Assignment No 2 – EE480
1.(2-1). Find the z-transform of the number sequence generated by sampling the time function e(t) = t every
T seconds, beginning at t = 0. Can you express this transform in closed form ?
Solution:
The sequence generated by sampling e(t) is simply replacing t = kT
k = 0,1,…
e(k) = kTu(k)
the z-transform is
Z[e(k)] = Z[kTu(k)] = TZ[ku(k)]
Using the derivative property (class slides pp. 12-13)
Z [ku(k )]   z
d
1
z

dz 1  z 1 ( z  1) 2
|z|<1
Finally:
Z [e(k )] 
Tz
( z  1) 2
|z|<1
2.(2-2-a). Write as a series, the z-transform of the number sequence generated by sampling the time
function e(t) = e-t every T seconds, beginning at t = 0. Can you express this transform in closed form ?
Solution:
The sequence generated by sampling e(t) is obtained using t=kT
k=0,1,…
e(k) = e-kTu(k)
the z-transform is, as a series
Z[e(k)] = e-0T + e-Tz-1 + e-2Tz-2 + e-3Tz-3 + e-4Tz-4 + ....
To express in closed form we use the decaying exponential function (class slide p. 6)


k 0
k 0
Z [e(k )]   e  kT z  k   (e T z 1 ) k 
1
1  e T z 1
|e-Tz-1|<1 or |z|>|e-T|
(2-2-b). Evaluate the coefficients in the series of part (a) for the case that T=0.05 s.
Solution:
e(0)=e-0=1.0
e(3)=e-0.15=0.860708
e(6)=e-0.30=0.740818
e(1)=e-0.05=0.951229
e(4)=e-0.20=0.818731
e(7)=e-0.35=0.704688
e(2)=e-0.10=0.904837
e(5)=e-0.25=0.778801
e(8)=e-0.40=0.670320 ....
3.(2-3-c). Find the z-transforms of the number sequences generated by sampling the following time
functions every T seconds, beginning at t = 0. Express these transforms in closed form.
(c) e(t) = e-(t-5T)u(t-5T)
Solution:
The number sequence generated by sampling e(t) is obtained using t=kT
e(kT) = e-(kT-5T)u(kT-5T) = e-(k-5)Tu((k-5)T)
e(k) = (e-T)(k-5)u(k-5)
the z-transform is, using the ‘m’ step delay property (class slides p. 10)
Z[e(k)] = Z[x(k-5)u(k-5)] = z-5Z[x(k)u(k)]
where x(k)=(e-T)k
Z[e(k)] = z-5Z[(e-T)ku(k)]
Finally we use the decaying exponential function (class slide p. 6):
Z [e(k )]  z 5
1
1  e T z 1
|e-Tz-1|<1 or |z|>|e-T|
4.(2-4). Find the z-transform, in closed form, of the number sequence generated by sampling the time
function e(t) every T seconds beginning at t=0. The function e(t) is specified by its Laplace transform,
E ( s) 
2(1  e 5 s )
s( s  2)
T=1s
Solution:
We need the time function e(t) before we can get the sequence of samples
E(s) = (1-e-5s)F(s)
where,
F ( s) 
2
( s  2)  s 1
1

 
s( s  2)
s( s  2)
s s2
Now we can get the inverse Laplace transform of F(s)
L-1[F(s)] = f(t)u(t) = (1-e-2t)u(t)
And using the time delay property of Laplace
L-1[e-5sF(s)]=f(t-5)u(t-5)
The time domain function e(t) then is
e(t)=f(t)u(t) – f(t-5)u(t-5)=(1-e-2t)u(t)-(1-e-2(t-5))u(t-5)
The number sequence generated by sampling the time function e(t) is found using t=kT=k, as T=1
e(k) = f(k)u(k)-f(k-5)u(k-5) = (1-e-2k)u(k)-(1-e-2(k-5))u(k-5)
The z-transform of f(k)u(k) is, using the decaying exponential function (class slide p. 6)
Z[f(k)u(k)]=Z[u(k)]-Z[(e-2)ku(k)]
Z [ f (k )u (k )] 
1
1
z (1  e2 )


1
 2 1
1 z
1 e z
( z  1)( z  e 2 )
|z-1|<1 and |e-2z-1|<1
The z-transform of f(k-5)u(k-5) is, using the ‘m’ step delay property (class slides p. 10)
Z[f(k-5)u(k-5)] = z-5Z[f(k)u(k)]
Finally:
Z [e(k )] 
(1  z 5 ) z (1  e 2 )
( z 5  1)(1  e 2 )

( z  1)( z  e 2 )
z 4 ( z  1)( z  e 2 )
|z|>1
5.(2-8). Find the inverse z-transform of each E(z) below by the four methods given in the text. Compare the
values of e(k) , for k=0,1,2,3, obtained by the four methods.
0.5 z
(2-8-a). E ( z ) 
Partial fraction expansion method only.
( z  1)( z  0.6)
Solution:
Get the function
E( z)
0.5

z
( z  1)( z  0.6)
Now get the coefficients
E' ( z) 
c1  ( z  1) E ' ( z ) | z 1 
0.5
 1.25
1  0.6
c2  ( z  0.6) E ' ( z ) |z  0.6 
0.5
 1.25
0.6  1
The function E(z) can be expanded as
E( z) 
1.25 z 1.25 z
1
1

 1.25(

)
z  1 z  0.6
1  z 1 1  0.6 z 1
The inverse z-transform can be obtained using the decaying exponential function (class slide p. 6)
Z-1[E(z)]=e(k)=1.25(1-0.6k)u(k)
k=0,1,2,...
(2-8-e). Use MATLAB to verify the partial-fraction expansions.
Solution:
>> [r,p,k]=residue([0.5],poly([1,0.6]));
>> [r,p]
% r:coefficients, p:poles
ans =
1.25000000000000
1.00000000000000
-1.25000000000000
0.60000000000000