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Assignment 3 solutions 11/3/
Assignment 3 Solutions
a)
0.5 0,1, 2,3........... ( ) 0 else
k (^) k x k
Z transform for the above can be found in series form using the basic definition of Z transform as shown below.
0 1 2 2 0 1 2
k k k k k
X Z x k z
z z z
z z
โ โ = โ โ โ โ = โ โ
๏ฅ
๏ฅ
In the closed form the above can be represented using the summation formula as below
2
1
1
1 ..... is the summation to formula 1 but in out case 0. 1 ( ) 1 0.5 0.
x x x x z z X Z z z
โ
โ
b) Plotting done in MATLAB. The plot of x ( k )is
Assignment 3 solutions 11/3/
Plot of the delayed function x k ( โ 3 ) is shown below.
c) Using the real shifting property of the Z transform we get,
( ) (^ )
0
0
0
3
2
where
but
3
k k k n (^) n k n m m n
Z f k n f k n z
f k n z z
z f m z k n m
z F Z
Z x k z F Z z F Z z z Z x k z
โ โ = โ (^) โ โ โ = โ โ โ = โ
โ
โ
๏ฅ
๏ฅ
๏ฅ
d) The given system is described as y k ( ) = 2 x k ( โ 3 ).
Taking Z transform on both sides we get
Y ( Z ) = 2 z โ^3 X ( Z )
Assignment 3 solutions 11/3/
The numbers at the sampling instances are tabulated in the table below.
0 1 1 0. 2 0. 3 0. 4 0. 5 1 6 0. 7 0. 8 0. 9 0. 10 0 11 0
b)
The discrete version of the function x t ( ) = e โ^2 t + e^ โ^2 (^^ t โ^5 )^ u t ( โ 5 )can be written as
( )
x k ( ) = e โ^2 kT + e^ โ^2^ k^ โ^5 Tu k โ 5 where T is the sampling time. Taking Z transform of the
expression above produces the closed form representation of the number sequence.
( )
4 2 T 2 T
z z
X Z
z e z e
โ
5 2 2 4 2
Taking 1 we get
( )
T
z z X Z z z e z e z z X Z z e
โ โ โ โ โ
a) Final value theorem states that
1
1
lim ( ) lim( 1) ( )
Applying final value theorem to ( ) we get
( ) 1
lim ( ) lim 1 1 As z tends to 1 the above function tends to 0 the final value can be said to be 0
k z
k z
e k z E Z
X Z z X Z z z e k z z
โโ โ
โโ โ
Assignment 3 solutions 11/3/
b) Inverse Z transform of the above function can be taken to get the discreet time domain function after which the final value theorem can be applied to that function. The working is as shown below.
1
When 1, we get
1 1
1 ( ) 1 Applying final value theorem to the above function we get lim ( ) lim 1
The above limit does not tend toward 0 as k increases to. Rathe
k
k
k
k k k
z Z a z a a z Z z z Z x k z
x k
โ
โโ โโ
r it oscillates between -1 and +1.
This inconsistency is due to the fact that the Z transform does not converge for the value of -1.
0 0 1 0 1 1
1
In the above function the magnitude of the term 1 must be less than 1
1 1
1 1
1
k k k k k k k k k
Z a a z
z z z z z z
โ (^) โ
= โ โ = โ โ = โ โ
For values of z > -1 the z transform converges And For values z < -1 it diverges For z = -1 it oscillates as obtained above.
