 1] + ( − 21 ) u[ − n − 1]
n 1 n
2

Interconnection of two LTI systems

parallel h[n ] = h1[n ] + h2[n ]

interconnection H (z ) = H 1(z ) + H 2 (z )

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 cascade h[n ] = h1[n ]  h2[n ]
interconnection H (z ) = H 1(z )H 2 (z )
4. Linear constant-coefficient difference equations
y[n ] + k =1 aky[n − k ] = k = 0 bkx[n − k ]  Y (z ) + k =1akz Y (z ) = k = 0 k X (z )
N M N −k M −k
b z

k = 0 k
M
Y (z ) b z −k
B (z ) b0z −M (z M + + bbM0 )
H (z ) = = = = −N N
X (z ) 1 +
k = 1 k A(z )
N
a z −k z (z + + aN )

k =1 (z − zk ) k = 1
M M −1
(1 − z k z )
H (z ) = b0z N − M = b0
k =1 (p − pk ) k = 1
N N −1
(1 − pk z )
where zk’s are the system zeros and pk’s are the system poles, and b0 is a
constant gain term.
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 Impulse response
▪ The transfer function H(z) with distinct poles can be expressed in the form:
M −N N
Ak
H (z ) =  C k z −k + 
1 − p z −1
where Ak = (1 − pk z −1) X (z ) z = p
k
k =0 k =1 k
and Ck = 0 when M  N, that is, when the rational function H(z) is proper.
If we assume that the system is causal, then the ROC is the exterior of a circle
starting at the outermost pole, and h[n ] = k = 0 C k [n − k ] + k =1 Ak (pk )nu[n ]
M −N N

▪ LTI systems can be classified into different classes based on:

1. Length of impulse response
• h[n] has infinite duration and the system is called an Infinite Impulse
Response (IIR) system.
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 • If N = 0, H(z) becomes a polynomial. The impulse response is given by:
h[n ] = k = 0 bk [n − k ] =
M bk , 0  n  M
0, otherwise 
the corresponding system is called Finite Impulse Response (FIR) system.
2. Feedback in implementation
• If N ≥ 1 the system is known as a recursive system.
• If N = 0 the output is a linear combination of the present and previous
inputs, only. Such a system is called nonrecursive system.
3. Poles and zeros
• If ak = 0, for k = 1, ..., N, the system has M zeros (all-zero systems).
• If bk = 0, for k = 1, ..., M, the system has N poles (all-pole systems).
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 5. Connections between pole-zero locations and time-domain behavior
M −N N
Ak
The TF H(z) with distinct poles: H (z ) =  C kz −k
+
1 − p z −1
k =0 k =1 k

where the first summation is included only if M ≥ N

▪ The roots of a polynomial with real coefficients either must be real or must
occur in complex conjugate pairs.
M −N
Ak K2 K1
bk 0 + bk1z −1
H (z ) =  C k z −k +
1 − p z −1
+
1 + a z −1
+ a z −2
k =0 k =1 k k =1 k1 k2

▪ Therefore, the behavior of a system with a rational transfer function can be

understood in terms of the behavior of a first-order system with a real pole and
a second-order system with complex conjugate poles.
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 First-order systems b
H (z ) = −1
, a , b real
1 − az
Assuming a causal system, the impulse response is given by the following
real exponential sequence: h[n ] = ba nu[n ]
Second-order systems
b0 + b1z −1 z (b0z + b1)
H (z ) = −1 −2
= 2
1 + a1z + a2z z + a1z + a2
There are three possible cases for poles: 1. Real and distinct, 2. Real and
equal, 3. Complex conjugate.
The impulse response of a causal system with a pair of complex conjugate
poles: h[n ] = 2 A r n cos(0n +  )u[n ] , where A is the PFE coefficient
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 Impulse responses associated with real poles in the z-plane
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 Impulse responses associated with a pair of complex conjugate poles in the z-plane
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 z +1
▪ Example 10: Given that causal system: H (z ) = , find
z − 0.9z + 0.81
2

a. its difference equation representation, and

b. its impulse response representation.
Y (z ) z +1 z −1 + z −2
a. H (z ) = = =
X (z ) z 2 − 0.9z + 0.81 1 − 0.9z −1 + 0.81z −2
Y (z ) − 0.9z −1Y (z ) + 0.81z −2Y (z ) = z −1X (z ) + z −2X (z )
y[n ] − 0.9y[n − 1] + 0.81y[n − 2] = x[n − 1] + x[n − 2]
y[n ] = 0.9y[n − 1] − 0.81y[n − 2] + x[n − 1] + x[n − 2]
−0.6173 + j 0.9979 −0.6173 − j 0.9979
b. H (z ) = 1.2346 + − j  / 3 −1
+ j  / 3 −1
, z  0.9
1 − 0.9e z 1 − 0.9e z
From z-transform table:
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 h[n ] = 1.2346 [n ] + [( − 0.6173 + j 0.9979)0.9ne − j  n / 3
+ ( − 0.6173 − j 0.9979)0.9ne j  n / 3 ]u[n ]
h[n ] = 1.2346 [n ] + 0.9n [ − 1.2346cos( n /3) + 1.9958sin( n /3)]u[n ]
h[0] = 0  h[n ] = 0.9n [ − 1.2346cos( n /3) + 1.9958sin( n /3)]u[n − 1]
6. The one-sided z-transform
one-sided or
 n = 0 x [n ]z
+ +  −n
X (z ) = Z {x [n ]} = Z{x [n ]u[n ]} = unilateral z-transform
▪ Almost all properties we have studied for the two-sided z-transform carry over
to the one-sided z-transform with the exception of the time shifting property.
Z + {x [n − 1]} = x [ − 1] + x [0]z −1 + x [1]z −2 + = x [ − 1] + z −1 (x [0] + x [1]z −1 + )
= x [ − 1] + z −1X + (z )
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 Z + {x [n − 2]} = x [ − 2] + x [ − 1]z −1 + z −2X + (z )
In general, for any k  0, we can show that
Case 1: Time delay: Z {x [n − k ]} = z X (z ) +  m =1 x [ − m ]z m −k
+ −k + k

In case x[n] is causal, then Z + {x [n − k ]} = z −k X + (z )

▪ This property makes possible the solution of linear constant-coefficient
difference equations with nonzero initial conditions.
Case 2: Time advance: Z {x [n + k ]} = z X (z ) −  m = 0 x [m ]z k −m ,
+ k + k −1
k 0
▪ Example 11: Determine the one-sided z-transform of the signals
x[n] = anu[n] x1[n] = x[n − 2] x2[n] = x[n + 2]
−2 2
1 z z
X + (z ) = −1
, X +
1 (z ) = −1
+ a −1 −1
z + a −2
, X +
2 (z ) = −1
− z 2
− az
1 − az 1 − az
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1 − az
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 ▪ Example 12: A linear time-invariant system
y[n ] = ay[n − 1] + x [n ], n  0 with y[−1] ≠ 0
ay[ − 1] 1
Y + (z ) = ay[ − 1] + az −1Y + (z ) + X + (z )  Y + (z ) = −1
+ −1
X +
(z )
1 − az 1 − az
zero input zero −state
If the input x[n] = 0 for all n, then: yzi[n] = ay[−1]an = y[−1]an+1, n ≥ −1
If the initial condition is zero then the system is at rest or at zero-state:
+ + 1 n
Y (z ) = H (z )X (z ), H (z ) = −1
or h [n ] = a u[n ]
1 − az
and hence the second term can be identified as the zero-state response yzs[n].
+ k = 0 h[k ]x[n − k ], n  −1
n +1 n
The complete response is given by: y[n ] = y[ − 1]a
If we set x[n] = u[n]
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 ay[ − 1] 1 1 ay[ − 1] 1/(1 − a ) a /(1 − a )
Y + (z ) = −1
+ −1 −1
= −1
+ −1
−
1 − az 1 − az 1 − z 1 − az 1−z 1 − az −1
n +1 1
y[n ] = y[ − 1]a + (1 − a n +1) , n  −1
1−a
yzi
yzs

▪ Example 13: Determine an expression for the nth term of the Fibonacci
sequence with initial conditions y[0] = 1 and y[1] = 1.
y[n] = y[n − 1] + y[n − 2]
y[1] = y[0] + y[−1] = 1 ⇒ y[−1] = 0
y[0] = y[−1] + y[−2] = 1 ⇒ y[−2] = 1
Y + (z ) = [z −1Y + (z ) + y[ − 1]] + [z −2Y + (z ) + y[ − 2] + y[ − 1]z −1 ]
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 1+ 5 1− 5 p1 p2
p1 = , p2 = , A1 = , A2 =
2 2 5 5
n n
1 + 5 1 + 5  1 − 5 1 − 5 
y[n ] =  2  −  2  , n0
2 5   2 5  
n +1
1 1
y[n ] = 2 (1 + 5 )n +1 − (1 − 5 )n +1  , n0
5   

▪ Example 15: Determine the unit step response of the system described by the
DE y[n] = 0.9y[n − 1] − 0.81y[n − 2] + x[n] with the ICs y[−1] = y[−2] = 1.
1 1
H (z ) = −1 −2
, X (z ) =
1 − 0.9z + 0.81z 1 − z −1
j / 3
H(z) has two complex-conjugate poles at p1 = 0.9e , p2 = 0.9e − j  / 3
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 1
Yzs (z ) =
(1 − 0.9e j  / 3z −1 )(1 − 0.9e − j  / 3z −1 )(1 − z −1 )
0.0496 − j 0.542 0.0496 + j 0.542 1.099
Yzs (z ) = j  / 3 −1
+ − j  / 3 −1
+
1 − 0.9e z 1 − 0.9e z 1 − z −1
yzs [n ] = 1.099 + 1.088 (0.9)n cos( 3 n − 5.2 ), n0
0.09 − 0.81z −1 0.045 + j 0.4936 0.045 − j 0.4936
Yzs (z ) = j  / 3 −1 − j  / 3 −1
= j  / 3 −1
+
(1 − 0.9e z )(1 − 0.9e z ) 1 − 0.9e z 1 − 0.9e − j  / 3z −1
yzi [n ] = 0.988 ( 0.9)n cos( 3 n + 87  ), n0
0.568 + j 0.445 0.568 − j 0.445 1.099
Y (z ) = Yzs (z )Yzi (z ) = j  / 3 −1
+ − j  / 3 −1
+
1 − 0.9e z 1 − 0.9e z 1 − z −1
yzs [n ] = 1.099 + 1.44 (0.9)n cos( 3 n + 38 ), n  0
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