EE422G Signals and Systems Laboratory Laplace and Z transforms Kevin D. Donohue

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EE422G
Signals and Systems Laboratory
Laplace and Z transforms
Kevin D. Donohue
Electrical and Computer Engineering
University of Kentucky
Laplace Transform
Fourier Series and Transforms are used for modeling signals in
terms of their frequency content, while systems and signal
interactions are modeled
with Laplace transforms, given by
∞
X  s= ∫ x t exp−st dt
−∞
To see relationship with Fourier let s = σ + jω:
∞
X  s= ∫ x t exp− t exp− j t dt
−∞
∞
X  s= ∫ x t exp− t   cos t− j sin  t  dt
−∞
Under what conditions does the Laplace Transform become the
Fourier Transform?
Convolution Relationships
Given a relaxed linear time invariant system, for input x(t) and
impulse response h(t) the output is computed through the
convolution integral:
∞
y t=∫−∞ ht− x d 
In the Laplace domain, convolution become multiplication:
Y s= X s H  s
where H  s is the system transfer function (TF).
  s
Y
H s= 
X  s
Differential Equation and TF
Given a relaxed system (zero for initial conditions) described by
differential equation:
ẋ t =6 ÿ t 2 ẏ t  y t 
Apply Laplace Transform relationships to find the Transfer
function and the impulse response.
Show:
1
s
6
 s=
H
1
1
s2 s
3
6
 
[
 
−t
h t =exp
6
cos
5 t
5
−
 ]
5 sin 5 t
6
6
3
The Z Transform
Consider a numerical integration of the Laplace integral
∞
X  s=
∑
x n T s exp−s n T s T s
n=−∞
Let z = exp(s Ts ), which implies z-n = exp(-n s Ts )
 
∞
ln  z
X
=
Ts
∑
x nT s  z−n T s
n=−∞
Now let Ts = 1 and define Z-transform as:
∞
X  z =
∑
n=−∞
x [n] z−n
z- Transform Example
Given impulse response h[n] = an u[n] (where u[] is the
unit step), find the TF (Use ZT of impulse response) and
comment on stability and convergence.
Show
z
H  z =
z−a
z- Transform Example
Find TF from difference equation for system in input x[n]
and output y[n]:
y [n]=x [n]−x [n−1]0.5 y [n−1]−0.25 y [n−2]
and comment on stability
Show
2
z
−z

H  z = 2
z −0.5 z0.25
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